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Which of the following species have bond order \(3 ?\) (a) \(\mathrm{CN}^{-}\) (b) \(\mathrm{O}_{2}^{-}\) (c) \(\mathrm{NO}^{+}\) (d) \(\mathrm{CO}\)

Short Answer

Expert verified
\( \mathrm{CN}^{-} \), \( \mathrm{NO}^{+} \), and \( \mathrm{CO} \).

Step by step solution

01

Understanding Bond Order

Bond order is defined as the difference between the number of bonding electrons and the number of antibonding electrons divided by two. The formula for bond order is given by \( \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2} \). A bond order of 3 indicates a triple bond.
02

Calculating Bond Order for \( \mathrm{CN}^{-} \)

The molecular orbital configuration of \( \mathrm{CN}^{-} \) is similar to that of \( \mathrm{CO} \), but we can also write the electron configuration as \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^2 \). The bond order is \( \frac{10 - 4}{2} = 3 \).
03

Calculating Bond Order for \( \mathrm{O}_{2}^{-} \)

For \( \mathrm{O}_{2}^{-} \), the electron configuration would be \( \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x, 2p_y)^4 \pi^*(2p_x, 2p_y)^3 \). The bond order is \( \frac{9 - 4}{2} = 2.5 \).
04

Calculating Bond Order for \( \mathrm{NO}^{+} \)

The electron configuration for \( \mathrm{NO}^{+} \) can be viewed as similar to \( \mathrm{N}_{2} \), \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x, 2p_y)^4 \sigma(2p_z)^2 \). The bond order is \( \frac{10 - 4}{2} = 3 \).
05

Calculating Bond Order for \( \mathrm{CO} \)

The electron configuration for \( \mathrm{CO} \) is \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x, 2p_y)^4 \sigma(2p_z)^2 \). The bond order is \( \frac{10 - 4}{2} = 3 \).
06

Conclusion

Comparing the bond orders calculated, \( \mathrm{CN}^{-} \), \( \mathrm{NO}^{+} \), and \( \mathrm{CO} \) each have a bond order of 3, indicating they are triple bonded.

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Key Concepts

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

Molecular Orbital Theory
Molecular Orbital Theory (MOT) is an essential concept in understanding how atoms combine to form molecules. Instead of viewing electrons in individual atoms, MOT suggests that electrons are within molecular orbitals that extend over the entire molecule. These molecular orbitals are formed by the linear combination of atomic orbitals. There are two types of molecular orbitals:
  • Bonding Orbitals: These are lower in energy than the atomic orbitals they come from and stabilize the molecule.
  • Antibonding Orbitals: These are higher in energy and can destabilize the molecule if occupied by electrons.
The concept of bond order derives from MOT and can be calculated using the formula: \[ \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2} \]This calculation helps predict the strength and number of bonds formed in a molecule. In a simple way, a higher bond order usually means a stronger, shorter bond.
Electron Configuration
Electron configuration refers to the distribution of electrons within an atomic or molecular orbital structure. Understanding electron configuration is crucial for determining the chemical properties and reactivity of molecules. For instance, when analyzing molecules such as \( \mathrm{CN}^{-} \), \( \mathrm{NO}^{+} \), and \( \mathrm{CO} \), different electron configurations explain their bond order. Electron configurations for these molecules are often expressed in terms of sigma (\( \sigma \)) and pi (\( \pi \)) orbitals:
  • For \( \mathrm{CN}^{-} \), \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^2 \), yielding a bond order of 3.
  • For \( \mathrm{NO}^{+} \), \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x, 2p_y)^4 \sigma(2p_z)^2 \), also with a bond order of 3.
  • For \( \mathrm{CO} \), \( \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x, 2p_y)^4 \sigma(2p_z)^2 \), again with a bond order of 3.
By comparing these configurations, we can identify the presence of bonding and antibonding interactions and the overall molecular stability. This understanding helps chemists predict how molecules will interact and react with one another.
Triple Bonds
Triple bonds are a type of chemical bond involving three shared pairs of electrons between two atoms, resulting in a bond order of 3. They are characterized by their strength and short length compared to single and double bonds. In the context of molecular orbital theory, a triple bond typically involves:
  • One sigma \( (\sigma) \) bond - a head-on overlap of orbitals.
  • Two pi \( (\pi) \) bonds - side-on overlaps of adjacent electron orbitals.
An example of a molecule with a triple bond is carbon monoxide \( (\mathrm{CO}) \). This triple bond results in a very stable molecule. It's crucial in various chemical processes, such as synthesis reactions and as a fundamental unit in organic chemistry governing reaction mechanisms. Understanding triple bonds helps visualize molecular interactions in both theoretical and practical chemistry.

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Most popular questions from this chapter

Using MO theory predict which of the following species has the shortest bond length? (a) \(\mathrm{O}_{2}^{+}\) (b) \(\mathrm{O}_{2}^{-}\) (c) \(\mathrm{O}_{2}^{2-}\) (d) \(\mathrm{O}_{2}^{2+}\)

Match the following:. List I (Species) 1\. \(\mathrm{O}_{2}^{2+}\) 2\. \(\mathrm{O}_{2}\) 3\. \(\mathrm{F}_{2}\) 4\. \(\mathrm{O}_{2}^{+}\) List II (Bond order) (i) \(1.0\) (ii) \(2.0\) (iii) \(2.5\) (iv) \(3.0\)

Among the following species, identify the isostructural pairs \(\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{HN}_{3}\) (a) \(\left[\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}\right]\)and \(\left[\mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\) (b) \(\left[\mathrm{NF}_{3}, \mathrm{HN}_{3}\right]\) and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (c) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (d) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{HN}_{3}, \mathrm{BF}_{3}\right]\)

A square planar complex is formed by hybridization of the following atomic orbitals (a) s, \(\mathrm{p}_{\mathrm{x}}, \mathrm{p}_{y}, \mathrm{p}_{z}\) (b) \(s, p_{x}, p_{y}, p_{z}, d\) (c) \(\mathrm{d}_{x^{2}-y^{2}}, \mathrm{~s}, \mathrm{p}_{x}, \mathrm{p}_{\mathrm{y}}\) (d) \(\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{y}, \mathrm{p}_{2}, \mathrm{~d}_{z^{2}}\)

Which of the pairs have identical values of bond order? (a) \(\mathrm{F}_{2}\) and \(\mathrm{Ne}_{2}\) (b) \(\mathrm{N}_{2}^{+}\)and \(\mathrm{O}_{2}^{+}\) (c) \(\mathrm{O}_{2}^{2-}\) and \(\mathrm{B}_{2}\) (d) \(\mathrm{C}_{2}\) and \(\mathrm{N}_{2}\)

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