Question

From the molecular orbital theory, one can show that the bond order in $F_{2}$ molecule as

• A. 2
• B. 1
• C. 3
• D. 4

Answer 1

Answer: (B)

1

Answer 2

According to molecular orbital theory,
$F_{2}(18)=\sigma 1 s^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}$
$\approx \pi 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{2} \approx \overset{*}{\pi} 2 p_{y}^{2}$
$\therefore$ Bond order in $F_{2}=\frac{N_{b}-N_{a}}{2}=\frac{10-8}{2}=1$