1

### JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)
The compound(s) with TWO lone pairs of electrons on the central atom is(are)
A
BrF5
B
ClF3
C
XeF4
D
SF4
2

### JEE Advanced 2016 Paper 2 Offline

MCQ (More than One Correct Answer)
According to Molecular Orbital Theory, which of the following statements is(are) correct?
A
$C_2^{2-}$ is expected to be diamagnetic
B
$O_2^{2+}$ expected to have a longer bond length than O2
C
$N_2^+$ and $N_2^-$ have the same bond order
D
$He_2^+$ has the same energy as two isolated He atoms

## Explanation

Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :

For $C_2^{2 - }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $O_2^{2 + }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ *$

Thus the bond order = (10 $-$ 4)/2 = 3;

Molecular orbital configuration of O2 (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ *$

So O$_2$ has 2 unpaired electrons. for O2, bond order = 2.

As bond order is inversely proportional to bond length, thus the bond length of $O_2^{2 + }$ is less than the bond length of O2.

Option (C) : For $N_2^ +$, the total number of electrons is 13. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Bond order of $N_2^ +$ = (9 $-$ 4)/2 = 2.5

$N_2^{ - }$ has 15 electrons.

Moleculer orbital configuration of $N_2^{ - }$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$

$\therefore\,\,\,\,$Na = 5

Nb = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right] = 2.5$

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $He_2^ +$ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.
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### IIT-JEE 1992

MCQ (More than One Correct Answer)
The molecules that will have dipole moment are
A
2, 2-dimethylpropane
B
trans-2-pentene
C
cis-3-hexene
D
2,2,3,3-tetramethylbutane
4

### IIT-JEE 1992

MCQ (More than One Correct Answer)
Which of the following have identical bond order?
A
CN-
B
$O_2^-$
C
NO+
D
CN+