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.
3

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+

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