Question

Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?

• A. $N_2^{2-}$
• B. $O_2$
• C. $N_2$
• D. $O_2^{2-}$

Answer 1

Answer: (C), (D)

$O_2^{2-}$

Electronic configuration of $N_2$ molecule
Number of electrons in $N_2=14$
Electronic configuration of $N_2$ molecule:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma p_2^2$

Number of unpaired electrons = 0

Electronic configuration of $N_2^{2-}$ molecule

Number of electrons in $N_2^{2-}=16$

Electronic configuratino of $N_2^{2-}$ molecule:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma p_2^2\sigma 2p_z^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

Number of unpaired electrons = 2

Electronic configuration of $0_2$ molecule

Number of electrons in $0_2=16$

Electronic configuration of $0_2$ molecule:

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

Number of unpaired electrons = 2

Electronic configuration of $0_2^{2-}$ molecule

Number of electrons in $0_2^{2-}=18$

Electronic configuration of $0_2^{2-}$ molecule:

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

Number of unpaired electrons = 0

So, molecules with zero unpaired electrons are $N_2\text{and}{0}_2^{2-}.$ So $N_2\text{and}{0}_2^{2-}$ are diamagnetic.

Hence, the correct options are (A) and (D).