Subtopic 4.2 MOLECULAR SHAPE AND POLARITY

Subtopic 4.2 MOLECULAR SHAPE AND POLARITY 1

LEARNING OUTCOMES (covalent bonding) 1. Draw the Lewis structure of covalent molecules (octet rule such as NH 3, CCl 4, H 2 O, CO 2, N 2 O 4, and exception to the octet rule such as BF 3, NO, NO 2, PCl 5, SF 6 ) 2. Explain the concept of overlapping and hybridisation of the s and p orbitals such as BeCl 2, BF 3, CH 4, N 2, HCN, NH 3, H 2 O molecules 3. Predict and explain the shapes of and bond angles in molecules and ions using the principle of valence valence shell electron pair repulsion, e.g. linear, trigonal planar, tetrahedral, trigonal bipyramid, octahedral, v-shaped, seesaw and pyramidal 4. Explain the existence of polar and non-polar bonds (including C- Cl, C-N, C-O, C-Mg) resulting in polar or/and non-polar molecules

LEARNING OUTCOMES (covalent bonding) 5. Relate bond lengths and bond strengths with respect to single, double and triple bonds 6. Explain the inertness of nitrogen molecule in terms of its strong triple bond and nonpolarity 7. Describe typical properties associated with ionic and covalent bonding in terms of bond strength, melting point and electrical conductivity 8. Explain the existence of covalent character in ionic compounds such as Al 2 O 3, All 3, and Lil 9. Explain the existence of coordinate (dative covalent) bonding such as H 3 O +, NH 4 +, Al 2 Cl 6, and [Fe (CN) 6 ]³ˉ

MOLECULAR GEOMETRY Molecular Geometry The 3D arrangement of atoms in a molecule Affects physical and chemical properties Is predicted by using Valence Shell Electron Pair Repulsion (VSEPR) model

VSEPR THEORY 1. Valence Shell Electron Pair Repulsion theory 2. It states that each group of valence electrons around a central atom is located as far away as possible from the others in order to minimize repulsion 3. To predict the molecular shape from the Lewis structure * Kalau dekat, akan repulse each other

VSEPR THEORY VSEPR THEORY Electron pairs around the central atom will repel one another and arrange themselves as far apart as possible from each other. Why? To minimise the electron pair-electron pair repulsion around the central atom Two type of electron pair around the central atom: bonding pairs and lone pairs

ELECTRON GROUP 1. Electrons pair of lone electron that occupy localized region around the central atom 2. The repulsion may occur either between : Bonding pair (single bond or double bond or triple) with bonding pair Lone pair (odd electron) with bonding pair Lone pair with lone pair

Bonding pair electron Lone pair electron

ELECTRON GROUP Lone pairlone pair repulsion > Lone pairbonding > pair repulsion Bonding pairbonding pair repulsion Decrease of the repulsion force Note: The electron pairs repulsion will determine the orientation of atoms in space

FIVE BASIC MOLECULAR SHAPES 1. Linear (180 ) 2. Trigonal planar (120 ) 3. Tetrahedral (109.5 ) 4. Trigonal bipyramid (90, 120 ) 5. Octahedral (90 )

ELECTRON GROUP ARRANGEMENT 1. Determined by number of electron groups around the central atom 2. Can be one of the 5 basic shapes : 2 electron group/bonding pair : linear 3 electron group/bonding pair : trigonal planar 4 electron group/bonding pair : tetrahedral 5 electron group/bonding pair : trigonal bipyramidal 6 electron group/bonding pair : octahedral

SOME RULES WHEN USING VSEPR THEORY 1. Double bond and triple bonds can be treated like single bond (approximation) One electron group 2. Order of electron group repulsion Lone pair lone pair > lone pair bonding pair > bonding pair bonding pair

SHAPE OF MOLECULE 1. Basic shapes are based on the repulsion between the bonding pairs. 2. Tips to determine the molecular shape : Step 1 : Calculate total number of valence electron Step 2 : Draw Lewis structure of the molecule Step 3 : Calculate the number of bonding pairs or electron groups (Place bonding pairs as far as possible to minimize repulsion) Step 4 : Predict the shape using VSEPR theory

A. MOLECULAR SHAPE WITH 2 BONDING PAIR Example: BeCl 2 Lewis structure Be : 2e 2Cl : 2 X 7e = 14e : Total : 16 e.... Cl Be Cl :.... shape 180 Linear AX 2 Electron group arrangement : linear Molecular shape : linear (bond angle : 180 )

