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In the following questions, a statement ...

In the following questions, a statement of assertion (A) is followed by a statement of reason (R )
A : `N_(2)` CO and `CN^(-)` are having same bond order.
R: Isoelectronic species always have same bond order .

A

If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1)

B

If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)

C

If Assertion is true statement but Reason is false, then mark (3)

D

If both Assertion and Reason are false statements , then mark (4).

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the assertion (A) and the reason (R) provided in the question. ### Step 1: Analyze Assertion (A) The assertion states that \( N_2 \), CO, and \( CN^- \) have the same bond order. 1. **Determine the number of electrons in each species:** - \( N_2 \): Each nitrogen atom has 7 electrons, so for \( N_2 \) (2 nitrogen atoms), the total is \( 7 \times 2 = 14 \) electrons. - CO: Carbon has 6 electrons and oxygen has 8 electrons, so for CO, the total is \( 6 + 8 = 14 \) electrons. - \( CN^- \): Carbon has 6 electrons, nitrogen has 7 electrons, and the negative charge adds 1 more electron, so the total is \( 6 + 7 + 1 = 14 \) electrons. Since all three species have 14 electrons, we can proceed to calculate their bond orders. ### Step 2: Calculate Bond Order The bond order can be calculated using the Molecular Orbital Theory (MOT) formula: \[ \text{Bond Order} = \frac{1}{2} (\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals}) \] 1. **Filling the molecular orbitals for 14 electrons:** - The order of filling for 14 electrons is: - \( \sigma_{1s}^2 \) - \( \sigma^*_{1s}^2 \) - \( \sigma_{2s}^2 \) - \( \sigma^*_{2s}^2 \) - \( \sigma_{2p_z}^2 \) - \( \pi_{2p_x}^2 \) - \( \pi_{2p_y}^2 \) The filling results in: - Bonding orbitals: \( 2 + 2 + 2 + 2 + 2 + 2 = 10 \) - Antibonding orbitals: \( 2 + 2 = 4 \) 2. **Calculate the bond order:** \[ \text{Bond Order} = \frac{1}{2} (10 - 4) = \frac{6}{2} = 3 \] Thus, the bond order for \( N_2 \), CO, and \( CN^- \) is 3. ### Step 3: Analyze Reason (R) The reason states that isoelectronic species always have the same bond order. 1. **Definition of Isoelectronic Species:** - Isoelectronic species are species that have the same number of electrons. Since \( N_2 \), CO, and \( CN^- \) all have 14 electrons, they are indeed isoelectronic. 2. **Conclusion on Reason:** - The statement is true; isoelectronic species can have the same bond order, as seen in this case. ### Final Conclusion Both the assertion (A) and the reason (R) are true, and the reason correctly explains the assertion. Therefore, the correct answer is: **Option 1: If both assertion and reason are true and reason is giving correct explanation.** ---
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