4. Chemical bonding and molecular structure — Chemistry STD 11 Science — Question

$\mathrm{O}=$ Each oxygen atom has $2+6=8$ electrons

$\mathrm{F}=$ Each fluorine atom has $2+7=9$ electrons

$\mathrm{O}==\sigma 1 \mathrm{s}^{2} \quad \sigma 1 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s} \quad \sigma 2 \mathrm{p}^{2} \mathrm{x} \quad\left\{\begin{array}{l}\pi 2 \mathrm{py}^{1} \\ \pi 2 \mathrm{pz}^{1}\end{array}\right.$

$\mathrm{F}==\sigma 1 \mathrm{s}^{2} \quad \sigma 1 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{px}^{2} \quad\left\{\begin{array}{l}\pi 2 \mathrm{py}^{2} \\ \pi 2 \mathrm{pz}^{1}\end{array}\right.$

Bond order $=\frac{\text { No. of bonding - No. of anti-bonding }}{2}$

$=\frac{10-7}{2}$

$=\frac{3}{2}=1.5$