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Chapter 10: Problem 38

What is the hybridization state of the central \(\mathrm{N}\) atom in the azide ion, \(\mathrm{N}_{3}^{-} ?\) (Arrangement of atoms: NNN.)

Short Answer

Expert verified
The hybridization state of the central Nitrogen (N) atom in the azide ion (\(\mathrm{N}_{3}^{-}\)) is 'sp'.

Step by step solution

01

Determine the central atom

Identify the central atom in the molecule. In this case, the central atom is the middle Nitrogen (N) atom in the azide ion (\(\mathrm{N}_{3}^{-}\)).
02

Determine the number of sigma bonds and lone pairs around the Nitrogen atom

The Nitrogen (N) atom in the center is bonded to the two other Nitrogen atoms. Each bond counts as 1 sigma bond, which gives us a total of 2 sigma bonds. Since there are no other atoms around the central N atom, there are no lone pairs.
03

Identify the Hybridization state

The Hybridization state is determined by the sum of sigma bonds and lone pairs. This gives 2 (from Step 2). So the hybridization state is 'sp', where 's' represents one of the electron orbitals and 'p' represents the other electron orbital linked by two sigma bonds.

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Most popular questions from this chapter

Consider the reaction $$ \mathrm{BF}_{3}+\mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B}-\mathrm{NH}_{3} $$ Describe the changes in hybridization (if any) of the \(B\) and \(N\) atoms as a result of this reaction

Antimony pentafluoride, \(\mathrm{SbF}_{5}\), reacts with \(\mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\) to form ionic compounds, \(\mathrm{XeF}_{3}^{+} \mathrm{SbF}_{6}^{-}\) and \(\mathrm{XeF}_{5}^{+} \mathrm{SbF}_{6}^{-}\). Describe the geometries of the cations and anion in these two compounds.

Draw a potential energy curve for the bond formation in \(\mathrm{F}_{2}\).

The molecules cis-dichloroethylene and transdichloroethylene shown on p. 324 can be interconverted by heating or irradiation. (a) Starting with cis-dichloroethylene, show that rotating the \(\mathrm{C}=\mathrm{C}\) bond by \(180^{\circ}\) will break only the pi bond but will leave the sigma bond intact. Explain the formation of trans- dichloroethylene from this process. (Treat the rotation as two, stepwise \(90^{\circ}\) rotations.) (b) Account for the difference in the bond enthalpies for the pi bond (about \(270 \mathrm{~kJ} / \mathrm{mol}\) ) and the sigma bond (about \(350 \mathrm{~kJ} / \mathrm{mol}\) ). (c) Calculate the longest wavelength of light needed to bring about this conversion.

Which of these pairs of atomic orbitals of adjacent nuclei can overlap to form a sigma bond? Which overlap to form a pi bond? Which cannot overlap (no bond)? Consider the \(x\) -axis to be the internuclear axis, that is, the line joining the nuclei of the two atoms. (a) \(1 s\) and \(1 s,\) (b) \(1 s\) and \(2 p_{x},\) (c) \(2 p_{x}\) and \(2 p_{y}\) (d) \(3 p_{y}\) and \(3 p_{y},\) (e) \(2 p_{x}\) and \(2 p_{x}\), (f) 1 s and 2 s.
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