Question

The internuclear distance in o-o bonds for o2+, o2, o2- and o2 2- resp are

Answer 1

here is ur and...hope it helps u

Answer 2

Answer:

The intermolecular distance in the O-O bonds as:

O₂⁺ < O₂ < O₂⁻ < O₂²⁻.

Explanation:

• Intermolecular distance between two atoms is the distance between their nuclei present in the same molecule. It is also called bond length.
• We know that bond length is inversely proportional to bond order. As the bond order decreases, the bond length increases.
• Bond order is the half of the difference of  number of electrons in bonding and in antibonding molecular orbitals.

• O₂ molecule :- The electronic configuration on the basis of molecular orbital theory: $\sigma 1s^2 \; \sigma 2s^2\; \sigma^*1s^2\; \sigma ^*2s^2\; \sigma \; 2p_z^2 \; \pi 2p_x^2\; \pi 2p_y ^2 \; \pi ^*2p_x^2$

       Bond order of O₂ molecule $=\frac{1}{2}[10-6]=2$

• O₂⁺ molecule :- The electronic configuration on the basis of molecular orbital theory: $\sigma 1s^2 \; \sigma 2s^2\; \sigma^*1s^2\; \sigma ^*2s^2\; \sigma \; 2p_z^2 \; \pi 2p_x^2\; \pi 2p_y ^2 \; \pi ^*2p_x^1$
•        Bond order of O₂ molecule $=\frac{1}{2}[10-5]=2.5$
• O₂⁻ molecule :- The electronic configuration on the basis of molecular orbital theory: $\sigma 1s^2 \; \sigma 2s^2\; \sigma^*1s^2\; \sigma ^*2s^2\; \sigma \; 2p_z^2 \; \pi 2p_x^2\; \pi 2p_y ^2 \; \pi ^*2p_x^2\; \pi ^*2p_y^1$
•        Bond order of O₂ molecule $=\frac{1}{2}[10-7]=1.5$
• O₂²⁻ molecule :- The electronic configuration on the basis of molecular orbital theory: $\sigma 1s^2 \; \sigma 2s^2\; \sigma^*1s^2\; \sigma ^*2s^2\; \sigma \; 2p_z^2 \; \pi 2p_x^2\; \pi 2p_y ^2 \; \pi ^*2p_x^2 \; \pi ^*2p_y^2$

       Bond order of O₂ molecule $=\frac{1}{2}[10-8]=1$

• Order of bond order of given molecules: O₂⁺ > O₂ > O₂⁻ > O₂²⁻.
• Order of intermolecular distance in the O-O bonds:

        O₂⁺ < O₂ < O₂⁻ < O₂²⁻.

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