Question

What is hybridisation in XeF5- ?​

Answer 1

Answer:

sp3d2

Explanation:

steric no = 1/2(no. of valence electrons + no. of singly bonded species with the atom whose hybridisation is to be determined - cationic charge on molecule + anionic charge on molecule)

Thus in XeF5+

steric no. = 1/2(8+5–1)=6

Hybridisation is sp3d2

Answer 2

The hybridization of Xe in $XeF_5^-$  is, $sp^3d^3$

Explanation :

Formula used  :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, $XeF_5^-$

$\text{Number of electrons}=\frac{1}{2}\times [8+5+1]=7$

The number of electron pair are 7 that means the hybridization will be $sp^3d^3$ and the electronic geometry of the molecule will be pentagonal bipyramidal.

But as there are 5 atoms around the central xenon atom, the 6th and 7th position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be different.

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