8.48 What can be inferred from the magnetic moment values of the following complex species ?

Example	Magnetic moment
[K4 [Mn(CN)6]	2.2
[Fe(H2O)6]2+.	5.3
K2 [MnCl4]	5.9

<p><strong>8.48 </strong>To calculate magnetic moment of the complex species, we use the spin formula:</p><p>μ =√n(n+2) BM</p><table><tbody><tr><td width="197"><p><u>When n= 1</u></p></td><td width="197"><p>⇒ μ = √1(1+2)</p><p>⇒ u= √3⇒</p><p>u=1.73 BM</p></td></tr><tr><td width="197"><p><u>When n=2</u></p></td><td width="197"><p>⇒ μ = √2(2+2)</p><p>⇒ μ = √8</p><p>⇒ μ = 2.83 BM</p></td></tr></tbody></table><table><tbody><tr><td width="197"><p><u>When n= 3</u></p></td><td width="197"><p>⇒ μ = √3(3+2)</p><p>⇒ μ = 15</p><p>⇒ μ = 3.87 BM</p></td></tr><tr><td width="197"><p><u>When n = 4</u></p></td><td width="197"><p>⇒ μ = √ 4(4+2)</p><p>⇒ μ = √24</p><p>⇒ μ = 4.899 BM</p></td></tr><tr><td width="197"><p><u>When n= 5</u></p></td><td width="197"><p>⇒ μ = √5(5+2)</p><p>⇒ μ = √35</p><p>⇒ μ = 5.92 BM</p></td></tr></tbody></table><p><strong>1.</strong> [K₄ [Mn(CN)₆]</p><p>⇒μ = 2.2 BM (given)</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1687174592phpoHxzCe_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1687174592phpoHxzCe.jpeg" alt="" width="482" height="187"></picture></div></div><p>&nbsp;</p><p>We can see from the above calculation that the given value(2.2) is close to n=1. It means that it has only one unpaired electron Also in this complex Mn is in +2 oxidation state,i.e., as Mn²⁺.Thus when CN^- ligands approach Mn²⁺ ion, The electrons in 3d do not pair up.</p><p>The atomic number of Manganese (Mn) is Z = 25</p><p>The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²</p><p>And, the electronic configuration of Mn²⁺=[Ar] 3d⁵</p><p>Thus CN^- is a strong ligand.</p><p>The hybridization involved is d²sp³ forming <u>inner orbital octahedral complex</u></p><p><strong>2.</strong> [Fe(H2O)₆]^2+.</p><p>⇒ μ = 5.3 BM (given)</p><p>We can see from the above calculation that the given value(5.3) is close to n = 4. It means that it has four unpaired electrons. In this complex, Fe is in +2 state, i.e., as Fe²⁺. This means that the electrons in 3d do not pair up when the ligands, H₂O molecules approach.</p><p>The atomic number of Iron(Fe)is Z = 26</p><p>The electronic configuration of ₂₆Fe= [Ar] 3d⁶ 4s²</p><p>And, the electronic configuration of Fe²⁺=[Ar] 3d⁶</p><p>Thus H₂O is a weak ligand. To accommodate the electrons donated by six H₂O molecules, the hybridization will be sp³d². Hence it will be an <u>outer orbital octahedral complex.</u></p><p><strong>3.</strong> K₂ [MnCl₄]</p><p>⇒μ = 5.9 BM (given)</p><p>We can see from the above calculation that the given value(5.9) is close to n=5. It means that it has five unpaired electrons</p><p>In this complex, Mn is in +2 state, i.e., as Mn²⁺. Hence, we can say that Cl- is a weak ligand and does not cause the pairing of electrons.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1687174880phpgg5oCo_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1687174880phpgg5oCo.jpeg" alt="" width="275" height="124"></picture></div></div><p>The atomic number of Manganese (Mn) is Z = 25</p><p>The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²</p><p>And, the electronic configuration of Mn²⁺=[Ar] 3d⁵</p><p>Thus, the hybridization involved will be sp³ and the complex will be <u>tetrahedral complex</u></p><p>Being Lewis bases (those who donate electrons) the ligands with less electronegativity will be stronger. Therefore, in general halogen or oxygen donors (F^-, Cl^-,Br^-, H₂O) are weak field ligands and the ones in which carbon or nitrogen atom is the donor (eg-CN^-,CO,NH₃) are strong field ligands.</p>

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