$2 B F_{3} (g) + 6 N a H 180^{o} C - --- \to P (g) + Q (s)$
Find out maximum number of atom(s) that can lie in a plane of covalent molecule 'P'.

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Solution

$2 B F_{3} + 6 N a H 180^{o} C - --- \to B_{2} H_{6} (P) + 6 N a F$

In $B_{2} H_{6}$ , two $B$ atoms and $4$ H atoms are present in same plane, but 1 bridge $H$ atom is present below the plane and another above the plane.

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