.. : B. MOLECULAR SHAPE WITH 3 BONDING PAIR Example: BCl 3 Lewis structure B: 3e 3Cl : 21e Total: 24e Repulsive forces between pairs are the same AX 3 :.. Cl.. Cl B.... Cl :.. 120 Trigonal planar Electron group arrangement : trigonal planar Molecular shape : trigonal planar (bond angle : 120 )

C. MOLECULAR SHAPE WITH 4 BONDING PAIR Example: CH 4 Lewis structure C: 4e H : 4e Total: 8e H Equal repulsion between bonding pairs equal angle 109.5 AX 4 H C H H Electron group arrangement : tetrahedral Molecular shape : tetrahedral (bond angle :109.5 ) Pyramid is not tetrahedral Tetrahedral

.... : D. Molecules with 5 bonding pairs Example: PCl 5 Lewis structure Shape: Cl Cl P Cl :...... Cl.... Cl.. : : 90 Trigonal bipyramidal 120 Electron group arrangement : trigonal bipyramidal Molecular shape : trigonal bipyramidal bond angle :120, 90

E. Molecules with 6 bonding pairs Example: SF 6 Lewis structure S : 6e 6F : 42e Total : 48e F F Octahedral 90 o 90 o F S F F F Electron group arrangement : octahedral Molecular shape : octahedral bond angle : 90

Shape of molecules which the central atom has one or more lone pairs (3 ELECTRON PAIRS) Class of molecules Number of bonding pairs Number of lone pairs Shape AX 2 E 2 1 e.g. SnCl 2 Sn: 4e 19 2Cl : 14e Total : 18e Bent / V-shaped Bond angle : < 120 o

: Example: SnCl 2 Lewis structure :.. Cl : Sn: 4e 2Cl : 14e Total: 18e Sn.. Cl :.. Electron group arrangement : trigonal planar Molecular shape : bent or v-shaped (bond angle < 120 ) AX 2 E 1. Calculate total number of valence electron 2. Draw the Lewis structure 3. Calculate the number of bonding pairs or electron groups (Place bonding pairs as far as possible to minimize repulsion) 4. Predict the shape using VSEPR theory

4 electron pairs in the valence shell of central atom: Class of molecules AX 3 E Number of bonding pairs Number of lone pairs 3 1 Shape 21 e.g. NH 3 N : 5e H : 3e Total : 8e Trigonal pyramidal Bond angle : < 109.5 o

4 electron pairs in the valence shell of central atom: Class of molecules AX 2 E 2 Number of bonding pairs Number of lone pairs 2 2 Shape e.g. H 2 O O : 6e H : 2e Total : 8e Bent / V-shaped Bond angle : < 109.5 o 22

5 electron pairs in the valence shell of central atom: Class of molecules AX 4 E Number of bonding pairs Number of lone pairs 4 1 Shape e.g. SF 4 S : 6e F : 28e Total : 34e Distorted tetrahedral (see-saw) Bond angle : < 90 o, < 120 o 23

5 electron pairs in the valence shell of central atom: Class of molecules AX 3 E 2 Number of bonding pairs Number of lone pairs 3 2 Shape eg. BrF 3 Br : 7e F : 21e Total : 28e 24 T-shaped Bond angle : < 90 o

5 electron pairs in the valence shell of central atom: Class of molecules AX 2 E 3 eg. I 3 Number of bonding pairs Number of lone pairs 2 3 Shape 25 I : 7e X 3 Total : 21e Linear Bond angle : 180 o

6 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AX 5 E 5 1 e.g. IF 5 I : 7e 26 F : 35e Total : 42e Square pyramidal Bond angle :90 o and 180 o

6 electron pairs in the valence shell of central atom: Class of molecules AX 4 E 2 Number of bonding pairs Number of lone pairs 4 2 Shape e.g. XeF 4 Xe : 8e F : 28e Total : 36e 27 Square planar Bond angle : 90 o

KEEP IN MIND! Elements period 3 or higher can form compounds that have both octet and expanded octet valence shell Example : PCl 3 + Cl 2 PCl 5 octet expanded octet *Period 3 : Na, Mg, Al, Si, P, S, Cl, Ar Never expand atoms of period 1 and 2 such as C and O. They must be octet! ********************************************************

