1. Set of species having same bond order is
1) C2, O2, N2-2 2) CN-, NO+, CO
3) N2+, N2-, NO 4) All the above
Ans: 4 (All the above)
Solution:
Total No. of e-→ 8 9 10 11 12 13 
15 16 17 18 19 20
Bond order → 0 0.5 1 1.5 2 2.5 2.5 2 1.5 1 0.5 0
2. Incorrect statement among the following is
1) Bond order: N2 > N2+ = N2- > N2-2
2) Bond order of H2+ & H2- is same
3) Stability of H2+ & H2- is same
4) Diamagnetic ions: NO+, O2-2
Ans: 3 (Stability of H2+ & H2- is same)
Solution:
Stability of H2- is less than H2+ as H2- has 1 e- in anti bonding orbital.
3. Highest dipole moment is observed in
1) CH3Cl3 2) CH3Cl 3) CH2Cl2 4) CCl4
Ans: 2 (CH3Cl)
Solution:
The resultant of 2 C - H bond moments adds to the resultant bond moments of C - H & C - Cl
4. Permanent dipole moment is observed in both the molecules is
1) NO2 & O3 2) NO2 & CO2 3) NO2 & SiF2 4) CO2 & SiF2
Ans: 1 (NO2 & O3)
Solution:
Both NO2 & O3 are bent shaped. Hence they have permanent dipole moment.
5. Iso structural pair(s) is / are
1) NO3- & CO3-2 2) BrO3- & XeO3 3) Both 1 & 2 4) BF3 & ClF3
Ans: 3 (Both 1 & 2)
Solution:
NO3- & CO3-2, BF3 are trigonal planar
BrO3- & XeO3 are pyramidal.
ClF3 is T - shape
6. Correct order of dipole moment of alkyl halides is
1) CH3F > CH3CI > CH3Br > CH3I
2) CH3I > CH3Br > CH3CI > CH3F
3) CH3Br > CH3CI> CH3F > CH3I
4) CH3CI > CH3F > CH3Br > CH3I
Ans: 4 (CH3CI > CH3F > CH3Br > CH3I)
Solution:
Very small atomic size of F dominates greater electronegativity of it.
CH3Cl (1.86 D) > CH3F (1.85 D) > CH3Br (1.83 D) > CH3I (1.6 D)
7. The formal charges present on P & H respectively in PH3 are
Ans: 0, 0
Solution:
F.C. on P = Group No. - no. of e- in lone pairs - No. of bonds = 5 - 2 - 3 = 0
F.C. on H = 1 - 0 - 1 = 0
8. Electronegativity of C is maximum in
1) C2H2 2) C2H2 3) C2H6 4) Same in all three molecules
Ans: 2 (C2H2)
Solution:
E.N. ∝ % of S character
9. Br atom in its 2nd excited state, reacts with excess of F2 to form a compound X. The formula and shape of X are
Ans: BrF5, Square pyramid
Solution:
BrF5 is AB5E Type molecule having one lone pair, so it is square pyramidal.
10. The ratio of total no. of Na+ & Cl- ions in 1 unit cell of NaCl lattice is
Ans: 4 : 4
Solution:
Na+ ions = 1 (body) + 12 × 
(edges) = 4Cl − ions = 6 × 
(face centres) + 8 × 
(corners) = 4
11. Number of lattice points occupied by Na+ & Cl- ions in 1 unit cell are respectively
Ans: 13 : 14
Solution:
Na+ ions = 1 (body) + 12 (edges) = 13 Cl- ions = 6 (face centres) + 8 (corners) = 14
12. The type of hybridization on each carbon in H2C = C = C = CH2 is
Ans: sp2 - sp - sp - sp2
Solution:
Terminal carbons = sp2
2 middle carbons have 2σ & 2π each so sp.
13. The types of hybridization of N in NO3−, NO2+ , NH4+ are respectively
Ans: sp2, sp, sp3
Solution:
No. of H.O.S = [V + M − C]
V = Group no. of central atom, M = No. of H or halogen atoms, C = charge on the ions.
