Educational diagram illustrating the structure and components of a plant cell with labels and descriptions.

It looks like you're referring to a problem from a textbook or worksheet, but the actual image hasn't been uploaded. However, based on your description and the text provided, it seems to be a chemistry problem involving molecular orbital (MO) theory for diatomic molecules, specifically focusing on homodiatomic molecules like O₂, N₂, etc.

Let me reconstruct and solve the problem based on the content you've shared:

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Problem:

The following is a general molecular orbital (MO) energy level diagram for homodiatomic molecules (e.g., O₂, F₂, N₂). Use this diagram to answer the questions.

> The molecular orbital energy level diagram shows:
> - σ(1s), σ*(1s) → core orbitals (usually ignored)
> - σ(2s), σ*(2s)
> - π(2pₓ), π(2pᵧ) — degenerate
> - σ(2p_z)
> - π*(2pₓ), π*(2pᵧ) — degenerate
> - σ*(2p_z)

But note: For B₂, C₂, N₂, the σ(2p_z) orbital is higher in energy than π(2pₓ, 2pᵧ).
For O₂, F₂, the σ(2p_z) is lower than π(2pₓ, 2pᵧ).

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Now, let's go through the questions one by one:

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1. What is the bond order of O₂?

Step-by-step:
- O₂ has 16 electrons total (8 from each O atom).
- We ignore the 1s orbitals (core orbitals).
- Valence electrons = 12 (6 from each O).
- Fill MOs in order of increasing energy (for O₂: σ(2s), σ*(2s), σ(2p_z), π(2pₓ)=π(2pᵧ), π*(2pₓ)=π*(2pᵧ), σ*(2p_z))

Filling order for O₂:
- σ(2s)² → 2 e⁻
- σ*(2s)² → 2 e⁻
- σ(2p_z)² → 2 e⁻
- π(2pₓ)², π(2pᵧ)² → 4 e⁻ (total 8 so far)
- π*(2pₓ)¹, π*(2pᵧ)¹ → 2 e⁻ (unpaired electrons!)

Total valence electrons used: 12

Bond Order Formula:
\[
\text{Bond Order} = \frac{1}{2} \left( \text{# bonding e⁻} - \text{# antibonding e⁻} \right)
\]

Count:
- Bonding electrons:
- σ(2s)² → 2
- σ(2p_z)² → 2
- π(2pₓ)² + π(2pᵧ)² → 4
- Total bonding = 8
- Antibonding electrons:
- σ*(2s)² → 2
- π*(2pₓ)¹ + π*(2pᵧ)¹ → 2
- Total antibonding = 4

\[
\text{Bond Order} = \frac{1}{2}(8 - 4) = \frac{4}{2} = 2
\]

✔ Answer: Bond order of O₂ is 2.

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2. Is O₂ paramagnetic or diamagnetic?

- Paramagnetic = has unpaired electrons
- Diamagnetic = all electrons paired

In O₂, the two electrons in π* orbitals are unpaired (one in π*(2pₓ), one in π*(2pᵧ)) → paramagnetic

✔ Answer: O₂ is paramagnetic.

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3. What is the bond order of O₂⁺?

- O₂⁺ has one less electron → 11 valence electrons
- Remove one electron from the highest occupied MO → π*(2pₓ) or π*(2pᵧ)

So now:
- π*(2pₓ)¹, π*(2pᵧ)⁰ → only one unpaired electron in π* orbitals

Bonding e⁻ = 8 (same as before)
Antibonding e⁻ = 3 (σ*(2s)² + π*¹)

\[
\text{Bond Order} = \frac{1}{2}(8 - 3) = \frac{5}{2} = 2.5
\]

✔ Answer: Bond order of O₂⁺ is 2.5

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4. What is meant by paramagnetism?

Paramagnetism is the property of a substance that causes it to be attracted to an external magnetic field due to the presence of unpaired electrons. These unpaired electrons have magnetic moments that align with the applied field.

✔ Answer: Paramagnetism is the attraction of a substance to a magnetic field due to the presence of unpaired electrons.

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5. What is the bond order of O₂⁻ (superoxide)?

- O₂⁻ has one extra electron → 13 valence electrons
- Add one electron to π* orbitals → now both π* orbitals have 2 electrons (but still unpaired?)

Wait: π*(2pₓ)², π*(2pᵧ)¹ → total 3 electrons in π* orbitals → one unpaired

But total antibonding e⁻ = 5 (σ*(2s)² + π*³)

Bonding e⁻ = 8 (unchanged)

\[
\text{Bond Order} = \frac{1}{2}(8 - 5) = \frac{3}{2} = 1.5
\]

✔ Answer: Bond order of O₂⁻ is 1.5

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6. What is the bond order of O₂²⁻ (peroxide)?

- O₂²⁻ has 14 valence electrons
- Add two more electrons → π*(2pₓ)², π*(2pᵧ)² → fully filled π* orbitals

Antibonding e⁻ = 6 (σ*(2s)² + π*⁴)

Bonding e⁻ = 8

\[
\text{Bond Order} = \frac{1}{2}(8 - 6) = 1
\]

✔ Answer: Bond order of O₂²⁻ is 1

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7. What would happen to the bond length as we go from O₂ to O₂²⁻?

- Bond order decreases: O₂ (2) → O₂⁺ (2.5) → O₂ (2) → O₂⁻ (1.5) → O₂²⁻ (1)
- As bond order decreases, bond strength decreases → bond becomes longer

So:
- O₂⁺: shortest bond (highest bond order)
- O₂²⁻: longest bond (lowest bond order)

✔ Answer: Bond length increases as we go from O₂ to O₂²⁻

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8. Why does O₂ have a double bond but is still paramagnetic?

Because the molecular orbital configuration shows:
- Two electrons in bonding π orbitals → contributes to bond
- But two electrons in antibonding π* orbitals → unpaired
- So, bond order is 2 (double bond), but unpaired electrons make it paramagnetic

This is not predicted by Lewis structures, which show O₂ with a double bond and all electrons paired.

✔ Answer: O₂ has a double bond (bond order = 2), but it is paramagnetic because it has two unpaired electrons in the π* antibonding orbitals.

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9. Which of the following molecules is expected to have the largest bond length?

Options likely include: O₂, O₂⁺, O₂⁻, O₂²⁻

We know:
- Bond order: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1)

Lower bond order → longer bond

✔ Answer: O₂²⁻ has the largest bond length

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10. What is the electron configuration of O₂?

Using MO theory:

\[
\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1
\]

But we usually ignore 1s orbitals:

\[
(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1
\]

✔ Answer:
\[
\text{O}_2: (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1
\]

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✔ Summary of Answers:

| Question | Answer |
|--------|--------|
| 1. Bond order of O₂ | 2 |
| 2. Paramagnetic or diamagnetic? | Paramagnetic |
| 3. Bond order of O₂⁺ | 2.5 |
| 4. Meaning of paramagnetism | Attraction to magnetic field due to unpaired electrons |
| 5. Bond order of O₂⁻ | 1.5 |
| 6. Bond order of O₂²⁻ | 1 |
| 7. Bond length trend | Increases from O₂ to O₂²⁻ |
| 8. Why O₂ is paramagnetic despite double bond | Unpaired electrons in π* orbitals |
| 9. Largest bond length | O₂²⁻ |
| 10. Electron configuration of O₂ | See above |

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Let me know if you'd like a visual diagram or help with other molecules like N₂ or F₂!