The state hybridization of phosphorus in phosphoric acid is:

s p

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{s p}^{2}

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{s p}^{3}

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{s p}^{3} d

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Solution

The correct option is C ${s p}^{3}$
Number of sigma bonds attached to atom is considered as its hybridization state.
If 2 ,3,4 sigma bonds attached to atom then the hybridization state of that atom is considered as $s p, s p^{2}, s p^{3}$ respectively.
So in $H_{3} P O_{4}$ phosporous has four sigma bonds attached to it.So its hybridization is $s p^{3}$ .
Hence option C is correct.

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Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

s p, s p^{2} a n d s p^{3}

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