Question

Arrange the following in increasing value of magnetic moments.
(i) ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$
(ii) ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3-}}$
(iii) ${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$
(iv) ${{\left[ Ni{{\left( {{H}_{2}}O \right)}_{4}} \right]}^{2+}}$

A. (I) < (II) < (III) < (IV)
B. (I) < (II) < (IV) < (III)
C. (II) < (III) < (I) < (IV)
D. (IV) < (III) < (II) < (I)

Answer 1

Hint: There is a formula to calculate the magnetic moment of the coordination complexes and it is as follows.
Magnetic moment = $\sqrt{n(n+2)}BM$
Here n = number of unpaired electron in the complex
BM = Bohr magneton (a unit to measure magnetic moment)

Complete step by step answer:
- In the question it is asked to find the magnetic moment of the given complexes and arrange them in an order.
- The first compound is ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$ .
- The electronic configuration of iron (II) in the complex is $[Ar]3{{d}^{6}}4{{s}^{0}}$ .
- We know that cyanide is a strong field ligand. So, all the electrons in the 3d orbital pairs up and form a low spin complex.
- So, the number of unpaired electrons in ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$ is zero.
- Means n = 0
- Then the magnetic moment of ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$ is as follows.
\[\begin{align}
&\implies \sqrt{n(n+2)}BM \\
&\implies \sqrt{0(0+2)}BM \\
&\therefore 0BM \\
\end{align}\]
- The second compound is ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3-}}$ .
- The electronic configuration of iron (III) in the complex is $[Ar]3{{d}^{5}}4{{s}^{0}}$ .
- We know that cyanide is a strong field ligand. So, all the electrons in the 3d orbital pairs up and form a low spin complex.
- So, the number of unpaired electrons in ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3-}}$ is one.
- Means n = 1
- Then the magnetic moment of ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3-}}$ is as follows.
\[\begin{align}
&\implies \sqrt{n(n+2)}BM \\
&\implies \sqrt{1(1+2)}BM \\
&\therefore\sqrt{3}BM \\
\end{align}\]
- The third compound is ${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$ .
- The electronic configuration of Chromium (III) in the complex is $[Ar]3{{d}^{3}}4{{s}^{0}}$ .
- We know that ammonia is a weak field ligand. So, all the electrons in the 3d orbital are unpaired and form a high spin complex.
- So, the number of unpaired electrons in ${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$ is three.
- Means n = 3
- Then the magnetic moment of ${{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}$ is as follows.
\[\begin{align}
&\implies \sqrt{n(n+2)}BM \\
&\implies \sqrt{3(3+2)}BM \\
&\therefore \sqrt{15}BM \\
\end{align}\]
- The fourth compound is ${{\left[ Ni{{\left( {{H}_{2}}O \right)}_{4}} \right]}^{2+}}$ .
- The electronic configuration of Nickel (II) in the complex is $[Ar]3{{d}^{8}}4{{s}^{0}}$ .
- We know that water is a weak field ligand. So, all the electrons in the 3d orbital are unpaired and form a high spin complex.
- So, the number of unpaired electrons in ${{\left[ Ni{{\left( {{H}_{2}}O \right)}_{4}} \right]}^{2+}}$ is two.
- Means n = 2
- Then the magnetic moment of ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$ is as follows.
\[\begin{align}
&\implies \sqrt{n(n+2)}BM \\
&\implies \sqrt{2(2+2)}BM \\
&\therefore \sqrt{8}BM \\
\end{align}\]
- Therefore the correct order of magnetic moment of the given molecules is as follows.
(I) < (II) < (IV) < (III)

So, the correct option is B.

Note: The magnetic moment of the coordination complexes is going to increase with the number of unpaired electrons. If the coordination complex has strong field ligands then the complex has less magnetic moment and vice versa.