Which of the following molecules contain odd electron bonds?

H_{2} O

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N O

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N O_{2}

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C l O_{2}

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Solution

The correct options are
B $N O$
C $N O_{2}$
D $C l O_{2}$
$N O_{2}$
Nitrogen dioxide is the chemical compound with the formula $N O_{2}$ . Again, nitrogen dioxide does not follow the octet rule for one of its atoms, namely nitrogen.
The total number of valence electrons is $5 + 2 (6) = 17$

There is the persistent radical character on nitrogen because it has an unpaired electron.

$C l O_{2}$

A Lewis structure shows $C l O_{2}$ has a total of $2 \times 6 + 7 = 19$ valence electrons, or 9 electron pairs and a lone electron.

Cl is the central atom.

The molecule $C l O_{2}$ has an odd number of valence electrons.

$N O$

14 valence electrons in your trial structure.

The valence electrons you have available are: $1 N + 1 O = 1 \times 5 + 1 \times 6 = 11$ .

The molecule $N O$ , $N O_{2}$ and $C l O_{2}$ has an odd number of valence electrons.

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K O_{2}, N O_{2}, S O_{2}, B a O_{2}, C l O_{2}, O_{2}, N O, N_{2} O, O_{3}, C O

C l O_{2}

C l - O

N O_{2}

C l O_{2}

C H_{3}

C F_{3}

What is (are) true about $N O$ and $C l O_{2}$ ?

K O_{2}, N O_{2}, S O_{2}, B a O_{2}, C l O_{2}, O_{2}, N O, N_{2} O, O_{3}, C O

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