Question

The bond order is maximum in:

• A. $ H_{2}$
• B. $ H_{2}^{+} $
• C. $ He_{2} $
• D. $ He_{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] $ He_{2}=(2+2)=\sigma 1s^{2},$
$\overset{*}{\sigma} 1 s ^{2}$ Bond order $=\frac{2-2}{2}=0$
[d] $H_{2}{ }^{+}(4-1=3)=\sigma 1 s^{2}$,
$\overset{*}{\sigma} 1 s^{2}$ Bond order $=\frac{2-1}{2}=\frac{1}{2}$
$\therefore H _{2}$ has highest bond order (1).