XH_3 + H^{+} longrightarrow XH_4^{+} (X = N, | vedclass

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Question

(D) The bond angle in $XH_3$ molecules depends on the hybridization of the central atom $X$. For $NH_3$,the bond angle is approximately $107^{\circ}$

Vedclass · Accepted Answer

$SbH_3$

Vedclass · Answer

(D) The bond angle in $XH_3$ molecules depends on the hybridization of the central atom $X$. For $NH_3$,the bond angle is approximately $107^{\circ}$ due to $sp^3$ hybridization with a lone pair. For $PH_3$,$AsH_3$,and $SbH_3$,the bond angle is close to $90^{\circ}$ because they involve almost pure $p$-orbitals for bonding (Drago's rule).In the formation of $XH_4^{+}$,the central atom undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.The increase in bond angle is calculated as $\Delta 	heta = 109.5^{\circ} - 	ext{Bond angle of } XH_3$.For $NH_3$: $\Delta 	heta = 109.5^{\circ} - 107^{\circ} = 2.5^{\circ}$.For $PH_3$: $\Delta 	heta = 109.5^{\circ} - 93.5^{\circ} = 16^{\circ}$.For $AsH_3$: $\Delta 	heta = 109.5^{\circ} - 91.8^{\circ} = 17.7^{\circ}$.For $SbH_3$: $\Delta 	heta = 109.5^{\circ} - 91.3^{\circ} = 18.2^{\circ}$.Thus,the maximum increase in bond angle occurs for $SbH_3$.

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

$XH_3 + H^{+} \longrightarrow XH_4^{+} (X = N, P, As, Sb)$. In this process,the maximum increase in bond angle will be in the case of $:-$

Similar Questions

The structure of $XeF_5^-$ is:

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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Reason : $Se$ in $SeCl_4$ has two lone pairs.

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