Question

The hybridization of atomic orbitals of nitrogen in $N{{O}_ {2}} ^ {+}, N{{O}_ {3}} ^ {-}, N{{H}_ {4}} ^ {+}$ are:
(A) $sp, s{{p}^ {3}}$ and $s{{p}^ {2}}$ respectively
(B) $sp, s{{p}^ {2}}$ and $s{{p}^ {3}}$ respectively
(C) $s{{p}^ {2}}, sp$ and $s{{p}^ {3}}$ respectively
(D) $s{{p}^ {2}}, s{{p}^ {3}}$ and $sp$ respectively

Answer 1

to solve this question, refer to the bond pairs, and the given charge in the given three ionic compounds: $N{{O}_{2}}^{+},N{{O}_{3}}^{-},N{{H}_{4}}^{+}$, we can easily find the hybridization of these compounds.

Complete answer:
We have been provided with three ionic compounds: $N{{O}_ {2}} ^ {+}, N{{O}_ {3}} ^ {-}, N{{H}_ {4}} ^ {+}$,
We need to find the hybridization of these compounds,
So, for that:
Firstly, we know that hybridization is the pairing of electrons to form chemical bonds in valence bond theory,
So, the first compound we have is $N{{O}_ {2}} ^ {+}$,
So, we know that electrons in the valence shell of nitrogen is 5 and that in oxygen is 6,
So, for finding the hybridization: 1/2 valence electron of central atom+ No. of surrounding monovalent atom- cationic charge+ anionic charge
So, for $N{{O}_ {2}} ^ {+}$: $\dfrac {1}{2} \left (5+0-0-1 \right) =2$,
So, the hybridization would be: $sp$,
The next compound we have is $N{{O}_ {3}} ^ {-}$,
So, we know that electrons in the valence shell of nitrogen is 5 and that in oxygen is 6,
So, for finding the hybridization: 1/2 valence electron of central atom+ No. of surrounding monovalent atom- cationic charge+ anionic charge
So, for $N{{O}_ {3}} ^ {-}$: $\dfrac {1}{2} \left (5+0-0+1 \right) =3$
So, the hybridization would be: $s{{p}^ {2}}$,
The next compound we have is $N{{H}_ {4}} ^ {+}$,
So, we know that electrons in the valence shell of nitrogen is 5 and that in hydrogen is 1,
So, for finding the hybridization: 1/2 valence electron of central atom+ No. of surrounding monovalent atom- cationic charge+ anionic charge
So, for $N{{H}_ {4}} ^ {+}$: $\dfrac {1}{2} \left (5+4-0-1 \right) =8$
Number of bonding orbitals=4,
Lone pair orbitals=0,
Total orbitals= bond orbitals + lone pair orbitals =4+0=4
So, the hybridization would be: $s{{p}^ {3}}$.

Note:
Hybridization does not accurately predict the observed spectroscopic energies of the species, which is the limitation of hybridized atomic orbitals, because of which it is little confusing.