Question

What is the hydridization of the central atom in XeF5+

Answer 1

Answer:

To find out the hybridization, first you need to find steric number of the central atom by the formula

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.

Xe valance e- = 8

Xe+ valance e- = 7 (we assume the overall charge on central atom)

bond pair= 5

lone pair=1

total bond pair= 5+ 1=6

hybridization= sp3d2

shape= square pyramidal

Hope it helps :)

Answer 2

The hydridization of the central atom in $XeF_5^+$ is $sp^3d^2$.

• The formula to determine hybridization state is -1/2(total valance electrons+number of single bonded atoms-charge of cation+charge of annion)
• Xe has total 8 valance electrons and number of single bonded atoms are 5. Charge on Xe is +1.
• So, hybridization is 1/2(8+5-1) =6
• So, it posses $sp^3d^2$ hybridization.