Test Index

BITSAT 2018 Paper

Section: Chemistry

Question : 46 of 150

Marks: +1, -0

O_2^{+}

Solution:

O_2

: σ

1s^{2}

, σ*

1s^{2}

, σ2

s^{2}

, σ*

2s^{2}

, σ

2p_{z}^{2}

\pi 2p_{x}^{2},\pi 2p_{y}^{2}

\pi^{*}2p_{x}^{1} , \pi^{*}2p_{y}^{1}

Bond order =

\frac{10-6}{2}

= 2

(two unpaired electrons in antibonding molecularorbital)

O_2^{+}

: σ

1s^{2}

, σ*

1s^{2}

, σ

2s^{2}

, σ*

2s^{2}

, σ

2p_{z}^{2}

\pi 2p_{x}62,\pi 2p_{y}^{2}

\pi^{*}2p_{x}^{1} , \pi^{*}2p_{y}^{0}

Bond order =

\frac{10-5}{2}

= 2.5

(One unpaired electron in antibonding molecularorbital)

Hence

O_2

as well as

O_2^{+}

both are paramagnetic,

and bond order of

O_2^{+}

is greater than that of

O_2

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