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Chapter 9: Problem 73

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

Short Answer

Expert verified
The linear and polar N2O molecule has the arrangement NNO. The Lewis structure is :N≡N-O:, and the formal charge on each atom is 0. The central nitrogen atom is sp hybridized. The triple bond between the nitrogen atoms consists of one σ bond from the overlap of sp-hybrid orbitals and two π bonds from the overlap of unhybridized p-orbitals. The single bond between the central nitrogen and the oxygen is formed due to the overlap of the other sp-hybrid orbital on the nitrogen and the hybrid orbital on the oxygen atom.

Step by step solution

01

(Determine the arrangement of N2O molecule)

We are given that the \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. The possible arrangements are NNO or NON. Since the molecule is polar, the more electronegative atom must be on one side, while the less electronegative atoms should be on the other side to maintain the polarity. Since oxygen is more electronegative than nitrogen, the correct arrangement is NNO.
02

(Drawing the Lewis structure of N2O)

To draw the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\), we need to count the total number of valence electrons available for bonding. Nitrogen has 5 valence electrons and oxygen has 6. There are two nitrogen atoms, so the total number of electrons is (2 x 5) + 6 = 16. The Lewis structure is: :N≡N-O:
03

(Determine the formal charges and hybridization of the central atom)

To find the formal charges, we use the formula: Formal charge = (valence electrons) - (non-bonding electrons) - 0.5 x (bonding electrons) For the leftmost nitrogen: Formal charge = 5 - 2 - 0.5 x (8) = 0 For the central nitrogen: Formal charge = 5 - 0 - 0.5 x (10) = 0 For the oxygen: Formal charge = 6 - 4 - 0.5 x (4) = 0 To determine the hybridization of the central nitrogen atom, we look at the electron regions around it. Since the central nitrogen is triple bonded to one nitrogen and single bonded to one oxygen, there are two regions of electrons around it. Thus, it is sp hybridized.
04

(Describe the multiple bonding in terms of orbitals)

The multiple bonding between the two nitrogen atoms (the triple bond) can be described in terms of orbitals as follows: 1. One \(\sigma\) bond is formed due to the overlap of the two sp-hybrid orbitals on the nitrogen atoms. 2. Two \(\pi\) bonds are formed due to the overlap of two pairs of unhybridized p-orbitals on the nitrogen atoms. The single bond between the central nitrogen and the oxygen atom is formed due to the overlap of the other sp-hybrid orbital on the nitrogen and the hybrid orbital on the oxygen atom.

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Most popular questions from this chapter

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) can be produced from the reaction of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion \(\left(\mathrm{C}_{2}^{2-}\right)\)

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\). Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\). Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and NO than the hybrid orbital model?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C} \equiv \mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43 .\) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.
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