Total number of electron lone pair on central atom of A, B and C.

Question

TestHub · Accepted Answer

Step 1: XeF₆ reacts with NaF to form an addition compound A:

$X e F_{6} + N a F \to N a^{+} [X e F_{7}]^{-} (A)$

In $[X e F_{7}]^{-}$ , the central Xe atom has 14 valence-shell electrons involved in 7 bonds, leaving 1 lone pair.

Step 2: At $50 ℃$ , NaXeF₇ undergoes controlled thermal decomposition to yield B and C:

$N a X e F_{7} 50 ℃ N a F (B) + X e F_{6} (C)$

Step 3: In the ionic compound NaF (B), there is 0 lone pair on a central atom.

Step 4: In XeF₆ (C), the central Xe atom has 8 valence electrons forming 6 bonds, leaving 1 lone pair.

Total number of lone pairs = $1 (in A) + 0 (in B) + 1 (in C) = 2.00$

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