To determine which of the given species is paramagnetic, we need to analyze the electron configuration of each species and check for the presence of unpaired electrons. Here’s a step-by-step solution:
### Step 1: Understand Paramagnetism and Diamagnetism
- **Paramagnetic species** have one or more unpaired electrons in their electron configuration.
- **Diamagnetic species** have all electrons paired.
### Step 2: Analyze Each Species
1. **O2^2- (Oxygen molecule with a -2 charge)**
- Total electrons = 8 (O) × 2 + 2 (for the -2 charge) = 18 electrons.
- Draw the Molecular Orbital (MO) diagram for 18 electrons (without sp-mixing).
- Filling the MO:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)², π*(2p_x)², π*(2p_y)².
- All electrons are paired.
- **Conclusion**: O2^2- is **diamagnetic**.
2. **NO (Nitric Oxide)**
- Total electrons = 7 (N) + 8 (O) = 15 electrons.
- Draw the MO diagram with sp-mixing.
- Filling the MO:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)², π*(2p_x)¹.
- There is one unpaired electron in π*(2p_x).
- **Conclusion**: NO is **paramagnetic**.
3. **O2 (Oxygen molecule)**
- Total electrons = 8 (O) × 2 = 16 electrons.
- Draw the MO diagram with sp-mixing.
- Filling the MO:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)², π*(2p_x)¹, π*(2p_y)¹.
- There are two unpaired electrons in π*(2p_x) and π*(2p_y).
- **Conclusion**: O2 is **paramagnetic**.
4. **CN^- (Cyanide ion)**
- Total electrons = 6 (C) + 7 (N) + 1 (for -1 charge) = 14 electrons.
- Draw the MO diagram with sp-mixing.
- Filling the MO:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)².
- All electrons are paired.
- **Conclusion**: CN^- is **diamagnetic**.
### Step 3: Final Conclusion
- The species that is paramagnetic among the options is **NO**.
### Answer
**The correct answer is option B: NO, which is paramagnetic due to the presence of one unpaired electron.**
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