Question

Among the following species
$N_2,N_2^{+},N_2^{-},N_2^{2-},O_2,O_2^{+},O_2^{-},O_2^{2-}$
the number of species showing diamagnetism is

• A. 2.00
• B. 2.0
• C. 2

Answer 1

Answer: (A), (B), (C)

According to molecules orbital theory. The electronic configurations of the given species are
$N_2:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2$

$N_2^{+}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^1$

$N_2^{-}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1$

$N_2^{2-}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

$O_2:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

$O_2^{+}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1$

$O_2^{-}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^1$

$O_2^{2-}:\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^2$

Diamagnetic species are $N_2\text{and}{O}_2^{2-}$
$\therefore$Number of species showing diamagnetism $=2$