Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Extra ion resonance energy is the extra bond energy gained due to difference in electro negativity which leads to ionic character

$${\mathrm{E}}_{\mathrm{H}-\mathrm{F}}=\frac{1}{2}[104+38]$$

Extra resonance energy = experimental ${\mathrm{E}}_{\mathrm{H}-\mathrm{F}}$-calculated ${\mathrm{E}}_{\mathrm{H}-\mathrm{F}}$

$$=135-71=64$$

If we calculated by taking ${\mathrm{E}}_{\mathrm{H}-\mathrm{F}}=\sqrt{{\mathrm{E}}_{\mathrm{H}-\mathrm{F}}\times {\mathrm{F}}_{\mathrm{F}-\mathrm{F}}}$

$$=\sqrt{104\times 38}=62.56$$

Then extra resonance energy $=35-62.56=72.44$

$$\begin{array}{rlr}& X_A-X_B=0.208\sqrt{\Delta }& \Delta =83\text{ so range is }61\text{ to }82\end{array}$$