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Chemistry Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn = 25; Fe = 26)

(A) ${[F e F_{6}]}^{3 -}$ (B) ${[F e {(C N)}_{6}]}^{3 -}$ (C) ${[M n C l_{6}]}^{3 -}$ (high spin) (D) ${[M n {(C N)}_{6}]}^{3 -}$

Choose the correct answer from the options given below

Option 1 -

A < B < D < C

Option 2 -

B < D < C < A

Option 3 -

A < C < D < B

Option 4 -

B < D < A < C

3 Views | Posted 3 months ago

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

A. ${[F e F_{6}]}^{3 -}$

Fe³⁺ - 3d⁵4s⁰

F^- is weak field ligand, so no pairing of electrons :

Number of unpaired electrons, n = 5

B. ${[F e {(C N)}_{6}]}^{3 -}$

$F e^{3 +} - 3 d^{5} 4 s^{0}$

CN^- is strong field ligand, so pairing of electrons takes place.

Number of unpaired electrons, n = 1

C. ${[M n C l_{6}]}^{3 -}$

...more

A. ${[F e F_{6}]}^{3 -}$

Fe³⁺ - 3d⁵4s⁰

F^- is weak field ligand, so no pairing of electrons :

Number of unpaired electrons, n = 5

B. ${[F e {(C N)}_{6}]}^{3 -}$

$F e^{3 +} - 3 d^{5} 4 s^{0}$

CN^- is strong field ligand, so pairing of electrons takes place.

Number of unpaired electrons, n = 1

C. ${[M n C l_{6}]}^{3 -}$

$M n^{3 +} - 3 d^{3} 4 s^{0}$

Number of unpaired electrons, n = 3

D. ${[M n {(C N)}_{6}]}^{3 -}$

$M n^{3 +} - 3 d^{3} 4 s^{0}$

Number of unpaired electrons, n = 3

Higher the number of unpaired electrons. Then higher be the magnetic moment as;

Magnetic moment, $μ = \sqrt[]{n (n + 2)} B . M$

Hence, order of magnetic moment is ;

B < D = C < A

less

<p><strong>A.</strong> <span contenteditable="false"> <math> <mrow> <msup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>F</mi> <mi>e</mi> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>6</mn> </mrow> </msub> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>3</mn> <mo>−</mo> </mrow> </msup> </mrow> </math> </span>           </p><p>Fe<sup>3+</sup> - 3d<sup>5</sup>4s<sup>0</sup></p><p>F<sup>-</sup> is weak field ligand, so no pairing of electrons :</p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752035063phpNjFBcI.jpeg" alt="" width="280" height="47" loading="lazy"></picture>                      </div></div><p>Number of unpaired electrons, n = 5</p><p><strong>B.</strong> <span contenteditable="false"> <math> <mrow> <msup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>F</mi> <mi>e</mi> <msub> <mrow> <mrow> <mo>(</mo> <mrow> <mi>C</mi> <mi>N</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>6</mn> </mrow> </msub> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>3</mn> <mo>−</mo> </mrow> </msup> </mrow> </math> </span>          </p><p><span contenteditable="false"> <math> <mrow> <mi>F</mi> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>3</mn> <mo>+</mo> </mrow> </msup> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>d</mi> </mrow> <mrow> <mn>5</mn> </mrow> </msup> <mn>4</mn> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> </mrow> </math> </span>                              </p><p>CN<sup>-</sup> is strong field ligand, so pairing of electrons takes place.</p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752035083phpInbKvQ.jpeg" alt="" width="296" height="54" loading="lazy"></picture>                            </div></div><p>Number of unpaired electrons, n = 1</p><p><strong>C.</strong> <span contenteditable="false"> <math> <mrow> <msup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>M</mi> <mi>n</mi> <mi>C</mi> <msub> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <mn>6</mn> </mrow> </msub> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>3</mn> <mo>−</mo> </mrow> </msup> </mrow> </math> </span>           </p><p><span contenteditable="false"> <math> <mrow> <mi>M</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>3</mn> <mo>+</mo> </mrow> </msup> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>d</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mn>4</mn> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> </mrow> </math> </span> </p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752035120phpgpxJIy.jpeg" alt="" width="287" height="49" loading="lazy"></picture><strong>                    </strong></div></div><p>Number of unpaired electrons, n = 3</p><p><strong>D.</strong> <span contenteditable="false"> <math> <mrow> <msup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>M</mi> <mi>n</mi> <msub> <mrow> <mrow> <mo>(</mo> <mrow> <mi>C</mi> <mi>N</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>6</mn> </mrow> </msub> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>3</mn> <mo>−</mo> </mrow> </msup> </mrow> </math> </span>           </p><p><span contenteditable="false"> <math> <mrow> <mi>M</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>3</mn> <mo>+</mo> </mrow> </msup> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>d</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mn>4</mn> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> </mrow> </math> </span>                              </p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752035141php4bCyas.jpeg" alt="" width="280" height="54" loading="lazy"></picture>                            </div></div><p>Number of unpaired electrons, n = 3</p><p>Higher the number of unpaired electrons. Then higher be the magnetic moment as;</p><p>Magnetic moment, <span contenteditable="false"> <math> <mrow> <mi>μ</mi> <mo>=</mo> <mroot> <mrow> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> <mi>B</mi> <mo>.</mo> <mi>M</mi> </mrow> </math> </span></p><p>Hence, order of magnetic moment is ;</p><p>B < D = C < A</p>

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Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn = 25; Fe = 26) (A) [ F e F 6 ] 3 − (B) [ F e ( C N ) 6 ] 3 − (C) [ M n C l 6 ] 3 − (high spin) (D) [ M n ( C N ) 6 ] 3 − Choose the correct answer from the options given below