Question

Match the following:$\begin{array}{cc}\text{Column-I}& \text{Column -II}\\ {\text{(a) O}}_2^{2-}& \text{(p)Bond order = 0.5}\\ \text{(b) CO}& \text{(q)Bond order = 3}\\ {\text{(c)NO}}^{+}& \text{(r)Bond order = 1}\\ {\text{(d)He}}_2^{+}& \text{(s)Paramagnetic}\\ & \text{(t) Diamagnetic}\end{array}$

• A. $(a-r,t)$
  $(b-q,t)$
  $(c-q,t)$
  $(d-p,s)$
• B. $(a-q,t)$
  $(b-r,t)$
  $(c-r,t)$
  $(d-p,s)$
• C. $(a-r,s)$
  $(b-q,s)$
  $(c-r,s)$
  $(d-r,t)$
• D. $(a-q,s)$
  $(b-p,s)$
  $(c-p,t)$
  $(d-r,t)$

Answer 1

The correct option is A $(a-r,t)$

$(b-q,t)$
$(c-q,t)$
$(d-p,s)$
Bond order:
$B.O=\frac{1}{2}(bonding{e}^{-}-antibonding{e}^{-})$
Magnetic nature:
$Unpaired{e}^{-}\Rightarrow Paramagnetic$
$Paired{e}^{-}\Rightarrow Diamagnetic$

For $O_2^{2-}$
total number of electron $=18$
Molecular configuration:
$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2$,
$\pi (2p_x^2=2p_y^2),{\pi }^{\ast }(2p_x^2=2p_y^2)$
$B.O=\frac{1}{2}(10-8)=1$
all electrons are paired so nature of $O_2^{2-}$ is diamagnetic.

For $H{e}_2^{+}$
total number of electrons $=3$
Molecular configuration:
$\sigma 1s^2,{\sigma }^{\ast }1s^1$
$B.O=\frac{1}{2}(2-1)=0.5$
Paramagnetic nature.

For $CO$
total number of electron $=14$
Molecular configuration:
$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,\pi (2p_x^2=2p_y^2),\sigma 2p_z^2,{\sigma }^{\ast }2s^2$
$B.O=\frac{1}{2}(10-4)=3$
Diamagnetic in nature.

For $N{O}^{+}$
total number of electrons $=14$
Molecular configuration:
$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi (2p_x^2=2p_y^2),\sigma 2p_z^2$
$B.O=\frac{1}{2}(10-4)=3$
Diamagnetic in nature.