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a
NH3
b
N(CH3)3
c
N(SiH3)3
d
NF3
answer is C.
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Detailed Solution
Since the conditions of back bonding is satisfied in N(SiH3)3, the lone pair on the N atom remains in pure p orbital so that it can get delocalised into the empty orbitals of Si. Therefore, the hybridisation of N is sp2. In the given molecule the bond angle is 1200.