- GeneralForum: 1General
- UNIT 1 : STRUCTURE OF AN ATOM AND MASS SPECTRUMUNIT 1 : STRUCTURE OF AN ATOM AND MASS SPECTRUM
Key unit competency
Interpret simple mass spectra and use them to calculate the relative atomic mass (R.A.M) of different elements.
By the end of this unit, I will be able to:
•Outline the discovery of the sub-atomic particles.
•Compare the properties of sub-atomic particles.
•Explain what is an isotope of an element.
•Assess the relationship between the number of protons and the number of electrons.
•Calculate the mass number knowing the number of protons and the number of neutrons.
•Understand the meaning of relative atomic mass and relative abundances
•Calculate the relative atomic mass of an element, given isotopic masses and abundances.
•Draw and label the mass spectrometer.
•Explain the fundamental processes occurring in the functioning of a mass spectrometer.
• Interpret different mass spectra.
•State the uses of the mass spectrometer.
•Calculate the relative atomic mass of an element, from a mass spectrum.
Each country has its own culture (language, traditions and norms, attitudes and values, etc.). Our culture defines our identity which is unique to each Rwandan citizen and differentiates us from foreigners; if one element of our culture is rejected or disappears, we become a different Rwandan people. When we introduce foreign cultures to replace ours, we can lose our identity. However, some of our cultural elements such as language can be shared with others to build the social relationship.
Similarly, in the atom, the number of protons within the nucleus defines the atomic number, which is unique to each chemical element; the atomic number or the number of protons of an atom defines its identity. If a proton is added or removed from an element, it becomes a different element. Electrons around the nucleus can be lost, gained, or shared to create bonds with other atoms in chemical reactions to produce useful substances, but this does not change the identity of the elements involved.
1.1. Outline of the discovery of the atom constituents and their properties
1. Regardless of some exceptions, all atoms are composed of the same com-ponents. True or False? If this statement is true why do different atoms have different chemical properties?
2. The contributions of Joseph John Thomson and Ernest Rutherford led the way to today’s understanding of the structure of the atom. What were their contributions?
3. Explain the modern view of the structure of the atom?
4. Using your knowledge about atom, what is the role each particle plays in an atom?
1.1.1. Constituents of atoms and their properties
Atoms are the basic units of elements and compounds. In ordinary chemical reactions, atoms retain their identity. An atom is the smallest identifiable unit of an element. There are about 91 different naturally occurring elements. In addition, scientists have succeeded in making over 20 synthetic elements (elements not found in nature but produced in Laboratories of Reasearch Centers).
An element is defined as a substance that cannot be broken down by ordinary chemical methods in simpler substances. Some examples of elements include hydrogen (H), helium (He), potassium (K), carbon (C), and mercury (Hg). In an element, all atoms have the same number of protons or electrons although the number of neutrons canvary. A substance made of only one type of atom is called also element or elemental substance, for example: hydrogen (H2), chlorine (Cl2), sodium (Na). Elements are the basic building blocks of more complex matter.
A compound is a matter or substance formed by the combination of two or more different elements in fixed ratios. Consider, Hydrogen peroxide (H2O2) is a compound composed of two elements, hydrogen and oxygen, in a fixed ratio (2:2).
During the early twentieth century, scientists discovered that atoms can be divided into more basic particles. Their findings made it clear that atoms contain a central portion called the nucleus. The nucleus contains protons and neutrons. Protons are positively charged, and neutrons are neutral. Whirling about the nucleus are particles called electrons which are negatively charged. The relative masses and charges of the three fundamental particles are shown in Table 1.1
The mass of an electron is very small compared with the mass of either a proton or a neutron.
The charge on a proton is equal in magnitude, but opposite in sign, to the charge on an electron.
1.1.2. Discovery of the atom constituents.
The oldest description of matter in science was advanced by the Greek philosopher Democritus in 400 BC.He suggested that matter can be divided into small particles up to an ultimate particle that cannot any more be divided, and called that particle atom. Atoms came from the Greek word atomos meaning indivisible.
The work of Dalton and other scientists such as Avogadro, etc., contributed more so that chemistry was beginning to be understood. They proposed new concept of atom, and from that moment scientists started to think about the nature of the atom. What are the constituents of an atom, and what are the features that make atoms of the various elements to differ?
In 1808 Dalton published A New System of Chemical Philosophy, in which he presented his theory of atoms:
a) Dalton’s Atomic Theory
1. Each element is made up of tiny particles called atoms.
2. The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways.
3. Chemical compounds are formed when atoms of different elements combine with each other. A given compound always has the same relative numbers and types of atoms.
4. Chemical reactions involve reorganization of the atoms—changes in the way they are bound together. The atoms themselves are not changed in a chemical reaction.
b)Discovery of Electrons and Thomson’s Atomic Model
In 1897 J. J. Thomson (1856–1940) and other scientists conducted several experiments, and found that atoms are divisible. They conducted experiments with gas discharge tubes. A gas discharge tube is shown in Figure 1.2.
The gas discharge tube is an evacuated glass tube and has two electrodes, a cathode (negative electrode) and an anode (positive electrode). The electrodes are connected to a high voltage source. Inside the tube, an electric discharge occurs between the electrodes.
The discharge or ‘rays’ originate from the cathode and move toward the anode, and hence are called cathode rays. Using luminescent techniques, the cathode rays are made visible and it was found that these rays are deflected away from negatively charged plates. The scientist J. J. Thomson concluded that the cathode ray consists of negatively charged particles, and he called them electrons.
Thomson postulated that an atom consisted of a diffuse cloud of positive charge with the negative electrons embedded randomly in it. This model, shown in Figure 1.3, is often called the plum pudding model because the electrons are like raisins dispersed in a pudding (the positive charge cloud), as in plum pudding.
In 1909 Robert Millikan (1868–1953) conducted the famous charged oil drop experiment and came to several conclusions: He found the magnitude of the charge of an electron equal to -1.602 x 10-19c. From the charge-to-mass ratio(e/m) determined by Thomson, the mass of an electron was also calculated.
c)Discovery of Protons and Rutherford’s Atomic Model
In 1886 Eugene Goldstein (1850–1930) observed that a cathode-ray tube also generates a stream of positively charged particles that move towards the cathode. These were called canal rays because they were observed occasionally to pass through a channel, or “canal,” drilled in the negative electrode (Figure 1.4). These positive rays, or positive ions, are created when the gaseous atoms in the tube lose electrons. Positive ions are formed by the process
Different elements give positive ions with different e/m ratios. The regularity of the e/m values for different ions led to the idea that there is a subatomic particle with one unit of positive charge, called the proton. The proton is a fundamental particle with a charge equal in magnitude but opposite in sign to the charge on the electron. Its mass is almost 1836 times that of the electron.
The proton was observed by Ernest Rutherford and James Chadwick in 1919 as a particle that is emitted by bombardment of certain atoms with α-particles.
Rutherford reasoned that if Thomson’s model were accurate, the massive α-particles should crash through the thin foil like cannonballs through gauze, as shown in Figure 1.6(a). He expected α-particles to travel through the foil with, at the most, very minor deflections in their paths. The results of the experiment were very different from those Rutherford anticipated. Although most of the α- particles passed straight through, many of the particles were deflected at large angles, as shown in Figure 1.6(b), and some were reflected, never hitting the detector. This outcome was a great surprise to Rutherford. Rutherford knew from these results that the plum pudding model for the atom could not be correct. The large deflections of the α-particles could be caused only by a center of concentrated positive charge that contains most of the atom’s mass, as illustrated in Figure 1.6(b). Most of the α-particles pass directly through the foil because the atom is mostly open space. The deflected α-particles are those that had a “close encounter” with the massive positive center of the atom, and the few reflected α-particles are those that made a “direct hit” on the much more
massive positive center.
In Rutherford’s mind these results could be explained only in terms of a nuclear atom—an atom with a dense center of positive charge (the nucleus) with electrons moving around the nucleus at a distance that is large relative to the nuclear radius.
d)Discovery of Neutrons
In spite of the success of Rutherford and his co-workers in explaining atomic structure, one major problem remained unsolved.
If the hydrogen contains one proton and the helium atom contains two protons, the relative atomic mass of helium should be twice that of hydrogen. However, the relative atomic mass of helium is four and not two.
This question was answered by the discovery of James Chadwick, English physicist who showed the origin of the extra mass of helium by bombarding a beryllium foil with alpha particles. (See figure 1.7).
In the presence of beryllium, the alpha particles are not detected; but they displace
uncharged particles from the nuclei of beryllium atoms. These uncharged particles cannot be detected by a charged counter of particles.
However, those uncharged particles can displace positively charged particles from another substance. They were called neutrons.The mass of the neutron is slightly greater than that of proton.
Figure 1.8 shows the location of the elementary particles (protons, neutrons, and electrons) in an atom. There are other subatomic particles, but the electron, the proton, and the neutron are the three fundamental components of the atom that are important in chemistry.
Atoms consist of very small, very dense nuclei surrounded by clouds of electrons at relatively great distances from the nuclei. All nuclei contain protons; nuclei of all atoms except the common form of hydrogen also contain neutrons.
Checking up 1.1
1. In an experiment, it was found that the total charge on an oil drop was 5.93 × 10-18 C. How many negative charges does the drop contain?
2. All atoms of the elements contain three fundamental particles. True or false? Give an example to support your answer.
3. Compare the atom constituents
a. in terms of their relative masses
b. in terms of their relative charges
4. Using the periodic table as a guide, specify the number of protons and electrons in a neutral atom of each of these elements.a. carbon (C) b. calcium (Ca) c. chlorine (Cl) d. chromium (Cr)
1.2. Concept of atomic number, mass number, and isotopic mass
The diagram below shows a representation of sodium isotopes . Observe it and answer to the questions below
1. Compare the two sodium isotopes in the figures above.
2. From your observation, how do you define the isotopes of an element?
3. How is the mass number, A, determined?
4. What information is provided by the atomic number, Z?
5. What is the relationship between the number of protons and the number of electrons in an atom?
6. Where are the electrons, protons, and neutrons located in an atom?
7. Why is the mass of an atom concentrated in the center?
8. Sodium-24 and sodium-23 react similarly with other substances. Explain the statement
9. Say which one(s) of the following statements is(are) correct and which one(s) is(are) wrong: (i) isotopes differ in their number of electrons, (ii) isotopes differ in their mass numbers, (iii) isotopes differ in their number of protons, (iv) isotopes differ by their number of neutrons, (v) all the statements are wrong.
The atomic number denotes the number of protons in an atom’s nucleus. The mass number denotes the total number of protons and neutrons. Protons and neutrons are often called nucleons. By convention, the atomic number is usually written to the left subscript of the elemental notation, and the mass number to the left superscript of the elemental notation as represented by the example below, where X represents any elemental symbol.
Some atoms of the same of element have the same atomic number, but different mass numbers. This means a different number of neutrons. Such atoms are called isotopes of the element.
Isotopes are atoms of the same element with different masses; they are atoms containing the same number of protons but different numbers of neutrons.
In a given atom, the number of protons, also called “atomic number” are equal to the number of electrons because the atom is electrically neutral. The sum of the number of protons and neutrons in an atom gives the mass number of that atom.
Mass number = number of protons + number of neutrons
= atomic number + neutron number
Checking up 1.2
1. How do you call the members of each of the following pairs? Explain.
2. Write, using the periodic table, the correct symbols to identify an atom that contains
a. 4 protons, 4 electrons, and 5 neutrons;
b. 23 protons, 23 electrons, and 28 neutrons;
c. 54 protons, 54 electrons, and 70 neutrons; and
d. 31 protons, 31 electrons, and 38 neutrons.
3. Use the list of the words given below to fill in the blank spaces. Each word will be used once.
Atomic number, Mass number, protons, Electrons, Isotopes, neutron
a) The atomic number tells you how many.................................. and ............................................................. are in an atom.......................................................is the number written as subscript on the left of the atomic symbol.
b) The total number of protons and neutrons in an atom is called the .......................................................................
c) Atoms with the same number of protons but different number of neu-trons are called .................................................
d) The subatomic particle that has no charge is called a .......................................................
1.3. Calculation of relative atomic mass of elements with isotopes
1. Argon has three naturally occurring isotopes: argon-36, argon-38, and argon-40. Based on argon’s reported relative atomic mass from the periodic table, which isotope do you think is the most abundant in nature? Explain.
2. Calculate the average atomic mass of an element with two naturally occurring isotopes: 85X (72.15%, 84.9118 amu) and 87X (27.85%, 86.9092 amu). Identify this element?
3. Boron has two naturally occurring isotopes. Find the percent abundances of 10B and 11B given the isotopic mass of 10B = 10.0129 amu and the isotopic mass of 11B = 11.0093 amu.
Relative atomic mass, symbolized as R.A.M or Ar, is defined as the mass of one atom of an element relative to 1/12 of the mass of an atom of carbon-12, which has a mass of 12.00 atomic mass units. The relative atomic mass, also known as the atomic weight or average atomic weight, is the average of the atomic masses of all of the element’s isotopes.
Relative isotopic mass is like relative atomic mass in that it deals with individual isotopes. The difference is that we are dealing with different forms of the same element but with different masses.
Thus, the different isotopic masses of the same elements and the percentage abundance of each isotope of an element must be known in order to accurately calculate the relative atomic mass of an element.
Notice: Remember that mass number is not the same as the relative atomic mass or isotopic mass! The mass number is the number of protons + neutrons; while relative atomic mass (or isotopic mass) is the mass if you were to somehow weigh it on a balance.
Let A1, A2, A3,..., An be an abundance of n isotopes of the same chemical element with atomic mass M1, M2, M3,..., Mn respectively, the relative atomic mass(R.A.M) is given by the following equation:
Example 1: Oxygen contains three isotopes 16O, 17O, and 18O. Their respective relative abundancesare 99.76%, 0.04%, and 0.20%. Calculate the relative atomic mass of oxygen.
Relative isotopic mass of 16O is 16 and its relative abundance is 99.76%;
Relative isotopic mass of 17O is 17, abundance 0.04%;
Relative isotopic mass of 18O is 18, abundance 0.20%.
By applying the same formula, the relative abundance of the isotopes may be calculated knowing the relative atomic mass of the element and the atomic masses of the respective isotopes.
Example 2: Chlorine contains two isotopes 35Cl and 37Cl, what is the relative abundance of each isotope in a sample of chlorine if its relative atomic mass is 35.5?
Checking up 1.3
1. Three isotopes of magnesium occur in nature. Their abundances and masses, determined by mass spectrometry, are listed in the following table. Use this information to calculate the relative atomic mass of magnesium.
2.The atomic weight of gallium is 69.72 amu. The masses of the naturally occurring Isotopes are 68.9257 amu for 69Ga and 70.9249 amu for 71Ga. Calculate the percent abundance of each isotope
1.4. Description and functioning of the mass spectrometer
By using the information in this book and other sources, attempt to answer the following questions
1. Define the mass spectrometer and state the main stages of its functioning.
2. Here below is given the block diagram of the mass spectrometer. Identify the unidentified marked component.
d) Ion transducer
3. Inlet system is also known as which of the following?
c) Sample handling system
d) Element injection system
The mass spectrometer is an instrument that separates positive gaseous atoms and molecules according to their mass-charge ratio and that records the resulting mass spectrum.
