SF₂, SF₄ and SF₆ have the hybridisation at sulphur atom respectively as :

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${sp}^{3}, {spd}^{2}, d^{2} {sp}^{3}$

${sp}^{3}, {sp}^{3}, {sp}^{3} d^{2}$

${sp}^{3}, {sp}^{3} d, {sp}^{3} d^{2}$

${sp}^{2}, {sp}^{3}, {sp}^{2} d^{2}$

answer is C.

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Detailed Solution

Hybridisation :
$1 . : {\ddot{SF}}_{2} \Rightarrow \frac{1}{2} (6 + 2) = 4 = {sp}^{3} 2. : {SF}_{4} \Rightarrow \frac{1}{2} (6 + 4) = 5 = {sp}^{3} d 3. {SF}_{6} \Rightarrow \frac{1}{2} (6 + 6) = 6 = {sp}^{3} d^{2}$

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