Question

Which of the following is diamagnetic?
A. \[{[Cu{\left( {N{H_3}} \right)_4}]^{2 + }}\]
B. \[{\left[ {NiC{l_4}} \right]^{2 - }}\]
C. \[{[PtC{l_4}]^{2 - }}\]
D. \[{\left[ {Cu{{\left( {H20} \right)}_\$ }} \right]^{2 + }}\]

Answer 1

Hint: Transition elements show a magnetic moment due to the presence of unpaired electrons to their d orbitals. With increasing the number of unpaired electrons, the spin magnetic moment value increases and vice-versa.If there is no unpaired electron then it is diamagnetic.

Complete step by step solution:
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. Based on interactions, magnetic moments can be divided into two parts.
A. Diamagnetism
B. Paramagnetism
In the 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. There should not be an unpaired electron to be diamagnetic.
Now, in case \[{[PtC{l_4}]^{2 - }}\] the electronic configuration is \[5{d^8}6{s^0}6{p^0}\] in a low spin. Therefore the
hybridization is \[ds{p^2}\]. The geometry is square planar. In this electronic configuration, there is no unpaired electron.
Therefore, \[{[PtC{l_4}]^{2 - }}\] is diamagnetic.

The correct option is C.

 Note: 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.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 the magnetic moment would be,\[\mu = \sqrt {\mathop L\limits^ \to \left( {\mathop L\limits^ \to + 1} \right) + 4\mathop S\limits^ \to \left( {\mathop S\limits^ \to + 1} \right)} \] .