Chemical Bonding Solved Examples

SOLVED EXAMPLES Ex. 1 Sol. Which of the following will be strongest Lewis base ? (A) CH3 CN (B) CH3 NH2 (C) N2 (D) None of these (B) In CH3 NH2, hybridisation of N is sp3 while in CH3 CN hybridisation of N is sp. N2 is also sp hybridised. We know that in hybridisation, as s-character increases the electronegativity of atom also increases. Due to this tendency the release of lone pair of electrons becomes some what difficult. In methyl amine the nitrogen is in sp 3 hybridisation and therefore the donation of the lone pair of electron will be quite easier; so it is strongest Lewis base. Ex. 2 Which type of hybridisation is found in H3O+ ? Sol. According to steric no. rule Steric number = Number of bond pair(s) + number of lone pair(s) at central atom H O + H So, steric number = 3 + 1 = 4. H Thus the hybridisation of oxygen in H3O+ is sp3. Ex. 3 Classify the following bonds as ionic, polar covalent or covalent and give your reasons : (A) SiSi bond in Cl3SiSiCl3 (B) SiCl bond in Cl3SiSiCl3 (C) CaF bond in CaF2 (D) NH bond in NH3 Sol. (A) Covalent, due to identical electronegativity. (B) Covalent, due to less electronegativity difference. (C) Ionic, due to more electronegativity difference. (D) Covalent, due to nearly similar electronegativity. Ex. 4 (A) Which one has highest and lowest melting point and why? NaCl KCl RbCl CsCl (B) Why melting points of cesium halide (CsX) decrease in the order given below ? CsF > CsCl > CsBr > CsI. Sol. (A) NaCl will have highest lattice energy on account of the smaller Na+ while CsCl has lowest lattice energy on account of the larger Cs+. Hence NaCl has highest melting point and CsCl has lowest melting point . (B) As size of anions increase their polarisability increases thus their covalent character increases and melting point decrease. CHEMICAL BONDING Ex. 5 Which is incorrect statement ? (A) In CF2=C=CF2 molecule all the four fluorine atoms are not in the same plane. (B) Ka2 of fumaric acid is more than Ka2 of maleic acid due to intra molecular hydrogen bonding in maleic acid. (C) The O–O bond length in O2[AsF4] is longer than KO2. (D) The bond angle order in halogen – S – halogen is OSF2 < OSCl2 < OSBr2 Ans. (C) F Sol. pz C (A) F C (s+p x+py) F py (s+px) C (s+px+pz) F As both carbon atoms are in different planes, all the four fluorine atoms are not in the same plane. (B) maleic acid fumaric acid O H C O – O C C H C H O– O H C H C C O H C O O Due to intramolecular hydrogen bonding the maleate ion gets stabilized and thus removal of second hydrogen becomes quite difficult. (C) O2[AsF4] = O2+ KO2 = O2 – Bond order  (D) B.O = 2.5 B.O = 1.5 1 ; so O2+ has smaller bond length than O2– bond length Greater the size of the halogen atoms greater will be the steric repulsions and thus larger will be the bond angles. Ex. 6 Which of the compounds MgCO3 and ZnCO3 is thermally more stable ? Explain. Sol. Mg+2 has less polarising power due to inert gas configuration while Zn+2 has higher polarising power due to pseudo inert gas configuration. A cation i.e. Zn2+ with a greater, polarising power exercise a strong pull on the electron cloud of the neighbouring O-atom of the CO32– ion and as such the metal carbonate (ZnCO3) gets readily decomposed into CO2 and the oxide of the metal, ZnO. Thus ZnCO3 is less stable than MgCO3. Ex. 7 Why is anhydrous HCl predominantly covalent in the gaseous state but is ionic in aqueous solution? Sol. It exists as HCl (bond formed by equal sharing of electrons) but in aqueous solution ionises as H+ (or H3O+) and Cl due to polarity of HCl. Ex. 8 Super oxides are