KEEP IN MIND! Elements period 3 or higher can form compounds that have both octet and expanded octet valence shell Example : PCl 3 + Cl 2 PCl 5 octet expanded octet *Period 3 : Na, Mg, Al, Si, P, S, Cl, Ar Never expand atoms of period 1 and 2 such as C and O. They must be octet! Example : ********************************************************

EXERCISE 1 Predict the geometry molecular shape for the following molecules and ions by using VSEPR theory: (a) PCl 5 (b) IF 4 + (c) ClF 3 (d) I 3 - (e) IOF 5 (f) BrF 5 (g) XeF 4 (h) O 3

ANSWER 1. Valence electron ClF 3 I 3-2. Lewis structure 3. Lone and bond pair 4. Predict the shape 32

GUIDELINES FOR APPLYING VSEPR THEORY Step 1 : Write the Lewis structure to see the number of electron groups (Bonding and lone pairs at the central atom) Step 2 : Determine the electron group arrangement Step 3 : Predict the ideal bond angle and any deviation caused by lone pairs Step 4 : Draw and name the molecular shape by counting the bonding groups

POLAR AND NON POLAR BONDS

BOND POLARITY 1. Atoms with different electronegative from polar bonds (difference in EN) 2. Depicted as polar arrow : 3. Example : C - Cl δ δ + C - Cl # Polar bond

POLAR BOND electron poor region electron rich region H F Polar bond 36

NON POLAR BOND F F Non polar bond 37

BOND POLARITY Example : C F δ + δ C - F # Polar bond C C (no difference of electronegativity) # Non polar bond

ELECTRONEGATIVITY Electronegativity of an atom is the ability of an atom that is covalently bonded to another atom to attract electrons to itself An atom with a high electronegativity will attract electrons towards itself and away from the atom with a lower electronegativity Electronegativity of an atom is inversely proportional to its size. The smaller the atomic size, the stronger the attraction for the bonding electrons and the higher electronegativity.

NON POLAR BONDS 1. For covalent bond that consists of two identical atoms, the bonding electrons are shared equally between the two atoms and are attracted equally to both the atoms. This type of bond is called non-polar bond. 2. Diatomic molecules such as H 2, O 2, Cl 2, N 2, have non polar bonds because the atoms in the molecules have same electronegativity. These molecules are called non polar molecules.

POLAR BOND 1. In covalent bond that contains two atoms that are not identical. The bonding electrons will be attracted more strongly by the more electronegative element and this result in unsymmetrical distribution of electrons. 2. Ex : H Cl. The more electronegative chlorine atom attracts the bonding electron pair more strongly than hydrogen does. 3. HCl is a polar molecule and it contains polar bond. 4. The separation of charge in polar covalent bond like H-Cl is called polarisation. 5. When two electrical charge of opposite signs are separated by small distance, a dipole is established. Thus, HCl has a dipole because the H-Cl is a polar.

DIPOLE MOMENT (μ) A quantitative measure of the polarity of a bond is its dipole moment ( µ ). µ = Q r Where : µ = dipole moment Q = the product of the charge from electronegativity r = distance between the charges Dipole moments are usually expressed in debye units(d) 1D (Debye) = 3.36 x 10 30 Cm

RESULTANT (NET) DIPOLE MOMENT 1. Determined by molecular shape and bond polarity Resultant dipole moment > 0 (polar molecule) Resultant dipole moment = 0 (non-polar molecule) 2. Example : I - I EN = 0 (charge in electronegativity) μ = 0 I 2 is a non polar molecule

RESULTANT (NET) DIPOLE MOMENT CI - I 3. Example : EN 0 (charge in electronegativity) μ 0 I 2 is a polar molecule

RESULTANT (NET) DIPOLE MOMENT CO 2 O = C = O 4. Example : CO 2 shape =linear The two bond dipoles cancel each other (molecule is symmetrical) Resultant dipole moment, μ = 0 CO 2 is a non polar molecule

RESULTANT (NET) DIPOLE MOMENT OCS O = C = S 5. Example : OCS shape =linear The two bond dipoles do not cancel each other Resultant dipole moment, μ 0 OCS is a polar molecule

RESULTANT (NET) DIPOLE MOMENT BF 3 F 6. Example : BF 3 shape =trigonal planar The two bond dipoles cancel each other Resultant dipole moment, μ = 0 BF 3 is a NON polar molecule F B F

RESULTANT (NET) DIPOLE MOMENT BF 2 Br F 7. Example : BF 3 Br shape =trigonal planar The three bond dipoles do not cancel each other Resultant dipole moment, μ 0 BF 3 Br is a polar molecule F B Br

Carbon tetrachloride, CCl 4 - molecular geometry : tetrahedral - Chlorine is more electronegative than carbon, - Dipole moment can cancel each other - has no net dipole moment (µ = 0) - therefore CCl 4 is a nonpolar molecule.

