`XeF_(4)` is square planar where as `C"C"l_(4)` is tetrahedral because

Options

A. in `XeF_(4), 'xe'" is "sp^(3)` hybridised and in `C"C"l_(4),'C'` is `sp^(3)` hybridised
B. in both `XeF_(4) and C"C"l_(4)` the central atom is `sp^(3)` hybridised
C. in `XeF_(4), Xe" is "sp^(3)d^(2)` hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in `C"C"l_(4)` 'C' is `sp^(3)` hybridised
D. Xe is noble gas, whereas C is a non-metal

Answer

C

Explanation

`XeF_(4)" is "sp^(3) d^(2)` hybridised due to `4 sigma +2lp` and shape is square planar.
`C"C"l_(4)" is "sp^(3)` hybridised due to `4 sigma`- bond.