Chemical Bonding Molecular Orbital Theory MCQs for CSIR UGC NET & GATE - Download PDF

Chemical Bonding | Assignment 1
Molecular Orbital Theory

QUIZ              PDF

Q1. The highest occupied molecular orbital in N₂ and O₂⁺ respectively are (take x as internuclear axis)

[NET June 2011]

(a) σ_2px, π*_2p
(b) π_2py, π_2pz
(c) σ*_2px, σ_2px
(d) π*_2py, π*_2pz

Q2. The number of anti bonding electrons in NO and CO according to MO theory are respectively

[NET Dec 2011]

(a) 1, 0
(b) 2, 2
(c) 3, 2
(d) 2, 3

Q3. The molecule in which the bond order increases upon addition of an electron is

[NET Dec 2013]

(a) O₂
(b) B₂
(c) P₂
(d) N₂

Q4. The correct statement among the following is

[NET Dec 2015]

(a) N₂ has higher bond order than N₂⁺ and hence has larger bond length compared to N₂⁺
(b) N₂⁺ has higher bond order than N₂ and hence has larger bond length compared to N₂
(c) N₂ has higher bond order than N₂⁺ and hence has larger dissociation energy compared to N₂⁺
(d) N₂ has lower bond order than N₂⁺ and hence has larger dissociation energy compared to N₂⁺

Q5. Correct combination of π and π* orbitals in B₂ molecule is

[NET June 2016]

(a) Gerade Ungerade
(b) Ungerade Gerade
(c) Gerade Gerade
(d) Ungerade Ungerade

Q6. Boron in BCl₃ has

[NET Dec 2017]

(a) sp hybridization
(b) sp² hybridization
(c) sp³ hybridization
(d) no hybridization

Q7. According to Bent's rule, for p-block elements the correct combination of geometry around the central atom and the position of more electronegative substituent is

[NET Dec 2017]

(a) trigonal bipyramidal and axial
(b) trigonal bipyramidal and equatorial
(c) square pyramidal and axial
(d) square pyramidal and equatorial

Q8. Removal of an electron from NO molecule results in

[NET June 2018]

(A) an increase in the (NO) in IR spectrum
(B) an EPR active species
(C) electrons in HOMO being closer to oxygen than to nitrogen 2p orbitals
(D) electrons in HOMO being closer to nitrogen than to oxygen 2p orbitals

The correct answer is
(a) A only
(b) A and C
(c) B and D
(d) A, B and C

Q9. Consider the nature of solvents in column-I and the corresponding _λmax for I2 in various solvents given in column-II. (for I₂ vapours, λ_max is 520 nm). Match column-I with column-II

[NET June 2018] 

Column I	Column II
(A) Non donor	(I) 520
(B) Weak donor	(II) 500
(C) Strong donor	(III) 450
(D) π donor	(IV) 360

The correct match is
(a) A-I, B-II, C-III, D-IV
(b) A-III, B-IV, C-II, D-I
(c) A-I, B-III, C-IV, D-II
(d) A-IV, B-III, C-II, D-I

Q10. Among the following diatomic molecules, the shortest bond length is to be found in

[GATE 2000]

(a) C₂
(b) N₂
(c) O₂
(d) F₂

Q11. The paramagnetic species among the following is

[GATE 2001]

(a) B₂
(b) C₂
(c) O₂^2-
(d) CO

Q12. The highest occupied molecular orbital of HF is

[GATE 2008]

(a) bonding
(b) antibonding
(c) ionic
(d) non bonding

Ans Key

1.a 2.a 3.b 4.c 5.b 6.b 7.a 8.b 9.c 10.b 11.a 12.d

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