The percentage of s-character in the orbital forming $P - S$ bonds in $P_{4} S_{3}$ is:

25 %

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33 %

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75 %

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50 %

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Solution

The correct option is A $25 %$
The percentage of s-character in the orbital forming $P - S$ bonds in $P_{4} S_{3}$ is $25 %$ due to $s p^{3}$ hybridization.

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Similar questions

The percentage of p-character in forming P-P bonds in P₄ molecule is:

The percentage of p-character in the orbitals forming p – p bonds in $P_{4}$ is

N H_{3}

S_{1} : S e F_{6}

P F_{5}

s p^{3} d^{2}

S_{2}

S_{8}

P_{4}

S_{3} :

[C o (N H_{3})_{6}]^{3 +}

P_{4}

P_{4}

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