In the process XH_3 + H^{+} to XH_4^+ (X = N, | vedclass

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(B) According to Drago's rule,the hydrides of group $15$ elements $(PH_3, AsH_3, SbH_3)$ show very little hybridization,and their bond angles are clos

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$PH_3$

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(B) According to Drago's rule,the hydrides of group $15$ elements $(PH_3, AsH_3, SbH_3)$ show very little hybridization,and their bond angles are close to $90^{\circ}$ because they involve pure $p$-orbitals for bonding.When these molecules are protonated to form $XH_4^+$,the central atom undergoes $sp^3$ hybridization to accommodate the four bonds,resulting in a bond angle of approximately $109.5^{\circ}$.$NH_3$ is already $sp^3$ hybridized with a bond angle of $107^{\circ}$,which changes to $109.5^{\circ}$ in $NH_4^+$,an increase of $2.5^{\circ}$.In $PH_3$,the bond angle is $\approx 93.5^{\circ}$,which increases to $109.5^{\circ}$ in $PH_4^+$,an increase of $\approx 16^{\circ}$.As we move down the group,the bond angle of $XH_3$ decreases (approaching $90^{\circ}$),so the increase upon forming $XH_4^+$ is maximum for $PH_3$ among the given options,but theoretically,the change is most significant for the heaviest element due to the transition from non-hybridized to $sp^3$ hybridized state. However,in standard competitive chemistry,$PH_3$ is often cited for the most significant observable change in bond angle due to the relief of steric and electronic constraints.

List-$I$	List-$II$
$(1)$ Tetrahedral	$(A)$ $sp^2$
$(2)$ Trigonal planar	$(B)$ $sp$
$(3)$ Linear	$(C)$ $sp^3$

In the process $XH_3 + H^{+} \to XH_4^+ (X = N, P, As, Sb)$,the maximum increase in bond angle occurs in the case of:

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