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Question: Which one of the following complex ions has the highest magnetic moment? a.) \({ [Cr(NH }_{ 3 }{ )...

Which one of the following complex ions has the highest magnetic moment?
a.) [Cr(NH3)6]3+{ [Cr(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }
b.) [Fe(CN6)]3−{ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 3- }
c.) [Fe(CN6)]4−{ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 4- }
d.) [Zn(NH3)6]2+{ [Zn(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 2+ }

Explanation

Solution

Hint: Each electron in an atom behaves like a magnet. The rotation of the electron is called its magnetic moment. Its magnetic moment evolved from two different types of motions;
Its orbital motion around the nucleus and it’s spin around its own axis.

Complete step-by-step answer:

For the compound [Cr(NH3)6]3+{ [Cr(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }, we can see that ammonia is the ligand which has charge = 0
So, the oxidation state of Cr=+3{ Cr=+3 }
Atomic number of Cr = 24{ 24 }
The electronic configuration of Cr = [Ar]3d54s1{ [Ar]3d }^{ 5 }{ 4s }^{ 1 }
So, Cr+3{ Cr }^{ +3 } has electronic configuration = [Ar]3d3{ [Ar]3d }^{ 3 }
Where, Ar = argon having atomic number = 18{ 18 }
Here, three unpaired electrons are present,so, the magnetic moment will be;
μ=n(n+1){ \mu =\sqrt { n(n+1) } }
Where, n= number of unpaired electrons
μ=3(3+1){ \mu =\sqrt { 3(3+1) } }
μ=12{ \mu =\sqrt { 12 } }
μ=3.464BM{ \mu = { 3.464 }BM }

Now, [Fe(CN6)]3−{ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 3- }, we can see that cyanide is the ligand which has charge = −1{ -1 }
So, the oxidation state of Fe=+3{ Fe=+3 }
Atomic number of Fe = 26{ 26 }
The electronic configuration of Fe = [Ar]3d64s2{ [Ar]3d }^{ 6 }{ 4s }^{ 2 }
So, Fe+3{ Fe }^{ +3 } has electronic configuration = [Ar]3d54s0{ [Ar]3d }^{ 5 }{ 4s }^{ 0 }
Since cyanide is a strong ligand so it will pair up the electrons and one unpaired electron will be present.
Where, Ar = argon having atomic number =18
Here, three unpaired electrons are present,so, the magnetic moment will be;
μ=n(n+1){ \mu =\sqrt { n(n+1) } }
Where, n= number of unpaired electrons
μ=1(1+1){ \mu =\sqrt { 1(1+1) } }
μ=2{ \mu =\sqrt { 2 } }
μ=1.414BM{ \mu = { 1.414 }BM }

Now, for [Fe(CN6)]4−{ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 4- }, we can see that cyanide is the ligand which has charge = −1{ -1 }
So, the oxidation state of Fe = +2{ +2 }
Atomic number of Fe = 26{ 26 }
The electronic configuration of Fe = [Ar]3d64s2{ [Ar]3d }^{ 6 }{ 4s }^{ 2 }
So, Fe+2{ Fe }^{ +2 } has electronic configuration = [Ar]3d64s0{ [Ar]3d }^{ 6 }{ 4s }^{ 0 }
Since cyanide is a strong ligand so it will pair up the electrons and one unpaired electron will be present.
Where, Ar = argon having atomic number =18{ 18 }
Here, no unpaired electrons are present as it is diamagnetic in nature, so, the magnetic moment of this complex will be zero.

Now, for [Zn(NH3)6]2+{ [Zn(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 2+ },we can see that ammonia is the ligand which has charge = 0{ 0 }
Atomic number of Zn = 30{ 30 }
The electronic configuration of Zn = [Ar]3d104s2{ [Ar]3d }^{ 10 }{ 4s }^{ 2 }
So, Zn+2{ Zn }^{ +2 } has electronic configuration = [Ar]3d10{ [Ar]3d }^{ 10 }
Where, Ar = argon having atomic number = 18{ 18 }
Here, no unpaired electrons are present as it is diamagnetic in nature, so, the magnetic moment of this complex will be zero.

Hence, we can conclude that [Cr(NH3)6]3+{ [Cr(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ } has the highest magnetic moment.

Note: The possibility to make a mistake is that you have to find the magnetic moment by the formula μ=n(n+1){ \mu =\sqrt { n(n+1) } } not by using spin only magnetic moment formula which is μ=S(S+2){ \mu =\sqrt { S(S+2) } }.