The state of hybridization of the central ato | vedclass

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Question

(A) In $BF_3$,the central atom $B$ has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridization.In $H_3O^{+}$,the central atom $O$ has $3$

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$B$ in $BF_3$

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(A) In $BF_3$,the central atom $B$ has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridization.In $H_3O^{+}$,the central atom $O$ has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization.In $NH_3$,the central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization.In $PCl_3$,the central atom $P$ has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization.Therefore,the hybridization of $B$ in $BF_3$ is different from the others.

List-$I$ (Species)	List-$II$ (Hybrid Orbitals)
$(a)$ $SF_{4}$	$(i)$ $sp^{3}d^{2}$
$(b)$ $IF_{5}$	$(ii)$ $d^{2}sp^{3}$
$(c)$ $NO_{2}^{+}$	$(iii)$ $sp^{3}d$
$(d)$ $NH_{4}^{+}$	$(iv)$ $sp^{3}$
	$(v)$ $sp$

The state of hybridization of the central atom is not the same as in the others.

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Match List-$I$ with List-$II$ :

List-$I$ (Species) List-$II$ (Hybrid Orbitals)

$(a)$ $SF_{4}$ $(i)$ $sp^{3}d^{2}$

$(b)$ $IF_{5}$ $(ii)$ $d^{2}sp^{3}$

$(c)$ $NO_{2}^{+}$ $(iii)$ $sp^{3}d$

$(d)$ $NH_{4}^{+}$ $(iv)$ $sp^{3}$

$(v)$ $sp$

Choose the correct answer from the options given below :

As the $s$-character of the hybrid orbital decreases,the bond angle ...........

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