Isoelectronic species are:

Options

A. `CO,CN^(-),NO^(+),C_(2)^(-2)`
B. `CO^(-),CN,NO,C_(2)^(-)`
C. `CO^(-),CN^(+),NO,C_(2)`
D. `CO,CN,NO,C_(2)`

Answer

To determine which of the given options contains isoelectronic species, we first need to understand what isoelectronic species are. Isoelectronic species are atoms or ions that have the same number of electrons.

Let's analyze each of the given options step by step:

### Step 1: Calculate the number of electrons in each species

1. **CO (Carbon Monoxide)**:
- Carbon (C) has 6 electrons.
- Oxygen (O) has 8 electrons.
- Total electrons in CO = 6 (C) + 8 (O) = **14 electrons**.

2. **CN⁻ (Cyanide Ion)**:
- Carbon (C) has 6 electrons.
- Nitrogen (N) has 7 electrons.
- The negative charge (−1) adds 1 electron.
- Total electrons in CN⁻ = 6 (C) + 7 (N) + 1 = **14 electrons**.

3. **NO⁺ (Nitric Oxide Ion)**:
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- The positive charge (+1) removes 1 electron.
- Total electrons in NO⁺ = 7 (N) + 8 (O) - 1 = **14 electrons**.

4. **C₂²⁻ (Dicarbon Ion)**:
- Each Carbon (C) has 6 electrons, and there are 2 Carbons.
- The negative charge (−2) adds 2 electrons.
- Total electrons in C₂²⁻ = 2 × 6 (C) + 2 = 12 + 2 = **14 electrons**.

### Step 2: Compare the total number of electrons
Now, we compare the total number of electrons in each species:

- CO: 14 electrons
- CN⁻: 14 electrons
- NO⁺: 14 electrons
- C₂²⁻: 14 electrons

Since all four species have the same number of electrons (14), they are all isoelectronic.

### Conclusion
The correct option containing isoelectronic species is **Option A: CO, CN⁻, NO⁺, and C₂²⁻**.

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