Question

The spin magnetic moment of cobalt in the compound \[{K_2}\left[ {Co{{\left( {SCN} \right)}_4}} \right]\] is:
A.\[\sqrt 3 B.M\]
B.\[\sqrt 8 B.M\]
C.\[\sqrt {15} B.M\]
D.\[\sqrt {24} B.M\]

Answer 1

Hint: Transition elements show magnetic moment due to presence of unpaired electrons to their d orbitals. With increasing the number of unpaired electrons, the spin magnetic moment value increases. The formula to calculate the spin magnetic moment of the transition elements is \[\mu = \sqrt {n\left( {n + 2} \right)} \]. Where n is the number of unpaired electrons.

Complete step by step answer:
When an atom is in a magnetic field it behaves like a magnet, this is called a magnetic moment. This is because the electrons of the atom interact with the magnetic field. On the basis of interactions, the magnetic moments can be divided into two parts.
- Diamagnetism
-Paramagnetism.
In solid structures due to the different kind of alignment of the magnetic moments with each other, there are some other kind of magnetism, which are,
-Ferromagnetism
-Antiferromagnetism
-Ferromagnetism
 In case of electrons there are two spins possible \[ + \dfrac{1}{2}\] and \[ - \dfrac{1}{2}\]. If unpaired electrons are present the transition metal becomes paramagnetic and Paramagnetism increases with increasing the number of unpaired electrons. The magnetic moment of transition metals depends upon the spin angular momentum of electrons and the orbital angular momentum. According to this the formula of magnetic moment would be,\[\mu = \sqrt {\mathop L\limits^ \to \left( {\mathop L\limits^ \to + 1} \right) + 4\mathop S\limits^ \to \left( {\mathop S\limits^ \to + 1} \right)} \]
For the 1st transition series metals the orbital contribution towards the magnetic moments can be neglected. Therefore, the magnetic moment can be calculated using the modified formula,\[{\mu _s} = \sqrt {4\mathop S\limits^ \to \left( {\mathop S\limits^ \to + 1} \right)} \]. Where, \[\mathop S\limits^ \to \] is the spin angular momentum of the electrons. Spin angular momentum is equal to the half of the unpaired electrons present in the d orbital of transition metals. Therefore, the spin only formula is \[{\mu _s} = \sqrt {n\left( {n + 2} \right)} \]. Where, n is the number of unpaired electrons.
 Now in the complex, \[{K_2}\left[ {Co{{\left( {SCN} \right)}_4}} \right]\] the oxidation state of Co is, +2. Therefore, the number of unpaired electrons in the d orbital is 3.
The spin only magnetic moment is,
 \[
  {\mu _s} = \sqrt {n\left( {n + 2} \right)} \\
  or,{\mu _s} = \sqrt {3\left( {3 + 2} \right)} \\
  or,{\mu _s} = \sqrt {3\left( 5 \right)} \\
  or,{\mu _s} = \sqrt {15} \\
   \\
 \]

So, the correct answer is C.

Note:
For 1st transition series the orbital contribution is negligible, so there we can calculate the magnetic moment by using the spin only formula \[\mu = \sqrt {n\left( {n + 2} \right)} \].