In the mass spectrometer, atoms and molecules are converted into ions. The ions are separated as a result of the deflection which occurs in a magnetic and electric field.
The basic components of a mass spectrometer are: vaporisation chamber (to produce gaseous atoms or molecules), ionization chamber (to produce positive ions), accelerating chamber (to accelerate the positive ions to a high and constant velocity), magnetic field (to separate positive ions of different m/z ratios), detector (to detect the number and m/z ratio of the positive ions) and the recorder (to plot the mass spectrum of the sample).
A mass spectrometer works in five main stages, namely vaporization, ionization, acceleration, deflection, and detection to produce the mass spectrum.
Stage 1: Vaporization
At the beginning the test sample is heated until it becomes vapour and is introduced as a vapour into the ionization chamber. When a sample is a solid with low vapour pressure, it can directly be introduced into the ionization chamber.
Stage 2: Ionisation
The vaporized sample passes into the ionization chamber (with a positive voltage of about 10,000 volts). The electrically heated metal coil gives off electrons which are attracted to the electron trap which is a positively charged plate.
The particles in the sample (atoms or molecules) are therefore bombarded with a stream of electrons (electrons gun), and some of the collisions are energetic enough to knock one or more electrons out of the sample particles to make positive ions. Mass spectrometers always work with positive ions.
Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion.
Most of the sample molecules are not ionized at all but are continuously drawn off by vacuum pumps which are connected to the ionization chamber (figure 1.9). Some of the molecules are converted to negative ions through the absorption of electrons.
The repeller plate absorbs these negative ions. A small proportion of the positive ions which are formed may have a charge greater than one (a loss of more than one electron). These are accelerated in the same way as the singly charged positive ions.
Stage 3: Acceleration
The positive ions are accelerated by an electric field so that they move rapidly through the machine at high and constant velocity.
Stage 4: Deflection
The ions are then deflected by a magnetic field according to their masses and charges ratio. Different ions are deflected by the magnetic field at different extents. The extent to which the beam of ions is deflected depends on four factors:
1. The magnitude of the accelerating voltage (electric field strength). Higher voltages result in beams of more rapidly moving particles to be deflected less than the beams of the more slowly moving particles produced by lower voltages.
2. Magnetic field strength. Stronger fields deflect a given beam more than weaker fields.
3. Masses of the particles. Because of their inertia, heavier particles are deflected less than lighter particles that carry the same charge.4. Charges on the particles. Particles with higher charges interact more strongly with magnetic fields and are thus deflected more than particles of equal mass with smaller charges
The two last factors (mass of the ion and charge on the ion) are combined into the mass/charge ratio. Mass/charge ratio is given the symbol m/z (or sometimes m/e).For example, if an ion had a mass of 28 and a charge of 1+, its mass/charge ratio would be 28. An ion with a mass of 56 and a charge of 2+ would also have a mass/charge ratio of 28.
In the figure 1.11 above, ion stream A is most deflected: it will contain ions with the smallest mass/charge ratio. Ion stream C is the least deflected: it contains ions with the greatest mass/charge ratio. Assuming 1+ ions, stream A has the lightest ions, stream B the next lightest and stream C the heaviest. Lighter ions are going to be more deflected than heavy ones.
Stage 5: Detection
The beam of ions passing through the machine is detected electrically. As they pass out of the magnetic field, ions are detected by an ion detector which records the position of the ions on the screen and the number of ions that hit the screen at each position. These two pieces of information are used to produce a mass spectrum for the sample.
A flow of electrons in the wire is detected as an electric current which can be amplified and recorded. The more ions arriving, the greater the current
Detecting the other ions
How might the other ions be detected (those in streams A and C which have been lost in the machine)?
Remember that stream A was most deflected. To bring them on to the detector, you would need to deflect them less by using a smaller magnetic field.
To bring those with a larger m/z value (the heavier ions if the charge is +1) to the detector you would have to deflect them more by using a larger magnetic field.If you vary the magnetic field, you can bring each ion stream in turn on the detector to produce a current which is proportional to the number of ions arriving. The mass of each ion being detected is related to the size of the magnetic field used to bring it on to the detector. The machine can be calibrated to record current (which is a measure of the number of ions) against m/z directly. The mass is measured on the 12C scale.
Note: The 12C scale is a scale on which the 12C isotope weighs exactly 12 units..
The detector of a typical instrument consists of a counter which produces a current that is proportional to the number of ions which strike it. Through the use of electron multiplier circuits, this current can be measured so accurately that the current caused by just one ion striking the detector can be measured. The signal from the detector is fed to a recorder, which produces the mass spectrum. In modern instruments, the output of the detector is fed through an interface to a computer. The computer can store the data, provide the output in both tabular and graphic forms, and compare the data to standard spectra, which are contained in spectra libraries also stored in the computer.
This is an example of an appearance of a mass spectrum of unknown element that has 2 isotopes.
Checking up 1.4
1. Use the list of the words given below to fill in the blank spaces. Each word will be used once.
Vaporization chamber, mass spectrum, velocity, ionization, deflection, detector, acceleration
A sample of the element is placed in the _________ chamber where it is converted into gaseous atoms. The gaseous atoms are ionized by bombardment of high energy electrons emitted by a hot cathode to become positive ions (in practice, the voltage in the ________chamber is set in such a way that only one electron is removed from each atom). The positive ions (with different masses) are then going faster to a high and constant _________by two negatively charged plates: the process is called_________. The positive ions are then deviatedby the magnet field. This process is called ____________ (ions with smaller mass will be deflected more than the heavier ones). These ions are then detected by the ion _________. The information is fed into a computer which prints out the________ of the element.
2. The correct order for the basic features of a mass spectrometer is...
a. acceleration, deflection, detection, ionization
b. ionisation, acceleration, deflection, detection
c. acceleration, ionisation, deflection, detection
d. acceleration, deflection, ionisation, detection
3. Which one of the following statements about ionisation in a mass spectrometer is incorrect?
a. gaseous atoms are ionised by bombarding them with high energy electrons
b. atoms are ionised so they can be accelerated
c. atoms are ionised so they can be deflected
d. it doesn’t matter how much energy you use to ionise the atoms
4. The path of ions after deflection depends on...
a. only the mass of the ion
b. only the charge on the ionc. both the charge and the mass of the iond. neither the charge nor the mass of the ion
5. Which of the following species will be deflected to the greatest extent?
6. Which of the following separates the ions according to their mass-to-charge?
a) Ion source
c) Magnetic sector
d) Electric sector
1.5. Interpretation of mass spectra.
The mass spectrum of zirconium looks like this:
a) What does m/z mean?
b) Explain as fully as possible what the mass spectrum shows about zirconium.
(I am not expecting you to read actual values from the relative abundance axis.)
c) The spectrum shows lines for 1+ ions. If there were also peaks for 2+ ions, where would you expect to find them, and what would you predict about their heights relative to the 1+ peaks?
Example 1: The mass spectrum of boron
The mass spectrum of boron may be used to know the number of boron isotopes and their relative abundances
The two peaks in the mass spectrum shows that there are 2 isotopes of boron with relative isotopic masses of 10 and 11 on the 12C scale
The mass spectrum of an element shows how you can find out the masses and relative abundances of the various isotopes of the element and use that information to calculate the relative atomic mass of the element.
The relative size of the peaks gives you a direct measure of the relative abundances of the isotopes. The tallest peak is often given an arbitrary height of 100 but you may find all sorts of other scales used; it doesn’t matter. You can find the relative abundances by measuring the lines on the stick diagram.
In this case, the two isotopes (with their relative abundances) are:
Example 2: The mass spectrum for zirconium
The mass spectrum of zirconium may be used to know the number of zirconium isotopes and their relative abundances
The 5 peaks in the mass spectrum show that there are 5 isotopes of zirconium with relative isotopic masses of 90, 91, 92, 94 and 96 on the 12C scale.
This time, the relative abundances are given as percentages. Again you can find these relative abundances by measuring the lines on the stick diagram. In this case, the 5 isotopes (with their relative percentage abundances) are:
Example 3: The mass spectrum of chlorine
Chlorine is taken as typical of elements with more than one atom per molecule. Chlorine has two isotopes, 35Cl and 37Cl, in the approximate ratio of 3 atoms of 35Cl to 1 atom of 37Cl. You might suppose that the mass spectrum would look like this:
But it is not true. The problem is that chlorine consists of molecules, not individual atoms. When chlorine is passed into the ionization chamber, an electron is knocked off the molecule to give a molecular ion, Cl2+. These ions won’t be particularly stable, and some will fall apart to give a chlorine atom and a Cl+ ion. The term for this is fragmentation.
If the Cl atom formed isn’t then ionized in the ionization chamber, it simply gets lost in the machine (neither accelerated nor deflected).
The Cl+ ions will pass through the machine and will give lines at 35 and 37, depending on the isotope and you would get exactly the pattern in the last diagram. The problem is that you will also record lines for the unfragmented Cl2+ ions.
At the end the spectrum will show peaks due to ionized atoms, Cl+at 35, and 37, and ionized molecule Cl2+ at 70, 72, 74 as below
Checking up 1.5
The mass spectrum of magnesium is given below:
a. How many isotopes does magnesium possess
b. Estimate the isotopic mass of each of the magnesium isotopes
c. Estimate the relative abundance for each of the isotopes of magnesium
1.6. Uses of the mass spectrometer and involving calculations
1. Mass spectrometers are used to determine which of the following?
a) The atomic mass
b)Composition in sample
c) Concentration of elements in sample
2. The mass spectrum of an element, A, contained four lines at mass/charge of 54; 56; 57 and 58 with relative intensities of 5.84; 91.68; 542.17; 0.31 respectively. Explain these data and calculate the relative atomic mass of A
1.6.1. Calculation of RAM using mass spectrum
When the mass spectrum of the element is given, you can calculate the relative atomic mass of that element by using the information from the mass spectrum.
Example 1: the mass spectrum of boron is given below
Determine the relative atomic mass of boron
From the mass spectrum given, we have123 typical atoms of boron (sum of relative abundances). 23 of these would be 10B and 100 would be 11B.
The total mass of these would be (23 x 10) + (100 x 11) = 1330
The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant figures).10.8 is the relative atomic mass of boron.
Example 2: The figure below represents the mass spectrum of zirconium.
Suppose you had 100 typical atoms of zirconium (sum of relative abundances). 51.5 of these would be 90Zr, 11.2 would be 91Zr and so on. The total mass of these 100 typical atoms would be
(51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + (2.8 x 96) = 9131.8
The average mass of these 100 atoms would be 9131.8 / 100 = 91.3 (to 3 significant figures).
91.3 is the relative atomic mass of zirconium.
1.6.2. Uses of mass spectrometer
In addition to the use of mass spectrometer in the determination of isotopes of elements and their relative abundances, the applications of mass spectrometry are found:
•Pharmaceutical: drug discovery, combinatorial chemistry, pharmacokinetics, drug metabolism.
•Clinical: neonatal screening, haemoglobin analysis, drug testing.
•Environmental: water quality, soil and groundwater contamination, food contamination, pesticides on foods.
•Geological: oil composition.
•Biotechnology: the analysis of proteins, peptides.
Checking up 1.6
1. Which of the following is not done through mass spectrometry?
a. Calculating the isotopic abundance of elements
b. Investigating the elemental composition of planets
c. Confirming the presence of O-H and C=O in organic compounds
d. Calculating the molecular mass of organic compounds
2. Mass spectra enable you to find relative abundances of the isotopes of a particular element.
a) What are isotopes?
b) Define relative atomic mass.
c) The mass spectrum of strontium contains the following lines for 1+ ions:
Calculate the relative atomic mass of strontium.
3. The mass spectrum for chlorine looks like this:
a) Explain why there are two separate groups of peaks.
b) State what causes each of the 5 lines.
c) Explain the approximate relative heights of the lines at 35 and 37.
d) Why cannot you predict the relative heights of the two clusters of lines (35/37 and 70/72/74)?
1.7. End unit assessment
I. Multiple choice questions
1.Which of the following is true regarding a typical atom?
a. Neutrons and electrons have the same mass.
b. The mass of neutrons is much less than that of electrons.
c. Neutrons and protons together make the nucleus electrically neutral.
d. Protons are more massive than electrons
2. Which of the following statements is(are) true? For the false statements, correct them.
a. All particles in the nucleus of an atom are charged.
b. The atom is best described as a uniform sphere of matter in which electrons are embedded.
c. The mass of the nucleus is only a very small fraction of the mass of the entire atom.
d. The volume of the nucleus is only a very small fraction of the total volume of the atom.
e. The number of neutrons in a neutral atom must equal the number of electrons.
3. Each of the following statements is true, but Dalton might have had trouble explaining some of them with his atomic theory. Give explanations for the following statements.
a. Atoms can be broken down into smaller particles.
b. One sample of lithium hydride is 87.4% lithium by mass, while another sample of lithium hydride is 74.9% lithium by mass. However, the two samples have the same chemical properties
4. In mass spectrometer, the sample that has to be analysed is bombarded with which of the following?
d. Alpha particles top of Form
5. Mass spectrometer separates ions on the basis of which of the following?
c. Molecular weight
d. Mass to charge ratio
6. In a mass spectrometer, the ions are sorted out in which of the following ways?
a. By accelerating them through electric field
b. By accelerating them through magnetic field
c. By accelerating them through electric and magnetic field
d. By applying a high voltage
7. The procedure for mass spectroscopy starts with which of the following processes?
a. The sample is bombarded by electron beam
c. The sample is converted into gaseous state
d. The ions are detected
8. Which of the following ions pass through the slit and reach the collecting plate?
a. Negative ions of all masses
b. Positive ions of all masses
c. Negative ions of specific mass
d. Positive ions of specific mass
9. Which of the following statements is not true about mass spectrometry?
a. Impurities of masses different from the one being analysed interferes with the result
b. It has great sensitivity
c. It is suitable for data storage
d. It is suitable for library retrieval
10. In a mass spectrometer, the sample gas is introduced into the highly evacuated spectrometer tube and it is ionised by the electron beam.a. Trueb. FalseII.
Short and long answer questions
11. What are the three fundamental particles from which atoms are built? What are theirelectric charges? Which of these particles constitute the nucleus of an atom? Which is the least massive particle of the fundamental particles?
12. Verify that the atomic weight of lithium is 6.94, given the following information:
6Li, mass = 6.015121 u; percent abundance = 7.50%
7Li, mass = 7.016003 u; percent abundance = 92.50%13. The diagram below shows the main parts of a mass spectrometer.
a. Describe the different steps involved in taking a mass spectrum of a sample
b. (i) Which two properties of the ions determine how much they are deflected by the magnetic field? What effect does each of these properties have on the extentof deflection?
(ii) Of the three different ion streams in the diagram above, why is the ion stream C least deflected?
