Water is a weak-field ligand, so the crystal field splitting paired in lower energy. Hence, electrons are not paired in lower energy d-orbitals (i.e., no pairing of electrons in d-orbitals occurs aganist the Hunds rule).

Species	Oxidation state of central metal ion	Configuration of central metal ion	Number of unpaired d-electrons
A. $\left[\mathrm{C}\mathrm{r}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$	+3	$\mathrm{C}{\mathrm{r}}^{3+}\left(3{\mathrm{d}}^3\right)$	3
B. ${\left[\text{Fe}{({\text{H}}_2\text{O})}_6\right]}^{2+}$	+2	$\mathrm{F}{\mathrm{e}}^{2+}\left(3{\mathrm{d}}^6\right)$	4
C. $\left[\mathrm{C}\mathrm{u}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$	+2	$\mathrm{C}{\mathrm{u}}^{2+}\left(3{\mathrm{d}}^9\right)$	1
D. $\left[\mathrm{Z}\mathrm{n}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$	+2	$\mathrm{Z}{\mathrm{n}}^{2+}\left(3{\mathrm{d}}^{10}\right)$	0

The more the number of unpaired electrons, the higher is the paramagnetic character of the metal ion, As a result, $\left[\mathrm{F}\mathrm{e}\right({\mathrm{H}}_2{\mathrm{O}}_{}{)}_6{]}^{2+}$ has the highest paramagnetism.