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Chapter 10: Problem 35

What hybrid orbitals are used by nitrogen atoms in the following species? (a) \(\mathrm{NH}_{3}\), (b) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{NH}_{2}\) (c) \(\mathrm{NO}_{3}^{-}\)

Short Answer

Expert verified
The Nitrogen atom in \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{NH}_{2}\) uses \(sp^{3}\) hybrid orbitals, while in \(\mathrm{NO}_{3}^{-}\) it uses \(sp^{2}\) hybrid orbitals.

Step by step solution

01

Analyze each species separately

Here we need to assess three different molecules/species: \(\mathrm{NH}_{3}\), \(\mathrm{H}_{2} \mathrm{N}-\mathrm{NH}_{2}\), and \(\mathrm{NO}_{3}^{-}\). We'll determine the hybrid orbitals for the nitrogen atom in each species individually.
02

Determine the hybrid orbitals for nitrogen in \(\mathrm{NH}_{3}\)

In \(\mathrm{NH}_{3}\) (Ammonia), the nitrogen atom forms three sigma bonds with three hydrogen atoms and has one lone pair. According to the VSEPR theory, its shape is trigonal pyramidal. When forming bonds, the nitrogen atom uses three sp3 hybrid orbitals. The extra lone pair of electrons is also in an sp3 hybrid orbital. This results in \(sp^{3}\) hybridization.
03

Determine the hybrid orbitals for nitrogen in \(\mathrm{H}_{2} \mathrm{N}-\mathrm{NH}_{2}\)

In \(\mathrm{H}_{2} \mathrm{N}-\mathrm{NH}_{2}\) (Hydrazine), each nitrogen atom forms two sigma bonds with two hydrogen atoms and one sigma bond with another nitrogen atom. Each nitrogen atom has one lone pair of electrons. The molecule has a bend shape ( according to VSEPR theory). Each Nitrogen atom uses three sp3 hybrid orbitals to form bonds, and the lone pair is also in an sp3 hybrid orbital. Hence, hybridization is \(sp^{3}\).
04

Determine the hybrid orbitals for nitrogen in \(\mathrm{NO}_{3}^{-}\)

In \(\mathrm{NO}_{3}^{-}\) (Nitrate Ion), the central nitrogen atom is surrounded by three oxygen atoms in a plane. Hence it exhibits a planar molecular geometry. The nitrogen atom forms three sigma bonds with three oxygen atoms and there are no lone pairs of electrons. Therefore, all three sp2 hybrid orbitals are used for bond formation. So, the hybridization is \(sp^{2}\).

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Most popular questions from this chapter

The stable allotropic form of phosphorus is \(\mathrm{P}_{4},\) in which each \(\mathrm{P}\) atom is bonded to three other \(\mathrm{P}\) atoms. Draw a Lewis structure of this molecule and describe its geometry. At high temperatures, \(\mathrm{P}_{4}\) dissociates to form \(\mathrm{P}_{2}\) molecules containing a \(\mathrm{P}=\mathrm{P}\) bond. Explain why \(\mathrm{P}_{4}\) is more stable than \(\mathrm{P}_{2}\).

Predict the geometries of the following species: (a) \(\mathrm{AlCl}_{3},\) (b) \(\mathrm{ZnCl}_{2}\), (c) \(\mathrm{ZnCl}_{4}^{2-}\).

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) has a tendency to lose two protons \(\left(\mathrm{H}^{+}\right)\) and form the carbide ion \(\left(\mathrm{C}_{2}^{2-}\right),\) which is present in a number of ionic compounds, such as \(\mathrm{CaC}_{2}\) and \(\mathrm{MgC}_{2}\). Describe the bonding scheme in the \(\mathrm{C}_{2}^{2-}\) ion in terms of molecular orbital theory. Compare the bond order in \(\mathrm{C}_{2}^{2-}\) with that in \(\mathrm{C}_{2}\).

In 2009 the ion \(\mathrm{N}_{2}^{3-}\) was isolated. Use a molecular orbital diagram to compare its properties (bond order and magnetism) with the isoelectronic ion \(\mathrm{O}_{2}^{-}\)

Explain in molecular orbital terms the changes in \(\mathrm{H}-\mathrm{H}\) internuclear distance that occur as the \(\mathrm{mo}-\) lecular \(\mathrm{H}_{2}\) is ionized first to \(\mathrm{H}_{2}^{+}\) and then to \(\mathrm{H}_{2}^{2+}\).
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