Which of the following represent(s) the correct sequence of indicated property ?

After losing three electrons, the outer electronic configuration of ${\mathrm{Sc}}^{3+}$ is [Ar], ${\mathrm{Fe}}^{3+}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$ and ${\mathrm{Mn}}^{3+}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^4.$ Therefore, ${\mathrm{Sc}}^{3+}$ is most stable, than ${\mathrm{Fe}}^{3+}$ and finally, ${\mathrm{Mn}}^{3+}.$ Hence, given trend is correct.

More is the unpaired electrons, more is the magnetic moment. ${\mathrm{Ni}}^{2+}$ has two unpaired electrons, ${\mathrm{Mn}}^{2+}$ has five unpaired electrons, ${\mathrm{Fe}}^{2+}$ has four unpaired electrons, and ${\mathrm{Co}}^{2+}$ has three unpaired electrons. Therefore, the correct order is:

${\mathrm{Mn}}^{2+}>{\mathrm{Fe}}^{2+}>{\mathrm{Co}}^{2+}>{\mathrm{Ni}}^{2+}$

Number of unpaired electrons in ${\mathrm{Cr}}^{3+}$ is three and in ${\mathrm{Cr}}^{2+}$, number of unpaired electrons in four. Therefore, the correct order is:

${\mathrm{Cr}}^{2+}>{\mathrm{Cr}}^{3+}$

On moving left to right, size of elements decreases because of the addition of electrons in the same subshell. Therefore the given ionic order present in option-4 is wrong and the correct order is:

${\mathrm{Cr}}^{2+}>{\mathrm{Fe}}^{2+}>{\mathrm{Mn}}^{2+}>{\mathrm{Ni}}^{2+}$