Statement 1: `N_(2) and NO^(+)` are both dia-magnetic.
Statement 2: `NO^(+)` is isoelectronic with `N_(2)`

Options

A. If assertion is true but the reason is false
B. If assertion is false but reason is true
C. If both assertion and reason are true and the reason is the correct explanation of assertion
D. If both assertion and reason are true and the reason is not the correct explanation of assertion

Answer

To solve the question, we need to evaluate both statements regarding the molecules \( N_2 \) and \( NO^+ \).

### Step 1: Analyze Statement 1
**Statement 1:** \( N_2 \) and \( NO^+ \) are both diamagnetic.

- **Definition of Diamagnetism:** A substance is diamagnetic if it has no unpaired electrons.
- **Determine the Electron Configuration of \( N_2 \):**
- \( N_2 \) has a total of 14 electrons.
- The electronic configuration can be filled as follows:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
- This results in: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \)
- Since all electrons are paired, \( N_2 \) is diamagnetic.

- **Determine the Electron Configuration of \( NO^+ \):**
- \( NO \) has 15 electrons, and \( NO^+ \) has 14 electrons (1 electron removed).
- The electronic configuration for \( NO^+ \) would be the same as \( N_2 \) because it also has 14 electrons.
- Thus, \( NO^+ \) also has the configuration: \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \)
- Therefore, \( NO^+ \) is also diamagnetic.

**Conclusion for Statement 1:** Both \( N_2 \) and \( NO^+ \) are indeed diamagnetic.

### Step 2: Analyze Statement 2
**Statement 2:** \( NO^+ \) is isoelectronic with \( N_2 \).

- **Definition of Isoelectronic Species:** Two species are isoelectronic if they have the same number of electrons.
- Since both \( N_2 \) and \( NO^+ \) have 14 electrons, they are isoelectronic.

**Conclusion for Statement 2:** \( NO^+ \) is isoelectronic with \( N_2 \).

### Final Evaluation
- **Both statements are true.**
- However, the reason given in Statement 2 (that \( NO^+ \) is isoelectronic with \( N_2 \)) is not the correct explanation for Statement 1 (which is about diamagnetism). The correct explanation involves the electronic configuration and the presence of unpaired electrons.

### Final Answer
Both statements are true, but the reason is not the correct explanation for the assertion.