coloured and paramagnetic why ? Sol. Super oxides contain one unpaired electron in anti bonding molecular orbital and are coloured due to transition of HOMO orbital electron within visible region. – Ex. 9 Of the species O 2 , O 2 , O 2 and O 22 which would have the maximum bond strength ? Sol. O2+ has higher bond order i.e. 2.5 than O2(2) and O2– (1.5) and bond strength is directly proportional to bond order. Ex. 10 Why BeF2 has zero dipole moment whereas H2O has some dipole moment ? Sol. BeF2 has linear molecule and H2O has bent molecule. F Be F =0 O H H 0 Ex. 11 A diatomic molecule has a dipole moment of 1.2 D. If its bond distance is equal to 1.0Å then the fraction of an electronic charge on each atom is : (A) 25% (B) 37% (C) 52% (D) 42% Ans. (A) Sol. Assuming complete charge transfer then dipole moment = (4.8 × 10–10 esu) (10–8 cm) = 4.8 D so % ionic character = 1.2 × 100 % = 25% 4.8 Ex. 12 Why crystals of hydrated calcium sulphate are soft and easily cleaved where as anhydrous calcium sulphate are very hard and very difficult to cleave ? Sol. With in the Ca2+ / SO42– layer the ions are held together by strong electrovalent bonds but these separated Ca2+ / SO42– layers are linked by relative weak H–bond. The weak H-bonds link SO42– ion in the intermediate region. Ex. 13 The dipole moment of KCl is 3.336 × 10–29 Coulomb meter. The interionic distance in KCl is 2.6Å. Find the % ionic character in KCl. Sol. The theoretical dipole moment in KCl = e × d = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 C meter % ionic character = Ex. 14 experimental dipole moment × 100 = (3.336 × 10–29/4.1652 × 10–29) × 100 theoretical dipole moment For an ionic compound AX3 (s) formed between a metal A and a non-metal X (outermost shell configuration of X = ns2 np5). Find the enthalpy of formation (magnitude) of AX3(s) in k cal mol–1 with the help of the following data. (Non-metal X is found to exist in nature as a diatomic gas) HSublimation A(s) = 100 Kcal/mol HE1 A(g) = 60 Kcal/mol HE2 A (g) = 150 Kcal/mol HE3 A(g) = 280 Kcal/mol Hdiss X2(g) = 80 Kcal/mol He.g X(g) = – 110 Kcal/mol HLattice energy A X3(s) = – 470 Kcal/mol CHEMICAL BONDING Sol. Hf = Hsub + H.E1 + H.E 2 + H.E3 + = 100 + 60 + 150 + 280 + 3 Hdiss + 3 He.g + HL.E 2 3 × 80 – 3 × 110 – 470 = 100 + 60 + 150 + 280 + 120 – 330 – 470 2 = – 90 Kcal/mol Ans. Ex. 15 CO forms weak bonds to Lewis acid such as BF3. In contrast CO forms strong bonds to transition metals. Why explain ? Sol. CO  BF3  Transition metal CO This is called synergic interaction and because of it the bond between CO and transition metal is stronger. Ex. 16 The gaseous potassium chloride molecule has a measured dipole moment of 10.0 D, which indicates that it is a very polar molecule. The separation between the nuclei in this molecule is 2.67 × 10 –8 cm. Calculate the percentage ionic character in KCl molecule. Sol. Dipole moment of compound would have been completely ionic = (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D so % ionic character = Ex. 17 10 .0 × 100% = 78.125 % ~ 78% 12 .8 Ans. There will be three different fluorine-fluorine distances in molecule CF2 (C)2 CF2. Assuming ideal bond angles for a particular hybridisation (assume no distortion due to double bonds) find out the two smaller fluorinefluorine distances and fill the result (in pm) in the increasing order in your answer sheet. Round off your answer to the nearest whole number. (Given that C–F bond length = 134 pm, C = C bond length = 134 pm Sol. d 1 = 2 × 134 × sin 60° pm = 231.8 pm = 232 pm d 2 = 134 × 3 + 2 × 134 cos 60° pm = 536 pm Ans. 3 = 1.73)