Chemical Bonding Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. Select the most ionic and most covalent compounds respectively from the following. CrO5, Mn2O7, PbO, P4O10, SnO2 (A) CrO5, Mn2O7 2. (B) Mn2O7, PbO (D) CrO5, SnO2 In which of the following sets central atom of each member involves sp 3 hybridisation ? (A) IO4–, ICl4–, IF4+ (B) SO3, SO , SO4 2– 3 3. (C) CrO5, P4O10 (B) XeO3, XeO4, XeF4 (D) PCl4+, BF4–,ClO4– 2– What is wrong about P4O10 molecule? (A) POP angle is 180° (B) Each ‘P’ atom can be considerd to be sp3 hybridised (C) There are two types of P–O bond lengths (D) There are 6 P–O–P bonds. 4. Which of the following statement is true for IO2F2– according to VSEPR theory? (A) The lone pair and two I– O double bonds occupy the equatorial positions of trigonal bipyramid. (B) It has sp3d hybridisation and is T-shaped. (C) Its structure is analogous to SF4. (D) (A) and (C) both 5. In the equation ;  C + 2D A + 2B + H2O  where A = HNO2 ; B = H2SO3 ; C = NH2OH Identify the geometry and hybridisation of (D) 6. (A) Trigonal planar and sp2 (B) Bent and sp2 (C) Tetrahedral and sp (D) Trigonal pyramidal and sp3. The structure of F2SeO is analogous to : (A) SO3 7. (C) SO32– (D) Both (B) and (C) A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10–10 m then the fraction of an electronic charge on each atom will be : (A) 42% 8. (B) ClO3– (B) 52% (C) 37% (D) 25% Which of the following statement is correct? (A) (SiH3)3N is a weaker Lewis base than (CH3)3SiOH. (B) Like CO, its analogue of Si is not stable (C) In phosgene, C–O bond length is longer than expected while C–Cl bond length is shorter. (D) All of these CHEMICAL BONDING 9. Which reactions involve a change in the electron-pair gemoetry for the under lined geometry ? 10.  BF4– (A) BF3 + F–   NH4+ (B) NH3 + H+   2 SO3 (C) 2 SO2 + O2   H3O+ (D) H2O + H+  According to Molecular orbital theory which of the following is incorrect ? (A) LUMO level for C2 molecule is  2px orbital (B) In C2 molecules both the bonds are  bonds (C) In C22– ion there is one  bonds and two  bonds. (D) C2 is paramagnetic but C22– is diamagnetic. SECTION - II : MULTIPLE CORRECT ANSWER TYPE 11. Which of the following statements is/are true for P4S3 molecule ? (A) It contains six P–S bonds and three P–P bonds. (B) It contains six P–S bonds and ten lone pairs. (C) It has all atoms sp3 hybridised. (D) It contains six P–P bonds and ten lone pairs. 12. Which of the following is polar ? (A) XeF4 13. (B) XeF6 (D) XeF5– Hydrogen bonding is present in which of the following species? (A) CH3NH2 14. (C) XeOF4 (B) CH3– (C) CH3COOH (D) CCl3CH(OH)2 Correct order about bond angle is : (A) H2O > H2S > H2Se > H2Te (B) C2H2 > C2H4 > CH4 > NH3 (C) SF6 < NH3 < H2O < OF2 (D) ClO2 > H2O > H2S > SF6 15. Which of the following molecule are polar sp3d2 hybridized (A) IF5 (B) XeF4 (C) XeOF4 (D) ICl4– SECTION - III : ASSERTION AND REASON TYPE 16. Statement-1 : Crystals of hydrated calcium sulphate (gypsum : (CaSO4 . 2 H2O) are soft and easily cleavedx. Statement-2 : Crystals of anhydrous calcium sulphate (anhydrite : CaSO4) are very hard and very difficult to cleave. (A) Statement-1 is True, Statement02 is True ; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True ; Statement02 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. 17. Statement-1 : Carbon has unique ability to form p – p multiple bonds with itself and with other atoms of small size and high electronegativity. Statement-2 : Heavier elements of group 14th do not form p – p multiple bonds with itself because their atomic orbitals are too large and diffuse to have effective overlapping. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. F 18. Statement-1 : C F – P F bond angle P = bond angle Q but not precisely 90° Q Statement - 2 : The molecule is slightly bent T-shaped and there is repulsion between lone pairs of electrons. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. F Q 19. Statement - 1 : S R P R Bond angle R is a normal triangular angle 120°. F F Statement - 2 : P  180 ; Q  90 and thus R  120 but less than 120° due to the presence of a lone pair.. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 20. Statement - 1 : Compounds oh Hg2+ ions having an ionic radius of 116 pm are more covalent in character than those of Ca2+ ions with almost identical size (114 pm) and the same charge. Statement - 2 : For two ions of same size and charge, one with an (n – 1)d10ns0, electronic configuration will be more polarizing than a cation with an (n – 1)s2 (n –1)p6 ns0, electronic configuration. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. CHEMICAL BONDING SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension # 1 When anions and cations approach each other, the valence shell of anions are pulled towards a cation, is known as polarisation and ability of a cation to polarize the anion is called as polarising power of cation. Due to polarisation, sharing of electrons occurs between two ions to some extent and bond shows some covalent character. Fajan’s suggested following factors on which polarisation depend. (i) As the charge on cation or anion increases polarisation increases. (ii) Size of cation decreases or size of anion increases, polarisation increases. (iii) Cation with pseudo noble gas configuration shows higher polarisation power than that with inert gas configuration. 21. Arrange the following compounds in increasing order of covalent character. (i) SiCl4 (ii) MgCl2 (iii) NaCl (iv) AlCl3 (A) NaCl < MgCl2 < AlCl3 < SiCl4 (B) SiCl4 < AlCl3 < MgCl2 < NaCl (C) NaCl < MgCl2 < SiCl4 < AlCl3 (D) SiCl4 < AlCl3< NaCl < MgCl2 22. Which one is the most ionic ? (A) P4O10 23. (B) TiO2 (C) SnO Highest polarisation is shown in the following compounds ? (A) MgCl2 (B) BaCl2 (C) AgCl (D) SO3 (D) AgI Comprehension # 2 The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule which provides us the following very important informations about the molecule. (A) Stability Of molecule : The molecule is stable if number of bonding molecular orbital electrons (Nb) is greater than the number of antibonding molecular orbitals electrons (Na). (B) Bond Order : Bond order = 1 (N – Na) 2 b A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule. (C) Nature of the Bond : Bond order 1, 2 and 3 corresponds to a single, double and triple bond respectively. (D) Bond Length : Bond length decreases as bond order increases (E) Magnetic Nature : Molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic. 24. Which one of the following statement is incorrect ? (A) Among O2+ , O2 and O2– the bond length decreases as O2– > O2 > O2+ (B) He2 molecule does not exist as the bonding and anti-bonding orbitals cancel each other. (C) C2, O22–, Li2 are diamagnetic (D) In F2 molecule, the energy of  2pz is more than  2px and  2py . 25. The following molecules / species have been arranged in the order of their increasing bond orders. Identify the correct order. (I) O2 ; (II) O2– ; (III) O22– ; (IV) O2+ (A) III < II < I < IV (B) IV < III < II < I (C) III < II < IV < I (D) II < III < I < IV 26. N2 has greater dissociation energy than N2+, where as O2 has a lower dissociation energy than O2+ because : (A) bond order is reduced when O2 is ionized to O2+ and bond order is increased when N2 is ionized to N2+. (B) bond order is increased when O2 is ionized to O2+ and bond order is decreased when N2 is ionized to N2+ (C) bond order is decreased when O2 is ionized to O2+ and bond order is decreased when N2 is ionized to N2+ (D) none of these Comprehension # 3 Hydrogen bonding : O C (X) 27. 28. H O O C OH O C H H OH H (Y) Which of the following species shows the same type of bonding as found in (X) ? (A) Crystalline potassium bicarbonate (B) Fromic acid (C) Anion of Caro’s acid (D) Sulphuric acid Which of the following can not be explained by (Y) type of bondig? (A) The hydrolysis of NCl3 to NH3. (B) The p-hydroxybenzaldehyde as high melting solid at room temperature (C) Extra stability of the complex of Ni(II) with dimethylglyoxime. (D) The hygroscopic nature of sodium peroxide. 29. Select the incorrect statement for (X) and (Y). (A) p-hydroxy benzoic acid the higher boiling point than o-hydroxy benzoic acid on account of (Y). (B) Boric acid crystal breaks quite easily into very fine particles because of (X). (C) HF exists as zig-zag chains, (HF)n in solid as well as in liquid because of (Y). (D) The higher k1 value than k2 for maleic acid is because of (X). SECTION - V : MATRIX - MATCH TYPE 30. Match the species listed in column I with hybridisation of central atom listed in column II. Column I Column II (A) IO2F2– (p) sp3d (B) F2SeO (q) sp3 (C) ClOF3 (r) sp2 + (D) XeF5 (s) sp3d2 CHEMICAL BONDING 31. Match the species listed in column I with shape(s) listed in column II. Column I Column II (A) ClF2–, ClF2+ (p) Square pyramidal (B) IO F , F2SeO (q) See-saw and pyramidal molecular geometry respectively (C) IOF4–, XeOF2 (r) Linear and bent respectively (D) BrF5, XeOF4 (s) Square pyramidal and T – shaped respectively. – 2 2 32. Match the compounds listed in column I with characteristic(s) listed in column II. Column I Column II (A) SF2 (p) sp3 and bent (B) KrF4 (q) lone pair(s) (C) NOCl (r) Bond angle < 109°28’ (D) N(Me)3 (s) sp2 and bent (t) sp3d2 and square planar 33. Match the compounds listed in column I with type of bond(s) listed in column II Column I Column II (A) Blue vitriol (p) Ionic bond (B) Gypsum (q) Covalent bond (C) Pure orthophosphoric acid (r) Hydrogen bond (D) Chloral hydrate (s) Co-ordinate bond MOCK TEST 1. B 2. D 3. A 4. D 5. C 6. D 7. D 8. D 9. A 10. D 11. A,B,C 12. B,C 13. A,C,D 14. A,B,D 15. A,C 16. B 17. B 18. A 19. D 20. A 21. A 22. C 23. D 24. D 25. A 26. B 27. C 28. C 29. B 30. A  (p), B  (q), C  (s), D  (p) 31. A (r), B  (q), C  (s), D (p) 32. A  (p, q, r), B  (q, r, t), C  (q, s), D  (r) 33. A  (p, q, r, s), B  (p, q, r, s), C  (q, r), D  (q, r)