Chloromethane, CH 3 Cl - molecular geometry : tetrahedral - Cl is more electronegative than C, C is more electronegative than H - Dipole moment cannot cancel each other - has a net dipole moment (µ 0) - therefore CH 3 Cl is a polar molecule.

Ammonia, NH 3 - molecular geometry : tetrahedral - N is more electronegative than H, - Dipole moment cannot cancel each other - has a net dipole moment (µ 0) - therefore NH 3 is a polar molecule.

BOND MOMENTS AND RESULTANT DIPOLE MOMENTS H

DIFFERENCE BETWEEN POLAR BOND AND POLAR MOLECULES 1. Polar molecules possess polar bond. 2. A bond is polar when the two atoms that are participating in the bond formation have different electronegativities. In polar molecule, all the bonds collectively should produce a polarity. 3. Though a molecule has polar bonds, it does not make the molecule polar. 4. If the molecule is symmetric and all the bonds are similar, then the molecule may become non polar. 5. Therefore, not all the molecules with polar bonds are polar.

KEEP IN MIND! The presence of polar bonds does not always lead to a polar molecule C O is a polar bond But, CO 2 is a non polar molecule O = C = O We have to CONSIDER BOTH (bond polarity and molecular shape)

A SUMMARY ON HOW TO DETERMINE MOLECULAR POLARITY 1. A molecule will be nonpolar if : a) The bonds are non-polar CI CI (non polar) b) No lone pair in the central atom and all the surrounding atoms are the same POLAR F B F B NON POLAR F Br F F

A SUMMARY ON HOW TO DETERMINE MOLECULAR POLARITY 1. A molecule will be nonpolar if : c) A molecule in which the central atom has lone pair electron will usually be polar with few exceptions N O H H POLAR H H POLAR H

A SUMMARY ON HOW TO DETERMINE MOLECULAR POLARITY 2. Exceptions : (NON POLAR MOLECULES) X X X F F A A Xe X X F F X LINEAR SQUARE PLANAR

Exercises : Predict the polarity of the following molecules: SO 2 ; HBr ; SO 3 ; CH 2 Cl 2 ; ClF 3 ; CF 4 ; H 2 O ; XeF 4 ; NF 3 60

HYBRID ORBITAL OVERLAP AND HYBRIDIZATION 1. VSEPR theory : predict molecular shapes by assuming that electron groups tend to minimize their repulsions 2. But, it does not tell how those shapes (which is observed experimentally), can be explained from the interactions of atomic orbitals.

VALENCE BOND (VB) THEORY 1. Covalent bonds are formed by sharing electrons from overlapping atomic orbitals 2. Two types of bonds : σ bond and π bond 3. Example :

DIRECT OVERLAP OF s AND p ORBITAL 1. Atoms in simple molecules or ions such as H 2, HF, N 2, etc. use pure s and/or p orbitals in forming covalent bonds. 2. Example : H 2 (hydrogen molecules)

DIRECT OVERLAP OF s AND p Example : HF (Hydrogen Flouride) H = 1s 1 F = 1s 2 2s 2 2p 5 ORBITAL Example : F 2 (Flourine molecules) F = 1s 2 2s 2 2p 5

HYBRIDIZATION 1. Mixing of two or more atomic orbitals to form a new set of equivalent hybrid orbitals in the same energy level 2. The spatial orientation of the new orbitals is cause more stable bonds and are consistent with the observed molecular shape. 3. 3 types of hybridization : sp, sp 2, sp 3 hybridization

4.3.2 Formation Hybrid orbitals Overlapping of hybrid orbitals and the pure orbitals occur when different type of atoms are involved in the bonding. Hybridization of orbitals: mixing of two or more atomic orbitals to form a new set of hybrid orbitals The purpose of hybridisation is to produce new orbitals which have equivalent energy Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

Hybridization Hybrid orbitals have different shapes from original atomic orbitals Types of hybridisation reflects the shape/geometry of a molecule Only the central atoms will be involved in hybridisation

