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Kelvi

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Which of the following options represents the correct bond order?

medium

Chemical bonding and Molecular structure

2015

chemistry

$O_2^- < O_2 < O_2^+$

$O_2^- + < O_2 < O_2^+$

$O_2^- < O_2 < O_2^+$

$O_2^- > O_2 > O_2^+$

Explanation

To determine the bond order of $O_2^-, O_2,$ and $O_2^+,$

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we will use molecular orbital theory.

\\

The molecular orbital configuration for

O_2

is:

\\ \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 \\

The bond order is given by:

\\

Bond order

= \frac{1}{2} (

Number of bonding electrons

-

Number of antibonding electrons

) \\

For

O_2: \\

Number of bonding electrons

= 10 \\

Number of antibonding electrons

= 6 \\

Bond order

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

For

O_2^+: \\

Remove one electron from the antibonding orbital:

\\

Number of bonding electrons

= 10 \\

Number of antibonding electrons

= 5 \\

Bond order

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

For

O_2^-: \\

Add one electron to the antibonding orbital:

\\

Number of bonding electrons

= 10 \\

Number of antibonding electrons

= 7 \\

Bond order

= \frac{1}{2} (10 - 7) = 1.5 \\

Comparing the bond orders:

\\ O_2^- (1.5) < O_2 (2) < O_2^+ (2.5) \\

Therefore, the correct option is Option 1:

O_2^- < O_2 < O_2^+.