Which of the following has a geometry different from others?

Options

A. `BF_4^(-)`
B. `SO_4^(2-)`
C. `PH_4^(+)`
D. `XeF_4`

Answer

D

Explanation

In `BF_4^(-)`, the central B has 4 valence electrons due to one unit of negative charge. All are used up to form 4 bonds with F atoms giving zero lone pair. In `SO_4^(2-)`, the central S atom has 8 valence electrons due to two units of `-ve` charge. All are used up to form bonds with 4 O atoms giving zero lone pair. In `PH_4^(+)`, the central P atom has 4 valence electrons due to one unit of positive charge. All are used up to form bonds with H atoms giving zero lone pair. In `XeF_4`, the central Xe has 8 valence electrons. Four `e^(-)s` are used up to form bonds with F atoms giving two lone pairs. Thus, `BF_4^(-)`, `SO_4^(2-)`, and `PH_4^(+)` all have octahedral geometry due to SN equal to 6.