The $$d$$-electron configurations of $$\text{Cr}^{2+}, \text{Mn}^{2+}, \text{Fe}^{2+}$$ and $$\text{Co}^{2+}$$ are $$d^4, d^5, d^6$$ and $$d^7$$ respectively. Which one of the following will exhibit the lowest paramagnetic behaviour? (Atomic no. Cr = 24, Mn = 25, Fe = 26, Co = 27).

Magnetic behaviour of a coordination complex depends on the number of unpaired electrons, $$n$$.
Spin-only magnetic moment is given by $$\mu_\text{so}= \sqrt{n(n+2)}\;\text{BM}$$, so a smaller $$n$$ means weaker (lower) paramagnetism.

All four complexes contain the weak-field ligand $$\text{H}_2\text{O}$$.
In an octahedral field produced by a weak ligand, the crystal-field splitting energy ($$\Delta_\text{o}$$) is small; hence all the complexes are high-spin.
Therefore we distribute the metal-ion $$d$$ electrons according to Hund’s rule in the pattern $$t_{2g}^{\uparrow\uparrow\uparrow}e_g^{\uparrow\uparrow}t_{2g}^{\downarrow\downarrow\downarrow}\ldots$$ until all electrons are placed.

Case 1: $$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$$ : $$\text{Cr}^{2+}$$ is $$3d^4$$
High-spin arrangement: $$t_{2g}^{\uparrow\uparrow\uparrow}e_g^{\uparrow}$$  →  $$n = 4$$ unpaired.

Case 2: $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$ : $$\text{Mn}^{2+}$$ is $$3d^5$$
High-spin arrangement: $$t_{2g}^{\uparrow\uparrow\uparrow}e_g^{\uparrow\uparrow}$$  →  $$n = 5$$ unpaired.

Case 3: $$[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$$ : $$\text{Fe}^{2+}$$ is $$3d^6$$
High-spin arrangement: $$t_{2g}^{\uparrow\uparrow\uparrow\downarrow}e_g^{\uparrow\uparrow}$$  →  $$n = 4$$ unpaired.

Case 4: $$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$$ : $$\text{Co}^{2+}$$ is $$3d^7$$
High-spin arrangement: $$t_{2g}^{\uparrow\uparrow\uparrow\downarrow\downarrow}e_g^{\uparrow\uparrow}$$  →  $$n = 3$$ unpaired.

Comparison of unpaired electrons:
$$n_{[\text{Mn}]} = 5 \gt n_{[\text{Cr}]} = n_{[\text{Fe}]} = 4 \gt n_{[\text{Co}]} = 3$$

Since $$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$$ has the smallest number of unpaired electrons, it exhibits the lowest paramagnetism.

Option A which is: $$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$$