The hybridization of the central atom of ClF_ | vedclass

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(B) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = 	ext{number of sigma bonds} + 	ext{number of lone

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$sp^3d, sp^3, sp^2$

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(B) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = 	ext{number of sigma bonds} + 	ext{number of lone pairs}$.$1$. For $ClF_3$: The central atom $Cl$ has $3$ bond pairs and $2$ lone pairs. $SN = 3 + 2 = 5$,which corresponds to $sp^3d$ hybridization.$2$. For $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair. $SN = 3 + 1 = 4$,which corresponds to $sp^3$ hybridization.$3$. For $SO_3$: The central atom $S$ has $3$ bond pairs and $0$ lone pairs. $SN = 3 + 0 = 3$,which corresponds to $sp^2$ hybridization.Thus,the hybridizations are $sp^3d, sp^3, sp^2$ respectively.

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 hybridization of the central atom of $ClF_3$,$NH_3$,and $SO_3$ are respectively:

Similar Questions

What is the type of hybridization of the central atom in the following molecules: $BeCl_2, CO_2, BH_3, BCl_3$?

What is the hybridization of the nitrogen atomic orbitals in $NO_2^+$,$NO_3^-$,and $NH_4^+$ respectively?

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 :

Which of the following compounds shows the hybridization mode $sp^2 - sp^2 - sp - sp$ from left to right?

The hybridisations of $N$,$C$ and $O$ shown in the following compound are:

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