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Question: The hybridisation of the orbitals of N atom in \[{\rm{NO}}_3^ - {\rm{, NO}}_2^ + \] and \[{\rm{NH}}_...

The hybridisation of the orbitals of N atom in NO3−,NO2+{\rm{NO}}_3^ - {\rm{, NO}}_2^ + and NH4+{\rm{NH}}_4^ + are respectively:
A) sp,sp2,sp3sp,s{p^2},s{p^3}
B) sp2,sp,sp3s{p^2},sp,s{p^3}
C) sp,sp3,sp2asp,s{p^3},s{p^2}a
D) sp2,sp,sps{p^2},sp,sp

Explanation

Solution

Determine the number of atoms that are joined to the central atom. Also take into account the lone pairs to determine the hybridization.

Complete step by step answer:
Write the lewis dot structure of NO3−{\rm{NO}}_3^ - :

The central nitrogen atom has three bonding domains and zero lone pairs of electrons. Three bonding domains are obtained when three hybrid orbitals of nitrogen overlap with orbitals of three oxygen atoms. Three hybrid orbitals on nitrogen atom will be obtained by the hybridisation of three atomic orbitals of nitrogen atom. Thus, the nitrogen atom is sp2s{p^2} hybridised. The geometry of the ion is trigonal planar.

Write the lewis dot structure of NO2+{\rm{NO}}_2^ + :

The central nitrogen atom has two bonding domains and zero lone pairs of electrons. Two bonding domains are obtained when two hybrid orbitals of nitrogen overlap with orbitals of two oxygen atoms. Two hybrid orbitals on nitrogen atom will be obtained by the hybridisation of two atomic orbitals of nitrogen atom. Thus, the nitrogen atom is spsp hybridised. The geometry of the ion is linear.

Write the lewis dot structure o NH4+{\rm{NH}}_4^ + :

The central nitrogen atom has four bonding domains and zero lone pairs of electrons. Four bonding domains are obtained when four hybrid orbitals of nitrogen overlap with orbitals of four oxygen atoms. Four hybrid orbitals on nitrogen atom will be obtained by the hybridisation of four atomic orbitals of nitrogen atom. Thus, the nitrogen atom is sp3s{p^3} hybridised. The geometry of the ion is tetrahedral.

Hence, the hybridisation of the orbitals of N atom in NO3−,NO2+{\rm{NO}}_3^ - {\rm{, NO}}_2^ + and NH4+{\rm{NH}}_4^ + are sp2,sp,sp3s{p^2},sp,s{p^3} respectively.

Hence, option B) is the correct answer.

Note:
When a double bond is present, do not count it as two bonding domains. Consider it as only one bonding domain.