◄UNIT 3: FORMATION OF IONIC AND METALLIC BONDSUNIT 5: VARIATION IN TRENDS OF THE PHYSICAL PROPERTIES►

UNIT 4: COVALENT BOND AND MOLECULAR STRUCTURE

  • UNIT 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.

    Learning objectives

    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

    Activity 4.1

    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

    Activity 4.2(a)

    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.

    (ii) HCl

    (iii) N2

    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

    Activity 4.2(b)

    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

    Bonding structure:

    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 = 24
    2. Draw connectivities

    Checking up 4.2(b)

    1. Draw the Lewis diagram for the following species(a)SOCl2 (b) ICl4-(c) SO42- (d) CH3OH

    2. 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 properties

    Activity 4.3

    1. 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+) ion

    4.3.1. Co-ordinate or dative covalent bonding and properties

    A 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 ions

    A 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 4.3

    1. 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) AlCl3
    3. 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 4.4

    1. Draw the diagram to show how the single and double bonds are formed insadi)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 other
    to 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 4.4

    1. 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 orbitals

    4.5 Valence Shell Electron Pair Repulsion Theory (VSEPR) theory

    Activity 4.5

    1. 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 activity
    i) Get students into groups of 2 or 3 and pass out balloons and strings and scissors
    ii) Ask students to create visuals of molecular orbitals with balloons Blow up 2 balloons and tie them together
    iii) 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 geometry

    Why 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 one
    nucleus. 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 pair

    The 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 Hybridisation

    Activity 4.6

    1. 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 Hybridization

    This 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 Hybridization

    This 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 orbitals


    This 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 orbitals


    The 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 Hybridization

    This 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 Hybridization

    This 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 Hybridization

    Inthis 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 hybridization


    Attention: 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 4.5

    1.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 hybridization
    3.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 molecule
    7. Draw dots-and-crosses diagrams (showing outer electrons only) for the following:
    a) Oxygen, O2
    b) Carbon dioxide, CO2
    c) 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, C2H4
    8.
    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 bonds

    Activity 4.7

    1. 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 dipole

    What 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 left

    to right across a period and decreases as you go down a group

    The 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 structures

    Activity 4.8

    1. 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 uses


    simple 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 points

    Simple 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 conductivity

    There 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) Solubility

    Simple 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 density

    Van 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 properties

    Sometimes 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 properties

    Diamond 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 below


    Diamond 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 properties

    Graphite 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 properties

    Silica, 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 4.8

    1. 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 Forces

    Activity 4.9

    1. 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 gas
    2. 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 unit

    4.9.1. London dispersion forces

    These 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 London

    dispersion 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 interaction

    This 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.

    4.9.3. Hydrogen bond

    For a hydrogen bond to be possible, there are necessary conditions:

    •The first condition is that the molecule contains one group where hydrogen is bonded to one of the three most electronegative atoms (F,O, N),i.e. the presence of one or another of the following bonds: H-F, O-H, N-H,
    •Because of high electronegativity of the 3 atoms, the bond they form with hydrogen is highly polar, with a partial positive charge on hydrogen atom and a partial negative charge of the other atom (see Fig.4.14, red arrows).
    •When two molecules containing that group approach each other, they rearrange themselves so that Hδ+end on one molecule attracts the Fδ-end on the other molecule as shown in Figure 4.14 (broken line).
    It is this attraction force between the two molecules, the broken line that is called “Hydrogen bond”. Hydrogen bond is a van der Waals force, but stronger than other van der Waals forces and that is why it has been given a specific name. The bond. energy of hydrogen bond is of the order of 20 kJmol-1


    Consequences of the presence of hydrogen bond in a compound

    Because of the presence in a compound of hydrogen bond, a strong force that requires more energy to overcome, this compound will possess higher melting and boiling points than the compounds of comparable molecular mass but without  hydrogen bond.