(iii) What would you have to do to focus the ion stream C on the detector?
c. Why is it important that there is a vacuum in the instrument?
d. Describe briefly how the detector works.Bottom of Form
14. A mass spectrum of a sample of indium shows two peaks at m/z = 113 and m/z = 115. The relative atomic mass of indium is 114.5. Calculate the relative abundances of these two isotopes. (b)The mass spectrum of the sample of magnesium contains three peaks with the mass-to-charge rations and relative intensities shown below
i. Explain why magnesium gives three peaks in mass spectrum?
ii. Use the information in the table above to calculate the accurate value for the relative atomic mass of magnesium
15. There exists 3 isotopes of oxygen that occur naturally with atomic mass 16, 17 and 18 with abundance 99.1% ; 0.89% and 0.01% respectively. Given that oxygen occurs naturally as diatomic molecule,
a. Predict the number of peaks that will be observed on the screen of mass spectrometer.
b. Show the molecular ions that are responsible of these peaks.UNIT 1 : STRUCTURE OF AN ATOM AND MASS SPECTRUM
- UNIT 2: ELECTRONIC CONFIGURATION OF ATOMS AND IONSUNIT 2: ELECTRONIC CONFIGURATION OF ATOMS AND IONS
Key unit Competence
To relate Bohr’s atomic model with atomic spectrum of Hydrogen, write electronic configuration of atoms and ions using s, p, d and f atomic orbitals and interpret graphical information related to ionization energy of elements.
By the end of this unit, I will be able to:
•Explain the stability of atoms using the concept of quantization of energy.
•Explain the achievements and limitations of Bohr’s atomic model.
•Explain the existence of energy levels using the data from emission spectra.
•Describe hydrogen spectral lines and spectral line series
•Explain the types of spectra in relation with the nature of light
•Explain the quantum theory of the atom using the quantum numbers.
•Determine the number and shapes of orbitals in given level or principal quantum number
•Explain the rules governing the electronic configuration: Aufbau principle and Hund’s rule
•Explain the relationship between the electronic configuration and the stability of the atoms
•Interpret the graphs of first ionisation energy against the atomic number.
•Describe the factors which influence the first ionisation energy.
2.1. Bohr’s atomic model and concept of energy levels
The potential energy of a person walking up ramp increases in uniform and continuous manner whereas potential energy of person walking up steps increases in stepwise and quantized manner. This can be explained by the values of energy which are continuous for the person walking up ramp while they are discrete (discontinued) for the person walking up steps (Figure 2.1(a) and Figure 2.1(b).
Niels Bohr (1885-1962) a young Danish physicist working in Rutherford’s laboratory, suggested a model for the hydrogen atom and predicted the existence of line spectra. In his model, based on Planck’s and Einstein’s ideas about quantized energy, Bohr proposed three postulates:
•An electron can rotate around the nucleus in certain fixed orbits of definite energy without emission of any radiant energy. Such orbits are called stationary orbits.
•An atom can make a transition from its stationary state of higher energy E2 to a state of lower energy E1 and emit a single photon of frequency ν. Conversely, an atom can absorb an energy at the lower level E1 and transit to the higher energy level E2. That is, the change in energy for a system which can be represented by the equation:
where n is an integer (1,2,3,...) and h is Planck’s constant
determined from experiment and has a value of 6.626x10-34J.s; ⱱ is the frequency of the electromagnetic radiation absorbed or emitted. Each of these small “packets” of energy is called photon also called quantum of energy. Energy can be gained or lost only in whole-number multiples of the quantityνh,That is, the change in energy for a system can be represented by the equation:, where n is an integer (1,2,3,...).
An atom does not release energy when it is in one of its stationary states. That is, the atom does not change energy while the electron moves on a given orbit. When an electron on a given orbit absorbs an appropriate quantum of energy, it jumps, i.e. is promoted to a higher energy orbit; this process is called “excitation”of electron. On the contrary, if the electron loses an appropriate quantum of energy, it falls on the lower energy orbit by emission of a light corresponding to the lost quantum of energy and the process is called “de-excitation” of electron. As there are many energy levels on which electrons can be excited and de-excited, an atom will have many lines of absorption, each corresponding to a quantum of energy absorbed: this appears as a series of lines called absorption spectrum. In the same way the series of emission lines will produce an emission spectrum (see Fig. 2.4).
2.1.1. Achievements of Bohr’s Atomic Model
• Explanation of the stability of an atom
Based on Rutherford’s atomic model, the electrons move around the nucleus in circular paths called orbits. According to the classical theory of electromagnetism, a charged particle revolving around a charged nucleus would release energy and end up by spiraling into the nucleus; thus the atoms would be unstable. The Bohr’s atomic model makes an assumption of discreet orbit (allowed orbit) to explain why an atom is stable; by doing so, Bohr introduces the concept of quantization of energy. Bohr’s atomic model explains the origin of atomic absorption and emission spectra.
• Explanation of the production of the absorption and emission spectra
The Bohr’s atomic model explains the origin of atomic absorption and emission spectra.
2.1.2. Limitations of Bohr Model
1.Bohr’s theory fails to explain the origin of the spectral lines of multi-electron atoms.
It only explains the origin of the spectrum of hydrogen-like species having only one electron such as H, He+, Li2+, Be2+, ........
The model fails to explain the spectral lines of atoms or species with more than one electron.
2. According to Bohr, the circular orbits in which electrons revolve are planar. However, modern research has shown that an electron moves around the nucleus in the three dimensional space.
3. Bohr’s theory fails to account for Zeeman Effect and Stark Effect. Zeeman Effect is the splitting of the spectral lines into thinner and closely- spaced lines when an excited atom is placed in a magnetic field. Stark Effect consists of the splitting of the spectral lines into thinner and closely-spaced lines in presence of electric field.
4. Bohr’s theory is in contradiction with Heisenberg’s uncertainty principle. Bohr assumes that the electron revolves around the nucleus in circular orbits at fixed distance from the nucleus and with a fixed velocity. However, according to W. Heisenberg, it is not possible to know simultaneously the accurate position and the velocity of a very small moving particle such as an electron.
Checking up 2.1
1. Find out two more examples that you can use to illustrating the concept of quantization.
2. Discuss the main weakness of Rutherford’s nuclear atom.
2.2. Hydrogen spectrum and spectral lines
Look at the picture of neon tube light below and do research about how this neon tube light works to produce light and present your findings
Bohr’s atomic model allows to explain the emission spectra of atoms. This happens when excited electrons lose energy in form of electromagnetic radiation and fall to lower energy levels.
The wave-particle nature of the light
Light as a waveThe light is a wave-like phenomenon as shown in Figure 2.2.
It is characterized by its wave length, generally symbolized by the Greek letter lambda, λ, and its frequency, represented by the Greek letter nu1, ν.
As shown in the Figure below, the wavelength represents the distance between two successive summits/peaks (or two successive troughs).The frequency represents the number of complete wavelengths made by the light per second, also called cycles per second.
Visible light is composed by different visible lights with different λ and ν.
But all those lights have the same speed, the speed of light, which, in a vacuum, is equal to: 3.00x108 m/s; although different types of light have different λ and ν, they move at the same speed c. This results in the relation between the speed of light and its wavelength and frequency: c = νλFrom this relation, and since c is constant, we can conclude that:
•Light with long wavelength has low frequency, whereas
•Lights with short wavelength has high frequency.
Let’s take an example to illustrate: light1 has λ1 equal to 105m whereas light2 has λ2 equal to 10-5 m. After 1 second, both would have travelled 3.00x108m, the speed of light, but their frequencies will be different:
ν1 = c/λ1 = 3.00x108ms-/105m = 3.00x103s-= 3,000 cycles/s
v2 = c/λ2= 3.00x108ms-/10-5m= 3.00x1013s-= 30,000,000,000,000 cycles/sLight is energy, represented by: E = hν where h is Plank’s constant (h=6.626x10-34Js)
Hence energy in the light is proportional to its frequency; higher the frequency of the light, higher is its energy and vice-versa.
The different colors of the visible light differ by their wavelength as shown in the
As illustrated in Figure 2.2 below, the right side of the spectrum consists of high-energy, high-frequency and short wavelength radiations. Conversely, the left side consists of low-energy, low-frequency and long wavelength radiations.
The letter gamma, γ, may also be used.
When an electron is excited or de-excited, the energy absorbed or emitted corresponds to the difference of energy, ΔE, between the final energy level of the electron, E2, and the starting energy level of the electron, E1: E2 – E1 = ΔE = hν. ΔE is positive when E2>E1, this is the case of absorption and excitation of electron; on the other hand ΔE may be negative when E2<E1, in case of emission and de-excitation of electron.
Figure 2.4 below shows the different series of emission spectra of hydrogen. As you can see, the difference between those series is the final energy level where the electron fall after de-excitation.
The series have been named according to the scientists who discovered them.
Ionization of an atom or loss of an electron corresponds to excitation of an electron to the level n=∞.
1. What is the meaning of infinity level in the hydrogen spectral lines?
2. Given a transition of an electron from n=5 to n=2. Calculate energy
2.3. Atomic spectra
Observe the picture above, discuss in groups and answer the following questions.
a. What do you see on the above photo?
b. State the physical phenomenon which is related to the above photo.
c. Think of any other means of producing the same pattern. List two of them.
d. What property can you attribute to light with reference to the above process?
Checking up 2.3
1. Different metals, when exposed to a flame, emit different flame colors. Explain the origin of that difference.
2. Would you expect to see the emission rays and the absorption rays of a given element to appear at the same place of a photographic plate or not. Explain your answer.
3. How do you explain the many spectral rays for the same element?
2.4 Orbitals and Quantum Numbers
1. a)Write the electronic configuration of aluminium atom(Z=13)
b) Indicate the number of electrons in each energy level/quantum shell
c) The shells are numbered from inside-outward starting from 1, 2, 3, 4 ... which other name is given to these shells?
d) How did you obtain the exact number of electrons in each energy level/quantum shell in (c) above?
We have seen the weakness and critics against the atomic Bohr’s model. In order to answer to the questions not answered by that model, other atomic models were proposed. One of those models is the Quantum model that has been developed by the Austrarian physicist Erwin Schrödinger (1887-1961). The model is based on a mathematical equation called Schrödinger equation. This model is based on the following assumptions or hypotheses:
•An electron is in continuous movement around the nucleus but cannot be localized with precision; only the high probability of finding it in a cartain region around the nucleus can be known.
•The region where the probability of finding electron is high, at more than 95%, is called “orbital”; in other words, the orbital is the volume or the space (tridimensional) around the nucleus where there is a high probability of finding the electron.
Without going into the mathematical development of the Schrödinger equation, we can say that the energy of the electron depends on the orbital where it is located. And an atomic orbital is described by a certain number of “quantum numbers”
according to the solution of Schrodinger equation, i.e. 3 whole numbers:
1) The principal quantum number (n) is a positive integer which varies from 1 to ∞.
The principal quantum number indicates the energy level in an atom where electrons can be located: the higher the n value, the higher the energy level. An electron in energy level n=1 has lowest energy in an atom. The principal quantum number, n, has been traditionally given names by the letters: K(n=1), L(n=2), M(n=3), N(n=4), O(n=5), P(n=6).
In the Bohr’s atomic model, K, L, M, ... were used to represent different orbits or shells of electrons. Later on, the term shell sometimes is used to describe a group of orbitals with the same principal quantum number. The term subshell describes a group of orbitals with the same principal and second quantum number. The maximum number of orbitals and electrons that can be found in an energy level n are n2and 2n2, respectively (Table 2.1). The maximum number of sub shells in an energy level n equals n.
Table 2.1 Relation between the principal quantum number, the number of orbitals and the maximum number of electrons.
2)The angular momentum quantum number (l)
The second quantum number is the angular quantum number represented by the letter, l: it is an integer which can take any value from zero or higher but less than n-1, i.e. equal to: 0,1, 2, 3,....up to n-1. For example if n= 1, l is equal to 0, if n= 2, l can be 0, 1. It is also called secondary or azimuthal quantum number. It indicates the shape of the orbital and is sometimes called the orbital shape quantum number. By tradition, those different shapes of orbitals have been given names or letter symbols: l = 0 = s, l =1 = p, l = 2 = d, l=3 = f
3) Magnetic quantum number (ml)
The magnetic quantum number describes the orientation of the orbital. It is an integer that varies from -l to +l. For example if: l = 0, ml can only be 0; if l = 1, ml = -1, 0, +1; if l=2, ml = -2, -1, 0, 1, 2. As you can see for each value of l there are (2l+ 1) values of ml corresponding to (2l + 1) orientations under the influence of magnetic field. The s orbital where l is zero and ml has no orientation; it has the shape of a sphere as shown in
Table 2.2: Relationship between the n, l and ml
The table 2.2 shows that, apart s sub-level that has only one orbital, other sub-levels have a certain number of different orbitals; those orbitals have the same energy but differ in their specific orientations. Example p orbitals are 3 with different orientations: pxpypz.
Different sub-levels belonging to the same principal quantum number have different energies as follows: s < p < d < f
4) The spin quantum number (S)
The fourth quantum number is the spin quantum number, represented by the symbol S (or ms in some books). The electron behaves as a spinning magnet.The spin quantum number is the property of the electron, not the orbital.This number describes the spinning direction of the electron in a magnetic field. The direction could be either clockwise or counterclockwise. The electron behaves as if it were spinning about its axis, thereby generating a magnetic field whose direction depends on the direction of the spin. The two directions for the magnetic field correspond to the two possible values for the spin quantum number, S (ms). Only two values are possible: s = +1/2 and -1/2 as shown in the Figure 2.7 below.
ms = +1/2 and ms = -1/2 are commonly represented by ↑ and ↓ respectively.In conclusion an electron in any given atom is decribed by 4 quantum numbers: (i) three quantum numbers which describe the orbital where the electron is located: n, l and ml and (ii) one quantum number describing the spin of the electron, S or ms.
2.5 Electronic configuration of atoms and ions
1. Write the electronic structure of the following chemical speciesK (Z=19), Ne (Z=10), Al3+ (Z=13), Cl (Z=17), O2- (Z=16)
2. Using information in question 1,
a. Determine the group and period of K and Cl.
b.Which species have a stable electronic configuration? Explain.
The electron configuration is the distribution of electrons of an atom in its atomic orbitals. The electronic configuration of an atom is governed by three main rules.
1) Aufbau Principle
The Aufbau principle explains how to build the electronic configuration of an atom.The Aufbau principle states that atomic orbitals of lower energy must be filled before the higher energy orbitals. The Aufbau principle is referred to as the “building-up” principle.
According to that principle, if an atom has only one electron, this electron will occupy the lowest principal quantum energy level, n=1, and the lowest energy orbital in that principal quantum energy, i.e. l = 0 = s. This is represented as: 1s1, meaning one electron in s orbital of the 1st energy level. If the atom has 2 electrons, the second electron will be filled in the same s orbital to give the structure 1s2.
Note the way of notation: the first number indicates the principal energy level or principal quantum number, followed by the letter indicating the orbital, followed by the number of electrons present in the orbital, as super-script.
2) Pauli Exclusion Principle
What happens if we have 3 electrons in an atom? Can we squeeze them in 1s orbital?
There is a principle called Pauli Exclusion Principle states: in an atom, two electrons cannot have the four quantum numbers n, l, ml, and S(ms) identical. This explains why the two electrons in the same orbital must have their spin opposite:
The Pauli exclusion principle doesn’t allow us to put a 3rd electron in the same orbital, since if we put 3 electrons in the same orbital, 2 electrons will have 4 identical quantum numbers and this is not allowed
In other words, the Pauli Exclusion Principle is telling us that you cannot put more than 2 electrons in an orbital. i.e. the maximum number of electrons in an orbital is 2.
