Statement-1:`NO^+ and CN^(-)` both have same order and magnetism (i.e. magnetic property).
Statement-2:`NO^+` and `CN^(-)` are isoelectronic species.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements regarding the species \( \text{NO}^+ \) and \( \text{CN}^- \).

### Step 1: Determine the number of electrons in \( \text{NO}^+ \) and \( \text{CN}^- \)

1. **For \( \text{NO}^+ \)**:
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- Since \( \text{NO}^+ \) has a positive charge, we subtract 1 electron.
- Total electrons in \( \text{NO}^+ \) = \( 7 + 8 - 1 = 14 \).

2. **For \( \text{CN}^- \)**:
- Carbon (C) has 6 electrons.
- Nitrogen (N) has 7 electrons.
- Since \( \text{CN}^- \) has a negative charge, we add 1 electron.
- Total electrons in \( \text{CN}^- \) = \( 6 + 7 + 1 = 14 \).

### Step 2: Draw the Molecular Orbital (MO) Diagram

Both species have the same number of electrons (14), so we can draw the MO diagram for both.

1. **MO Diagram Structure**:
- The \( \sigma \) and \( \sigma^* \) orbitals from the 1s level will be filled first.
- Then, the 2s and 2p orbitals will be filled according to the energy levels and Hund's rule.

### Step 3: Fill the Molecular Orbitals

- **Filling Order**:
- Fill the \( \sigma_{1s} \) and \( \sigma_{1s}^* \) first with 2 electrons each.
- Next, fill \( \sigma_{2s} \) and \( \sigma_{2s}^* \) with 2 electrons each.
- Then fill the \( \sigma_{2p_z} \) with 2 electrons.
- Next, fill \( \pi_{2p_x} \) and \( \pi_{2p_y} \) with 4 electrons (2 in each).
- Finally, fill \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \) with 0 electrons.

### Step 4: Calculate Bond Order

The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

- For both \( \text{NO}^+ \) and \( \text{CN}^- \):
- Bonding electrons = 10 (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Antibonding electrons = 4 (from \( \sigma_{1s}^*, \sigma_{2s}^* \))

Thus, the bond order is:
\[ \text{Bond Order} = \frac{(10 - 4)}{2} = \frac{6}{2} = 3 \]

### Step 5: Determine Magnetic Properties

- **Magnetism**:
- Both \( \text{NO}^+ \) and \( \text{CN}^- \) have all electrons paired in their molecular orbitals.
- Therefore, both species are diamagnetic.

### Step 6: Check for Isoelectronicity

- Since both \( \text{NO}^+ \) and \( \text{CN}^- \) have the same number of electrons (14), they are isoelectronic.

### Conclusion

- **Statement 1**: True (Both have the same bond order and magnetic properties).
- **Statement 2**: True (Both are isoelectronic species).

### Final Answer

Both statements are correct, and statement 2 provides the correct explanation for statement 1. Therefore, the answer is option A.

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