    This explain why water has abnormally high melting and boiling points compared to other hydrides of Group 16 elements (Fig.4.15).


    The presence of O-H and N-H groups in a compound makes that compound very soluble in water because of strong interactions between water molecules and the molecules of the solute containing such groups; e.g. alcohols, ammonia, amines, etc....

    Hydrogen bond is responsible in maintaining the two chains of DNA together. It is found also in many other molecules of biological importance such as proteins.


    Hydrogen bonding in ice

    Each water molecule is hydrogen-bonded to 4 others in a tetrahedral formation. The ice molecule has a “diamond-like” structure. When liquid water freezes, the hydrogen bonds become more rigid, and the volume becomes larger than the liquid because of empty space generated by the rigidity of solid water. This explains why ice floats over liquid water because its density is lower than the density of liquid water.

    This explains also why, if a closed glass bottle full water is left in the fridge for long time, it will crack under the pressure of increasing volume of solid water, the ice.

    Checking up 4.9

    1. Ice floats on water and yet ice is in solid state. Explain the statement
    2. The melting and boiling points of alcohols are much higher than the corresponding alkanes of almost the same molecular mass. Explain the statement.
    3. Water boils at 100oC while H2S boils at -64oC; yet water has lower molecular mass than hydrogen sulfide.Explain why.
    4. Arrange the following compounds in order of increasing boiling points HCl, HBr and HI giving reasons.
    5. Arrange the following compounds in the order of increasing boiling points. Explain your answer. n- Pentane, 2-methylbutane and 2, 2-dimethylpropane has boiling points 360, 280, and 10oC respectively. Explain
    6. (a)What is a hydrogen bond?
    (b) Explain the origin of hydrogen bond in different molecules.

    4.10. End unit assessment

    PART 1: MULTIPLE CHOICE QUESTIONS

    1. In the periodic table, electronegativity genmerally decreases: A From right to left in a period, B Upwards in a group; C From left to right in a period.

    2. Which structure would sulphur, S8, have?A simple covalent molecules B simple covalent lattice C giant covalent lattice Dgiant ionic lattice
    3. Which statement(s) is/are true?
    1) Water has hydrogen bonds which increase the boiling point.
    2) Water as a solid is denser than as a liquid.
    3) Water has bond angles of 180o.
    A 1, 2 and 3
    B 1 and 2
    C 2 and 3
    D only 1

    4. Which molecules have strong intermolecular forces?
    A BF3            B H2S           C NH3               D H2

    5. Which of the following molecules would have a permanent dipole?
    A H2O          B CO2             C NH4+                  D CH4
    6. What is the bond angle of a H3O+ ion?
    A104.5°               B107°             C109.5°          D120°

    7.Which is the better representation of ammonium ion?

    8. Which of the following susbstances: has the highest melting point, can conduct electricity, is soluble in water.
    A Al(s)      B carbon (graphite)     C CH4(g)        D NaCl(aq)
    9. Among the following molecules, which one has a trigonal pyramidal shape?
    A H2O              B CO2                  C NH3           D BF3

    10.Which of the following statements is false about VSEPR theory?

    a) The geometry of a molecule is determined by the number of electron groups on the central atom.
    b) The geometry of the electron groups is determined by minimizing repulsions between them.
    c) A lone pair, a single bond, a double bond, a triple bond and a single electron - each of these is counted as a single electron group.
    d) Bond angles may depart from the idealized angles because lone pairs of electrons take up less space than bond pairs.
    e) The number of electron groups can be determined from the Lewis structure of the molecule.

