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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Chemistry

Question : 126 of 160

Marks: +1, -0

The correct sequence of bond order is

Solution:

Bond order (BO)

(Number of electrons in antibonding molecular orbital) - (Number of electrons

= \frac{\text{ in bonding molecular orbital }}{2}

Here, antibonding molecular orbital electrons denoted by

(*)

\mathrm{O}_2^+ = \sigma 1s^2 \sigma^{*} 1s^2 \sigma 2s^2 \sigma^{*} 2s^2 \sigma 2p_x^2 (\pi 2p_y^2 = \pi 2p_z^2)

(\pi^{*} 2p_y^1 = \pi^{*} 2p_z^0)

\mathrm{BO} = \frac{1}{2}(10-5)=2.5

\mathrm{O}_2 = \sigma 1s^2 \sigma^{*} 1s^2 \sigma 2s^2 \sigma^{*} 2s^2 \sigma 2p_x^2 (\pi 2p_y^2 = \pi^1 2p_z^2)

\mathrm{BO} = \frac{1}{2}(10-6)=2

\mathrm{O}_2^- = \sigma 1s^2 \sigma^{*} 1s^2 \sigma 2s^2 \sigma^{*} 2s^2 \sigma 2p_x^2 (\pi 2p_x^2 = \pi 2p_y^2)

\mathrm{BO} = \frac{1}{2}(10-7)=1.5 \;\; (\pi^{*} 2p_y^2 = \pi^{*} 2p_z^1)

\mathrm{O}_2^{2-} = \sigma 1s^2 \sigma^{*} 1s^2 \sigma 2s^2 \sigma^{*} 2s^2 \sigma 2p_x^2 (\pi 2p_x^2 = \pi 2p_y^2)

\;\;\; \mathrm{BO} = \frac{1}{2}(\pi^{*} 2p_y^2 = \pi^{*} 2p_z^2)

\therefore

Correct order of bond order is

\mathrm{O}_2^+ > \mathrm{O}_2 > \mathrm{O}_2^- > \mathrm{O}_2^{2-} \; \text{". "}

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