Hence, the 3rd electron must go in energy level n = 2, since energy level n=1 if full; it has only one orbital s. Then for that atom with 3 electrons, the electronic structure is: 1s22s1
3) Hund’s rule
Atom with 4 electrons: 1s22s2
Atom with 5 electrons: 1s22s22p1
What about an atom with 6 electrons? Are we putting the 6th electron in the same orbital as the 5th electrons? Remember that there are 3 p orbitals pxpypz of the same energy!
The Hund’s rule answers to that question.
It states that orbitals of equal energy are each occupied by one electron before electrons begin to pair up into the same orbital. By convention, all the unpaired electrons are given the same orientation spin. A slight preference for keeping electrons in separate orbitals helps to minimize the natural repulsive forces that exist between two electrons.
Therefore, atom with 6 electrons: 1s22s22px1py1pz0
When building the electronic configuration of elements, you must be guided by the principles and rules seen above and: writing the principal quantum number in Arabic number, followed by the orbitals immediately followed by the number of electrons in the orbital as superscript.
An atom X: 1s2: has only two electrons in s orbital at the 1st energy level
An atom Y: 1s22s22p3: has electrons in 2 levels of energy: level n=1, and level n= 2. In level 1, it has 2 electrons in s orbital. In level 2, it has 2 electrons in s orbital and 3 electrons in p orbitals.
Figure 2.8 is a useful and simple aid for keeping track of the order in which electrons are first filled for each atomic orbital. The different orbitals are filled in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.Notice that as energy levels increase starting from n=3, 4s orbital is filled before 3d, 5s before 4d, etc... as shown in the diagram below. But when ionized, 4s electrons are ionized before 3d, and 5s before 4d.
Checkup 2.5 (a)
1. Build the electronic configuration of the following atoms: 1H, 3Li, 5B, 11Na, 18Ar,19K, 21Sc, 24Cr, 26Fe, 29Cu
2. Write the electronic configuration for each of the following pairs of ions. State the more stable ion in gaseous state and explain your choice.
a. cu+ and cu2+
b.Fe2+ and Fe3+
Expanded notation is another method of writing the s, p, d and f notation. The method uses the same concept as s, p, d and f notation except that each individual orbital of a sub-level having many orbitals is represented with a subscript letter indicating the orientation of the orbital. This applies for p, d, and f orbitals. Considering that p-orbital has three componentsxp,ypand pz, the expanded electronic configuration of some elements is given hereafter.
Checkup 2.5 (b)
Write the expanded electronic configuration for each of the following atom/ions. S(z=16), P3-(z=15), Mg2+(z=12)
Orbital box representation
An orbital box representation consists of a box for each orbital in a given energy level, grouped by sublevel, with an arrow indicating an electron and its spin.
Note that two electrons in the same orbital have necessarily opposite spins as indicated in the examples below.
The table 2.4 shows the electronic configuration of some elements using orbital box representation and applying Hund’s rule.
Table 2.4: Electronic configuration using orbital box representation
Using boxes to represent orbitals, draw the electronic configuration ofN3- (z=7), Ti4+(z=22), Mg2+(z=12), Ar(z=18)Identify the isoelectronic species that are present.
Noble Gas Notation
All noble gases have completely filled subshells and can be used as a shorthand way of writing electron configurations for subsequent atoms.When using this method, the following steps are respected.
a. Identify the noble gas whose electronic configuration is included in that of the concerned element.
b.Write the chemical symbol of the identified noble gas within square brackets. We call this the noble gas core.
c. Add electrons beyond the noble gas core. Note that electrons that are add-ed to the electronic level of the highest principal quantum number (the outermost level or valence shell) are called valence electrons.
Example: Given the electronic configurations of the noble gases Ne and Ar, one can write the electronic configuration of some elements in noble gas notation of some elements as:
Checking up 2.5 (d)
Using the noble gas notation, write the electronic configuration of the following atoms/ions.
a. Ge (Z=32)
b. S (Z=16)
c. Co2+ (Z=27)
d. Br- (Z=35)
e. Sr (Z=38)
2.6. Relationship between ionization energy, energy levels and factors influencing ionization energy
Write the electronic configuration of the following elements/ions, use s, p, d, ...)
Sodium, magnesium, magnesium ion (Mg2+), aluminium, aluminium ion (Al3+), oxygen ion (O2-)
Identify the common feature of ions in (1) and why do they have such feature
Suggest what happened to aluminium atom when it changed to aluminium ion (Al3+)
Identify the group and the period of aluminium, sodium and oxygen atom
2.6.1. Concept of Ionization energy
The ionization energy is a measure of the energy needed for an atom, in gaseous state, to lose an electron and become positive ion.
The first ionisation energy is the energy required to remove one electron from an atom in its gaseous state. The example below shows how to represent the successive ionization energies of an atom M.
Second ionisation energy and nth ionisation energy: Two or more electrons can be removed and we have successive ionization energies.
The ionization energy is usually expressed in kiloJoules per mole (kJ.mol-1). This energy is required to overcome the attractive force between the nucleus and the electron and then remove the electron. Theoretically there are as many successive ionisation energies as there are electrons in the original atom. In figure 2.9, someone can make an interpretation of successive ionization energies of an atom
2.6.2. Interpretation of a graph of successive ionization energies of an atom
The graph shows that the energy to remove electron increases as more electrons are successively removed.
a) The energy required to remove the first electron is relatively low. This corresponds to the loss of one 3s electron.
b) To remove the second electron needs greater energy because this electron is closer to the nucleus in a 2p orbital. There is a steady increase in energy required as elec-trons are removed from 2p and then 2s-orbitals.
c) The removal of the tenth and eleventh electrons requires much greater amounts of energy, because these electrons are closer to the nucleus in 1s orbital.
2.6.3. Factors influencing the extent of ionization energy
The ionization energy is a physical property of elements that can be influenced by some factors:
1) Size of atom
The atomic size is the distance between the nucleus and valence shell. As the number of energy levels (shells) increases, the force of attraction between nucleus and valence electron decreases. Therefore, the valence electrons are loosely held to the nucleus and lower energy is required to remove them, i.e. ionization energy decreases with increase in atomic size and vice versa.This is what happens when you go down a Group.
2) Nuclear charge
The nuclear charge is the total charge of all the protons in the nucleus. As the nuclear charge increases, the force of attraction between nucleus and valence electrons on the same valence energy level increases and hence makes it difficult to remove an electron from the valence shell. The higher the nuclear charge, the higher the ionization energy. This is what happens when you cross a period from left to right.
3) Screening effect or Shielding effect
The Screening effect or Shielding effect is due to the presence of inner electrons which have a screening or shielding effect against the attraction of the nucleus towards the outermost electrons. The electrons present in inner shells between the nucleus and the valence shell reduce the attraction between nucleus and the outermost electrons. This shielding effect increases with the increasing number of inner electrons. A strong Shielding effect makes it easier to remove an external electron and hence lowers the ionisation energy.
2.6.4. Importance of ionization energy in the determination of the chemistry of an element
Ionization energy provides a basis to understand the chemistry of an element. The following information is provided.
Determination of metallic or non- metallic character.
The I.E informs us how the atom will behave chemically: a low I.E indicates that the element behaves as metal whereas a high I.E indicates that the element behaves as non-metal.
The first ionization energies of metals are all nearly below 800kJ mol-1 while those of non- metals are all generally above 800 kJ mol-1.
Down the group ionization energies decrease so that the elements became more metallic. In groups 14 and 15 there is change from non metallic to metallic character. Across a period from left to right 1st I.E. increases. The elements become less metallic to non- metallic.
Example: The first three ionization energies for elements A, B, C, and D are given in the table below
This table shows that for a given element: 1st IE < 2nd IE < 3rd IE.
From the 1st I.E. of the elements it can be predicted that elements B and C have typical metallic properties since their 1st ionization energies are low.
D is expected to be non-metal because of its high 1st IE.
Checking up 2.6
1. Given elements: Cl, Ca, and Na and the following 1st IE: 456, 578.8 and 1251 KJ/mol. Respectively.Match those 1st IE with the three element and justify.
2. Given elements: bromine and iodine and 1st IE: 1008 and 1140 KJ/mol. Which 1st IE correspond to which element? Explain.
3. Explain why 2nd IE is always greater than the 1st IE?
2.7. End unit assessment
1. Which of the following is the correct representation of the ground-state electron configuration of molybdenum? Explain what is wrong with each of the others.
2. Which of the following electron configurations are correct and which ones are wrong? Explain.
3. Photosynthesis uses 660 nm light to convert CO2 and H2O into glucose and O2. Calculate the frequency of this light.
4. Which of the following orbital designations are incorrect: 1s, 1p, 7d, 9s, 3f, 4f, 2d?
5. The data encoded on CDs, DVDs, and Blu-ray discs is read by lasers. What is the wavelength in nanometers and the energy in joules of the following lasers?
6. Concerning the concept of energy levels and orbitals,
a. How many subshells are found in 3=n?
b. What are the names of the orbitals in 3=n?
c. How many orbitals have the values 4=nand3=l?
d. How many orbitals have the values 23==,lnand2−=lm?(e) What is the total number of orbitals in the level4=n?
7. A hypothetical electromagnetic wave is pictured here. What is the wavelength of this radiation?
8. Consider the following waves representing electromagnetic radiation:
a. Which wave has the longer wavelength?
b. Calculate the wavelengths of the two radiations
c. Which wave has the higher frequency and larger photon energy?
d. Calculate these values.
9. Order the orbitals for a multielectron atom in each of the following lists according to increasing energy:a. p5 ,d5b. s4, p3c. s6,d4
10. According to the Aufbau principle, which orbital is filled immediately after each of the following in a multielectron atom?
11. According to the Aufbau principle, which orbital is filled immediately be-fore each of the following?
12. Four possible electron configurations for a nitrogen atom are shown be-low, but only one represents the correct configuration for a nitrogen atom in its ground state. Which one is the correct electron configuration? Which configurations violate the Pauli Exclusion Principle? Which configurations violate Hund’s rule?
13. Explain the variation in the ionization energies of carbon, as displayed in this graph.
14. The first seven ionization energies of an element W are shown below
What factors determine the magnitude of the first ionization energyUNIT 2: ELECTRONIC CONFIGURATION OF ATOMS AND IONS
- UNIT 3: FORMATION OF IONIC AND METALLIC BONDSUNIT 3: FORMATION OF IONIC AND METALLIC BONDS
Key unit competence
Describe how properties of ionic compounds and metals are related to the nature of their bonding
By the end of this unit, I will be able to:
•Explain why atoms bond together;
•Explain the mechanisms by which atoms of different elements attain stability;
•Explain the formation of ionic bonds using different examples;
•Represent ionic bonding by dot-and-cross diagrams;
•Describe the properties of ionic compounds based on observations;
•Perform experiments to show properties of ionic compounds;
•Assemble experimental set up appropriately and carefully;
•State the factors that influence the magnitude of lattice energy ;
•Relate the lattice structure of metals to their physical properties;
•Describe the formation of metallic bonds;
•State the physical properties of metals and forces of attraction that hold atoms of metal.
People like to bond with each other for many reason such as: to unite their forces and be stronger, to exchange idea and produce big things, to found a family, etc. we cannot live in isolation. People can have strong connection .Similarly some atoms can also have strong bonds between them. Some atoms have weak connections, just like two people can have connections.
Some atoms may not need to bond with others; they are self-sufficient as some people, a small number, may be self-sufficient.
Connections between atoms are called chemical bonds. Solids are one of the three fundamental states of matter. In molecules, atoms or ions are held together by forces called chemical bonds.There are 3 types of chemical bonds: Ionic, Covalent and Metallic bonds.
The type of a bond in molecules is determined by the nature and properties of the bonding atoms. However, in this unit we will only emphasize on ionic and metallic bonding.
3.1. Stability of atoms and why they bind together
1. In pairs discuss and write electronic configuration of sodium , neon, argon, magnesium, aluminium, oxygen and chlorine
2. What happens when oxygen and chlorine gain electrons?
3. What happens when sodium, magnesium and aluminium lose electrons?
4. Discuss on how atoms of elements can gain their stabilities by either loosing or gaining electron(s) on the valence shells and show with evidence that an atom is stable?
5. How does the formation of an ionic bond between sodium and chlorine reflect the octet rule?
Like people always relate and connect to others depending on their values, interests and goals so does unstable atoms. They are combine together to achieve stability. We know that noble gases are the most stable elements in the periodic table. The noble gases are extremely unreative: they do not tend to form compounds or combine to themselves
What do the noble gases have in common? They have a filled outer electron energy level. When an atom loses, gains, or shares electrons through bonding to achieve a filled outer electron energy level, the resulting compound is often more stable than individual separate atoms. Neutral sodium has one valence electron. When it loses this electron to chlorine, the resulting Na+ cation has an outermost electron energy level that contains eight electrons.
It is isoelectronic (same electronic configuration) with the noble gas neon. On the other hand, chlorine has an outer electron energy level that contains seven electrons. When chlorine gains sodium’s electron, it becomes an anion that is isoelectronic with the noble gas argon.
Below are examples of how magnesium bonds with oxygen and calcium with chlorine:
As you can see, the noble gases have the most stable electronic configuration, characterized by the presence of 8 electrons in the outermost shell: this structure is called the “Octet Structure” or “Octet Rule”; octet meaning eight in Latin.Any atom that does not have that structure, will combine with another atom (of the same element or different element) to achieve the octet structure.
Checking Up 3.1
Write the shorthand electronic configuration for copper.Predict the ions that will be formed from atoms in the table, name the isoelectronic noble gas and establish the electronic configuration (E.C) of the ion formed.
3.2. Ionic bonding
Atoms have many ways of combining together to achieve the octet structure, and one of them is the formation of an ionic bond.
In an ionic bond, electrons are transferred from one atom to another so that they form oppositely charged ions; in other words, one atom loses electron(s), the other gains electrons(s).The resulting strong force of attraction between the oppositely charged ions is what holds them together. Ionic bonding is the electrostatic attraction between positive and negative ions in an ionic crystal lattice.
3.2.1. Formation of ionic bond
Draw diagrams to illustrate the formation of ionic compounds in magnesium oxide, magnesium chloride, sodium peroxide, and sodium sulphide.
The transfer of electrons from one atom to another followed by attraction between positive and negative ions is called ionic bonding. This type of bonding occurs between metals and non-metals. The compounds formed are called ionic compounds. As stated previously, metals try to lose their outer electrons while non metals look to gain electrons to obtain a full outer shell. When metals lose their outer electrons they form positively charged ions called cations. When non-metals gain electrons they form negatively charged ions called anions. An example is shown below:
The curved arrow between sodium and fluorine represents the transfer of an electron from a sodium atom to a fluorine atom to form opposite ions. These 2 ions are strongly attracted to each other because of their opposite charges. A bond is now formed and the resulting compound is called Sodium Fluoride
Another example of ionic bonding is the bonding of Beryllium and Fluorine to form beryllium fluoride.
