The bond angle in NF_{3} (102.3^{circ}) is sm | vedclass

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(C) In $NH_{3}$, the electronegativity of $N$ $(3.04)$ is higher than that of $H$ $(2.20)$, so the bonding electron pairs are shifted towards $N$. Thi

Vedclass · Accepted Answer

opposite polarity of $N$ in the two molecules

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(C) In $NH_{3}$, the electronegativity of $N$ $(3.04)$ is higher than that of $H$ $(2.20)$, so the bonding electron pairs are shifted towards $N$. This increases the repulsion between the bonding pairs, leading to a larger bond angle $(107.2^{\circ})$.In $NF_{3}$, the electronegativity of $F$ $(3.98)$ is higher than that of $N$ $(3.04)$, so the bonding electron pairs are shifted away from $N$ towards $F$. This reduces the repulsion between the bonding pairs, resulting in a smaller bond angle $(102.3^{\circ})$.Therefore, the difference in the polarity of the $N-X$ bond (where $X = H$ or $F$) causes the difference in bond angles.

Column $I$	Column $II$
$(a) \ XeO_6^{4-}$	$(P)$ Tetrahedral
$(b) \ ClO_2$	$(Q) \ V\text{-shape}$
$(c) \ NH_4^+$	$(R)$ Trigonal Bipyramidal
$(d) \ XeO_3F_2$	$(S)$ Octahedral

The bond angle in $NF_{3}$ $(102.3^{\circ})$ is smaller than $NH_{3}$ $(107.2^{\circ})$. This is because of

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In which molecule are all bonds not equal?

Match the column for molecular shape?

Column $I$ Column $II$

$(a) \ XeO_6^{4-}$ $(P)$ Tetrahedral

$(b) \ ClO_2$ $(Q) \ V\text{-shape}$

$(c) \ NH_4^+$ $(R)$ Trigonal Bipyramidal

$(d) \ XeO_3F_2$ $(S)$ Octahedral

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