The increasing d-character in hybridisation of Xe in `XeF_(2), XeF_(4), XeF_(6)` is

Options

A. `sp^(2) , sp^(3) d, sp^(3) d^(2)`
B. `sp^(3) d, sp^(3) d^(2), sp^(3) d^(3)`
C. `sp^(3) d^(2), sp^(3) d, sp^(3) d^(3)`
D. `sp^(2), sp^(3), sp^(3) d`

Answer

B

Explanation

`XeF_(2) - sp^(3) d`,
Total no. of valence electrons = 22
` 22/8 = 2 (Q) + 6(R ) , 6/2 = 3 (Q)`
` X = 2 + 3 = 5`
Hybridisation is `sp^(3)` d.
`XeF_(4) - sp^(3) d^(2)`
Total no. of electrons in outermost shells `= 8 + 28 = 36 36/8 = 4 (Q) + 4(R) , 4/2 = 2 (Q) + 0 (R)`
` X = 4 + 2 + 0 = 6`
Hybridisation is `sp^(3) d^(2)`.
` XeF_(6) - sp^(3) d^(3)`
Total no. of valence electrons = ` 8 + 42 = 50 50/8 = 6 (Q) + 2 2 (R) , 2/2 = 1 (Q) , X = 6 + 1 = 7`
Hybridisation is `sp^(3) d^(3)`.