Test: Molecular Orbital Theory
23 Questions MCQ Test Topic-wise MCQ Tests for NEET | Test: Molecular Orbital Theory
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Direction (Q. Nos. 1-14) This section contains 14 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is
[IIT JEE 2010]
B2 (10 electrons)
MO electronic configuration is 
Bonding electrons = 6
Anti-bonding electrons = 4
No unpaired electron-diamagnetic
In which of the following pairs of molecules/ions both the species are not likely to exist?
Species with (zero) bond order will not exist.
Electrons in orbital
Among the following the maximum covalent character is shown by the compound
Assuming (2s-2p) mixing is not operative, the paramagnetic species among the following is
[JEE Advanced 2014]
If (2s-2p) mixing is not operative, then
and and O2.
The common features among the species CN-, CO, NO+ and N2 are
Thus, all these are isoelectronic. MO electronic configuration is 
No electron unpaired - diamagnetic
Bonding electrons = 10
Anti-bonding electrons = 4
Thus, bond- order = 
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?
Bond energy of O — F (1.5) > F — F (1)
Thus, OF is more stable than F2 thus, (c).
Which of the following diatomic molecules would be stabilised by the removed of an electron?
A molecule is stabilised if bond-order increases, an anti-bonding electron is lost
Probability (electron charge density) of bonding and anti-bonding molecular orbitals are given.
Select the correct probability,
Electron-charge density in a bonding molecular orbital is high in the internuclear region as shown in II.
In an anti-bonding molecular orbital, it is high in parts of the molecule away from the internuclear region.
If one of the electrons in the He2 molecule is taken to the next excited state, then bond order in He2
Electron is taken to next excited state that is 
hen electronic configuration is
Number of electrons in bonding molecular orbital = 3
and in anti-bonding molecular orbital = 1
Thus, bond order = (3 - 1) /2 =1
Thus, bond order increases by 1 unit.
If one of the electrons (1s2) of helium is taken in excited state then bond order of He2 is
Number of electrons in bonding orbital = 4 and in anti-bonding orbitai = 0
The bond energy of H2 is 436 kJ mol -1. Thus, bond energy of 
is
Consider the following oxidation/reduction process,
Q. Magnetic moment does not change in
Direction (Q. Nos. 15) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : N2 has a greater dissociation energy than 
, where as O2 has lower dissociation energy than 
.
Statement II : N2 has 14 electrons while O2 has 16 electrons .
Electron from bonding molecular orbital of higher stability is lost (requires higher energy).
Electron from antibonding molecular orbital of lower stability is lost (requires lower energy).
N2 has 14 electrons and O2 has 16 electrons.
Thus, both Statements I and II are correct but Statement II is not the correct explanation of Statement I
Direction (Q. Nos. 16-18) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.
Q. Set of species with identical bond order is/are
isoelectronic species will generally have identical bond order:
One of the electrons of the highest energy level is taken to next excited state in the following diatomic species. Select the species which undergoes change in bond order?
In which of the following processes, does the value of magnetic moment change ?
(a) CO (14e), no unpaired electron, magnetic moment = 0
CO+ (13e), One electron is lost from bonding melecular orbital unpaired electron = 1, magnetic moment = 
(b) 
(15e) Unpaired electron = 1, magnetic moment =
(13e) Unpaired electron = 1, magnetic moment =
(c) Zn, Zn2+, unpaired electron = 0
(d) O2 (16e), unpaired electron = 2, magnetic moment = 
, unpaired electron = 1, magnetic moment =
Direction (Q. Nos. 19-20) This section contains a paragraph, each describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)
Valence shell MO electronic configuration of a diatomic species is shown
* is for anti-bonding molecular orbital (MO).
Q. Bond order of this species is
Total number of electrons = 16
Number of bonding electrons = 10,
Number of anti-bonding electrons = 6,
Valence shell MO electronic configuration of a diatomic species is shown
* is for anti-bonding molecular orbital (MO).
Q. Divalent cation of this species
Total number of electrons = 16
Number of bonding electrons = 10, Number of anti-bonding electrons = 6,
Direction (Q. Nos. 21) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q. Match the conversion in Column l with the type of effect given in Column II.
Thus, (i) - (q , r) (ii) - (q.t),
(iii) - (q,r), (iv)- (P.s).
(v) - (q,s).
Direction (Q. Nos. 22 and 23) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).
Q. Total number of electrons in anti-bonding MO in 
(superoxide ion) is .......
(17 electrons) has molecular orbital electronic configuration
Underlined are anti-bonding molecular orbital Thus, seven electrons are in anti-bonding molecular orbitals.
How many bonding MO are used in the formation of NO?
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