Question

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

a) The strongest C-N bond is in CN⁻, as it has the highest bond order of 1.5. b) Among the given species, only CN has unpaired electrons.

Step-by-step solution

Determine electron configurations of C and N

First, we need to determine the electron configurations for Carbon (C) and Nitrogen (N). Carbon (C) has 6 electrons, so its electron configuration is: \(1s^2 2s^2 2p^2\) Nitrogen (N) has 7 electrons, so its electron configuration is: \(1s^2 2s^2 2p^3\)

Determine the molecular orbital configurations and bond orders

Now, we will determine the molecular orbital (MO) configurations for CN⁺, CN, and CN⁻. CN⁺ has 12 electrons (6 from C and 7 from N minus 1): MO configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4\) CN has 13 electrons (6 from C and 7 from N): MO configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^5\) CN⁻ has 14 electrons (6 from C and 7 from N and one additional electron): MO configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^6\) Next, we will use the bond order formula to determine the strength of the C-N bond in each species: \(Bond\:Order = \frac{1}{2} (Electrons\:in\:bonding\:orbitals - Electrons\:in\:antibonding\:orbitals)\) Bond order for CN⁺: \(= \frac{1}{2}(6 - 4) = 1\) Bond order for CN: \(= \frac{1}{2}(6.5 - 4) = 1.25\) Bond order for CN⁻: \(= \frac{1}{2}(7 - 4) = 1.5\) A higher bond order indicates a stronger bond.

Identify species with unpaired electrons

We will now identify species having unpaired electrons by examining their molecular orbital configurations: CN⁺: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4\) - No unpaired electrons. CN: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^5\) - One unpaired electron. CN⁻: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^6\) - No unpaired electrons. a) The strongest C-N bond is in CN⁻, as it has the highest bond order of 1.5. b) Among the given species, only CN has unpaired electrons.

Key concepts

Electron Configuration

In the realm of Molecular Orbital Theory, understanding electron configurations is a fundamental step. Electrons are arranged in specific orbitals, and this configuration helps determine a molecule's chemical properties and behavior. For individual atoms like carbon (C) and nitrogen (N), electron configurations are straightforward. Carbon, with 6 electrons, follows the configuration: \(1s^2 2s^2 2p^2\). Nitrogen, having 7 electrons, fits into \(1s^2 2s^2 2p^3\).

When these atoms form molecules like \(\text{CN}^+, \text{CN},\) and \(\text{CN}^-\), the electron configuration becomes more complex. These molecules swim in a shared electron pool, and the total number of electrons changes according to their charge. For example, \(\text{CN}^+\) has 12 electrons because it loses one electron compared to neutral \(\text{CN}\), which has 13 electrons. On the other hand, \(\text{CN}^-\) gains an extra electron, leading to a total of 14 electrons. Each of these configurations impacts how the molecules bond, react, and behave chemically.

Bond Order

Bond order is a crucial concept in molecular orbital theory, telling us about the strength and stability of a bond between atoms. The bond order can be determined using the formula:

• \(\text{Bond Order} = \frac{1}{2} (\text{Electrons in bonding orbitals} - \text{Electrons in antibonding orbitals})\)

For \(\text{CN}^+, \text{CN},\) and \(\text{CN}^-\), the bond orders are calculated based on their molecular orbital configurations.

- **\(\text{CN}^+\):** With a bond order of 1, calculated from its MO configuration \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4\), it's the weakest among these species.

- **\(\text{CN}\):** Exhibits a modest bond order of 1.25, indicating moderate strength. This is because it has one and a half pairs of bonding electrons (\(6.5\) bonding electrons minus 4 antibonding) using its MO.

- **\(\text{CN}^-\):** The strongest C-N bond with a bond order of 1.5, attributed to having the highest number of electrons in bonding orbitals. In general, a higher bond order means a stronger, shorter, and more stable bond, as demonstrated by \(\text{CN}^-\).

Unpaired Electrons

Unpaired electrons in a molecular orbital diagram significantly affect the magnetic and reactive properties of the molecule. They are like lone dancers without a partner, often leading to interesting chemical behaviors. In the sequence of climates as \(\text{CN}^+, \text{CN},\) and \(\text{CN}^-\), we can observe these electrons by focusing on their molecular orbital (MO) configurations.

In \(\text{CN}^+\), all the electrons are paired, making it diamagnetic, which means it does not possess a net magnetic moment.

Conversely, \(\text{CN}\) has an odd number of electrons, leading to one unpaired electron depicted in the MO configuration \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^5\) - this last electron in the \(\pi_{2p}\) orbital has no partner. As a consequence, \(\text{CN}\) is paramagnetic, which means it is attracted by a magnetic field.

Finally, \(\text{CN}^-\) has all electrons paired due to the additional electron, rendering it diamagnetic as well. This formation highlights that the presence of unpaired electrons can be revealed through careful examination of the molecular orbital configurations, crucial for understanding the essence of molecular reactivity and interactions.