Question

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

Answer 1

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

Now, ${ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 3- }$, we can see that cyanide is the ligand which has charge = ${ -1 }$
So, the oxidation state of ${ Fe=+3 }$
Atomic number of Fe = ${ 26 }$
The electronic configuration of Fe = ${ [Ar]3d }^{ 6 }{ 4s }^{ 2 }$
So, ${ Fe }^{ +3 }$ has electronic configuration = ${ [Ar]3d }^{ 5 }{ 4s }^{ 0 }$
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;
${ \mu =\sqrt { n(n+1) } }$
Where, n= number of unpaired electrons
 ${ \mu =\sqrt { 1(1+1) } }$
${ \mu =\sqrt { 2 } }$
${ \mu = { 1.414 }BM }$


Now, for ${ [Fe(CN }_{ 6 }{ ) }_{ ] }^{ 4- }$, we can see that cyanide is the ligand which has charge = ${ -1 }$
So, the oxidation state of Fe = ${ +2 }$
Atomic number of Fe = ${ 26 }$
The electronic configuration of Fe = ${ [Ar]3d }^{ 6 }{ 4s }^{ 2 }$
So, ${ Fe }^{ +2 }$ has electronic configuration = ${ [Ar]3d }^{ 6 }{ 4s }^{ 0 }$
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 =${ 18 }$
Here, no unpaired electrons are present as it is diamagnetic in nature, so, the magnetic moment of this complex will be zero.

Now, for ${ [Zn(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 2+ }$,we can see that ammonia is the ligand which has charge = ${ 0 }$
Atomic number of Zn = ${ 30 }$
The electronic configuration of Zn = ${ [Ar]3d }^{ 10 }{ 4s }^{ 2 }$
So, ${ Zn }^{ +2 }$ has electronic configuration = ${ [Ar]3d }^{ 10 }$
Where, Ar = argon having atomic number = ${ 18 }$
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 ${ [Cr(NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }$ has the highest magnetic moment.

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