Pick out the incorrect statement for X e F 4

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According to VSEPR theory, six electron pairs form an octahedron with two positions occupied by lone pairs. This results in square planar shape of $X e F_{4}$ .
A. $\overset{(+ 4)}{3 X e F_{4}} + 6 H_{2} O \to \overset{(0)}{2 X e} + \overset{(+ 6)}{X e O_{3}} + 12 H F + \frac{3}{2} O_{2}$
$X e F_{4}$ is reduced to Xe and also oxidized to $X e O_{3}$ .
B. (i) $X e F_{4} + 2 S F_{4} \to X e + 2 S F_{6}$
(ii) $3 X e F_{4} + 2 P t \to 3 X e + 2 P t F_{6}$
(iii) $X e F_{4} + C_{6} H_{6} \to X e + C_{6} H_{5} F + H F$
D. $X e F_{4} + 4 K I \to 4 K F + X e + 2 I_{2}$

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