Question

Which of these presents the correct order of their increasing bond order?

• A. $C_2^{2-}<H{e}_2^{+}<NO<O_2{}^{-}$
• B. $NO<O_2{}^{-}<C_2^{2-}<H{e}_2^{+}$
  
• C. $H{e}_2{}^{+}<O_2{}^{-}<NO<C_2{}^{2-}$
• D. $NO<C_2{}^{2-}<H{e}_2{}^{+}<O_2{}^{-}$
  

Answer 1

The correct option is C $H{e}_2{}^{+}<O_2{}^{-}<NO<C_2{}^{2-}$

Bond Order Calculation:

$H{e}_2^{+}$:

Number of electron= 3$\le$14

so,

$\sigma 1S^2\sigma \ast 1S^1$

Bond Order(B.O)= $\frac{Bondingelectron-NonBondingElectron}{2}$

=$\frac{2-1}{2}$

=0.5

$C_2^{2-}$

Number f electron=14$\le$14

so,

$\sigma 1S^2\sigma \ast 1S^2\sigma 2S^2\sigma \ast 2S^2\begin{array}{c}\Pi 2P_x^2\\ \Pi 2P_y^2\end{array}\sigma 2P_z^2$

Bond Order(B.O)=$\frac{10-4}{2}$ = $\frac{6}{2}=3$

$NO$

Number f electron=15$>$14

so,

$\sigma 1S^2\sigma \ast 1S^2\sigma 2S^2\sigma \ast 2S^2\sigma 2P_z^2\begin{array}{c}\Pi 2P_x^2\\ \Pi 2P_y^2\end{array}\sigma 2P_z^2$

Bond Order(B.O)=$\frac{10-5}{2}$ = $\frac{5}{2}=2.5$

$O_2^{-}$

Number f electron=17$>$14

so,

$\sigma 1S^2\sigma \ast 1S^2\sigma 2S^2\sigma \ast 2S^2\sigma 2P_z^2\begin{array}{c}\Pi 2P_x^2\\ \Pi 2P_y^2\end{array}\begin{array}{c}{\Pi }^{\ast }2P_z^2\\ {\Pi }^{\ast }2P_y^1\end{array}$

Bond Order(B.O)=$\frac{10-7}{2}$ = $\frac{3}{2}=1.5$