Question 29 of 32advanced🔍 AnalyzeLong Answer5 marks

Compare the relative stability of the following species and indicate their magnetic properties; O₂, O₂⁺, O₂⁻ (superoxide), O₂²⁻ (peroxide)

Correct Answer

To compare the relative stability and magnetic properties of O2\text{O}_2, O2+\text{O}_2^+, O2−\text{O}_2^-, and O22−\text{O}_2^{2-}, we use molecular orbital theory. The energy level order for these species is σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z.

For O2\text{O}_2 (16 electrons), there are 10 electrons in bonding molecular orbitals and 6 in antibonding molecular orbitals. The bond order is 12[10−6]=2\frac{1}{2} [10 - 6] = 2. It has two unpaired electrons in π∗2px\pi^* 2p_x and π∗2py\pi^* 2p_y orbitals, making it paramagnetic. For O2+\text{O}_2^+ (15 electrons), the bond order is 12[10−5]=2.5\frac{1}{2} [10 - 5] = 2.5. It has one unpaired electron, making it paramagnetic.

For O2−\text{O}_2^- (superoxide, 17 electrons), the bond order is 12[10−7]=1.5\frac{1}{2} [10 - 7] = 1.5. It has one unpaired electron, making it paramagnetic. For O22−\text{O}_2^{2-} (peroxide, 18 electrons), the bond order is 12[10−8]=1\frac{1}{2} [10 - 8] = 1. All electrons are paired, making it diamagnetic.

Since stability increases with an increase in bond order (as bond enthalpy increases and bond length decreases), the relative order of stability is: O2+>O2>O2−>O22−\text{O}_2^+ > \text{O}_2 > \text{O}_2^- > \text{O}_2^{2-}.

Exercise: EXERCISES | Q: 4.36 | (Page 34)
For More Understanding

Explanation

The answer uses the molecular orbital energy level sequence provided in the context for O2 and F2 to determine the electronic configurations of the given oxygen species. By applying the bond order formula, Bond order = 1/2 [Nb - Na], the bond orders are calculated as 2.5, 2, 1.5, and 1 for O2+, O2, O2-, and O22O2^{2}- respectively. The context states that with an increase in bond order, bond enthalpy increases and bond length decreases, which dictates the stability order. Magnetic properties are determined by the presence of unpaired electrons in the antibonding pi orbitals, as illustrated for O2 in the context.

Solution Steps

  1. Step 1: Identify the molecular orbital energy level order for O2: σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z.

  2. Step 2: Calculate bond order using the formula Bond order=12[Nb−Na]\text{Bond order} = \frac{1}{2} [\text{N}_b - \text{N}_a]. For O2\text{O}_2 (16e-), BO = 12[10−6]=2\frac{1}{2} [10 - 6] = 2. For O2+\text{O}_2^+ (15e-), BO = 12[10−5]=2.5\frac{1}{2} [10 - 5] = 2.5. For O2−\text{O}_2^- (17e-), BO = 12[10−7]=1.5\frac{1}{2} [10 - 7] = 1.5. For O22−\text{O}_2^{2-} (18e-), BO = 12[10−8]=1\frac{1}{2} [10 - 8] = 1.

  3. Step 3: Determine magnetic properties by checking for unpaired electrons. O2\text{O}_2 has 2 unpaired electrons (paramagnetic), O2+\text{O}_2^+ has 1 (paramagnetic), O2−\text{O}_2^- has 1 (paramagnetic), and O22−\text{O}_2^{2-} has 0 (diamagnetic).

  4. Step 4: Compare stability using the rule that stability increases with bond order. The stability order is O2+>O2>O2−>O22−\text{O}_2^+ > \text{O}_2 > \text{O}_2^- > \text{O}_2^{2-}.

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