Question

Assuming $(2s-2p)$ mixing is not operative, the paramagnetic species among the following is :

• A. $B{e}_2$
• B. $B_2$
• C. $C_2$
• D. $N_2$

Answer 1

The correct option is C $C_2$

1. $B{e}_2$:

valence electrons $=4$

${\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2$

2. $B_2$:

valence electrons $=6$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p_z\right)}^2$

3.$C_2$:

valence electrons $=8$

${\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^1{\left(\pi 2p_y\right)}^1$

4. $N_2$:

valence electrons=$10$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2$

$C_2$ has two unpaired electrons, thus its paramagnetic

Therefore, the option is C.