From: $S F_{4}, B r F_{3}, X e F_{2}, I F_{5}, S b F_{4}^{-} . S F_{5}^{-}, S e F_{3}^{+}, C H_{3}^{+}, P C l_{4}^{+}$ , the total number of planar species is/are:

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Solution

From the given compounds, only 3 are planar:
The easiest one among them is the methyl carbocation $C H_{3}^{+}$ . This is clearly planar.

$X e F_{2}$ is linear and hence planar. $B r F_{3}$ - like $C l F_{3}$ , and $I F_{3}$ - are T-shaped and planar. Use VSEPR.

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