Question

Which hybridization scheme occurs about nitrogen when nitrogen forms a double bond?

Answer 1

The electronic configuration of nitrogen is

1s2 2s2 2p3

Now during formation of double bond there will be formation of a pi bond

The nitrogen will undergo sp2 hybridization. One extra p orbital will form pi bond by side ways overlapping

the two sp2 hybridized orbital will form sigma bond with two atoms

the third sp2 hybridized orbital will have one lone pair of electrons .

Answer 2

The hybridization around nitrogen when it forms a double bond is $\boxed{s{p^2}}$ .

Further explanation:

Prediction of hybridization:

The hybridization can be determined by calculating the number of hybrid orbitals (X) which is formed by the atom. The formula to calculate the number of hybrid orbitals (X) is as follows:

 $\boxed{{\text{X = }}\frac{1}{2}\left[ {{\text{VE}} + {\text{MA}} - c + a}\right]}$

Where,

• VE is a total number of valence electrons of the central atom.

• MA is the total number of monovalent atoms/groups surrounding the central atom.

• c is the charge on the cation if the given species is a polyatomic cation.

• a is the charge on the anion if the given species is a polyatomic anion.

Note: In MA only monovalent species should be considered and for divalent atoms or groups MA is equal to zero.

Generally, the least electronegative atom is considered as the central atom. Calculate the hybridization as follows:

1. If the value of X is 2 then it means two hybrid orbitals are to be formed and thus the hybridization is sp.

2. If the value of X is 3 then it means three hybrid orbitals are to be formed and thus the hybridization is $s{p^2}$ .

3. If the value of X is 4 then it means four hybrid orbitals are to be formed and thus the hybridization is $s{p^3}$.

4. If the value of X is 5 then it means five hybrid orbitals are to be formed and thus the hybridization is $s{p^3}d$.

5. If the value of X is 6 then it means six hybrid orbitals are to be formed and thus the hybridization is $s{p^3}{d^2}$ .

The ground state electronic configuration for nitrogen (N) is,

 $1{s^2}2{s^2}2{p^3}$

Therefore, the valence electrons associated with nitrogen (N) atom are 5.

The two electrons of nitrogen are involved in the formation of the double bond and the electron present in 2s subshell will remain as the lone pair on nitrogen atom. There is only one electron left on the nitrogen atom out of five, and therefore the total number of monovalent atoms that can surrounding the central atom (MA) is 1.

Since the molecule is a neutral species and thus the value of a and c is 0.

Substitute these values in the above formula.

 $\begin{aligned}\text{X}&=\dfrac{1}{2}[5+1-0+0]\\&=\dfrac{1}{2}[6]\\&=\boxed{3}\end{aligned}$

Since the value of X is 3, it means 3 hybrid orbitals are to be formed and therefore the hybridization of nitrogen is ${\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}$ .

Learn more:

1. Molecular shape around the central atoms in the amino acid glycine: https://brainly.com/question/4341225

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: hybridization, nitrogen, ground state electronic configuration, sp2, valence electrons, monovalent atoms, VE, MA, a, c, X, 5, 0, N, least electronegative, central atom.