Chemical Bonding Exercise # 3

Exercise # 3 1. 2. Part # I [Matrix Match Type Questions] Match the compounds listed in column-I with characteristic(s) listed in column-II. Column-I Column-II (A) O2F2– (p) sp3d (B) F2SeO (q) sp3 (C) ClOF3 (r) sp2 (D) XeF5+ (s) sp3d2 Column-I (A) SF2 (B) KrF4 (C) NOCl (D) NF3 Column-II (p) sp3 and bent (q) two lone pairs (r) bond angle < 109028’ (s) sp2 and bent (t) sp3d2 and square planar 3. Column – I (A) ClF2– , CIF2+ (B) IO2F2– , F2SeO (C) IOF4–, XeOF2 (D) BrF5 , XeOF4 Column –II (p) Square pyramidal. (q) See – saw and pyramidal shaped respectively. (r) Linear and bent shaped respectively. (s) Square pyramidal and T-shaped respectively. (t) Both sp3d2. 4. Column () (A) P4 (B) SO42– (C) C2 H6 (D) P4O10 Column () (p) 7  - bond (q) central atom is in sp3 hybridisation (r) No, ‘P–P’ bond (s) No, ‘O–O’ bond 5. Column-I (A) H3P3O9 (B) H2S2O7 (C) H2S4O6 (D) H4P2O5 Column-II (p) S–O–S bond is present (q) Di-basic acid (r) P–O–P bond is present (s) Central atom (S or P) in maximum oxidation state. 6. Column-I (A) N2+ is stable than N2– (B) NO can easily lose its electron than N2 (C) NO have large bond length than NO+ Column-II (p) due to one have greater number of electrons in antibonding molecular orbitals than other (q) one has B.O. 3 and other has 2.5 (r) both are paramagnetic with same bond order (D) He2+ exists, but is less stable than H2+ (s) one is paramagnetic and other diamagnetic Column - I (A) BF3 (B) (SiH3)3 N (C) B2H6 (D) SiO2 Column - II (p) sp3 hybridization (q) p – p back bond (r) p – d back bond (s) 3c – 2e bond 7. CHEMICAL BONDING 8. Column – I (A) O2 and NO– (B) O2+ and NO (C) CO and CN– (D) C2 and CN+ Column – II (p) Same magnetic property and bond order as that in N2+ (q) Same bond order but not same magnetic property as that in O2 (r) Same magnetic property and bond order as that N22 – (s) Same magnetic property and bond order as that in NO+ 9. Column – I (A) SO3 (gas) (B) OSF4 (C) SO3F – (D) ClOF3 Column – II (p) Polar with p–d bonds and identical S–O bond, lengths. (q) One lone pair and p – d bond. (r) Non–polar with p – p and p – d bonds. Identical S–O bond lengths. (s) Polar with p – dbond. 10. Column – I (A) IF2– (B) ClF3 (C) XeO3F2 (D) SF4 Column – II (p) sp3d (q) polar (r) one of the bond angles is 1800 . (s) one lone pair 11. Column-I (A) Blue vitriol (B) Gypsum (C) Pure orthophosphoric acid (D) Chloral hydrate Column-II (p) Ionic bond (q) Covalent bond (r) Hydrogen bond (s) Resonance stabilisation 12. Column-I (A) Liquid bromine (B) Solid hydrogen fluoride (C) Solution of sodium fluoride in water (D) Liquid methylamine (E) Noble gas clathrate. Column-II (p) Hydrogen bond (q) Ion-dipole force (r) Dispersion force. (s) Dipole induced dipole interaction. 13. Column-I (A) HCl < HF (B) PH3 < NH3 (C) H2O < D2O (D) F2 < Cl2 Column-II (p) Strength of hydrogen bonding (q) Dipole moment (r) Boiling point (s) Bond energy Part # II [Comprehension Type Questions] Comprehension # 1 VSEPR Theory The trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore follows that the corners are not equivalent in ClF3 molecule. Lone pairs occupy two of the corners, and F atoms occupy the other three corners. These different arrangements are theoretically possible, as shown in figure. (i) The most stable structure will be the one of lowest energy, that is the one with the minimum repulsion between the five orbitals. The greatest repulsion occurs between two lone pairs. Lone pair bond pair repulsions are next strongest, and bond pair-bond pair repulsions the weakest. E a E – Equatorial position a – Axial position A E E a A rule of thumb can be theorised, that the position having maximum repulsion amongst them are occupied at equatorial points. Therefore (3) structure is right. (ii) Since double bond occupies more space compared to single bond therefore it will prefer equatorial position. (iii) More electronegative element will occupy axial position in case of trigonal bipyramidal geometry (iv) In case of sp3d2 hybridisation lone pairs should be placed opposite to each other because all the corners are identical. 