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  • Consider the ions/molecule <img alt="\mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BO%7D_%7B2%7

Consider the ions/molecule
\mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}
For increasing bond order the correct option is :

Option: 1

\mathrm{O}_{2}^{2-}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}<\mathrm{O}_{2}^{+}


Option: 2

\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{2-}<\mathrm{O}_{2}<\mathrm{O}_{2}^{+}


Option: 3

\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{2-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}


Option: 4

\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2-}<\mathrm{O}_{2}


Answers (1)

best_answer

From MOT i.e molecular orbital Theory,

\mathrm{Bond \: order= \frac{N_{b}-N_{a}}{2} }
\mathrm{N_{b} } = number of bonding electrons .
\mathrm{N_{a} } = number of antibonding electrons.

\mathrm{for \: O_{2}^{+} ,N_{b}= 10,\: N_{a}= 5}
       \mathrm{Bond \: order= \frac{10-5}{2}= 2.5}

\mathrm{for\: \: O_{2},\: N_{b}= 10 , \, N_{a}= 6}
        \mathrm{Bond\: order= \frac{10-6}{2}= 2}

\mathrm{for\: \: O_{2}^{-},\: N_{b}= 10 , \, N_{a}= 7}
         \mathrm{B\cdot O= \frac{N_{b}-N_{a}}{2}= \frac{10-7}{2}= 1.5}

\mathrm{for\: \: O_{2}^{2-},\: N_{b}= 10 , \, N_{a}= 8}
           \mathrm{B\cdot O= \frac{N_{b}-N_{a}}{2}= \frac{10-8}{2}= \frac{2}{2}= 1}

\mathrm{\therefore \: Bond\; order: O_{2}^{2-}< O_{2}^{-}< O_{2}< O_{2}^{+}}

Option (1) is correct

Posted by

HARSH KANKARIA

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