    11. Bond angle of SF6 is, explain
    A90°        B180°        C120°        D87.5°

    12. Lone pairs in CO2 are, explain
    A =1         B= 2         C= 3          D =4

    13. Number of bonding pairs of electrons in water H2O is,
    A1,      B 2,      C 3,       D 4

    14. Molecular structure of SF6 is;
    A linear,
    B tetrahedral,
    C hexagonal,
    D octahedral
    15. Which of the following pairs of substances have the same shape? Explain
    A CO2 and SiO2      B BCl3 and PCl3       C BeH2 and H2S      D CH4 and AlCl4-

    PART 2: Filling in questions

    16. Use the words listed below to fill in the correct appropriate word(s) in the spaces below in the text.

    Bigger, covalent bond, diamond, free electrons, halogens, hard crystals, high electrical conductivity, high melting points, increase, intermolecular forces, low electrical conductivity, low melting points, non-metals, sharing, soft crystals, strong, strong bond, weak, weak force.

    A ..................... is formed by two atoms................. one or more pairs of electrons to make a .......................between the two atoms in a molecule. However, between small molecules, only a ..................holds them together in the bulk liquid or solid. This results in small covalent molecules having ........................... and ......................................if solid. Small covalent molecules have no ...................... and so have a

     The Group 7 ............................ collectively known as the.......................form diatomic molecules of two atoms. The ................................between the molecules are ................ giving them relatively low melting points and boiling points. This also explains why they are gases, liquids or solids with ......................... As you go down Group 7 the melting boiling points and boiling points .......................because the molecules get .............. and the intermolecular forces ...................... In giant covalent structures the forces between all the atoms are .................forming ..................... like diamond or silica . In the atomic giant structure metals there are free electrons which allow ....................................................

    PART3:

    17. Fill in the table by putting a check mark in the compare-and-contrast matrix under the column(s) that each physical attribute describes.

    b) Using the hybridization theory, explain the formation of the CH3Cl molecule.
    (c) Predict the shape of the following molecules, and state whether they are polar or non-polar
    (i)NCl3
    (ii) BCl3
    (iii) CO2
    (d) Carbon and silicon are in group 14 of the periodic table. Both react with oxygen to form their respective dioxides. Carbon dioxide is a gas at room conditions but silicon dioxide is a hard solid. Explain their differences in terms of bonding and structure.
    25. The atomic number of sulphur is 16.
    (a) Write the electronic configuration of sulphur in the ground state using s,p,d... notation.
    (b) Under suitable conditions, sulphur reacts with chlorine to form SCl2.
    (i)Draw the Lewis structure of SCl2
    (ii) Predict the shape of SCl2
    26. For the following covalently bonded molecules please fill in the following missing information in the chart. Write the structural formula in the next column being sure to include unshared electronsWrite the name of the shape in the “Geometry name” column (ex: linear)Write whether the molecule is polar or non-polar by writing it in that column

    PART 4: MATCHING TYPE QUESTIONS

    27. Please match the term with the correct definition. There will be extra choices and some choices may be used more than once.

    28. Use the given information to complete the crossword puzzle belowAcross

    11. A covalent bond between atoms in which the electrons are shared unequally
    12. A covalent bond in which the electrons are shared equally by the two atoms
    14. Intermolecular forces resulting from the attraction of oppositely charged regions of polar molecules

    15. A bond formed when two atoms share a pair of electrons
    16. The two weakest intermolecular attractions - dispersion interactions and dipole forces
    18. A covalent bond in which one atom contributes both bonding electrons
    20. A chemical formula that shows the arrangement of atoms in a molecule or polyatomic ion
    21. A covalent bond in which three pairs of electrons are shared by two atoms
    22. A bond in which two atoms share two pairs of electrons
    23. A compound that is composed of molecules
    26. A molecule consisting of two atoms
    28. One of the two or more equally valid electron dot structures of a molecule or polyatomic ion
    31. valence-shell electron-pair repulsion theory; because electron pairs repel, molecules adjust their shapes so that valence electron pairs are as far apart as possible

    Down

    1. An orbital that applies to the entire molecule
    2. A bond angle of 109.5 degrees that results when a central atom forms four bonds directed toward the center of a regular tetrahedron
    3. The mixing of several atomic orbitals to form the same total number of equivalent hybrid orbitals

