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NEET 2015
Exam
NEET
Year
2015
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

NEET 2015 · Chemistry · Chemical Bonding and Molecular Structure

Which of the following options represents the correct bond order?

Which of the following options represents the correct bond order?

Options

  1. AO2−>O2>O2+{O}_{2}^{-} > {O}_{2} > {O}_{2}^{+}
  2. BO2−<O2<O2+{O}_{2}^{-} < {O}_{2} < {O}_{2}^{+}
  3. CO2−>O2<O2+{O}_{2}^{-} > {O}_{2} < {O}_{2}^{+}
  4. DO2−<O2>O2+{O}_{2}^{-} < {O}_{2} > {O}_{2}^{+}

Correct answer

B. O2−<O2<O2+{O}_{2}^{-} < {O}_{2} < {O}_{2}^{+}

Step-by-step solution

B.O=Nb−Na2B . O = \dfrac{{N}_{b} - {N}_{a}}{2}

O2−⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<π2px2=π2py2<π∗2px2=π∗2py1{O}_{2}^{-} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{2} = \overset{*}{\pi} 2{p}_{y}^{1}

B.O for O2−=1.5⇒10−72=1.5{O}_{2}^{-} = 1.5 ⇒ \dfrac{10 - 7}{2} = 1.5

O2⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<π2px2=π2py2<π∗2px1=π∗2py1{O}_{2} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{1} = \overset{*}{\pi} 2{p}_{y}^{1}

B.O for O2=2.0⇒10−62=2{O}_{2} = 2.0 ⇒ \dfrac{10 - 6}{2} = 2

O2+⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<π2px2<π2py2<π∗2px1<π∗2py0{O}_{2}^{+} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} < \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{1} < \overset{*}{\pi} 2{p}_{y}^{0}

B.O for O2+=2.5⇒10−52=2.5{O}_{2}^{+} = 2.5 ⇒ \dfrac{10 - 5}{2} = 2.5

As per M.O.T electronic configurations

O2−<O2<O2+{O}_{2}^{-} < {O}_{2} < {O}_{2}^{+}
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