The transfer of electrons from beryllium results in the formation of an ionic bond. Beryllium now has a positive (+2) charge and Fluorine now has a negative charge. The resulting compound is called Beryllium Fluoride (BeF2).
Other examples showing the formation of ionic compounds using dot and cross diagrams.
a. Formation of calcium chloride
b. Formation of magnesium oxide
c. Formation of bonds in sodium fluoride
Checking Up 3.2
1. For each of the following ionic bonds: Sodium + Chlorine, Magnesium + Iodine, Sodium + Oxygen, Calcium + Chlorine and Aluminium + Chlorine
a. Write the symbols for each element.
b. Draw a Lewis dot structure for the valence shell of each element.
c. Draw an arrow (or more if needed) to show the transfer of electrons to the new element.
d. Write the resulting chemical formula.
e. Write the electron configurations for each ion that is formed. Ex. H1+ = 1s0
2. Solid sodium chloride and solid magnesium oxide are both held together by ionic (electrovalent) bonds. a. Using s,p and d notation write down the symbol for and the electronic configuration of (i) a sodium ion; (ii) a chloride ion; (iii) a magnesium ion; (iv) an oxide ion.
b. Explain what holds sodium and chloride ions together in the solid crystal
c. Sodium chloride melts at 1074 K; magnesium oxide melts at 3125 K. Both have identical structures. Why is there such a difference in their melting points?
3.2.2. Physical properties of ionic compounds
Determination of Relative Melting Point of different substances
1. Cut a square of aluminum foil that is about5 by 5 cm
2. Set up a ring stand with an iron ring attached.
3. Place the aluminum square on the iron ring, as shown at right in Figure 3.6
4. Obtain a small pea-sized sample of NaCl. Place the sample on the aluminum foil, about 5cm from the center of the square.
5. Obtain a small pea-sized sample of table sugar. Place the sample on the aluminum foil, about 1 cm from the center of the square, but in the opposite direction from the salt.
6. Your square of aluminum foil should look like in Figure 3.7.
7. Light the Bunsen burner and adjust the flame height so that the tip of the flame is just a cm or so below the height of the aluminum foil.
8. Observe as the two compounds heat up.
9. Set up another sheet of aluminum foil and determine the relative melting points (low vs. high) of the four unknowns.
10. Record your results in the table 3.1 belowCaution: if the compounds burn with sparks do not panic.
1. Which compound melts first?
2. Giving reasons compare the melting points of the two compounds.
Table: 3.1 relative melting points of different substances
The melting points of ionic compounds are higher than those of covalent compounds; this is due to strong electrostatic forces between opposite charges in the ionic substances compared to the week forces of attraction between molecules in covalent substances . This also explains why all ionic compounds are solid at room temperature.
Conductivity in Solution
1. Dissolve a spoonful of NaCl in water.
2. Connect the apparatus as shown in figure 3.8
3. Make an observation and record your results as in table 3.2 below
4. Repeat the procedure 1 to 3 above using sugar solution, ethanol and copper(II) sulfate solution
5. Record your results in the table below.
Table 3.2: relative conductivity of different substances
1. Give reasons for your observations above.
2. Solid sodium chloride does not conduct electricity whereas an aqueous solutionof sodium chloride does. Explain
Based on our tests with salt and sugar, the ability to conduct electricity in solution of ionic compounds is much higher than in covalent compounds
Activity 3.3(c) Solubility test
1. Using forceps, place 5-8 crystals of each of sodium chloride, magnesium chloride, copper sulphate, calcium carbonate, copper carbonate, sodium sulphate (a small pinch) of the compound into one of the test tubes in test tube rack.
2. Half-fill the test tube with distilled water and stir with a clean stirring rod.
3. Observe if the crystals dissolve in water.
4. Record your findings in a suitable table.
Table 3.3: Solubility of different substances
1. Classify substances as soluble and slightly soluble in water.
2. Giving reasons, suggest an explanation for your observation.
Water is a good solvent for many ionic compounds but not a solvent for covalent compounds, apart few exceptions (you will learn about later on).
Shattering: Why are Ionic compounds generally hard, but brittle?
It takes a large amount of mechanical force, such as striking a crystal with a hammer, to force one layer of ions to shift relative to its neighbour. However, when that happens, it brings ions of the same charge next to each other (Figure3.9). The repulsive forces between like-charged ions cause the crystal to shatter. When an ionic crystal breaks, it tends to do so along smooth planes because of the regular arrangement of the ions.
Checking Up 3.3
1. The diagrams below show the electric conductivity of distilled water, solid sodium chloride and a solution of sodium chloride respectively. Use the diagrams to explain the observations from the set up.
i) no light is given out by bulb in A
ii) no light is given out by bulb in B
iii) light is given out in C
2.Why are ionic compounds brittle?
3.Why do ionic compounds have high melting points?
4.What happens when an electric current is passed through a solution of an ionic compound?
3.2.3. Lattice energy
By using information in this student’s chemistry book and other books from the school library, attempt to answer the following questions.
1. Define lattice energy
2. Explain how the lattice energy is used to describe high melting points of ionic compounds.
3. What is the bonding force present in ionic compounds?
4. Why is the melting temperature of magnesium oxide higher than that of magnesium chloride, even though both are almost 100% ionic?
5. How is lattice energy of ionic compounds related to their high melting points?
It is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component ions into gaseous ions (Endothermic process). On the other hand lattice energy is the energy released when gaseous ions bind to form an ionic solid (Exothermic process). Its values are usually expressed with the units’ kJ/mol.
Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favourable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.
Checking Up 3.4
Define lattice energy
Which one of the following has the greatest lattice energy?
MgO or NaCl
LiCl or MgCl2
Which one of the following has the greatest Lattice Energy?
NaCl or CaCl2
AlCl3 or KCl
How do the ionic radius and ion charge affect the lattice energy of an ionic substance?
Factors affecting lattice enthalpy
1. Basing on the information obtained from the definition of lattice energy, suggest the factors that affect the lattice energy.
2. Which has the larger lattice energy: NaCl or CsI?
Basing on the information obtained from the definition of lattice energy, suggest the factors that affect the lattice energy.
Which has the larger lattice energy: NaCl or CsI?
There are two main factors that affect lattice enthalpy.
a) The charges on the ions
Sodium chloride and magnesium oxide have exactly the same arrangements of ions in the crystal lattice, but the lattice enthalpies are very different.
From the above diagram the lattice enthalpy of magnesium oxide is much greater than that of sodium chloride. This is because in magnesium oxide, +2 ions are attracting -2 ions; in sodium chloride, the attraction is only between +1 and - 1ions.
b. The radius or the size (volume) of the ions
The lattice enthalpy of magnesium oxide is also increased relative to sodium chloride because magnesium ions are smaller than sodium ions, and oxide ions are smaller than chloride ions.
It means that the ions are closer together in the lattice, and that increases the strength of the attractions.
For example, as you go down Group 17 of the Periodic Table from fluorine to iodine, you would expect the lattice enthalpies of their sodium salts to fall as the negative ions get bigger - and that is the case:
As the negative ion gets bigger in size, the distance between the centres of oppositely charged ions increase and attractions so decrease. For instance as you go down group 1 halides, the enthalpy decreases.
3.3. Formation of metallic bonds and physical properties of metals
The figure above shows materials commonly used at home. If you reflect back around your house/home you will see hundreds of objects made from different kinds of materials.
1. Observe the objects (in picture) and classify them according to the materials they are made of.
2. Have you ever wondered why the manufacturers choose the material they did for each item?
3. Why are frying saucepans made of metals and dishes, cups and plates often made of glass and ceramic?
4. Could dishes be made of metal? And saucepans made of ceramic and glass
3.3.1. Formation of metallic bond
Another way of combining is the combination between metal atoms to form metallic bond.
When metal atoms combine together, there is no transfer of electrons since the combining atoms are of the same nature, i.e. all are metals and no one is ready to give up or to capture electrons.In metallic bonding, all metal atoms put together their valence electrons in a kind of pool of electrons where positive metallic cations seem to bathe. This model is called “Elecron Sea Model” (Fig. 3.15)
Metals have a sea of delocalized electrons within their structure. These electrons have become detached and the remaining atoms have a positive charge. This positive charged is attracted to the delocalized sea of electrons due to electrostatic forces of attraction (forces which result from unlike charges), and as a result has a strong interaction. It is this interaction which makes the metals so hard and rigid. Figures 3.13 and 3.14 are representation of metallic bond.
3.3.2. Physical properties of metals
Activity 3.7: Looking at metals
1. Collect a number of metal items from your home or school. Some exam-ples are listed below: hammer, electrical wiring, cooking pots, jewellery, burglar bars and coins, nails,
2. What is the function of each of these objects?
3. Discuss why you think metal was used to make each object. You should consider the properties of metals when you are answering this question.
a. Electrical conductivity
1. Take a dry cell/battery, a torch bulb/ bulb, connecting wires, crocodile clips and connect them. As in the figure 3.17
2. Repeat the experiment above using different metals
3. Record your results in a suitable table.
1. Compare the relative conductivity of the metals used in the above experi-ment.
2. Suggest the purpose of the resistor in the experimental set up.
Due to the mobile valence electrons of metals, electricity can pass through the metals easily. So they are conductors of electricity. Silver and copper are the best conductors of electricity.
Note: mercury is a poor conductor of electricity.Thermal conductivity
Experiment to demonstrate the ability of different substances to conduct heat
Apparatus/apparatus: two cups (made from the same material e.g. plastic); a metal
1. Pour boiling water into the two cups so that they are about half full.
2. At the same time, place a metal spoon into one cup and a plastic spoon in the other.
3. Note which spoon heats up more quickly.
4. Record your observations.
1. Which one heats faster plastic spoon or metallic spoon and why?
2. Why do we use plastic cups?
3. Why are cooking pots made of metallic materials not plastics?
Results: The metal spoon heats up more quickly than the plastic spoon. In other words, the metal conducts heat well, but the plastic does not.
Conclusion: Metals are good thermal conductors, while plastic is a poor thermal conductor. The reason is due to the mobility of electrons with transfer of kinetic energy between electrons. This explains why cooking pots are metallic, but their handles are often plastic or wooden. The pot itself must be metal so that heat from the cooking surface can heat up the pot to cook the food inside it, but the handle is made from a poor thermal conductor so that the heat does not burn the hand of the person who is cooking.
c. Malleability and ductility
Experiments to demonstrate the malleability and ductility of metals
Materials: wires, nails, hammer, piece of cloth.
1. Wrap the material to be test in a heavy plastic or cloth to avoid pieces flying from the material.
2. Place the material on a flat hard surface
3. Use a harmer to pound the material flat
4. Record your observations as malleable or non-malleable.
Metals can have their shapes changed relatively easily in two different ways i.e.
Malleable: can be hammered into sheets or
Ductile: can be drawn into rods and wires
As the metal is beaten into another shape the delocalised electron cloud continues to bind the “ions” together.
The positive ions in the lattice are all identical. So the planes of ions can slide easily over one another attractive forces in the lattice are the same whichever ions are adjacent and remain the same throughout the lattice
Some metals, such as gold, can be hammered into sheets thin enough to be translucent.
Note: If a material is pounded into a flat shape it is called malleable but if it breaks or does not change it is called non-malleable.
d) Metal appears hiny/lustrous
Demonstration of shininess in metals
1. Hold a small piece of sodium metal using forceps.
2. Place it on a hard surface and cut it into two parts
3. Observe the cut surface. What do you observe?
4. Look at the surface of aluminium sheets, how does it appear?
Explain what makes metal surfaces appear shiny/luster.
Light is composed of very small packages of electromagnetic energy called photons. We are able to see objects because light photons from the sun (or other light source) reflect off of the atoms within the object and some of these reflected photons reach the light sensors in our eyes and we can see the objects.
When photons of light hit the atoms within an object three things can happen:
The photons can bounce back from the atoms in the object, can pass through an object such as glass or can be stopped by the atoms within the object.Objects that reflect many photons into our eyes make the objects appear shiny. Objects that absorb photons and reflect less photons appear dull or even dark black to our eyes.
Did you know? Of all of the metals, aluminium and silver are the shiniest to our eyes. Gold is also one of the more shiny metals. However, gold is not as shiny as silver and aluminium. Mercury, a liquid metal, is also shiny and special telescope mirrors have been made of mercury.
(e)Melting and boiling points
1. Why do metals have variable melting points?
2. Why do metals have high melting points compared to non-metals?
Melting point is a measure of how easy it is to separate individual particles. In metals it is a measure of how strong the electron cloud holds the positive ions. The ease of separation of ions depends on the electron density and Ionic / Atomic size.
Melting point increases across the period (from Na---Al), the electron cloud density increases due to the greater number of valence electrons contributed per atom. As a result the ions are held more strongly. Hence more energy will be needed to separate the ions as shown in the table 3.4.
Table 3.4: Valence electron density as a factor that influence melting and boil-ing points of metal
Conversely, as the size of the atom/ion increases down the group the melting and boiling points decrease as shown in the table 3.5
Table 3.5. Atomic/ionic size as a factor that influences melting and boiling points of metals
(*) Sevenair & Burkett(1997), Introductory Chemistry, Investigating the Molecular Nature of Matter; WCB.
N.B: You may find, in another book, different values of atomic/ionic sizes, due to different methods used for the determination the sizes.
3.3.3. Factors affecting the strength of metallic bonds.
1. Explain different strengths of metallic bonds in different metals?
2. Compare the metallic strength of the following metals:
(i) Sodium and magnesium
(ii) Sodium and potassium
The three main factors that affect the metallic bond are:
•Number of protons/ Strength of nuclear attraction: The more protons the stronger the force of attraction between the positive ions and the delocalized electrons
•Number of delocalized electrons per atom: The more delocalized electrons the stronger the force of attraction between the positive ions and the delocalized electrons
Size of atom: The smaller the atom, the stronger the force of attraction between the positive ions and the delocalized electrons and vice-versa, the larger the atom, the weaker the force of attraction between the positive ions and the delocalised electrons.
The strength increases across a period from left to right because:
The atoms have more protons. There are more delocalized electrons per atom.Electrons are added to the same energy level. Group1 elements have 1 electron in their outer shells and so contribute 1 electron to the sea of electrons, Group 2 elements contribute 2 electrons per atom, and Group 3 elements contribute 3 electrons per atom
If the atoms/ions are smaller; there is therefore a greater force of attraction between the positive ions and the delocalized electrons.
In group 1 elements, the melting and boiling points decrease as the size increases hence attraction between the delocalized electrons and metal cations decreases down the group as shown table 3.6
Table 3.6: Variation of melting and boiling points of group 1 elements
Checking Up 3.6
1. Look at the table below, which shows the thermal conductivity of a number of different materials, and then answer the questions that follow:
The higher the number in the second column, the better the material is at conducting heat (i.e. it is a good thermal conductor). Remember that a material that conducts heat efficiently will also lose heat more quickly than an insulating material. Use this information to answer the following questions:
1. Name two materials that are good thermal conductors.
2. Name two materials that are good insulators.
3. Explain why:
a. cooler boxes are often made of polystyrene
b. Homes that are made from wood need less internal heating during the cold months.
c. Igloos (homes made from snow) are so good at maintaining warm temperatures, even in freezing conditions.
d. Houses covered by iron sheets and houses covered by tiles can be compared in their capacity of keeping the interior of the house hot of fresh during a sunny and hot day.