1. Geometry (i.e. arrangement of electron pairs around central atom) of ClOF3 is similar to the : (A) XeF4 (B) SOCl2 (C) 3¯ (D) ClO4¯ 2. The shape of SF5– can be : – F F S F S F F •• (A)  only 3. (B)  and  only – F F S F F II I – F •• F •• F – F F S F F III (C) V only Actual shape of the molecule BrF5 is similar to the molecule : (A) PCl5 (B) XeF4 (C) PCl4+ F F F F IV (D) , , &  (D) None of these 4. Which of the following statements is/are correct for CIO4 – oxoanion ? (A) It does not have any tendency of polymerisation. (B) It has strong p–p bonding between chlorine and oxygen. (C) All Cl–O bonds are identical and chlorine atom is sp3 hybridised. (D) The chemical bonding takes place in ground state and charge dispersion is more than CIO3 – oxo anion. 5. Which of the following do not exist ? (A) SH6 (B) HFO4 (C) SI6 (D) HClO3 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 orbital electrons (Na) and vice- versa. CHEMICAL BONDING (B) Bond order : 1 (N - N ) 2 b a 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,or 3 corresponds to single, double or triple bonds 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. Bond order = 1. Which of the following statements is incorrect ? (A) Among O2+, O2 and O2- the stability decreases as O2+ > O2 > O2(B) He2 molecule does not exit as the effect of bonding and anti-bonding molecular orbitals cancel each other (C) C2 ,O22 - and Li2 are diamagnetic (D) In F2 molecule, the energy of 2 Pz is more than  2px and  2 Py 2. The bromine (Br2) is coloured because: (A) the difference in energy (E) between HOMO and LUMO is large and the electronic excitation take place by absorption of light which falls in ultra violet region. (B) the difference in energy (E) between HOMO and LUMO is small and the electronic excitation take place by absorption of light which falls in infrared region. (C) the bromine molecule is paramagnetic and the difference in energy (E) is such that the electronic excitation take place in visible light. (D) the difference in energy (E) between HOMO and LUMO is such that the electronic excitation take place by absorption of light which falls in visible region and bromine molecule is diamagnetic. 3. N2 has greater bond dissociation energy than N2+ , where as O2 has a lower bond 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 deceased when O2 is ionized to O2+ and bond order is decreased when N2– is ionized to N2+ (D) None of these. Comprehension # 3 A covalent bond in which electrons are shared unequally and the bonded atoms acquire a partial positive and negative charge, is called a polar covalent bond. Bond polarity is described in terms of ionic character. Similarly in ionic bond, some covalent character is introduced because of the tendency of the cation to polarise the anion. The magnitude of covalent character in the ionic bond depends upon the extent of polarization caused by cations. In general : (i) Smaller the size of cation, larger is its polarizing power. (ii) Larger the anion, more will be its polarisability. (iii) Among two cations of similar size, the polarizing power of cations with pseudo - inert gas configuration (ns2np6nd10) is larger than cation with noble gas configuration (ns2np6) e.g. polarizing power of Ag+ is more than K+. 1. 2. 3. Which of the following will be most covalent ? (A) NaCl (B) Na2S (C) MgCl2 (D) MgS Which of the following is least ionic ? (A) Be2 (B) BeCl2 (C) BeBr2 Arrange the following compounds in increasing order of their ionic character : (D) BeF2 SnCl2, SnCl4, SiCl4, SnF4, SnF2 (A) SnF2 < SnCl2 < SnF4 < SnCl4 < SiCl4 (B) SnF2 < SnCl2 < SnF4 < SiCl4 < SnCl4 (C) SiCl4 < SnCl4 < SnF4 < SnCl2 < SnF2 (D) SnCl4 < SnF4 < SnCl2 < SnF2 < SiCl4 4. Which is the correct order of covalent character (A) BeF2 < BeCl2 < BeBr2 < Bel2 (B) BeCl2 < BeF2 < Bel2 < BeBr2 (C) Bel2 < BeBr2 < BeCl2 < BeF2 (D) Bel2 < BeCl2 < BeBr2 < BeF2 5. Which of the following combination of cation and anion has maximum covalent character. (A) K+ , Cl– (B) Na+, Cl– (C) Cs+ , Cl– (D) Mg+2, Cl– Comprehension # 4 The degree of polarity of a covalent compound is measured by the dipole moment (  bond) of the bond defined as:  bond = Charge on one of the poles  bond length  bond is a vector quantity. The dipole moment of a molecule is the vector addition of all the bond dipole moments present in it. For a triatomic molecule, containing two bond's like H2O,  molecule is given by  2 molecule =  2 bond +  2 bond + 2  bond .  