Total number of vacant orbitals in the valenc | vedclass

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

Question

(B) The atomic number of sulphur $(S)$ is $16$. Its ground state electronic configuration is $[Ne] 3s^2 3p^4 3d^0$. In the formation of $SF_4$,sulphur

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The atomic number of sulphur $(S)$ is $16$. Its ground state electronic configuration is $[Ne] 3s^2 3p^4 3d^0$. In the formation of $SF_4$,sulphur undergoes $sp^3d$ hybridization. To form $4$ bonds with fluorine atoms,sulphur needs $4$ unpaired electrons. One electron from the $3p_z$ orbital is excited to the $3d_{z^2}$ orbital. The excited state configuration becomes $[Ne] 3s^2 3p^3 3d^1$. In $sp^3d$ hybridization,one $3s$,three $3p$,and one $3d$ orbital participate. The valence shell of sulphur has $3s, 3p,$ and $3d$ orbitals. Total orbitals in the valence shell are $1 (3s) + 3 (3p) + 5 (3d) = 9$ orbitals. After hybridization,$5$ orbitals are used for bonding ($4$ for $S-F$ bonds and $1$ for the lone pair). The remaining orbitals in the valence shell are $9 - 5 = 4$ vacant $3d$ orbitals.

Total number of vacant orbitals in the valence shell of sulphur when it undergoes the formation of $SF_4$.

Similar Questions

Which molecule has $sp^3$ hybridization of the central atom?

Which type of overlap is involved in the formation of $O-H$ bonds in a water molecule?

Find the hybridisation of the $C_1$,$C_2$,and $C_3$ labelled carbon atoms in the given molecule: $H-C \equiv C_1-C_2H-C_3H=CH_2$

Which of the following molecules has unequal bond lengths?

What is the hybridization state of the chlorine atom in $ClO_4^-$,$ClO_3^-$,and $ClO_2^-$ ions,respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module