A simplified application of MO theory to the& | vedclass

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Question

$\mathrm{OF}$

$\mathrm{O}=$ Each oxygen atom has $2+6=8$ electrons

$\mathrm{F}=$ Each fluorine atom has $2+7=9$ electrons

$\mathrm{O}==\sigma

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

$\mathrm{OF}$

$\mathrm{O}=$ Each oxygen atom has $2+6=8$ electrons

$\mathrm{F}=$ Each fluorine atom has $2+7=9$ electrons

$\mathrm{O}==\sigma 1 \mathrm{s}^{2} \quad \sigma 1 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s} \quad \sigma 2 \mathrm{p}^{2} \mathrm{x} \quad\left\{\begin{array}{l}\pi 2 \mathrm{py}^{1} \ \pi 2 \mathrm{pz}^{1}\end{array}ight.$

$\mathrm{F}==\sigma 1 \mathrm{s}^{2} \quad \sigma 1 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{s}^{2} \quad \sigma 2 \mathrm{px}^{2} \quad\left\{\begin{array}{l}\pi 2 \mathrm{py}^{2} \ \pi 2 \mathrm{pz}^{1}\end{array}ight.$

Bond order $=\frac{	ext { No. of bonding - No. of anti-bonding }}{2}$

$=\frac{10-7}{2}$

$=\frac{3}{2}=1.5$

A simplified application of $MO$ theory to the hypothitical molecule $'OF'$ would give its bond order as :-

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