The number of species from the following in w | 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 steric number $(SN)$ = (Number of sigma bonds) + (Number of lone pairs).$1. NH

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) To determine the hybridization of the central atom,we calculate the steric number $(SN)$ = (Number of sigma bonds) + (Number of lone pairs).$1. NH_3$: $SN = 3 + 1 = 4 \rightarrow sp^3$$2. SO_2$: $SN = 2 + 1 = 3 \rightarrow sp^2$$3. SiO_2$: $SN = 4 + 0 = 4 \rightarrow sp^3$$4. BeCl_2$: $SN = 2 + 0 = 2 \rightarrow sp$$5. CO_2$: $SN = 2 + 0 = 2 \rightarrow sp$$6. H_2O$: $SN = 2 + 2 = 4 \rightarrow sp^3$$7. CH_4$: $SN = 4 + 0 = 4 \rightarrow sp^3$$8. BF_3$: $SN = 3 + 0 = 3 \rightarrow sp^2$The species with $sp^3$ hybridization are $NH_3, SiO_2, H_2O,$ and $CH_4$.Therefore,the total number of such species is $4$.

The number of species from the following in which the central atom uses $sp^3$ hybrid orbitals in its bonding is . . . . . . .
$NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3$

Similar Questions

The $d-$ orbitals which are involved in the hybridization of the central atom in $ICl_4^-$ are:

$H - C \equiv C - C \equiv C - CH_3$
Compare the bond lengths $a$ and $b$.

The type of hybridization of sulfur in $SO_2$ is ......

Discuss the hybridization and shape of $PCl_5$.

Two $p$-block elements $X$ and $Y$ form fluorides of the type $EF_3$. The fluoride compound $XF_3$ is a Lewis acid and $YF_3$ is a Lewis base. The hybridization of the central atoms of $XF_3$ and $YF_3$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The number of species from the following in which the central atom uses $sp^3$ hybrid orbitals in its bonding is . . . . . . .$NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3$