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Q.

In hydrazine (N2H4) the hybridization of nitrogen is

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a

sp

b

sp2

c

sp3

d

dsp2

answer is C.

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Detailed Solution

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The Lewis structure of N2H4 is

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Every N atom has one lone pair of electrons and is bonded to three other atoms (one N and two H atoms). Thus, steric number of every N  is 3 + 1 = 4 . This implies that four hybrid orbitals are needed by every N which is possible only through sp3  hybridization.

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