    4. A tightly bound group of atoms that behaves as a unit and has a positive or negative charge

    5. A pair of valence electrons that is not shared between atoms

    6. A molecule in which one side of the molecule is slightly negative and the opposite side is slightly positive.

    7.a covalent bond in which the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms

    8. A covalent bond between atoms in which the electrons are shared unequally

    9. A neutral group of atoms joined together by covalent bonds

    10. A molecule that has two poles, or regions, with opposite charges

    12. The energy required to break the bond between two covalently bonded atoms

    17. A chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of a compound

    PART4: SHORT AND LONG ANSWER QUESTIONS

    18. (a)What is a covalent bond and how does it differ from ionic and metallic bonds.(b) Using dot and cross diagrams show how bonding takes place in carbon dioxide (CO2) and methane (CH4 ).
    19. The proton numbers of X, Y, and Z are 12, 7 and 17 respectively. (a) Write the electronic configuration of X, Y, and Z.(b) Give the formula of the compound formed between Y and Z and draw its Lewis structure
    20. X is a hard solid at room temperature .X does not conduct electricity in the solid state or when molten. Describe the type of bonding in X
    21. All the sulphur-oxygen bonds in the SO42- ions are exactly identical. Explain the observation.
    22

    Explain the trend of the melting point of the above substances in relation to their structure and bonding.
    23. Nitrogen and phosphorus are in the same group in the periodic table. Nitrogen can form the NCl3 molecule only, but phosphorus can form both the PCl3 and PCl5 molecule.
    (a) Write the electronic configuration of the nitrogen and phosphorus atoms.
    (b) Draw the Lewis diagram for the NCl3 and PCl3 and predict their shapes.
    (c) Based on your answer in (a) and (b), explain why phosphorus can form the PC5 molecule but nitrogen cannot form NCl5 molecule.
    24(a) Wth the aid of suitable diagrams, explain the bond formation process in:
    (i) AlCl3 and (ii) Al2Cl6

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    Covalent bonds and Lewis theory

    1. Lewis diagrams: Molecules
    2. Lewis diagrams
    3. Lewis diagrams
    4. Properties of compounds
    5. Lewis diagrams of covalent molecules
    6. Definitions: Covalent and ionic bonding
    7. Lewis diagrams
    8. Electron transfer in covalent, ionic, and metallic bonds

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    Covalent bonds and Lewis theory

    1. Lewis diagrams: Molecules
    2. Lewis diagrams
    3. Lewis diagrams
    4. Properties of compounds
    5. Lewis diagrams of covalent molecules
    6. Definitions: Covalent and ionic bonding
    7. Lewis diagrams
    8. Electron transfer in covalent, ionic, and metallic bonds

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    Valence Shell Electron Pair Repulsion Theory (VSEPR) theory

    1. Molecular shape
    2. Multiple choice: Molecule type
    3. Multiple choice: Molecular shape
    4. Multiple choice: Electron geometry
    5. Multiple choice: Working out molecular shape
    6. Multiple choice: Molecular and electron geometry

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    Polar covalent bonds

    1. Electronegativity difference and polarity
    2. Molecular polarity
    3. Polarity of bonds and molecules
    4. Electronegativity difference
    5. Electronegativity and bonding
    6. Multiple choice: Types of bonding
    7. Electronegativity difference

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    Intermolecular Forces

    1. Intermolecular forces: definitions
    2. Intermolecular forces
    3. The relationship between physical properties and intermolecular forces
    4. Experimental investigations of boiling point
    5. Intermolecular forces and melting/boiling points
    6. Intermolecular forces and boiling points
    7. Boiling points and rates of evaporation
    8. Multiple choice: Intermolecular forces between small molecules
    9. Intermolecular forces
    ◄UNIT 3: FORMATION OF IONIC AND METALLIC BONDSUNIT 5: VARIATION IN TRENDS OF THE PHYSICAL PROPERTIES►