$$\text{MnCl}_2\cdot 4\text{H}_2\text{O}$$ (molar mass 198 g $$\text{mol}^{-1}$$) when dissolved in water forms a complex of $$\text{Mn}^{2+}$$. An aqueous solution containing 0.400 g of $$\text{MnCl}_2\cdot 4\text{H}_2\text{O}$$ was passed through a column of a cation exchange resin and the acid solution coming out was neutralized with 10 mL of 0.20 M NaOH. The formula of the complex formed is

A

$$[\text{Mn(H}_2\text{O})_4\text{Cl}_2]$$

B

$$[\text{Mn(H}_2\text{O})_6]\text{Cl}_2$$

C

$$[\text{Mn(H}_2\text{O})_5\text{Cl}]\text{Cl}$$

D

$$\text{Na}[\text{Mn(H}_2\text{O})_3\text{Cl}_3]$$