4. Magnesium has a higher melting and boiling point than sodium. This can be explained in terms of the electronic structures, the packing, and the atomic radii of the two elements.
a. Explain why each of these three things causes the magnesium melting and boiling points to be higher. b. Explain why metals are good conductors of electricity.
c. Explain why metals are also good conductors of heat.
5. Pure metals are usually malleable and ductile.
a. Explain what those two words mean. If a metal is subjected to a small stress, it will return to its original shape when the stress is removed. However, when it is subjected to a larger stress, it may change shape permanently. Explain, with the help of simple diagrams why there is a different result depending on the size of the stress.When a piece of metal is worked by a blacksmith, it is heated to a high temperature in a furnace to make it easier to shape. After working it with a hammer, it needs to be re-heated because it becomes too difficult to work. Explain what is going on in terms of the structure of the metal. Why is brass harder than either of its component metals, copper and zinc?
3.4. End Unit Assessment
1. Choose from a list of words and fill in the missing words in the text below:
List of words:
Conduct electricity, electrodes, electrolysis, electrostatic attraction, free electrons, good conductivity, great malleability, high density, high melting points, ionic bond,metal, negative ion, non-metal, positive ion,regular crystal shape and attractive forces.
Te x t :
Metals have a layered structure of ................... in fixed positions but between them are oppositely charged ............... that can move around at random between the metal atoms. There is a strong ...................................... between these oppositely charged particles which gives them ..........................The strong forces also give a ....................... making the average ........... heavier than an average ................ The presence of ........................... in the structure keeps the bonding intact when metals are bent or hammered giving them.......................................... Also, these ....................give metals ..............as regards heat and electricity.When electrons are transferred from (usually) ............ atom (e.g. sodium) to................. atom (e.g. chlorine) an ionic bond is formed. Sodium loses an electron to form a singly charged ........................... and chlorine gains an electron to form a singly charged negative ion. In an ionic compound, the ionic bond is the electrostatic attraction between the neighbouring positive ions and negative ions.
The strong forces holding this giant ionic lattice together give these ionic compounds............................ and...................................................................When ionic compounds are melted they are found to ................ in a process called ...........using electrical contacts called............ In this process, move to the negative electrode (cathode) and metalsare released. At the same time, ....................move to the positive electrode (anode) and ................ are formed. Research from the internet or text books to find out other physical properties of metals and ionic compounds that are not mentioned above.
Answer these questions by choosing the best alternative repre-sented by letters from A, B, C and D.
1. Metals lose electrons from their lattice to become
a. positive ions
b. negative ions
d. non- metals
2. Neither ions nor electrons are free to move in
c. ionic solids
d. All of the above
3. Attractive forces between metal ions and delocalized electrons can be weakened or overcome by
b. high temperature
d. All of the above
4. Metals are good conductors due to
a. ionic lattices
b. crystalline lumps
c. mostly solids
d. delocalized electrons
5. Most atoms adopt one of three simple strategies to achieve a filled shell. Which of the following is NOT one of these strategies?
a. They accept electrons
b. They share electrons
c. They give away electrons
d. They keep their own electrons
6. Which of the following is NOT a type of chemical bond?
7. In metallic bonding...
a . One atom takes the outer shell electrons from another atom.
b. A couple of atoms share their electrons with each other.
c. Some electrons are shared by all the atoms in the material.
d. Bonding takes place between positively charged areas of one atom with a negatively charged area of another atom.
8. Which of the following is NOT a characteristic of metals?
A. Shiny /lustre
B. Brittle/Shatters easily
C. Conducts electricity
9. When two or more metal elements are combined they form an...
c. Covalent bond
10. Sulphur is a solid non-metallic element at room tempera-ture, so it is?
a. A good conductor of heat
b. A substance with a low melting point
c. Easily bent into shape
d. A good conductor of electricity
11. Copper is a metallic element so it is likely to be a?
a. substance with a low boiling point
b. poor conductor of electricity
c. good conductor of heat
d. substance with a low melting point
12. Sodium chloride is a typical ionic compound formed by combining a metal with a non-metal. Sodium chloride will?
a. have a low melting point
b. consist of small NaCl molecules
c. conduct electricity when dissolved in water
d. not conduct electricity when molten
13. Copper is a metallic element so it is likely to be a?
a. Substance with a shiny surface
b. Poor conductor of electricity
c. Poor conductor of heat
d. Substance with a low melting point
14. When an ionic bond is formed between atoms of different elements?
a. Protons are transferred
b. Electrons are transferred
c. Protons are shared
d. Electrons are shared15. Sodium chloride has a high melting point because it has:
a. Many ions strongly attracted together
b. Strong covalent double bonds
c. A giant covalent 3-dimentional structure
d. Molecules packed tightly together
16. Which substance is likely to have a giant ionic structure:
a. Melts at 1400oC, insoluble in water, good conductor of electricity either when solid or molten
b. Melts at 2800oC, insoluble in water, non-conductor of electricity when molten or solid
c. Melts at 17oC, insoluble in water, non-conductor of electricity either when solid or molten
d. Melts at 2600oC, dissolves in water, non-conductor of electricity when solid,undergoes electrolysis in aqueous solution
17. Sodium chloride conducts electricity when:
a. Solid or molten
b. Solid or in solution
c. Molten or in solution
d. Non of the above
18. The structure of magnesium oxide is a
a. Giant covalent lattice
b. Giant ionic lattice
c. Simple ionic lattice
d. All the above19. What is the formula for magnesium chloride (contains Mg2+ and Cl? ions)?a. MgClb. Mg22Clc. MgCl2d. MgCl320. Why does sodium chloride have a lower melting point than magnesium chloride?
a. Its positive ions are smaller and have a smaller charge
b. Its positive ions are larger but have a smaller charge
c. Its positive ions are smaller but have a larger charge
d. All the above
21. Explain the conductivity of sodium chloride
a. It conducts electricity when molten because it contains free electrons
b. It conducts electricity when molten because sodium has metallic bonding
c. It conducts electricity when molten because its ions are free to move.
d. None of the above
Short and long answer questions
22.(a) Explain why the lattice dissociation enthalpy of NaBr is a bit less than that of NaCl.
(b) Explain why the lattice dissociation enthalpy of MgO is about 5 times greater than that of
23.a) The table (using figures for lattice energies from gives experimental and theoretical values for the silver halides.(The values are listed as lattice dissociation energies.) compare the values and give a detailed explanation.
(b) For AgF, the experimental and theoretical values are very close. What does that show?UNIT 3: FORMATION OF IONIC AND METALLIC BONDS
- UNIT 4: COVALENT BOND AND MOLECULAR STRUCTUREUNIT 4: COVALENT BOND AND MOLECULAR STRUCTURE
Key Unit Competence
Demonstrate how the nature of the bonding is related to the properties of covalent compounds and molecular structures.
By the end of this unit, I will be able to:
•Define octet rule as applied to covalent compounds.
•Explain the formation of covalent bonds and describe the properties of covalent compounds. •Describe how the properties of covalent compounds depend on their bonding.
•Explain the rules of writing proper Lewis structures
•Draw different Lewis structures
•State the difference between Lewis structures from other structures.
•Apply octet rule to draw Lewis structures of different compounds.
•Make the structures of molecules using models.
•Write the structures of some compounds that do not obey octet rule.
•Explain the formation of dative covalent bonds in different molecules.
•Compare the formation of dative covalent to normal covalent bonding.
•Describe the concept of valence bond theory.
•Relate the shapes of molecules to the type of hybridization.
•Differentiate sigma from pi bonds in terms of orbital overlap and formation.
•Explain the VSEPR theory.•Apply the VSEPR theory to predict the shapes of different molecules/ions.
•Predict whether the bonding between specified elements will be primarily covalent or ionic.
•Relate the structure of simple and giant molecular covalent compounds to their properties.•Describe simple and giant covalent molecular structures.
•Describe the origin of inter-molecular forces.
•Describe the effect of inter and intra molecular forces on the physical properties of certain molecules. •Describe the effect of hydrogen bonding in the biological molecules.
•Relate the physical properties to type of inter and intra molecular forces in molecules.
•Compare inter and intra molecular forces of attraction in different molecules.
4.1. Overlap of atomic orbitals to form covalent bonds
1. Using dot and cross diagrams, show how a covalent bond is formed in:
(a) Hydrogen molecule (H2)
(b) Hydrogenchloride molecule (HCl)
(c) Chlorine molecule (Cl2).
In UNIT 3, you have learnt that atoms have different ways of combination to achieve the stable octet electronic structure; two of those ways of combination led to the formation of ionic bond and metallic bond. But what happens where the two combining atoms need electrons to complete the octet structure and no one is willing to donate electrons? For example the combination of 2 hydrogen atoms or the combination of 2 chlorine atoms?
When this happens, the combining atoms share a pair of electrons where each atom brings or contributes one electron. In other words there is an overlapping of two orbitals, one orbital from one atom, each orbital containing one electron (see Fig.4.1): this bond is called “Covalent bond”. The attraction between the bonding pair of electrons and the two nuclei holds the two atoms together.
The covalent bond is a bond formed when atoms share a pair of electrons to complete the octet. Similarly, people need each other irrespective of their race, economic, political and social status for the success of human race. Some compounds that exist in nature such as hemoglobin in our blood, chlorophyll in plants, paracetamol,
esponsible for transport of oxygen, green color in plants and as pain killer respectively are made of the covalent bond. The covalent bonds mostly occur between non-metals or between two of the same (or similar) elements.Two atoms with similar electronegativity do not exchange an electron from their outermost shell; the atoms instead share electrons so that their valence electron shell is filled.
In general, covalent bonding occurs when atoms share electrons (Lewis model), concentrating electron density between nuclei. The build-up of electron density between two nuclei occurs when a valence atomic orbital of one atom combines with that of another atom (Valence bond theory).In Valence bond theory, the bonds are considered to form from the overlap of two atomic orbitals on different atoms, each orbital containing a single electron.
The orbitals share a region of space, i.e. they overlap. The overlap of orbitals allows two electrons of opposite spin to share the common space between the nuclei, forming a covalent bond.
As example, we can consider the case of hydrogen molecule. The molecule of hydrogen is made of two hydrogen atoms bonded together, this is made by the two spherical 1s orbitals overlap, and contains two electrons with opposite spins (Figure 4.1)
These two electrons are attracted to the positive charge of both the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
The figure (Figure 4.1) shows the distance between the two nuclei. If the two nuclei are far apart, their respective 1s-orbitals cannot overlap and no covalent bond is formed. As they move closer each other, the orbital overlapping begins to occur, and a bond starts to form.
The examples below represent different atoms overlapping in order to form covalent bonds.
Example 1: Water formation
Water molecule is formed by two atoms of hydrogen and one atom of oxygen, its formula is H2O.
The steps of water formation are indicated in the Figure 4.2
The oxygen atom has 6 electrons in the valence shell and needs two electrons to complete its outer shell. Similarly, each hydrogen atom has one electron valenceand needs one electron to complete its outer shell. Therefore, oxygen can only share 2 of its 6 electron valence otherwise it will exceed the maximum of 8; the two lone pairs remain.
Example 2: Formation of hydrogen chloride (Figure 4.3)
The molecule of hydrogen chloride is made by covalent bond between one chloride atom and a one hydrogen atom as shown below.
The chlorine atom has 7 electrons in the valence shell and needs one electron to complete its outer shell. Similarly, the hydrogen atom has one electron valence and needs one electron to complete its outer shell. The two atoms share one single electron in their outer shell to make a covalent bond in order to become stable.
Example 3: Formation of ammonia (NH3) (Figure 4.4)
The ammonia is a compound formed by one atom of nitrogen and three atoms of hydrogen as indicated below.
•Nitrogen atom needs three electrons to complete its outer shell and each hydrogen atom needs one electron to complete its outer shell. Then, nitrogen can only share 3 of its 5 electrons in order not to exceed the maximum of 8. A lone pair remains.
4.1.1 Properties of covalent molecules
Covalent molecules are chemical compounds in which atoms are all bonded together through covalent bonds. The covalent compounds possess different properties and some are emphasized below.•Covalent compounds exist as individual molecules, held together by weak van der Waals forces.
•Due to the weak van der Waals forces that hold molecules together, covalent compounds have low melting and boiling points; because the weak forces between molecules can be broken easily to separate the molecules. That is why covalent compounds can be solid, liquid and gaseous at room temperature.
•Covalent compounds do not display the electrical conductivity either in pure form or when dissolved in water. This can be explained by the fact that the covalent compounds do not dissociate into ions when dissolves in water.
•Generally non-polar covalent compounds do not dissolve in water; but
many polar covalent compounds are soluble in water( a polar solvent)
Non-polar covalent compounds are soluble in organic solvents (themselves non-polar covalent).
The two statements above are at the origin of the say by chemists: “Like dissolves like”.
Checking up 4.1
Using dot and cross diagram draw the diagrams to show the formation of bonds in the following molecules:
1 a) Ammonia (NH3) (b) carbon dioxide gas ( CO2) (c) Nitrogen molecule (N2) (d) tetra chloromethane (CCl4)
2. Explain the properties of covalent molecules compared to ionic compounds.
4.2. Theories on the formation of covalent bond
4.2.1. Lewis theory and structures
1. Draw the diagrams indicating only the valence electrons of the following :Chlorine molecule (Cl2), Carbon atom (C), Phosphorus atom (P)
2. Draw the diagram to show how all electrons are shared in a molecule of
(i) NH3 indicating all unshared electrons.
3. Identify the common feature possessed by the diagrams drawn above in 2.
4. Suggest the name given by those diagrams drawn above.
The Lewis structure is a representation of covalent molecules or polyatomic ions where all valence electrons are distributed to the bonded atoms as shared electron pair (bond pairs) or unshared electron (lone pair).We then combine electrons to form covalent bonds until we come up with a Lewis structure in which all of the elements (with the exception of the hydrogen atoms) have an octet of valence electrons.
The representation of Lewis structures follows different steps. Let us use the nitrate ion (NO3-) as a typical example. Determine the total number of valence electrons in a molecule or ion.
1) Determine the total number of valence electrons in a molecule or ion.
2) Draw a skeleton for the molecule or ion which connects all atoms using only single bonds. In simple molecules, the atom with the most available sites for bonding is usually placed at the center. The number of bonding sites is determined by considering the number of valence electrons and the ability of an atom to expand its octet. As you will progress in your study of chemistry, you will be able to recognise that certain groups of atoms prefer to bond together in a certain way!
(3) Of the 24 valence electrons in NO3-, 6 are required to make the skeleton. Consider the remaining 18 electrons and place them so as to fill the octets of as many atoms as possible (start with the most electronegative atoms first then proceed to the more electropositive atoms).
(4) Are the octets of all the atoms fulfilled? If not then fill the remaining octets by making multiple bonds (make a lone pair of electrons, located on a more electronegative atom, into a bonding pair of electrons that is shared with the atom that is electron deficient).
Check that you have the lowest formal charge (F.C.) possible for all the atoms, without violating the octet rule; F.C. = (valence e-) - (1/2 bonding e-) - (lone electrons).
