The correct option is D NH−2<NH3<NH+4 number of s and p - orbitals used in hybridisation
a) Bond angle is maximum in NH+4 due to absence of any lone pair. So, (a) is correct.
b) Number of lone pair maximum in NH−2 thus (b) is incorrect.
c) As the number of lone pair increases, p-character in the NH bond increases and hence NH bond length increases. Hence the correct order of bond length,
NH−2>NH3>NH+4(N−H) bond length
hence c is incorrect.
d)All the three species are sp3 hybridized and hence d is incorrect