The compounds with sp^2 hybridized central at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To determine the hybridization of the central atom,we calculate the number of electron pairs (steric number) around it: $(A)$ In $H_2CO_3$,the cen

Vedclass · Accepted Answer

$A$ and $C$ only

Vedclass · Answer

(A) To determine the hybridization of the central atom,we calculate the number of electron pairs (steric number) around it: $(A)$ In $H_2CO_3$,the central carbon atom is bonded to three oxygen atoms (one double bond and two single bonds). Steric number = $3$ (sigma bonds) + $0$ (lone pairs) = $3$,which corresponds to $sp^2$ hybridization.$(B)$ In $SiF_4$,the central silicon atom is bonded to four fluorine atoms. Steric number = $4$ (sigma bonds) + $0$ (lone pairs) = $4$,which corresponds to $sp^3$ hybridization.$(C)$ In $BF_3$,the central boron atom is bonded to three fluorine atoms. Steric number = $3$ (sigma bonds) + $0$ (lone pairs) = $3$,which corresponds to $sp^2$ hybridization.$(D)$ In $HClO_2$,the central chlorine atom is bonded to two oxygen atoms and one hydrogen atom,with two lone pairs. Steric number = $3$ (sigma bonds) + $2$ (lone pairs) = $5$,which corresponds to $sp^3d$ hybridization.Thus,only $H_2CO_3$ $(A)$ and $BF_3$ $(C)$ have $sp^2$ hybridized central atoms.

The compounds with $sp^2$ hybridized central atom among the following are:
$(A)$ $H_2CO_3$
$(B)$ $SiF_4$
$(C)$ $BF_3$
$(D)$ $HClO_2$

Similar Questions

What type of covalent bond is formed by hybrid orbitals?

The trigonal bipyramidal geometry results from the hybridization:

The hybridisation of xenon atom in $XeF_4$ is

Which compound of sulphur is formed in the second excited state?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The compounds with $sp^2$ hybridized central atom among the following are:$(A)$ $H_2CO_3$$(B)$ $SiF_4$$(C)$ $BF_3$$(D)$ $HClO_2$