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Question: Which of the following pairs of species have largest difference in spin only magnetic moment?...

Which of the following pairs of species have largest difference in spin only magnetic moment?

A

O2_2,O2+_2^+

B

O2_2,O2−_2^-

C

O2+_2^+,O22−_2^{2-}

D

O2−_2^-,O2+_2^+

Answer

O2+_2^+,O22−_2^{2-}

Explanation

Solution

To solve this problem, we need to determine the number of unpaired electrons (nn) for each given oxygen species (O2O_2, O2+O_2^+, O2−O_2^-, O22−O_2^{2-}) using Molecular Orbital Theory (MOT). The spin-only magnetic moment (μs\mu_s) is then calculated using the formula μs=n(n+2)\mu_s = \sqrt{n(n+2)} B.M.

The molecular orbital (MO) configuration for O2O_2 and its ions is based on the filling of molecular orbitals formed from atomic orbitals of oxygen atoms. The relevant MOs and their energy order for O2O_2 are: σ1s<σ1s∗<σ2s<σ2s∗<π2p<σ2p<π2p∗<σ2p∗\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \pi_{2p} < \sigma_{2p} < \pi^*_{2p} < \sigma^*_{2p}

Let's determine the electronic configuration and the number of unpaired electrons for each species:

  1. O2O_2:

    • Total electrons = 16 electrons.
    • MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2p)4(σ2p)2(π2p∗)2(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2 (\pi^*_{2p})^2
    • Number of unpaired electrons (nn) = 2.
    • Spin-only magnetic moment (μO2\mu_{O_2}) = 2(2+2)=8\sqrt{2(2+2)} = \sqrt{8} B.M.

  2. O2+O_2^+:

    • Total electrons = 15 electrons.
    • MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2p)4(σ2p)2(π2p∗)1(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2 (\pi^*_{2p})^1
    • Number of unpaired electrons (nn) = 1.
    • Spin-only magnetic moment (μO2+\mu_{O_2^+}) = 1(1+2)=3\sqrt{1(1+2)} = \sqrt{3} B.M.

  3. O2−O_2^-:

    • Total electrons = 17 electrons.
    • MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2p)4(σ2p)2(π2p∗)3(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2 (\pi^*_{2p})^3
    • Number of unpaired electrons (nn) = 1.
    • Spin-only magnetic moment (μO2−\mu_{O_2^-}) = 1(1+2)=3\sqrt{1(1+2)} = \sqrt{3} B.M.

  4. O22−O_2^{2-}:

    • Total electrons = 18 electrons.
    • MO configuration: (σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2p)4(σ2p)2(π2p∗)4(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2 (\pi^*_{2p})^4
    • Number of unpaired electrons (nn) = 0.
    • Spin-only magnetic moment (μO22−\mu_{O_2^{2-}}) = 0(0+2)=0\sqrt{0(0+2)} = 0 B.M.

Now, let's calculate the difference in spin-only magnetic moments for each pair:

  • Pair (1): O2O_2, O2+O_2^+ Difference = ∣μO2−μO2+∣=∣8−3∣|\mu_{O_2} - \mu_{O_2^+}| = |\sqrt{8} - \sqrt{3}| B.M. ≈1.096\approx 1.096 B.M.

  • Pair (2): O2O_2, O2−O_2^- Difference = ∣μO2−μO2−∣=∣8−3∣|\mu_{O_2} - \mu_{O_2^-}| = |\sqrt{8} - \sqrt{3}| B.M. ≈1.096\approx 1.096 B.M.

  • Pair (3): O2+O_2^+, O22−O_2^{2-} Difference = ∣μO2+−μO22−∣=∣3−0∣=3|\mu_{O_2^+} - \mu_{O_2^{2-}}| = |\sqrt{3} - 0| = \sqrt{3} B.M. ≈1.732\approx 1.732 B.M.

  • Pair (4): O2−O_2^-, O2+O_2^+ Difference = ∣μO2−−μO2+∣=∣3−3∣=0|\mu_{O_2^-} - \mu_{O_2^+}| = |\sqrt{3} - \sqrt{3}| = 0 B.M.

Comparing the differences, the largest difference is 3\sqrt{3} B.M., which occurs for the pair O2+O_2^+ and O22−O_2^{2-} (Option 3).