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Chemistry-

General

Easy

Question

The number of anti-bonding electron pairs in $O subscript 2 end subscript superscript 2 minus end superscript$ molecular ion on the basis of molecular orbital theory is (atomic number of O is 8)

5
2
4
6

The correct answer is: 4

$O subscript 2 end subscript superscript 2 minus end superscript equals 18$ electrons
$open parentheses sigma 1 s close parentheses to the power of 2 end exponent open parentheses sigma to the power of asterisk times end exponent 1 s close parentheses to the power of 2 end exponent open parentheses sigma 2 s close parentheses to the power of 2 end exponent open parentheses sigma to the power of asterisk times end exponent 2 s close parentheses to the power of 2 end exponent$
$open parentheses sigma 2 p subscript x end subscript close parentheses to the power of 2 end exponent open parentheses pi 2 p subscript x end subscript close parentheses to the power of 2 end exponent open parentheses pi 2 p subscript y end subscript close parentheses to the power of 2 end exponent open parentheses pi to the power of asterisk times end exponent 2 p subscript x end subscript close parentheses to the power of 2 end exponent open parentheses pi to the power of asterisk times end exponent 2 p subscript y end subscript close parentheses to the power of 2 end exponent$
Underlined orbitals are antibonding M.O
Total = 4

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