Which of the following is correct order of bo | vedclass

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Question

Hybridization of centre atom in $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$ is $\mathrm{sp}^{3}$ with one lone pair and three bond pairs of electrons henc

Vedclass · Accepted Answer

$NH_3 > NF_3 > PH_3$

Vedclass · Answer

Hybridization of centre atom in $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$ is $\mathrm{sp}^{3}$ with one lone pair and three bond pairs of electrons hence these are pyramidal in shape. However $\mathrm{F}$ is more electronegative than $\mathrm{H}$ the electron pair is attracted more towards $\mathrm{F}$ in $\mathrm{NF}_{3}$

So the repulsion between bond pairs are more in $\mathrm{NH}_{3}$ hence bond angle of $\mathrm{NH}_{3}$ is more than $\mathrm{NF}_{3}$ while $\mathrm{PH}_{3}$ molecule is formed by mutual overlapping of orbitals in which bond angle is nearly $92^{o} .$

Which of the following is correct order of bond angle ?

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$\begin{array}{{20}{c}} {N \equiv C} \\ {N \equiv C}\end{array}\,\,\left. {} \right\rangle C = C\,\left\langle {} \right.\,\begin{array}{{20}{c}} {C \equiv N} \\ {C \equiv N} \end{array}$