The change in hybridization (if any) of the A | vedclass

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Question

(B) In $AlCl_3$,the $Al$ atom is bonded to $3$ chlorine atoms with $3$ $\sigma$ bonds and has no lone pair. The steric number is $3$,which corresponds

Vedclass · Accepted Answer

$sp^2$ to $sp^3$

Vedclass · Answer

(B) In $AlCl_3$,the $Al$ atom is bonded to $3$ chlorine atoms with $3$ $\sigma$ bonds and has no lone pair. The steric number is $3$,which corresponds to $sp^2$ hybridization.In $AlCl_4^{-}$,the $Al$ atom is bonded to $4$ chlorine atoms with $4$ $\sigma$ bonds and has no lone pair. The steric number is $4$,which corresponds to $sp^3$ hybridization.Therefore,the hybridization of the $Al$ atom changes from $sp^2$ to $sp^3$.

List-$I$	List-$II$
$(1)$ $SF_4$	$(A)$ $sp^3d^2$
$(2)$ $IF_5$	$(B)$ $d^2sp^3$
$(3)$ $NO_2^+$	$(C)$ $sp^3d$
$(4)$ $NH_4^+$	$(D)$ $sp^3$
	$(E)$ $sp$

The change in hybridization (if any) of the $Al$ atom in the following reaction is $AlCl_3 + Cl^{-} \rightarrow AlCl_4^{-}$

Similar Questions

Which hybrid orbitals are used by carbon atoms in the following molecules?
$(a)$ $CH_3-CH_3$
$(b)$ $CH_3-CH=CH_2$
$(c)$ $CH_3-CH_2-OH$
$(d)$ $CH_3-CHO$
$(e)$ $CH_3COOH$

Hybridisation of the central atom in $NF_3$ is

Match the species in List-$I$ with their correct hybrid orbitals in List-$II$.

List-$I$ List-$II$

$(1)$ $SF_4$ $(A)$ $sp^3d^2$

$(2)$ $IF_5$ $(B)$ $d^2sp^3$

$(3)$ $NO_2^+$ $(C)$ $sp^3d$

$(4)$ $NH_4^+$ $(D)$ $sp^3$

$(E)$ $sp$

Explain bond formation in $BCl_{3}$. It is a symmetrical trigonal molecule.

Explain $sp^{3}d$ hybridization by a suitable example.

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