Question

Which of the following has maximum %s character in N-H bond?
A. ${{N}_{2}}{{H}_{2}}$
B. $N{{H}_{4}}^{+}$
C. $N{{H}_{3}}$
D. ${{N}_{2}}{{H}_{4}}$

Answer 1

When two atomic orbitals combine to form hybrid orbital in a molecule then redistribution of the energy of orbitals of individual atoms gives orbitals of equivalent energy. This process is called hybridization. The new orbitals thus formed are known as hybrid orbitals.

Complete step by step answer:
Hybrid orbitals are involved in mixing and hybridization is further classified in to $sp,s{{p}^{2}},s{{p}^{3}},s{{p}^{3}}d,s{{p}^{3}}{{d}^{2}},s{{p}^{3}}{{d}^{3}}$. The s$\%$ character generally tells us about the involvement of s species in the compound.
In case of sp s$\%$ character is 50$\%$ as there are only two orbitals i.e. s and p are present so they have same contribution character while in case of $s{{p}^{2}}$ three orbitals are there out of which two are p orbitals and one is s orbital so the s character is now decreases with increase in p orbitals here s character is 33.3$\%$ so we can say that with the increase in number of orbitals s character goes on decreasing.
In case of ${{N}_{2}}{{H}_{2}}$ hybridization is $s{{p}^{2}}$, so s character is 33.33$\%$
$N{{H}_{4}}^{+}$ shows $s{{p}^{3}}$ hybridization now s character is 25$\%$
$N{{H}_{3}}$ also shows $s{{p}^{3}}$ hybridization and s character remains 25$\%$
${{N}_{2}}{{H}_{4}}$ also shows $s{{p}^{3}}$ hybridization and s character remains 25$\%$

Hence the maximum s character is shown in ${{N}_{2}}{{H}_{2}}$ compound, thus we can say that option A is the correct answer.

Note: Only atomic orbitals will undergo hybridization which have equal energies and the number of hybrid orbitals formed is equal to the number of atomic orbitals mixing. Also we can say that it is not necessary that only the half-filled orbitals participate in hybridization even completely filled orbitals with slightly different energies can also participate.