Q60E
Question
For each of the following, indicate the hybridization of the nitrogen atom (for\({{\bf{N}}_{\bf{3}}}\),the central nitrogen).
(a)\({{\bf{N}}_{\bf{2}}}{{\bf{F}}_{\bf{4}}}\)
(b)\(\;{\bf{N}}{{\bf{H}}_{\bf{2}}}^ - \)
(c) \({\bf{N}}{{\bf{F}}_{\bf{3}}}\)
(d) \({{\bf{N}}_{\bf{3}}} - \)
Step-by-Step Solution
VerifiedHybridization of the nitrogen atom is the following:
a) \({N_2}{F_4} - s{p^3}\)
b) \({\rm{NH}}_2^ - - s{p^3}\)
c) \(N{F_3} - s{p^3}\)
d) \(N_3^ - - sp\)
The nitrogen has four \(s{p^3}\) hybrids orbital since it is \(s{p^3}\) hybridized. The two \(N - H\) sigma bonds are formed when two of the \(s{p^3}\) hybridized orbital collide with \(s\) orbital from hydrogen.
a.
\({N_2}{F_4}\) Possesses \(s{p^3}\) hybridization and bond angles smaller than \(109.5\)degrees due to a lone pair present in Nitrogen.
b.
- The hybridization type of \(N{H_2} - \) is \(s{p^3}\).
- Four areas on the central Nitrogen atom are responsible for detecting hybridization.
- Two unbond electron pairs and two sigma bonds are present in N.
- \(N{H_2} - \) exhibits \(s{p^3}\) hybridization because of these four areas around the core nitrogen atom.
c.
- \(N{F_3}\) has a pyramidal structure (the core atom is \(s{p^3}\) hybridized) and a lone pair of electrons in the fourth orbital.
d.
- The central atom hybridization in \({N_3},{\rm{ }}NOCl,{\rm{ }}and{\rm{ }}{N_2}O{\rm{ }}is{\rm{ }}sp,{\rm{ }}s{p^2},{\rm{ }}sp,\) respectively.
- In the azide ion, the core N atom has two bonding domains and no lone pairs of electrons. In \(NOCl\), the core \(O\) atom has two bonding domains and one lone pair of electrons.
Hence, Hybridization of the nitrogen atom is the following:
a) \({N_2}{F_4} - s{p^3}\)
b) \({\rm{NH}}_2^ - - s{p^3}\)
c) \(N{F_3} - s{p^3}\)
d) \(N_3^ - - sp\)