HYBRIDIZATION s orbital p orbital sp orbital

TYPES OF HYBRID ORBITALS Type Examples Electron group Electron group arrangement sp BeCl 2 2 Linear sp 2 BF 3 3 Trigonal planar sp 3 CH 4 4 Tetrahedral

DETERMINING HYBRID ORBITALS 1. Draw Lewis structure 2. Predict electron group arrangement using VSEPR model 3. Deduce the hybridization of the central atom by matching the arrangement of the electron groups with the hybrid orbitals 4. Use partial the orbital diagram to explain the mixing of atomic orbitals

Molecular formula Lewis Structure Molecular shape and electron group arrangement Hybrid orbitals

SIGMA (σ) BOND 1. Resulting from end to end overlap 2. Has highest electron density along the bond axis 3. Allow free rotation 4. All single bonds are σ bond

bond It formed when orbitals overlap from end to end Example: i. overlapping s orbitals H + H H H 73 bond

ii. Overlapping of s and p orbitals P x orbital H + x H x bond 74

iii. Overlapping of p orbitals x + x x bond 75

Pi (π) Bond 1. Resulting from side to side overlap 2. Has two regions of electron density One above and one below the σ bond axis 3. One π bond hold two electrons that move through both regions of the bond 4. π bond restricts rotation

bond It formed when two p-orbitals of the same orientation overlap sideways Double bond consists of one σ bond and one π bond Example : y y y y + bond

bond Example : π π O = C = O σ σ CO 2 has two π bond and one σ bond Triple bond always consists of one σ bond and two π bond N N π σ π

How do I predict the hybridization of the central atom? Count the number of lone pairs AND the number of atoms bonded to the central atom No of Lone Pairs + No of Bonded Atoms Hybridization Examples 2 3 4 sp sp 2 sp 3 BeCl 2 BF 3 CH 4, NH 3, H 2 O

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FORMATION OF sp HYBRIDIZATION Two equivalents sp hybrid orbitals that lie 180 apart 2 electron groups (from VSEPR theory) Electron group arrangement : linear Molecular shape : linear sp sp 180

FORMATION OF sp HYBRIDIZATION Lewis structure BeCl 2 σ σ Cl Be Cl Valence electron configuration Be O 1 : O 2 : 2s 2s 2p 2p Hybridisation sp sp 2p 2p sp orbitals Empty 2p orbitals One s orbital + one p orbital two equivalent sp orbitals

FORMATION OF sp 2 HYBRIDIZATION Three equivalents sp 2 hybrid orbitals that lie 120 apart 3 electron groups (from VSEPR theory) Electron group arrangement : trigonal planar Molecular shape : trigonal planar sp 2 sp 2 sp 2 120

Lewis structure FORMATION OF sp 2 HYBRIDIZATION Valence electron configuration B O 1 : O 2 : BF 3 2s 2s 2p 2p F σ σ F B σ Br Hybridisation sp 2 sp 2 sp 2 2p sp 2 orbitals Empty 2p orbitals One s orbital + two p orbital three equivalent sp 2 orbitals

TRY DRAW THEM sp 2 hybrid!!

FORMATION OF sp 3 HYBRIDIZATION Bond angle : 109.5 4 electron groups (from VSEPR theory) Electron group arrangement : tetrahedral Molecular shape : tetrahedral sp 3 sp 3 sp 3 sp 3 109.5

Lewis structure FORMATION OF sp 3 HYBRIDIZATION Valence electron configuration B O 1 : O 2 : CH 4 2s 2s 2p 2p H σ σ σ H C H σ H Hybridisation sp 3 sp 3 sp 3 sp 3 sp 3 orbitals Empty 2p orbitals One s orbital + three p orbital four equivalent sp 3 orbitals

TRY DRAW THEM sp 3!!

