Given below are two statements;

Statement I: Ferromagnetism is considered as an extreme form of paramagnetism

Statement II: The number of unpaired electrons in a Cr2+ ion (Z=24) is the same as that of a Nd3+ ion (Z= 60).

In the light of the above statements, choose the correct answer from the options given below:

The correct answer is Statement I is true but Statement II is false.

Analysis of Statement I:
Paramagnetism is a property shown by substances that are weakly attracted by a magnetic field due to the presence of unpaired electrons, which align themselves in the direction of the magnetic field. In ferromagnetic materials, the magnetic moments of the atoms are grouped into domains. In the presence of an external magnetic field, these domains align strongly in the direction of the magnetic field, resulting in a very strong magnetization that can persist even after the external field is removed. Thus, ferromagnetism represents a highly ordered, spontaneous, and extreme alignment of magnetic moments, and it is considered an extreme form of paramagnetism. Therefore, Statement I is true.

Analysis of Statement II:
Let us find the number of unpaired electrons in ${\mathrm{Cr}}^{2+}$ and ${\mathrm{Nd}}^{3+}$ ions:
1. Chromium (Cr) has an atomic number of Z = 24. Its ground-state electronic configuration is:
$\left[\mathrm{Ar}\right]3d^{5}4s^1$
To form the ${\mathrm{Cr}}^{2+}$ ion, we remove two electrons (one from the 4s orbital and one from the 3d orbital):
${\mathrm{Cr}}^{2+}:\left[\mathrm{Ar}\right]3d^4$
In the 3d subshell, there are 5 degenerate orbitals. According to Hund's rule, these 4 electrons will occupy 4 separate orbitals singly. Hence, the number of unpaired electrons in ${\mathrm{Cr}}^{2+}$ is 4.

2. Neodymium (Nd) has an atomic number of Z = 60. Its ground-state electronic configuration is:
$\left[\mathrm{Xe}\right]4f^{4}6s^2$
To form the ${\mathrm{Nd}}^{3+}$ ion, we remove three electrons (two from the 6s orbital and one from the 4f orbital):
${\mathrm{Nd}}^{3+}:\left[\mathrm{Xe}\right]4f^3$
In the 4f subshell, there are 7 degenerate orbitals. The 3 electrons will occupy 3 separate orbitals singly. Hence, the number of unpaired electrons in ${\mathrm{Nd}}^{3+}$ is 3.

Since the number of unpaired electrons in ${\mathrm{Cr}}^{2+}$ (which is 4) is not equal to that of ${\mathrm{Nd}}^{3+}$ (which is 3), Statement II is false.