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The percentage of p -character in the orbital forming P-P bond in $P_4$ is

$\begin{array}{1 1}(a)\;25\\(b)\; 33\\(c)\; 50\\(d)\; 75\end{array}$

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$P_4$ has $sp^3$ hybridization. Therefore 75 % p-character and 25 % S-character
Hence (d) is the correct answer.
answered Dec 9, 2013 by sreemathi.v

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