The bond order and magnetic behaviour of ion are, respectively

Option 1 -

2 and diamagnetic

Option 2 -

1 and paramagnetic.

Option 3 -

1.5 and diamagnetic

Option 4 -

1.5 and paramagnetic

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Electronic configuration of $O_{2}^{-}$ anion is;

${(σ 1 s)}^{2} {(σ_{1 s}^{*})}^{2} {(σ_{2 s})}^{2} {(σ_{2 s}^{*})}^{2} {(σ 2 p z)}^{2} {(π 2 p x)}^{2} = {(π 2 p y)}^{2} {(π * 2 p x)}^{2} = {(π * 2 p y)}^{1}$

In anion total 10 e^- resides in Bonding Molecular Orbitals and 7e^- resides in anti-Bonding Molecular orbitals. Hence bond order $= \frac{10 - 7}{2} = \frac{3}{2} = 1.5 .$

It is paramagnetic due to one unpaired electron.

Electronic configuration of <math> <mrow> <msubsup> <mrow> <mi>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mo>−</mo> </mrow> </msubsup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815812861"> </o:OLEObject></xml><! [endif]->anion is; <math> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>σ</mi> <mn>1</mn> <mi>s</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> </mrow> </msubsup> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> </mrow> </msubsup> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>σ</mi> <mn>2</mn> <mi>p</mi> <mi>z</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>π</mi> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>π</mi> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>π</mi> <mo>*</mo> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>π</mi> <mo>*</mo> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> </mrow> </msup> </mrow> </math> In anion total 10 e- resides in Bonding Molecular Orbitals and 7e- resides in anti-Bonding Molecular orbitals. Hence bond order <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mo>−</mo> <mn>7</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>.</mo> <mn>5</mn> <mo>.</mo> </mrow> </math> It is paramagnetic due to one unpaired electron.

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

The bond order and magnetic behaviour of ion are, respectively

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question