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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Chemistry

Question : 126 of 160

Marks: +1, -0

\mathrm{C}_{2}, \mathrm{B}_{2}^{2-}, \mathrm{N}_{2}^{2+}, \mathrm{CN}^{+}, \mathrm{NO}^{-}, \mathrm{O}_{2}, \mathrm{C}_{2}^{+}

Solution:

Molecule	Total number of electrons	Electronic configuration	Bond order (BO) = $\frac{a-b}{2}$
$\mathrm{CN}^{+}$	12	$\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{1}=\pi 2p_{y}^{1} \pi^{} 2p_{x}^{0}=\pi^{} 2p_{y}^{0}$	$\;\frac{8-4}{2}=2$
$\mathrm{NO}^{-}$	16	$\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2}=\pi 2p_{y}^{2} \pi^{} 2p_{x}^{1}=\pi^{} 2p_{y}^{1}$	$\;\frac{10-6}{2}=2$
$\mathrm{N}_{2}^{2+}$	12	$\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{1}=\pi 2p_{y}^{1} \pi^{} 2p_{x}^{0}=\pi^{} 2p_{y}^{0}$	$\;\frac{8-4}{2}=2$
$\mathrm{O}_{2}$	16	$\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2}=\pi 2p_{y}^{2} \pi^{} 2p_{x}^{1}=\pi^{} 2p_{y}^{1}$	$\;\frac{10-6}{2}=2$
$\mathrm{C}_{2}$	12	$\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{1}=\pi 2p_{y}^{1} \pi^{} 2p_{x}^{0}=\pi^{} 2p_{y}^{0}$	$\;\frac{8-4}{2}=2$
$\mathrm{C}_{2}^{+}$	11	$\sigma l s^{2} \sigma^{2} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{1}=\pi 2p_{y}^{0} \pi^{} 2p_{x}^{0}=\pi^{*} 2p_{y}^{0}$	$\;\frac{7-4}{2}=1.5$
$\mathrm{B}_{2}^{2-}$	12	$\sigma l s^{2} \sigma^{+} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{1}=\pi 2p_{y}^{1} \pi^{} 2p_{x}^{0}=\pi^{*} 2p_{y}^{0}$	$\;\frac{8-4}{2}=2$

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