Question

The bond order is maximum in:

• A. $H_2$
• B. $H_2^{+}$
• C. $H{e}_2$
• D. $H{e}_2^{+}$

Answer 1

Answer: (A)

Key Idea: Write the configuration of each molecule according to molecular orbital theory and then find the bond order.
Compare the bond orders to find correct answer.
[a] $H_2=(1+1=2)\sigma 1s^2$
no. of electrons - No. of electrons
B. $O.=\frac{\text{ in bonding }M.O.\text{ in and bonding }M.O}{2}$
$=\frac{2-0}{2}$ = 1
[b] $H_2^{+}(1)=\sigma 1s^1$
Bond order $=\frac{1-0}{2}=\frac{1}{2}$
[c] $H{e}_2=(2+2)=\sigma 1s^2,$
$\stackrel{\ast }{\sigma }1s^2$ Bond order $=\frac{2-2}{2}=0$
[d] $H_2{}^{+}(4-1=3)=\sigma 1s^2$,
$\stackrel{\ast }{\sigma }1s^2$ Bond order $=\frac{2-1}{2}=\frac{1}{2}$
$\therefore H_2$ has highest bond order (1).