Consider the species Cl2+, Cl2, and Cl2–. Which of these species will be paramagnetic?Only Cl2Cl2+ and Cl2Cl2 and Cl2–Cl2+ and Cl2–all three are paramagnetic

To determine which of the species $\text{Cl}_2^+, \text{Cl}_2,$ and $\text{Cl}_2^-$ are paramagnetic, we must analyze the electron configuration and the presence of unpaired electrons in each species.

1. Break Down the Problem

• Identify the electron configurations for $\text{Cl}_2^+, \text{Cl}_2,$ and $\text{Cl}_2^-$.
• Determine if each species has unpaired electrons.

2. Relevant Concepts

• A molecule is paramagnetic if it has unpaired electrons.
• The electron configuration of molecular chlorine $\text{Cl}_2$ can be built from the individual chlorine atoms.

3. Analysis and Detail

• Electron configuration of $\text{Cl}_2$:

  • The total number of electrons in $\text{Cl}_2$ is 14 (7 from each Cl atom). The appropriate molecular orbital configuration will have:
    • $\sigma(1s)^2$
    • $\sigma^*(1s)^2$
    • $\sigma(2s)^2$
    • $\sigma^*(2s)^2$
    • $\sigma(2p_z)^2$
    • $\pi(2p_x)^2$
    • $\pi(2p_y)^2$
    • $\pi^*(2p_x)^1$
  • This shows $\text{Cl}_2$ has 1 unpaired electron, making it paramagnetic.

• Electron configuration of $\text{Cl}_2^+$:

  • Removing one electron (from the highest occupied molecular orbital) gives:
    • $\sigma(1s)^2$
    • $\sigma^*(1s)^2$
    • $\sigma(2s)^2$
    • $\sigma^*(2s)^2$
    • $\sigma(2p_z)^2$
    • $\pi(2p_x)^2$
    • $\pi(2p_y)^2$
    • $\pi^*(2p_x)^0$
  • This shows $\text{Cl}_2^+$ has no unpaired electrons, making it diamagnetic.

• Electron configuration of $\text{Cl}_2^-$:

  • Adding one electron:
    • $\sigma(1s)^2$
    • $\sigma^*(1s)^2$
    • $\sigma(2s)^2$
    • $\sigma^*(2s)^2$
    • $\sigma(2p_z)^2$
    • $\pi(2p_x)^2$
    • $\pi(2p_y)^2$
    • $\pi^*(2p_x)^1$
    • This indicates that $\text{Cl}_2^-$ has one unpaired electron, making it paramagnetic.

4. Verify and Summarize

• Summary of findings:
  • $\text{Cl}_2^+$: diamagnetic (no unpaired electrons).
  • $\text{Cl}_2$: paramagnetic (1 unpaired electron).
  • $\text{Cl}_2^-$: paramagnetic (1 unpaired electron).

Final Answer

The species that will be paramagnetic are $\text{Cl}_2$ and $\text{Cl}_2^-$.