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On the basis of molecular orbital theory, would you expect \(\mathrm{NH}_{2}^{+}\) to be linear or bent?

Short Answer

Expert verified
According to the molecular orbital theory, \(\mathrm{NH}_{2}^{+}\) would be expected to have a linear shape.

Step by step solution

01

Determine the total number of valence electrons

Firstly, count the total number of valence electrons in the molecule. Nitrogen has 5 valence electrons and hydrogen has 1 valence electron. Since there is a positive charge on the molecule, it means one electron has been removed. Summing it up, \(\mathrm{NH}_{2}^{+}\) has \(5 + 2*1 - 1 = 6\) valence electrons.
02

Ascertain the type of molecular orbitals

Molecular orbitals combine from atomic orbitals. Two atomic orbitals combine to form two molecular orbitals, one bonding and the other antibonding. The types of molecular orbitals important to forming \(\mathrm{NH}_{2}^{+}\) include sigma \(\mathrm{s}\), sigma \(\mathrm{s}*\), sigma \(\mathrm{p}\), pi \(\mathrm{p}\) and pi \(\mathrm{p}*\). In this case, nitrogen contributes 2s and 2p atomic orbitals and hydrogen contributes 1s atomic orbitals.
03

Fill electrons in molecular orbitals according to the order

Electrons are filled in ascending order of energy levels in molecular orbitals. The first two electrons will occupy the sigma \(\mathrm{s}\) orbital, the next two will occupy the sigma \(\mathrm{s}*\) orbital, and the last two electrons will occupy the sigma \(\mathrm{p}\) orbital.
04

Decide the shape of molecule

From the molecular orbital theory, it is known the sigma \(\mathrm{p}\) orbital leads to linear geometry. After filling the electrons, we found that the six electrons of \(NH_{2}^{+}\) occupy sigma \(\mathrm{s}\), sigma \(\mathrm{s}*\) and sigma \(\mathrm{p}\) orbitals. There are no leftover lone pair electrons to cause a bend in the molecule, so it would be expected that \(\mathrm{NH}_{2}^{+}\) is linear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in determining how atoms interact and bond with each other. In the context of molecular orbital theory, knowing the number of valence electrons is essential because it helps predict the molecular structure and bonding characteristics of a molecule. For \(\mathrm{NH}_{2}^{+}\), we start by calculating the number of valence electrons for each atom. Nitrogen (N) in its neutral form has 5 valence electrons, while each hydrogen (H) atom contributes one valence electron. Because the molecule carries a positive charge, one electron is removed, resulting in 6 valence electrons total: 5 from nitrogen, 2 from hydrogens, minus 1 for the charge. This configuration of valence electrons lays the foundation for determining the molecular orbitals that will form.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule, which is crucial in determining how the molecule interacts with other molecules. In molecular orbital theory, the geometry is largely influenced by the distribution of electrons in the molecular orbitals. For instance, in the case of \(\mathrm{NH}_{2}^{+}\), the molecular geometry is determined by how the electrons fill the available molecular orbitals.
In general, linear molecular geometry occurs when there are no lone pairs to bend the molecule. In \(\mathrm{NH}_{2}^{+}\), after filling the molecular orbitals according to their energy levels, the electrons occupy the sigma-s, sigma-s*, and sigma-p orbitals. There are no additional lone pairs that could cause bending, so the molecule is predicted to be linear. Understanding the molecular geometry helps in predicting the behavior and interactions of the molecule in various chemical reactions.
Sigma and Pi Bonds
Sigma (\(\sigma\)) and pi (\(\pi\)) bonds are types of covalent bonds that form between atoms in a molecule. Each serves a distinct purpose and contributes to the molecule's overall stability and form. In molecular orbital theory, these bonds are described in terms of the overlap of atomic orbitals.
- **Sigma Bonds (\(\sigma\)):** These bonds occur when atomic orbitals overlap head-on. In the MO of \(\mathrm{NH}_{2}^{+}\), the first molecular orbitals filled are sigma orbitals (sigma-s, followed by sigma-s*), which generally determine the backbone and basic framework of the molecule.- **Pi Bonds (\(\pi\)):** These bonds form when two atomic orbitals overlap sideways. Although pi bonds are not significant in the formation of \(\mathrm{NH}_{2}^{+}\) as a linear molecule, they play an essential role in other molecules where additional bonding is needed.Understanding the distinction between these bonds helps in knowing not just the shape but also the strength and properties of a molecule. In \(\mathrm{NH}_{2}^{+}\), however, sigma bonds are more pronounced given its linear geometry.

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Most popular questions from this chapter

Without doing detailed calculations, explain in which of the following materials you would expect to find the greatest mass percent of hydrogen: seawater, the atmosphere, natural gas \(\left(\mathrm{CH}_{4}\right),\) ammonia.

The oxides of the phosphorus(III), antimony(III), and bismuth(III) are \(\mathrm{P}_{4} \mathrm{O}_{6}, \mathrm{Sb}_{4} \mathrm{O}_{6},\) and \(\mathrm{Bi}_{2} \mathrm{O}_{3} .\) Only one of these oxides is amphoteric. Which one? Which of these oxides is most acidic? Which is most basic?

Figure \(15-1\) (page 656 ) shows that \(I_{2}\) is considerably more soluble in \(\mathrm{CCl}_{4}(1)\) than it is in \(\mathrm{H}_{2} \mathrm{O}(1) .\) The concentration of \(I_{2}\) in its saturated aqueous solution is \(1.33 \times 10^{-3} \mathrm{M},\) and the equilibrium achieved when \(\bar{I}_{2}\) distributes itself between \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CCl}_{4}\) is $$\mathrm{I}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) \quad K_{\mathrm{c}}=85.5$$ (a) \(\mathrm{A} 10.0 \mathrm{mL}\) sample of saturated \(\mathrm{I}_{2}(\mathrm{aq})\) is shaken with \(10.0 \mathrm{mL} \mathrm{CCl}_{4} .\) After equilibrium is established, the two liquid layers are separated. How many milligrams of \(I_{2}\) will be in the aqueous layer? (b) If the \(10.0 \mathrm{mL}\) of aqueous layer from part (a) is extracted with a second \(10.0 \mathrm{mL}\) portion of \(\mathrm{CCl}_{4}\) how many milligrams of \(\mathrm{I}_{2}\) will remain in the aqueous layer when equilibrium is reestablished? (c) If the 10.0 mL sample of saturated \(I_{2}(\) aq) in part (a) had originally been extracted with \(20.0 \mathrm{mL} \mathrm{CCl}_{4}\) would the mass of \(I_{2}\) remaining in the aqueous layer have been less than, equal to, or greater than that in part (b)? Explain.

Explain why the volumes of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) obtained in the electrolysis of water are not the same.

A 55 L cylinder contains \(A r\) at 145 atm and \(26^{\circ}\) C. What minimum volume of air at STP must have been liquefied and distilled to produce this Ar? Air contains \(0.934 \%\) Ar, by volume.

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