Example : Methane, CH 4 Ground state : C : 1s 2 2s 2 2p 2 Excitation: to have 4 unpaired electrons 1s 2s 2p Lewis Structure H H C H H H Excited state : 1s 2s 2p sp 3 sp 3 hybrid H sp 3 C sp 3 sp 3 H H shape: tetrahedral

sp 3 -Hybridized C atom in CH 4 sp 3 1s sp 3 sp 3 sp 3 1s 1s 91

sp 3 hybrid Mixing of s and three p orbitals sp 3 sp 3

Other Example: 1. BF 3 Lewis structure : Valence Electron configuration : F : B ground state : B hybrid : Molecular geometry Orbital overlap/hybridisation:

Example: BF 3 Pure p orbital sp2 sp 2 sp 2 F : 1s 2 2s 2 2p 5 Shape : trigonal planar

2. NH 3 Lewis structure : Other Example : Valence orbital diagram ; H : N ground state : N hybrid : Orbital Overlap : Molecular Geometry :

sp 3 1s sp 3 sp 3 sp 3 1s 1s 96

3. H 2 O Other Example: Lewis structure : Valence orbital diagram; O ground state : O hybrid : Molecular geometry Orbitals overlap: 97

HYBRIDISATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS

FORMATION OF HYBRIDIZATION Lewis structure Valence electron configuration C O 1 : O 2 : C 2 H 4 2s 2s (ETHANE) 2p 2p H H σ C = C π H H Hybridisation sp 2 sp 2 sp 2 2p sp 2 orbitals UNHYBRIDIZED 2p orbitals One s orbital + two p orbital three equivalent sp 2 orbitals

HOW TO DRAW?

bonds bond 101

102

FORMATION OF HYBRIDIZATION Lewis structure Valence electron configuration C O 1 : O 2 : C 2 H 2 2s 2s (ACETYLENE) 2p 2p H σ C C π π H Hybridisation sp sp 2p 2p sp orbitals UNHYBRIDIZED 2p orbitals One s orbital + one p orbital two equivalent sp orbitals

HOW TO DRAW?

105

FORMATION OF HYBRIDIZATION Lewis structure C 6 H 6 (BENZENE) Valence electron configuration C O 1 : O 2 : 2s 2s 2p 2p Hybridisation sp 2 sp 2 sp 2 2p sp 2 orbitals UNHYBRIDIZED 2p orbitals One s orbital + two p orbital three equivalent sp 2 orbitals

BENZENE???? (Look at the notes!)

ANSWER: For each of the following, draw the orbital overlap to show the formation of covalent bond a) H 2 O b) N 2 c) CH 3 Cl d) AlCl 3 108

INERTNESS OF NITROGEN MOLECULE 1. Nitrogen is a very electronegative element. 2. It is an inert (unreactive) element. 3. Inertness due to 2 factors : a) Strong triple bond b) Non polarity of N 2 4. Bond energy N N is very high due to triple bond. 5. This strong bond must be broken first, then it can form with other compounds. 6. A lot of energy needed to break 7. Only at high temperature, nitrogen can react with other elements to form compounds. 8. Nitrogen molecules is non-polar. The absence of polarity on the molecule explains why nitrogen is unreactive.

COVALENT CHARACTER IN IONIC COMPOUNDS 1. Not all compounds are ionic, and not all compounds are covalent. 2. Polarisation of chemical bonds occur not only in covalent compound but also in ionic compound. 3. Most ionic compounds have covalent character due to incomplete transfer of electrons. 4. If a small cation with high electric charge approaches a large anion, the cation will attract the electron cloud from the negative ion. 5. Polarisation : The distortion of electron cloud of the anion by a neighbouring cation. 6. Polarising power : The extent to which a cation (positive ion) can polarise an anion (negative ion)

COVALENT CHARACTER IN IONIC COMPOUNDS 7. Polarisation (distortion of electron cloud) of the anion produces covalent character in the ionic bond because the valence electron is partially shared between the cation and anion. 8. The greater the degree of polarisation of the anion, the greater the amount of covalent character in the ionic bond. 9. The polarising power of a cation towards an anion is proportional to the charge density. charge charge density = ionic radius (size) 10. Small and highly charged cations such as Li + and Al 3+ have high charge density, so, high polarising power.

Question 1 Arrange the following chlorides in order of increasing covalent character. Explain your answer. NaCl, MgCl 2, AlCl 3

Answer 1 NaCl < MgCl 2 < AlCl 3 The charge density of the cations increases in the order : Na + < Mg 2+ < Al 3+ Hence, polarising power of the cations towards the Cl _ ion increases in the same order. Polarising power of Al 3+ is high so, it is a covalent compound.

Question 2 Arrange two compounds of beryllium order of increasing covalent character. BeF 2, BeI 2

ANSWER 2 BeI 2 is expected to have a high degree of covalent character. The size of I ion is larger than F ion, hence it is easier to be polarised by the Be 2+ ion