Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them

Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

\mathbf{A}\) \quad \mathrm{O}_{2}^{-}<\mathrm{NO}<\mathrm{C}_{2}^{2-}<\mathrm{He}_{2}^{+}

\mathbf{B}\) \quad \mathrm{C}_{2}^{2-}<\mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{NO}

\mathbf{C}\) \quad \mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{N} \mathrm{O}<\mathrm{C}_{2}^{2-}

\mathbf{D}\) \quad \mathrm{N} \mathrm{O}<\mathrm{O}_{2}^{-}<\mathrm{C}_{2}^{2-}<\mathrm{H} \mathrm{e}_{2}^{+}

Answers (1)

\\\text{First we will calculate the bond order for all elements. } \\\\ \text{Bond order } =\frac{N_{b}-N_{a}}{2} \\ \mathrm{N}_{\mathrm{b}}= \text{ Number of electrons in bonding molecular orbital } \\ \mathrm{N}_{\mathrm{a}}=\text{ Number of electrons in antiobonding molecular orbital} \\

\begin{array}{l} \mathrm{NO}=15 \text { electrons } \\ \sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{y}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{4}}^{4} \\ \mathrm{B.O}=\frac{8-3}{2}=2.5 \end{array}

\begin{aligned} &\mathrm{O}_{2}^{-}=17 \text { electrons }\\ &\sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{\mathrm{y}}^{2}},, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}^{*}=\pi_{2 \mathrm{p}_{\mathrm{r}}^{2}}^{*}\\ &\mathrm{B} . \mathrm{O}=\frac{8-5}{2}=1.5 \end{aligned}

\begin{array}{l} \mathrm{C}_{2}^{2-}=14 \text { electrons } \\ \sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{y}^{2}} \\ \mathrm{B.O}=\frac{8-2}{2}=3 \end{array}

\begin{aligned} &\mathrm{He}_{2}^{2+}=3 \text { electrons }\\ &\sigma_{1 s^{2}}, \sigma_{1 s^{1}}^{*}\\ &\mathrm{B.O}=\frac{2-1}{2}=0.5\\ &\text { Therefore order is } \mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{NO}<\mathrm{C}_{2}^{2-} \end{aligned}

Posted by

lovekush

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

When will NTA issue JEE Main 2026 city slip for over 14.5 lakh? previous years’ trends
anam-khan

JEE Main 2026: NTA directs non-Aadhaar applicants to submit photo verification certificate by January 7
anam-khan

Jharkhand CM inaugurates free coaching institute for tribal medical, engineering students

Latest Question