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CGPET 2017 Solved Paper

Section: Chemistry

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Question : 65 of 150

Marks: +1, -0

The bond order in CN

^{-}

ion is

$2.5$
3
2
$3.5$

Solution:

Bond order (BO) Number of Bonding electrons

=\frac{-\text{Number of non - bonding electrons}}{2}

Total number of electrons in

\mathrm{CN}^{\ominus}

ions

=6+7+1=14

MO configuration of

\mathrm{CN}^{\ominus}

ion

=\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2}

\approx \pi 2p_{y}^{2}, \sigma 2p_{z}^{2}

and number of bonding electrons

=10

Number of non-bonding Electrons

=4

\therefore B.O. = \frac{10-4}{2}=3

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