Which of the following orders are correct against the property given? \(\begin{array}{lll} \hline \text {…

1. I, II only
2. II, III only
3. I, III only
4. I, II, III

Solution

(I) In \(\mathrm{BF}_3\), electronegativity of \(\mathrm{F}\)-atom is greater than \(B\).
In \(\mathrm{NH}_3\), electronegativity of \(\mathrm{N}\)-atom is greater than \(\mathrm{H}\). Therefore, direction of all dipole moment with lone pairs lies in same direction, hence net dipole moment increases.
\(\mu_{\text {net }} \simeq 1.46 \mathrm{D}\) In \(\mathrm{NF}_3\), electronegativity of \(\mathrm{F}\) is greater than \(\mathrm{N}\). Therefore, direction of dipole moment are opposite from the lone pair, hence, net dipole decreases.
\(\mu_{\mathrm{net}} \simeq 0.24 \mathrm{D}\) Correct order of dipole moment will be \(\mathrm{NH}_3 > \mathrm{NF}_3 > \mathrm{BF}_3 \text {. }\) (II) Covalent bond length \(\mathrm{C}-\mathrm{O} > \mathrm{N}-\mathrm{O} > \mathrm{O}-\mathrm{H}\) as electronegativity difference increases then bond length decreases but lone pair-lone pair repulsion increases the bond length. (III) Bond order \(=\) Number of bonding electron \(\left(N_b\right)\) \(\frac{- \text { Number of antibonding electron }}{2}\) \(\begin{gathered} \mathrm{C}_2=\sigma 1 s^2 < \sigma^{\star} 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \pi 2 p_x^2 \simeq \pi 2 p_y^2 \\ \mathrm{BO}=\frac{8-4}{2}=2 \\ \mathrm{~B}_2=\sigma s^2 < \sigma^{\star} 1 s^2 < \sigma 2 s^2 < \sigma^{\star} 2 s^2 < \pi 2 p_x^1 \simeq \pi 2 p_y^1 \\ \mathrm{BO}=\frac{6-4}{2}=1 \\ \mathrm{He}_2=\sigma l s^2 < \sigma^* 1 s^2 \\ \mathrm{BO}=\frac{2-2}{2}=0 \text { (does not exist.) } \end{gathered}\) II and III statement are correct.

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

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