bond cos   = bond angle The % ionic character of a bond is calculated using the equations % ionic character =  obs  100  ionic ionic = dipole moment when the molecule is assumed to be completely ionic. 1. Which of the following molecule has non-zero dipole moment : (A) XeF2 2. The dipole moment of (B) ClF3 (C) XeO2F4 is 1.5 D. The dipole moment of (D) XeF4 will be - CHEMICAL BONDING (A) 0 D 3. (B) 1.5 D (C) 2.86 D (D) 2.25 D Which of the following compound has Zero dipole moment (A) PCl3 (B) PCl2F3 (C) PCl3F2 (D) PClF4 Comprehension # 5 Molecular geometry is the general shape of a molecule as determined by the relative positions of the atomic nuclei. VSEPR model predicts the shape of the molecules & ions in which valence shell electron pairs are arranged about the atom as far away from one another as possible, thus minimizing pair repulsion information about the geometry of a molecule can sometimes be obtained from an experimental quantity called dipole moment. 1. The dipole moment of a triatomic molecule AX2 was found to be equal to the bond moment of A – X bond. Which of the following information regarding geometry of the molecule can be drawn from the above observation. (A) Molecule is linear (B) Molecule is V shaped with X – A – X = 90° (C) Molecule is V shaped with  X–A–X = 120° (D) Molecular geometry can not be predicted with the given information 2. Which of the following inter-halogen compounds is non-polar in nature: (A) ClF3 (B) BrF5 (C) IF7 (D) BrCl Comprehension # 6 1. Nitrogen, oxygen and fluorine are the highly electronegative elements. When they are tied to a hydrogen atom to form covalent bond , the electrons of the covalent bond are shifted towards the more electronegative atom. This partially positively charged hydrogen atom forms a bond with the other electronegative atom. This bond is called as hydrogen bond and is weaker than covalent bond. For example, in HF molecule, the hydrogen bond exists between hydrogen atom of one molecule and fluorine atom of another molecule as depicted :– – – H+ – F – – – – H+ – F– – – – H+ – F– Here, hydrogen bond acts as a bridge between atoms which holds one atom by covalent bond and the other by hydrogen bond. Hydrogen bond is represented by a dotted line (– – –) while a solid line represents the covalent bond. Thus, hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. What would be correct about bonding in chloral hydrate CCl3CH(OH)2 ? (A) Only intra molecular H-bonding (B) Only inter molecular H-bonding (C) Both inter and intra molecular H-bonding (D) No hydrogen bonding 2. If boiling point of ortho–nitro phenol and para nitro phenol are t1 and t2 respectively and mixture of these two are heated at temperature 't' where 't' is between these two temperature. Vapours will contains (A) Only ortho nitro phenol (B) Only para nitro phenol (C) Both ortho and para nitro phenol (D) None of these 3. What statement is incorrect about D2O (solid) (A) It will float in D2O liquid (B) It will sink in H2O liquid (C) It has maximum density at 4°C (D) It has less volume than H2 314 CHEMICAL BONDING EXERCISE - 3 : PART # I 1. A  (p), B  (q), C  (p), D  (s) 2. A  (p, q, r), B  (q, r, t), C  (s), D  (r) 3. A  (r), B  (q), C  (s), D  (p, t) 4. A  (q, s), B - (q, r, s), C  (p, q), D  (q, s) 5. A  (r, s), B  (p, q, s), C  (q), D - (q, r) 6. A  (p, r), B  (p, q, s), C  (q, s), D  (p, r) 7. A  (q), B  (r), C  (s, p), D - (p) 8. A  (r), B  (p), C  (s), D  (q) 9. A  (r), B  (s), C  (p), D - (q) 10. A  (p, r), B  (p, q), C  (p, r), D  (p, q, s) 11. A  (p, q, r, s), B  (p, q, r, s), C  (q, r), D  (q, r) 12. A  (r), B  (p, q, r), C  (p, q, r), D  (p, r), E  (p, r, s) 13. A  (p, q, r, s), B  (p, q, r, s), C  (p, q, r, s), D  (r, s) PART # II Comprehension # 1 : 1. C 2. D 3. D Comprehension # 2 : 1. D 2. D 3. B Comprehension # 3 : 1. D 2. A 3. C Comprehension # 4 : 1. B 2. B 3. C Comprehension # 5 : 1. C 2. C Comprehension # 6 : 1. A 2. A 3. C 4. A,C 5. A,B,C 4. A 5. D