Thus the Lewis structure of NO3- ion can be written in the following ways:
The 4 steps are used to find Lewis structures. If the octets are incomplete, and more electrons remain to be shared, move one electron per bond per atom to make another bond. Note that in some structures there will be empty orbitals (example: the B of BF3), or atoms which have ten electrons (example: P in PF5) or twelve electrons (example: S in SF6).
The first step in this process involves the calculation of the number of valence electrons in the molecule or ion. In the case of a neutral molecule, this is nothing more than the sum of the valence electrons on each atom. If the molecule carries an electric charge, we add one electron for each negative charge or subtract an electron for each positive charge.
In Lewis structure, the least electronegative element is usually the central element, except H that is never the central element, because it forms only one bond.
Another way of finding Lewis structure
1. Calculate n (the number of valence (outer) shell electrons needed by all atoms in the molecule or ion to achieve noble gas configurations for instance,NO3-, n=1× 8(for N atom) + 3×8 (for O atom) = 32 electrons.
2. Calculate A, number of electrons available in the valence (outer) shells of all the atoms. For negatively charged ions, add to this number the number of electrons equal to the charge of the anions. For cations you subtract the number of electrons equal to the charge on the cation.
For instance: NO3-,
A= 1×5(for N) +3×6 (for O atom) +1(for -1 charge) = 5+18+1=24 electrons.
3. Calculate S, total number of electrons shared in the molecule or ion, using the relationship
S = n-A
S= n-A= 32-24 =8 electrons shared (4pairs of electron shared)
4. Place S electrons into the skeleton as shared pair’s. Use double and triple bonds only when necessary. Lewis formulas may be shown as either dot formula or dash formulas.
5. Place the additional electrons into the skeleton as unshared (lone) pairs to fill the octet of every A group element (except H, which can share only 2 electrons) check that the total number of valence electrons is equal to A from step 2.
Check that you have the lowest formal charges (f.c.) possible for all the atoms, without violating the octet rule; F.C. = (valence e-) - (1/2 bonding e-) - (lone electrons).
Thus the Lewis structure of NO3- ion can be written in the following ways:
Checking up 4.2(a)
1 .What do you consider when drawing Lewis structures of different molecules?
2. Draw the Lewis diagram for the following compounds(a)PCl3 (b) H2S (c) CH3Cl (d) C2H2
3.What is meant by the term Lewis structure?
4. Using water differentiates between Lewis structures from other structures.
5. Deduce the importance of Lewis structures in covalent molecules.What critics can you make to those strcutures?
Exceptions to the octet rule
1. Draw the Lewis structure of aluminium chloride (AlCl3)?
2. How many electrons surround the aluminium atom in the centre?
3. Does it obey the octet rule? And why?
There are three general ways in which the octet rule doesn’t work:
•Molecules with an odd number of electrons
•Molecules in which an atom has less than an octet
•Molecules in which an atom has more than an octet
a. Odd number of electrons
Consider the example of the Lewis structure for the molecule nitrous oxide (NO):Total electrons: 6+5=11
There are currently 5 valence electrons around the nitrogen. A double bond would place 7 around the nitrogen, and a triple bond would place 9 around the nitrogen. We appear unable to get an octet around each atom.(b) Less than an octet (most often encountered with elements of Boron and Beryllium)Consider the example of the Lewis structure for boron trifluoride (BF3):1. Add electrons (3 x 7) + 3 = 242. Draw connectivitiesChecking up 4.2(b)1. Draw the Lewis diagram for the following species(a)SOCl2 (b) ICl4-(c) SO42- (d) CH3OH2. During sharing of electrons some elements can accommodate more than eight electrons in their outer shell while other becomes comfortable with less than eight electrons on the observation with reference to BeCl2 and PCl5.4.3 Coordinate or dative covalent bonding and propertiesActivity 4.31. Write the electronic configuration of (i) nitrogen atom (ii) aluminium atom (ii) oxygen atom using s, p, d.... notation.2.Draw the Lewis structures of water molecule(H20), Aluminium Chloride (AlCl3), ammonia molecule (NH3) and carbon dioxide (CO2)3. During formation of hydroxonium ion (H3O+) Water (H2O) combine with Hydrogen ion (H+). Draw the electronic structure to show how the electrons are shared.4. Aluminium chloride in vapor phase exists as Al2Cl6. Draw the Lewis structure indicating all the bonds.5.Using diagrams, suggest how sharing of electrons take place during the formation of ammonium ion(NH4+) from ammonia (NH3) and (H+) ion4.3.1. Co-ordinate or dative covalent bonding and propertiesA dative covalent bond, or coordinate bond is another type of covalent bonding. In this case, the shared electron pair(s) are completely provided by one of the participants in the union, and not by contributions from the two of them. The contributors of the shared electrons are either neutral molecules which contain lone pair(s) of electrons on one of their atoms, or negatively charged groups with free pairs of electrons to donate for sharing.Note: Once the coordinate bond is formed, the four N-H bonds in NH4+ ion are identical and cannot be distinguished.Another example comprising co-ordinate bond is the reaction between ammonia and boron trifluoride. Boron trifluoride (BF3) is an electron deficient (Lewis acid) because it has 3 pairs of electrons at its bonding level instead of four pairs. For that kind of reaction, the ammonia is used to supply this extra lone pair(Lewis base). A coordinate bond is formed where the lone pair from the nitrogen moves towards the boron.4.3.2. Dative covalent bonding in complex ionsA complex ion consists of a central metal cation bonded to a group of atoms, molecules or ions which are known as ligands through co-ordinate bonds.The solid copper (II) hydroxide which was initially formed reacts with the excess ammonia (which acts as ligands) to form the water soluble tetra ammine copper (II) complex as shown below.In complex ion formation, the central metal cation must have empty orbitals while the ligands must have a lone pair of electrons for the co-ordinate bond to be formed as indicated below.Checking up 4.31. Draw diagrams to show how the dative covalent bond when water combines with Hydrogen ion to form hydroxonium ion (H3O+)2. Draw the Lewis structures for the following speciesNO2 (b) N2O4 (c) AlCl33. Suggest the properties of dative covalent bond and explain in which way it differs from normal covalent bonds.4. Use the Lewis diagram for aluminium chloride AlCl3 drawn in 2(c) above to identify and label the dative covalent bonds and normal covalent bonds.5. What do you understand by the term co-ordinate (dative covalent bond)?6. Describe with examples how a dative covalent bond is formed using a specificexample of your choice.7. Biological molecules such as hemoglobin contain iron cation in the center surrounded by oxygen ligands. Suggest reasons why is this important to us?4.4 Valence bond theory (VBT)Activity 4.41. Draw the diagram to show how the single and double bonds are formed in:(i)Oxygen molecule (O2) (ii) nitrogen molecule (N2) (iii) ethene molecule(C2H4)2. Use valence bond theory to explain the formation of the above molecules.3. Explain how sigma and pi bonds are formed using valence bond theory.This theory of covalent bond is based on the concept that electrons are located around the atomic nucleus in orbitals. Then when two atoms approach each otherto share electrons, their two orbitals, each containing one electron, overlap in the region between the two nuclei to form a pair of electrons. That pair of electron is attracted by each nucleus and this force of attraction maintains the two atoms together; it is this force that is called chemical bond and in this case, it is qualified as ” covalent”.The two examples above have in common that the concentration of the bonding electrons are on the inter-atomic (inter-nuclei) axis; such bonds are called “sigma bond”, represented by the symbol “σ”.As you can observe, p orbitals overlap head-to-head or axially, they form a σ bond.(3) Formation of O2 molecule (O=O)When O2 forms, two orbitals in the same orientation, e.g. px, overlap head-to-head to form a σ bond. The other orbitals, e.g. py, will overlap side-by-side or laterally:As you notice, the density of bonding electrons is not on the inter-nuclei axis, it is rather located outside the axis but surrounding it. This kind of covalent bond is called “ Pi bond”, represented by the symbol “π”. Hence the double bond O=O is made of two covelent bonds: a σ bond and a π bond.Due to the position of their electrons density in relation with the two nuclei, σ bond participates in maintaining the two nuclei together more strongly than the π bond; that is why σ bond is stronger than π bond. In addition, π bond cannot exist alone, it exists only where there is a double or triple bond. Hence, in a double or triple bond, there is one σ bond and one or two π bonds respectively.Checking Up 4.41. Describe the aspects and postulates of valence bond theory(VBT)2. Use VBT to explain the formation of single(sigma) and double (pi)bonds(a) Explanation of lateral overlap of atomic orbitals and(b) Explanation of head-to-head overlap of atomic orbitals4.5 Valence Shell Electron Pair Repulsion Theory (VSEPR) theoryActivity 4.51. What is VSEPR in full?2. Draw the Lewis structures of water (H2O) , ammonia (NH3) and identify the areas with high density.3. In groups carry out the activityi) Get students into groups of 2 or 3 and pass out balloons and strings and scissorsii) Ask students to create visuals of molecular orbitals with balloons Blow up 2 balloons and tie them togetheriii) Do the same thing with 3 balloons, 4 balloons, 5 balloons, and 6 balloons iv) Meet back as a class and have each group show their models and name the molecular geometryWhy VSEPR Theory?There is no direct relationship between the formula of a compound and the shape of its molecules; in other words, looking at chemical formula of compound does not allow you to know the shape or geometry of the molecule.The Lewis structures of molecules show molecules as having planar or bidimensional shapes; yet it has been proven that some molecules are tridimensional. Even the VBT theory that shows that tridimensional molecules are possible, it can’t justifiy some molecular structures such as for example the tetrahedral shape of methane molecule CH4.In order to predict the geometry of molecules, Nyholm and Gillespie developed a qualitative model known as Valence Shell Electron Pair Repulsion Theory (VSEPR Theory)The basic assumptions of this theory are summarized below.1) The electron pairs in the valence shell around the central atom of a mol-ecule repel each other and tend to orient in space so as to minimize the repulsions and maximize the distance between them.2) There are two types of valence shell electron pairs such as Bond pairs and Lone pairs.Bond pairs are shared by two atoms and are attracted by two nuclei. Hence they occupy less space and cause less repulsion.Lone pairs are not involved in bond formation and are in attraction with only onenucleus. Hence they occupy more space. As a result, the lone pairs cause more repulsion.The order of repulsion between different types of electron pairs is as follows:Lone pair - Lone pair > Lone Pair - Bond pair > Bond pair - Bond pairThe bond pairs are usually represented by a solid line, whereas the lone pairs are represented by a lobe with two electrons.3) In VSEPR theory, the multiple bonds are treated as if they were single bonds. The electron pairs in multiple bonds are treated collectively as a single su-per pair. The repulsion caused by bonds increases with increase in the number of bonded pairs between two atoms i.e., a triple bond causes more repulsion than a double bond which in turn causes more repulsion than a single bond.4) The shape of a molecule can be predicted from the number and type of valence shell electron pairs around the central atom.The principle of the VSEPR is based on the idea that: the most stable structure of a molecule is the one where the electron pairs are far away one from another in order to minimize the repulsions between the pairs of electrons surrounding the central atom.The VSEPR theory assumes that each atom in a molecule will achieve a geometry that minimizes the repulsion between electrons in the valence shell of that atom. The use of VSEPR involves the following steps:•Draw a Lewis structure for the ion or molecule in question.•The shape is based on the location of the nuclei in a molecule, so double and triple bonds count as one shared pair when determining the shape of the molecule•Locate the shared pairs and lone pairs on the central atom•Determine the shape based on the above considerations.4.6. Hybridisation and types of HybridisationActivity 4.61. Write the electronic configuration of carbon and hydrogen using s,p, d.. Notation.2. Use the electronic configurations above to identify the orbitals that contain electrons used during the formation of methane.3. Use the knowledge of overlap of atomic orbitals to indicate how orbitals overlap in formation of hydrogen chloride, methane and beryllium chloride and predict the shapes of the molecules.But why hybridization?The concept of hybridization has been developed by Linus Pauling to explain shapes of molecules that cannot be explained by overlapping of pure atomic orbitals.Hybridization is a mathematical operation consisting in mixing atomic orbitals (which are mathematical functions) to generate new orbitals called “hybridized atomic orbitals”. This textbook will not develop the mathematical aspect of hybridization, only the results of that operation will be used.1. sp HybridizationThis type of hybridization involves mixing of one ‘s’ orbital and one ‘p’ orbital of comparable energy to give a new hybrid orbital known as an sp hybridized orbital. When two orbitals are hybridized, they generate two new sp hybridized orbitals. sp hybridization forms a linear shape or structure where the two new orbitals form an angle of 180o.Example 1. The element Be has 4 electrons and its electronic configuration in ground state is 1s2 2s2. BeCl2 is linear, Cl-Be-Cl; this cannot be explained by using pure atomic orbital of Be.But through sphybridisation on Be atom, two new atomic hybridized orbitals, sp, are generated. The two s2 electrons can now be distributed one in each hybridized orbital and this allows Be to form two linear sigma bonds around, by overlapping its 2 sp hybridized orbital with p orbitals from 2 chlorine atoms as shown below.In this way, the linear shape of Cl-Be-Cl (BeCl2) can be explained.2. sp2 HybridizationThis type of hybridization involves mixing of one ‘s’ orbital and two ‘p’ orbitals to give 3 new hybrid orbitals known as sp2.This kind of hybridization results into a trigonal planar structure (equilateral triangle), with an angle of 120o between the hybridized orbitals, as shown below.This is the case of ethylene (C2H4). During the formation of ethylene molecule, each carbon atom undergoes sp2 hybridization by mixing one 2s orbital and two 2p orbitals to give three half-filled sp2 hybrid orbitals oriented in trigonal planar symmetry. There is also one half-filled un-hybridized 2pz orbital on each carbon perpendicular to the plane of sp2 hybrid orbitalsThis is also found in the case of ethylene (C2H4).During the formation of ethylene molecule, each carbon atom undergoes sp2 hybridization by mixing one 2s orbital and two 2p orbitals to give three half-filled sp2 hybrid orbitals oriented in trigonal planar symmetry. There is also one half-filled un-hybridized 2pz orbital on each carbon perpendicular to the plane of sp2 hybrid orbitalsThe carbon atoms form a σsp2-sp2 bond with each other by using sp2 hybrid orbitals. A πp-p bond is also formed between them due to lateral overlapping of un-hybridized 2pz orbitals. Thus there is a double bond (σsp2-sp2 and πp-p) between two carbon atoms. Each carbon atom also forms two σsp2-s bonds with two hydrogen atoms. Therefore, ethylene molecule is planar with angles equal to 120o.3. sp3 HybridizationThis type involves mixing of one ‘s’ orbital and three ‘p’ orbitals to give four new hybrid orbitals known as sp3. Hybridization of one s and three p orbitals forms a tetrahedral with an angle of 109.50 between the new orbitals; this angle is often called tetrahedral angle.After hybridization, the four valence electrons are distributed in the four orbitals one by one and this allow the carbon atom to form 4 identical C-H sigma bonds. That is how the tetrahedral structure of methane, CH4, can be explained.During the formation of methane molecule, the carbon atom undergoes sp3 hybridization by mixing one ‘2s’ and three 2p orbitals to form four half-filled sp3 hybrid orbitals, which are oriented in tetrahedral symmetry in space around the carbon atom. Each of these sp3 hybrid orbitals forms a σsp3-s bond with one hydrogen atom. Thus carbon forms four σsp3-s bonds with four hydrogen atoms. Methane molecule is tetrahedral in shape with 109o50’ bond angle.4. sp3d HybridizationThis type of hybridization involves mixing of one ‘s’ orbital, three ‘p’ orbital’s and one ‘d’ orbital to give a 5 new hybrid orbitals known as sp3d