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Chapter 9: Problem 71

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

Short Answer

Expert verified
(a) The strongest C-N bond is present in CN鈦�, as it has the least number of antibonding electrons (4) compared to CN (5) and CN鈦� (6). (b) Out of the three species, only CN has unpaired electrons, with one unpaired electron in the 蟺*(2py) orbital. CN鈦� and CN鈦� have no unpaired electrons.

Step by step solution

01

Determine the total number of electrons for each ion

The number of electrons for each ion is determined by considering the atomic number of each element and their charge. Carbon (C) has 6 electrons and nitrogen (N) has 7 electrons. CN鈦�: Carbon (6) + Nitrogen (7) - 1 (positive charge) = 12 electrons CN: Carbon (6) + Nitrogen (7) = 13 electrons CN鈦�: Carbon (6) + Nitrogen (7) + 1 (negative charge) = 14 electrons
02

Determine the electron configuration using the molecular orbital theory

The molecular orbital theory uses atomic orbitals from the elements to create molecular orbitals. There are five molecular orbitals: 蟽(2s), 蟽*(2s), 蟽(2pz), 蟺(2px), and 蟺(2py). The order of these orbitals in terms of their energy is: 蟽(2s) < 蟽*(2s) < 蟺(2px) = 蟺(2py) < 蟽(2pz) Now, we fill these orbitals with the electrons of CN鈦�, CN, and CN鈦� to find their electron configurations.
03

Electron configurations of CN鈦�, CN, and CN鈦�

Using the molecular orbitals, we fill the electrons from the lowest energy level up. CN鈦� (12 electrons): 蟽(2s)虏, 蟽*(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏, 蟺*(2px)虏 CN (13 electrons): 蟽(2s)虏, 蟽*(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏, 蟺*(2px)虏, 蟺*(2py)鹿 CN鈦� (14 electrons): 蟽(2s)虏, 蟽*(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏, 蟺*(2px)虏, 蟺*(2py)虏
04

Identify the species with the strongest C-N bond

The bond strength is determined by the difference between bonding and antibonding electrons. The fewer antibonding electrons and the greater bonding electrons, the stronger the bond. From the electron configurations: CN鈦�: Bonding electrons = 8 (蟽(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏) Antibonding electrons = 4 (蟽*(2s)虏, 蟺*(2px)虏) CN: Bonding electrons = 8 (蟽(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏) Antibonding electrons = 5 (蟽*(2s)虏, 蟺*(2px)虏, 蟺*(2py)鹿) CN鈦�: Bonding electrons = 8 (蟽(2s)虏, 蟺(2px)虏, 蟺(2py)虏, 蟽(2pz)虏) Antibonding electrons = 6 (蟽*(2s)虏, 蟺*(2px)虏, 蟺*(2py)虏) Among the three species, CN鈦� has the strongest C-N bond as it has the least number of antibonding electrons.
05

Determine which species has unpaired electrons

The species with unpaired electrons should have an odd number of electrons in its antibonding orbitals. From the electron configurations: CN鈦�: All electrons are paired (none unpaired) CN: One unpaired electron in the 蟺*(2py) orbital CN鈦�: All electrons are paired (none unpaired) Hence, the CN species has one unpaired electron.

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Most popular questions from this chapter

Draw a picture that shows all three \(2 p\) orbitals on one atom and all three \(2 p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of \(2 p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of \(2 p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of \(2 p\) orbitals?

Why are there no \(s p^{4}\) or \(s p^{5}\) hybrid orbitals?

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Shown below are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type or types of hybridization that could lead to hybrid orbitals at the specified angle. |Section 9.5]

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