QJEEM19Chdb61: JEE Main 2019 Chemistry - Class 11 JEEM19Chdb61 Solution: Among the following molecules / ions, which one is diamagnetic…

Note :

(1) $\,\,\,\,$ Bond strength $\propto$ Bond order

(2) $\,\,\,\,$ Bond length $\propto$ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $= {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bending molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

(A) In O atom 8 electrons present, so in O₂, 8 $\times$ 2 = 16 electrons present.

Then in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

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

$\therefore\,\,\,\,$ N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons are present so it is paramagnetic.

(D) Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

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

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

Here no unpaired electrons are present so it is diamagnetic.

(B)   $C_2^{2 - }$ has 14 electrons.

Moleculer orbital configuration of $C_2^{2 - }$ is

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

$\therefore\,\,\,\,$ N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here no unpaired electron present, so it is diamagnetic.

(C)   $N_2^{2 - }$ has 16 electrons.

Moleculer orbital configuration of $N_2^{2 - }$ is

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

$\therefore\,\,\,\,$ N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electron present, so it is paramagnetic.

As Bond length $\propto$ ${1 \over {Bond\,\,order}}$

so among two diamagnetic ions $C_2^{2 - }$ and $O_2^{2 - }$ , bond order of $C_2^{2 - }$ is more so it will have shorter bond length.