hybridized orbitals.The 5 new hybridisedorbitals form atrigonalbipyramidalgeometry as shown in the example below. Three of the hybrid orbitals lie in horizontal plane at angle of 120° to one another.Example: Formation of Phosphorus pentachloride (PCl5)The ground state electronic configuration of phosphorus atom is: 1s2 2s22p63s23px13py13pz1. The formation of PCl5 molecule requires 5 unpaired electrons. Hence the phosphorus atom undergoes excitation to promote one electron from 3s orbital to one of empty 3d orbital.Thus the electronic configuration of ‘P’ in the excited state is 1s2 2s22p6 3s13px13py13pz13d1.5. sp3d2 HybridizationInthis type of hybridization one s, three p and two d-orbitals undergo intermixing to form six identical sp3 d2 hybrid orbitals. These six orbitals are directed towards the corners of an octahedron and lie in space at an angle of 90° to one another.Example: Formation of Sulfur hexaflouride (SF6)The electronic configuration of ‘S’ in ground state is 1s2 2s22p6 3s23px23py13pz1. In SF6 molecule, there are six bonds formed by sulfur atom. Hence there must be 6 unpaired electrons. However there are only 2 unpaired electrons (px2py1pz1) in the ground state of sulfur. Through sp3d2 hybridation, the 6 electrons in 3s23px23py13pz1can be redistributed in the new 6 hybridized orbitals to get one electron in each new orbital:After hybridization of its orbitals, Sulfur atom forms six σ bonds with 6 fluorine atoms by using these sp3d2 orbitals. Each fluorine atom uses its half-filled 2pz orbitals for the bond formation. SF6 is octahedral in shape with bond angles equal to 90o.Summary about hybridizationAttention: The regular shapes or geometries in this table are found when the central atom is surrounded by identical atoms. But when the entities surrounding the central atom are different, we get deforemed shapes.Examples:(i) Ammonia molecule, :NH3, results from sp3 hybridisation. This molecule should be tetrahedral. But since one tretrahedral summit is occupied by a lone pair of electrons, this causes ammonia molecule to adopt a pyramidal shape, which is a deformed tetrahedral. The angle between N-H bonds becomes 107.3O(ii) Water molecule, H2Ö:, is also a results of sp3 hybridisation of oxygen atom. But in the molecule, two tetrahedral summits are occupied by lone pairs of electrons and this causes the deformation of tetrahedral shape into a bent shape with the characteristic angle of 104.5O between the two O-H bonds.Checking up 4.51.What is the importance of hybridization of atomic orbitals?Using carbon atom describe the three types of hybridization it can undergo.Give some examples of molecules, other than the ones illustrated above as examples that are sp, sp2, sp3 and sp3d hybridized.2.What criteria do you consider when predicting the type of hybridization3.What are hybrid orbitals?4.What are characteristics of hybrid orbitals?What is meant by the term “hybridising” of atomic orbitals?5. Why atomic orbitals in a given atom undergo hybridization?6. Explain how the type of hybridization is related to the shape of the molecule7. Draw dots-and-crosses diagrams (showing outer electrons only) for the following:a) Oxygen, O2b) Carbon dioxide, CO2c) Methanal (formaldehyde). You are unlikely to be familiar with this compound yet. Its molecular formula is CH2O. You will have to play around to work out how it is joined up. Make sure that you make the maximum possible number of bonds.d) Ethene, C2H48.a) Write down the electronic structure of carbon in s and p notation.b) Before the carbon atoms bond to each other and to hydrogen atoms to make ethene, they undergo hybridization. What does this mean, and how are the electrons arranged once this has happened?c) We think of the next stage as the formation of sigma bonds between the two carbon atoms andthe carbons and hydrogens. What are sigma bonds, and how are they formed?d) Finally, we picture the formation of a pi bond between the two carbon atoms. How does thisform, and how is it different from a sigma bond?4.7 Polar covalent bondsActivity 4.71. Can you define the term electronegativity?2. How is electronegativity related to polarity of the compound?3. How does the polarity of a given molecule affect its physical properties?4. Can you describe the general trends of electronegativity across and down the groups in the periodic table?5. What is meant by the term dipole and Net dipoleWhat happens if shairing of the bonding pair of electrons between the two atoms forming the bond is not equal? For instance when two different non-metal elements such as hydrogen and bromine combine?In this case, there is unequal sharing where the more electronegative element takes a bigger share of the bonding pair of electrons (Fig. 4.7)In a polar covalent bond, binding pair of electrons is unequally shared between two atoms. The power of an atom to attract the pair of electrons that constitutes the bond in a molecule is called “electronegativity”.The ‘electronegativity’ can be used to determine whether a given bond is non-polar covalent, polar covalent or ionic bond. The electronegativity increases from leftto right across a period and decreases as you go down a groupThe larger the electronegativity, the greater is the strength to attract a bonding pair of electrons; and the larger the difference in electronegativites of the atoms, the more polar the covalent bond between the two atoms.If we consider the molecule of F2, the electrons are shared equally between the atoms and the bond is called a non-polar covalent. However, in HF the fluorine atom has greater electronegativity than the hydrogen atom. The sharing of electrons in HF is unequal: the electron pair that is bonding HF together shifts towards the fluorine atom because it has a larger electronegativity value and there is a polar covalent bond. The fluorine end becomes partially negatively charged and the hydrogen end becomes partially positively charged. The H-F bond can be represented as follows.The ‘ δ+’ and ‘ δ-’ symbols indicate partial positive and negative charge respectively.The arrow indicates the “pull” of electrons off the hydrogen and towards the more electronegative atom, fluorine.The following is the general thumb rule for predicting the type of bond based upon the electronegativity differences:•If the electronegativities are equal and the difference in electronegativity difference is less than 0.5, the bond is non-polar covalent.•If the difference in electronegativities between the two atoms is greater than 0.4, but less than 2.0, the bond is polar covalent.•If the difference in electronegativities between the two atoms is 2.0, or greater, the bond is ionic.Notice here that, although individual C-Cl bonds in CCl4 are polar, the whole molecule is non-polar because the four dipoles cancel each other. The same for a molecule like O=C=O where each C=O bond is polar but cancel each other in the molecule since they are directed in opposite directions.N.B: The limit 2.0 is a little bit arbitrary and indicative, even some books give 1.7 as the limit.In order to simplify things, some bonds with very small defferences in elec-tronegativities are considered as pure covalent; e.g. with the difference in elec-tronegativity of 0.4, C-H bond, in organic chemistry, is considered as a pure covalent bond.Checking up 4.7(a)Predict whether the following bonds in compounds are polar, non polar or ionic.(i)BeCl2 (ii) BF3 (iii) CH4 (iv) PCl3 (v) H2S (vi) SnCl2(vii) CO2 (viii) SO2 (ix) SO3 (x) SF6 (xi) PCl5 (xii) Cl2(b)Some molecules above have polar bonds but overall the molecules are non polar. Identify the molecules and explain the statement.(c )Explain how the nature of the bonds (polar, non polar or ionic) affect the physical properties of the different compounds. Use specific examples.(d)Hydrogen chloride and ammonia are very soluble in water and yet they are covalent compounds. Explain(e)Draw the shapes of the molecules above in (a) and predict whether the molecule is overall polar, or non polar.4.8. Simple and giant covalent structuresActivity 4.81. With reference to CO2, C4H10, diamond and graphite explain how their sizes can affect their physical properties.2. Diamond is said to be the hardest natural substance known. Explain the origin of that hardness.3. Graphite is soft and can be used as a lubricant. Explain why.4. Graphite conducts electricity while diamond does not. Explain.5.Relate the physical properties of diamond and graphite to their usessimple molecular compounds display some physical properties as follows.(i) Low melting and boiling points Simple molecular compounds display some physical properties as follows.(i) Low melting and boiling pointsSimple Molecular Structures tend to have low melting and boiling points since the forces between molecules (intermolecular forces, which are van der Waals forces) are quite weak.Little energy is required to separate the molecules.e.g.of boiling points: CH4: -161°C, C2H6: -88°C, C3H8 : -42°C(ii) Poor electrical conductivityThere are no charged particles (ions or electrons) delocalized throughout the molecular crystal lattice to conduct electricity. They cannot conduct electricity in either the solid or molten state.(iii) SolubilitySimple structures tend to be quite insoluble in water, but this depends on how the polarized molecule is. The more polar the molecules, the more water molecules will be attracted to them (some may dissolve in water as a result of forming hydrogen bonds within it). Molecular crystals tend to dissolve in non-polar solvents such as alcohol.(iv) Soft and low densityVan der Waals forces are weak and non-directional. The lattice is readily destroyed and the crystals are soft and have low density.b. Giant covalent structures and their physical propertiesSometimes covalently bonded structures can form giant networks, known as Giant Covalent Structures. In these structures, each network of bonds connects all the atoms to each other.These structures are usually very hard and have high melting and boiling points. This is because of the strong covalent bonds holding each atom in place. In general, Giant Covalent Structures cannot conduct electricity due to the fact that there are no free charge carriers. One notable exception is Graphite. This is a structure composed of ‘sheets’ of carbon atoms on top of each other. Electrons can move between the sheets and carry the electricity. The main giant covalent molecular structures are the two allotropes of carbon (diamond and graphite), and silica (silicon dioxide).(i) Diamond structure and the physical propertiesDiamond is a form of carbon in which each carbon atom is joined to four other carbon atoms, forming a giant covalent structure with four single bonds. As a result, diamond is very hard and has a high melting point. It does not conduct electricity. Diamond is tetrahedral face-centered cubic as shown in the figure below belowDiamond has a very high melting point (almost 4000°C): the carbon-carbon covalent bonds are very strong and have to be broken throughout the structure before melting occurs.The compound is very hard due to the necessityto break very strong covalent bonds operating in 3-dimensions.Diamond does not conduct electricity: All the electrons are held tightly between the atoms, and are not able to move freely.The compound is insoluble in water and other organic solvents due to no possible attractions which could occur between solvent molecules and carbon atoms which could outweigh the attractions between the covalently bound carbon atoms.(ii) Graphite and the physical propertiesGraphite is another form of carbon in which the carbon atoms form layers. These layers can slide over each other and graphite is much softer than diamond. Each carbon atom in a layer is joined to only three other carbon atoms in hexagonal rings as shown in the figure below.Concerning the properties of graphite:•It has a high melting point, similar to that of diamond: In order to melt graphite, you have to break the covalent bonding throughout the whole structure.• You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper.•Graphite has a lower density than diamond because of the relatively large amount of space between the sheets.•It is insoluble in water and organic solventsdue to the same reason as that of diamond.The attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds in graphite.•Graphite is a conductor of electricity due to the delocalized electrons which are free to move throughout the sheets. If a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet and be replaced with new ones at the other end.(iii) Silicon dioxide (SiO2) and the physical propertiesSilica, which is found in sand, has a similar structure to that of diamond. It is also hard and has a high melting point, but contains silicon and oxygen atoms, instead of carbon atoms. It is tetrahedral (Figure 4.10). The fact that it is a semiconductor makes it immensely useful in the electronics industry: most transistors are made of silica.Silicon dioxide exhibits some physical properties such as:•It has a high melting point (around 1700°C) which varies depending on what the particular structure is (remember that the structure given is only one of three possible structures).The silicon-oxygen covalent bonds are very strong and have to be broken throughout the structure before the melting occurs.•Silicon dioxide is hard due to the need to break the very strong covalent bonds.•Silicon dioxide is not displaying the property of electrical conductivity because all the electrons are held tightly between the atoms, and are not able to move freely. No any delocalized electrons are observed.•It is insoluble in water and organic solvents because no possible attractions occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure.Checking up 4.81. Diamond, graphite and silicon dioxide are all examples of giant covalent structures. What does the word giant mean in this context?a) Draw a diagram to show the arrangement of carbon atoms in a diamond crystal.b) Draw a diagram or diagrams to show the arrangement of carbon atoms in a graphite crystal.2. Answer the following questions by referring to the diagrams you have drawn in question 2.a) Explain why diamond is very hard, whereas graphite is so soft that it can be used in pencils or as a lubricant.b) The densities of diamond and graphite are: diamond 3.51 g cm-3; graphite 2.25 gcm-3 . Explain why graphite is less dense than diamond.c) Although graphite is very much softer than diamond, both substances have very high melting points. Explain why that is.d) Explain why graphite conducts electricity whereas diamond doesn’t.e) Explain why neither material is soluble in water or any other solvent under normal conditions.4.a) Draw a diagram to show the structure of silicon dioxide.b) Explain why silicon dioxide(i) is hard;(ii) has a high melting point;(iii) Doesn’t conduct electricity;(iv) is insoluble in water and other solvents.4.9. Intermolecular ForcesActivity 4.91. Make a research and describe why:i) Ice floats over water and the bottle full of water breaks on cooling(freezing)ii) Water is a liquid at room temperature while Hydrogen sulfide is a gas2. Trichloromethane (ii) ethanol (iii) aluminium fluoride. Arrange these compounds in order of increasing boiling points.Intermolecular forces are electrostatic forces which may arise from the interaction between partial positively and negatively charged particles. Intermolecular forces exist between two molecules while intramolecular forces hold atoms of a molecule together in a molecule (Figure 4.11).Intermolecular forces are much weaker than the intramolecular forces of attraction but are important because they determine the physical properties of molecules such as their boiling point, melting point, density, and enthalpies of fusion and vaporization.Intramolecular forces hold the atoms in the molecule together; they are called chemical bonds. Intermolecular forces hold covalent molecules together and are responsible of a certain number of properties of the substance such as the melting and boiling temperatures of covalent substances. They can be grouped in a category forces called van der Waals forces. There are three main kinds of intermolecular interactions such as London dispersion forces, dipole-dipole interactions and hydrogen bonding later in the unit4.9.1. London dispersion forcesThese are the weakest of the intermolecular forces and exist between all types of molecules, whether ionic or covalent—polar or non-polar. The more electrons a molecule has, the stronger the London dispersion forces are. For example, bromine, Br2 , has more electrons than chlorine, Cl2, so bromine will have stronger Londondispersion forces than chlorine, contributing to increasing the boiling point of bromine, 59 oC, compared to chlorine, –35oC. Those London forces are very weak for non-polar covalent compounds; hence breaking them does not require much energy, which explains why non-polar covalent compounds such as methane and nitrogen which only have London dispersion forces of attraction between the molecules have very low melting and boiling points.4.9.2. Dipole-dipole interactionThis kind of interaction occurs between molecules containing polar bonds and acts in addition to the basic van der Waals’ forces. The extra attraction between dipoles means that more energy must be added to separate molecules. The boiling points are higher than the expected for a given mass. The figure 4.13 below shows the dipole-dipole attraction between molecules of HCl.