The shape of XeF_{5}^{-} will be | vedclass

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

Question

(C) To determine the shape of $XeF_{5}^{-}$, we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac

Vedclass · Accepted Answer

planar

Vedclass · Answer

(C) To determine the shape of $XeF_{5}^{-}$, we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac{1}{2} [V + M - C + A] = \frac{1}{2} [8 + 5 - 0 + 1] = 7$. Here, $V = 8$ (valence electrons of $Xe$), $M = 5$ (monovalent atoms), and $A = 1$ (anionic charge). With $7$ electron pairs, the electron geometry is pentagonal bipyramidal ($sp^{3}d^{3}$ hybridization). Out of these $7$ pairs, $5$ are bond pairs and $2$ are lone pairs. The two lone pairs occupy the axial positions to minimize repulsion, leaving the $5$ $Xe-F$ bonds in a single equatorial plane. Therefore, the molecular shape of $XeF_{5}^{-}$ is pentagonal planar.

The shape of $XeF_{5}^{-}$ will be

Similar Questions

Which of the following set of species have planar structures?

What is the $H-S-H$ bond angle in $H_2S$ (in $^{\circ}$)?

Which of the following species has tetrahedral geometry?

Which of the following is not linear?

In $ClF_3$ molecule,if the lone pair of the central atom occupied the axial position,then which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module