Question

The correct bond order in the following species is: [2015]
(a) O₂²⁺ < O₂⁻ < O₂⁺ (b) O₂⁺ < O₂⁻ < O₂²⁺
(c) O₂⁻ < O₂⁺ < O₂²⁺ (d) O₂²⁺ < O₂⁺ < O₂⁻

Answer 1

Answer:

Options c Should ne corrected

Explanation:

He'llo this is maanik follow me for more anwers .... Search Google for it....

Plzzzz support me.....

Answer 2

The correct bond order of given species is O₂⁻ < O₂⁺ < O₂²⁺

Explanation:

Electronic configurations and bond order of the given species are

$O_2^{-} - \sigma1s^2,\sigma^{*}1s^2,\sigma2s^2,\sigma^{*}2s^2,\sigma2p_x^2,\pi2p_y^2=\pi2p_z^2,\pi^{*}2p_y^2=\pi^{*}2p_z^1$

$Bond \,order=\frac{N_b-N_a}{2} =\frac{10-7}{2} =1.5$

$O_2^{+} - \sigma1s^2,\sigma^{*}1s^2,\sigma2s^2,\sigma^{*}2s^2,\sigma2p_x^2,\pi2p_y^2=\pi2p_z^2,\pi^{*}2p_y^1=\pi^{*}2p_z^0$

$Bond \,order=\frac{N_b-N_a}{2} =\frac{10-5}{2} =2.5$

$O_2^{2+} - \sigma1s^2,\sigma^{*}1s^2,\sigma2s^2,\sigma^{*}2s^2,\sigma2p_x^2,\pi2p_y^2=\pi2p_z^2$

$Bond \,order=\frac{N_b-N_a}{2} =\frac{10-6}{2} =3$

$\therefore$ The correct bond order is O₂⁻ < O₂⁺ < O₂²⁺

Hence the correct answer is option c