Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. a. \(\mathrm{CN}^{+}\) b. \(\mathrm{CN}\) c. \(\mathrm{CN}^{-}\)

Short Answer

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The electron configurations for \(\mathrm{CN}^{+}\), \(\mathrm{CN}\), and \(\mathrm{CN}^{-}\) are 1σ² 2σ² 1π⁴ 3σ¹, 1σ² 2σ² 1π⁴ 3σ², and 1σ² 2σ² 1π⁴ 3σ² 2π¹, respectively. The bond orders are 1.5, 1, and 0.5 respectively, with \(\mathrm{CN}^{+}\) and \(\mathrm{CN}^{-}\) being paramagnetic. The order of increasing bond length is \(\mathrm{CN}^{-}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}\), and the order of increasing bond energy is \(\mathrm{CN}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}^{-}\).

Step by step solution

01

Determine Electron Configurations

Firstly, write down the electron configurations of each element and count the total number of valence electrons in the molecule: Carbon (C): 1s² 2s² 2p² Nitrogen (N): 1s² 2s² 2p³ a. \(\mathrm{CN}^{+}\) (C and N have 9 valence electrons because of +1 charge): The molecular orbital model predicts the following electron arrangements: 1σ: 2 electrons 2σ: 2 electrons 1π: 4 electrons 3σ: 1 electron b. \(\mathrm{CN}\) (C and N have 10 valence electrons): The molecular orbital model predicts the following electron arrangements: 1σ: 2 electrons 2σ: 2 electrons 1π: 4 electrons 3σ: 2 electrons c. \(\mathrm{CN}^{-}\) (C and N have 11 valence electrons because of -1 charge): The molecular orbital model predicts the following electron arrangements: 1σ: 2 electrons 2σ: 2 electrons 1π: 4 electrons 3σ: 2 electrons 2π: 1 electron
02

Calculate Bond Orders and Identify Paramagnetic Species

The bond order is given by the formula: (number of electrons in bonding orbitals - number of electrons in anti-bonding orbitals)/2. We will use this formula to calculate bond orders for each diatomic species: a. \(\mathrm{CN}^{+}\): Bond Order (BO) = (6 - 3)/2 = 1.5 - Paramagnetic due to unpaired electron in 3σ orbital b. \(\mathrm{CN}\): BO = (6 - 4)/2 = 1 - Diamagnetic no unpaired electrons c. \(\mathrm{CN}^{-}\): BO = (6 - 5)/2 = 0.5 - Paramagnetic due to unpaired electron in 2π orbital
03

Place the Species in Order of Increasing Bond Length and Bond Energy

Higher bond order corresponds to shorter bond length and higher bond energy. So, \<bond length and energy of \( BO \) can be arranged as follows: Increasing Bond Length: \(\mathrm{CN}^{-}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}\) Increasing Bond Energy: \(\mathrm{CN}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}^{-}\) In conclusion, the electron configurations and bond orders have been determined for \(\mathrm{CN}^{+}\), \(\mathrm{CN}\) and \(\mathrm{CN}^{-}\). \(\mathrm{CN}^{+}\) and \(\mathrm{CN}^{-}\) were found to be paramagnetic species. The order of increasing bond length is \(\mathrm{CN}^{-}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}\), while the order of increasing bond energy is \(\mathrm{CN}\) < \(\mathrm{CN}^{+}\) < \(\mathrm{CN}^{-}\).

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Most popular questions from this chapter

Which of the following statements concerning \(\mathrm{SO}_{2}\) is(are) true? a. The central sulfur atom is \(s p^{2}\) hybridized. b. One of the sulfur-oxygen bonds is longer than the other(s). c. The bond angles about the central sulfur atom are about 120 degrees. d. There are two \(\sigma\) bonds in \(\mathrm{SO}_{2}\). e. There are no resonance structures for \(\mathrm{SO}_{2}\).

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