Observe the following statements:Statement-I: | vedclass

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

Question

(C) For ${SF}_6$: The central atom $S$ has $6$ valence electrons and forms $6$ bonds with $F$ atoms. The steric number is $6 + 0 = 6$,which correspond

Vedclass · Accepted Answer

Statement $I$ is not correct,but statement $II$ is correct

Vedclass · Answer

(C) For ${SF}_6$: The central atom $S$ has $6$ valence electrons and forms $6$ bonds with $F$ atoms. The steric number is $6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridisation and an octahedral geometry.For ${BrF}_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridisation. The geometry is square pyramidal due to the presence of one lone pair.Statement-$I$ says hybridisation is not the same,which is incorrect because both are $sp^3d^2$.Statement-$II$ says ${BrF}_5$ is square pyramidal and ${SF}_6$ is octahedral,which is correct.

Observe the following statements:
Statement-$I$: Hybridisation is not the same in both ${SF}_6$ and ${BrF}_5$.
Statement-$II$: ${BrF}_5$ is square pyramidal while ${SF}_6$ is octahedral in shape.
The correct answer is:

Similar Questions

The shape of the $C_2H_4$ (ethene) molecule is.......

Structure of $IF_4^+$ and hybridization of iodine in this structure are

The hybridizations of the central atom in the molecules $BF_3, BeF_2, BrF_3$ are respectively

Explain the tetrahedral structure of $CH_4$ (methane) using its bonding structure.

The order of electronegativity of carbon in $sp$,$sp^{2}$ and $sp^{3}$ hybridised states follows:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Observe the following statements:Statement-$I$: Hybridisation is not the same in both ${SF}_6$ and ${BrF}_5$.Statement-$II$: ${BrF}_5$ is square pyramidal while ${SF}_6$ is octahedral in shape.The correct answer is: