Question

Calculate the magnetic moment of a divalent ion in the aqueous solution if its atomic number is 25?

Answer 1

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 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. Based on interactions, magnetic moments can be divided into two parts.
Diamagnetism
Paramagnetism
In solid structures due to a different kind of alignment of the magnetic moments with each other, there are some other kinds of magnetism, which are,
Ferromagnetism
Antiferromagnetism
Ferrimagnetism
In the case of electrons, two spins are 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 the magnetic moment would be, $\mu = \sqrt {\mathop L\limits^ \to \left( {\mathop L\limits^ \to + 1} \right) + 4\mathop S\limits^ \to \left( {\mathop S\limits^ \to + 1} \right)}$
The atomic number of 25 is from the 1st transition series of metals, 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 equals to 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 the electronic configuration of the divalent ion of atomic number 25 is, $3{d^5}6{s^0}$ .
Therefore, the number of unpaired electrons in the d orbital is 5.
The spin only magnetic moment is,

$${\mu _s} = \sqrt {n\left( {n + 2} \right)} \\\ or,{\mu _s} = \sqrt {5\left( {5 + 2} \right)} \\\ or,{\mu _s} = \sqrt {5\left( 7 \right)} \\\ or,{\mu _s} = \sqrt {35} \\\$$

Therefore, the magnetic moment of a divalent ion in the aqueous solution if its atomic number is 25 is, $\sqrt {{\text{35}}} \,\,B.M$

Note:
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.