Solveeit Logo
Login

Question

Question: The species in which the N-atom is in a state of sp hybridization A. \(N{{O}_{2}}^{-}\) B. \(N{{...

The species in which the N-atom is in a state of sp hybridization
A. NO2−N{{O}_{2}}^{-}
B. NO3−N{{O}_{3}}^{-}
C. NO2N{{O}_{2}}
D. NO2+N{{O}_{2}}^{+}

Explanation

Solution

. Hybridization is defined as the concept of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals. The atomic orbitals of the same energy level can only take part in hybridization and both full filled and half-filled orbitals can also take part in this process if they have equal energy.

Complete step by step answer:
Redistribution of the energy of orbitals of individual atoms to give orbitals of equivalent energy. Hybridization is a process in when two atomic orbitals combine to form hybrid orbital in a molecule and the new orbitals formed are known as hybrid orbitals. Based on the types of orbitals involved in mixing hybridization can be classified in to sp,sp2,sp3,sp3d,sp3d2,sp3d3sp,s{{p}^{2}},s{{p}^{3}},s{{p}^{3}}d,s{{p}^{3}}{{d}^{2}},s{{p}^{3}}{{d}^{3}}.

Atomic orbitals with equal energies undergo hybridization. The number of hybrid orbitals formed is equal to the number of atomic orbitals mixing. This hybridization is based on the combination of molecules i.e. if s orbital is combine with p orbital then it is said to have sp hybridized molecule in similar manner if two p orbitals combined with s orbital it is said to have sp2s{{p}^{2}} configuration and so on.
Hybridization of a molecule or ion can be calculated by the formula:
½[Group number central atom + number of atoms attached to central metal atom (like F, H, Cl except O + magnitude of negative charge - magnitude of positive charge]

Hybridization of NO2N{{O}_{2}}= 12[5+0+0−0]=52=2.5\dfrac{1}{2}[5+0+0-0]=\dfrac{5}{2}=2.5 will not give stable hybridization.
NO2−N{{O}_{2}}^{-}= 12[5+0+1−0]=62=3\dfrac{1}{2}[5+0+1-0]=\dfrac{6}{2}=3 i.e. sp2s{{p}^{2}} configuration.
NO3−N{{O}_{3}}^{-}= 12[5+0+1−0]=62=3\dfrac{1}{2}[5+0+1-0]=\dfrac{6}{2}=3 in this case hybridization remains same as in previous case as oxygen atom will not involve.
NO2+N{{O}_{2}}^{+}= 12[5+0+0−1]=42=2\dfrac{1}{2}[5+0+0-1]=\dfrac{4}{2}=2 i.e. sp configuration.
Hence we can say in NO2+N{{O}_{2}}^{+} species N-atom is in a state of sp hybridization
So, the correct answer is “Option D”.

Note: It is not necessary that all the half-filled orbitals must participate in hybridization. Even completely filled orbitals with slightly different energies can also participate. Hybridization happens only during the bond formation and not in an isolated gaseous atom. The shape of the molecule can be predicted if hybridization of the molecule is known. The bigger lobe of the hybrid orbital always has a positive sign, while the smaller lobe on the opposite side has a negative sign.