Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following: 

(i) [CoF₆]^3- ,[Co(H₂O)₆]²⁺ ,[Co(CN)₆]^3-

^{(ii) [FeF₆]3- ,[Fe(H₂O)₆]2 + ,[Fe(CN)₆]4 -}

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment =   μ= $\sqrt[]{n(n+2}$   =  $\sqrt[]{4(4+2}$

   =  $\sqrt[]{24}$   = 4.9 BM 

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3(3+2}$ = $\sqrt[]{15}$ = 3.87BM

 

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Ener

...more

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment =   μ= $\sqrt[]{n(n+2}$   =  $\sqrt[]{4(4+2}$

   =  $\sqrt[]{24}$   = 4.9 BM 

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3(3+2}$ = $\sqrt[]{15}$ = 3.87BM

 

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Energy level diagram:

Number of unpaired electrons =  μ= $\sqrt[]{n(n+2}=\mathrm{}\sqrt[]{5(5+2}$ =  μ= $\sqrt[]{35}$ = 5.95BM [Fe(H₂O)₆]²⁺

Fe²⁺=[Ar]3 d⁶

Energy level diagram:

Magnetic moment

Since, number of unpaired electrons (n)=4, therefore magnetic moment is-

$\sqrt[]{4(4+2}$ = $\sqrt[]{24}$ = 4.9BM

[Fe(CN)₆]⁴⁻

Electronic configuration: Fe²⁺=[Ar]3 d⁶

Energy level diagram

Since CN⁻ is strong field ligand, it forces all the electrons to get paired up Magnetic moment.

In the absence of unpaired electrons it behaves as diamagnetic ( ie. no magnetic behaviour ).

So magnetic moment = zero

 

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<p>This is a Long Answer Type Questions &nbsp;as classified in NCERT Exemplar</p><p><strong>Ans: (i) </strong>Electronic cnfiguration: Co³⁺ =[Ar]3d⁶</p><p>Energy level diagram:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992179phpWhTIOr_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992179phpWhTIOr.jpeg" alt="" width="400" height="225"></picture></div></div><p><strong>Magnetic moment:</strong></p><p>Number of unpaired electrons (n)=4</p><p>Magnetic moment = &nbsp;&nbsp;μ= <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mi>n</mi> <mo>(</mo> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>&nbsp;&nbsp;=&nbsp; <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>4</mn> <mo>(</mo> <mn>4</mn> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p>&nbsp;&nbsp;&nbsp;=&nbsp; <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>&nbsp;&nbsp;= 4.9 BM&nbsp;</p><p>[Co(H₂O)₆]²⁺</p><p>Electronic cnfiguration: Co²⁺=[Ar]3&nbsp;d⁷</p><p>Energy level diagram:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992226php003sZK_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992226php003sZK.jpeg" alt="" width="234" height="114"></picture></div></div><p><strong>Magnetic moment:</strong> Since ,number of unpaired electrons (n)=3, therefore magnetic moment =<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>3</mn> <mo>(</mo> <mn>3</mn> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>= <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>= 3.87BM</p><p>&nbsp;</p><p><strong>[Co(CN)₆]³⁻</strong></p><p>Electronic configuration: [Ar]Co³⁺=3&nbsp;d⁶</p><p>Energy level diagram:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992263phpp7Xf0Q_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992263phpp7Xf0Q.jpeg" alt="" width="190" height="90"></picture></div><div><strong>(ii) </strong><p><strong>Ans: </strong><strong>[FeF₆]³⁻</strong></p><p>Electronic configuration: Fe³⁺=[Ar]3&nbsp;d⁵</p><p>Energy level diagram:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992304phpYykxnB_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992304phpYykxnB.jpeg" alt="" width="230" height="104"></picture></div></div><p>Number of unpaired electrons = &nbsp;μ=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mi>n</mi> <mo>(</mo> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mi mathvariant="normal"> </mi> <mroot> <mrow> <mrow> <mn>5</mn> <mo>(</mo> <mn>5</mn> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi> </mi> </math> </span> = &nbsp;μ= <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>35</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi> </mi> </math> </span>= 5.95BM [Fe(H₂O)₆]²⁺</p><p>Fe²⁺=[Ar]3&nbsp;d⁶</p><p>Energy level diagram:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992367phpisJy4z_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992367phpisJy4z.jpeg" alt="" width="266" height="117"></picture></div></div><p>Magnetic moment</p><p>Since, number of unpaired electrons (n)=4, therefore magnetic moment is-</p><p><span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>4</mn> <mo>(</mo> <mn>4</mn> <mo>+</mo> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi> </mi> </math> </span>=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi> </mi> </math> </span>= 4.9BM</p><p><strong>[Fe(CN)₆]⁴⁻</strong></p><p>Electronic configuration: Fe²⁺=[Ar]3&nbsp;d⁶</p><p>Energy level diagram</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745992419phpJXN0uG_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745992419phpJXN0uG.jpeg" alt="" width="244" height="112"></picture></div></div><p>Since CN⁻is strong field ligand, it forces all the electrons to get paired up Magnetic moment.</p><p>In the absence of unpaired electrons it behaves as diamagnetic ( ie. no magnetic behaviour ).</p><p>So magnetic moment = zero</p>&nbsp;</div></div>

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