NO3− = H.O.S. = [5 + 0 −(−1)] = 3 ( ∵ sp2)
NO2+ = H.O.S. = [5 + 0 −(+1)] = 2 ( ∴ sp)
NH4+ = H.O.S. = [5 + 4 −(+1)]= 4 ( ∴ sp3)
14. The correct order of increasing covalent character is
1) NaCl > LiCl > BeCl2
2) BeCl2< LiCl < NaCl
3) NaCl < LiCl < BeCl2
4) LiCl < BeCl2< NaCl
Ans: 3 (NaCl < LiCl < BeCl2)
Solution:
Covalent character Polarising power ∝ 
∴ Na+ > Li+ > Be+2 − Order of size of cation
15. The dipole moment of HBr is 1.6 × 10−30 coulomb metre. The bond length of HBr bond is 1A°. The % ionic character of HBr is
Ans: 10
Solution:
% of ionic character = 
16. Number of sigma & pi bonds in P4O10 is
Ans: 16, 0
Solution:
Each P is having 4 sigma bonds with O & P.
∴ No. of sigma bonds = 4 × 4 = 16
No. of pi bonds = 0
17. Total number of paired electrons and unpaired electrons present in anti bonding orbitals in O2 molecule on the basis of molecular orbital theory are respectively
Ans: 4, 2
Solution:
Paired electrons = (π2Px)2 (π2Py)2 = 2 + 2 = 4
unpaired electrons = (π2*px)1 (π2py*)1 = 1 + 1 = 2
18. Which of the following statement is not correct when N2 & O2 are converted into N2+ & O2+
1) In N2+, N − N bond strength decreases.
2) In O2+, O − O bond order increases
3) N2+ becomes diamagnetic
4) In O2+, paramagnetism decreases
Ans: 3 (N2+ becomes diamagnetic)
Solution:
Bond order in N2+ is 2.5, it is paramagnetic.
19. Incorrect statement(s) among the following is/ are
1) HFH bond angle 120.1°
2) The magnitude of H− bonding is maximum in solid state and minimum in gaseous state
3) Most of the hydrogen bonds are symmetric
4) All are incorrect statements
Ans: 3 (Most of the hydrogen bonds are symmetric)
20. Which of the following is a incorrect set with respect to molecule / ion, hybridisation and shape?
1) BrF5, sp3d, Trigonal bipyramidal
2) [CO (NH3)6]+3, d2 sp3, octahedral
3) [Pt(Cl)4]−2, dsp2, square planar
4) [CoF6]−3, sp3 d2, Octachedral
Ans: 1 (BrF5, sp3d, Trigonal bipyramidal)
Solution:
BrF5, sp3d2, square pyramidal
21. CsI3
Ans: Contains Cs+ and I3− ions
Solution:
CsI3→ Cs+ + I3−
22. Strongest bond is present in the given ions
1) O2-2 2) O2- 3) O2+2 4) O2+
Ans: 3 (O2+2)
Solution:
Bond order of O2+2 is 3 (maximum)
B.O. = (10 − 4) = 3
As O2+2 has shortest bond length, bond is strongest
23. Least volatile compound among the following is
Ans: H2O
Solution:
H2O has high boiling point due to presence of strong inter molecular hydrogen bonding. So its volatile nature is least.
24. Shortest hydrogen bond is observed in
1) F - H .... O 2) F - H .... F 3) Both 1 & 2 4) N - H ... O
Ans: 2 (F - H .... F)
Solution:
F is highly electronegative, forms strongest H − Bond. So the bond is shortest.
25. Type of repulsion(s) observed in H2O is/are H2O
1) l.p. − l.p 2) b.p. − b.p. 3) l.p. − b.p. 4) All the above
Ans: 4 (All the above)
Solution:
has 2 l.p.s & 2 b.p.s
26. The possible bond angles in the molecule in which 20% of 'S' character in the hybrid orbital formed by central atom is
Ans: 90° & 120°
