Stability of transition metals in aqueous solution
Both elements are stable under certain conditions in various oxidation states. In chromium they range from -2 to +6 and in manganese from 0 to +7.

However, some of them are more common than others, especially in aqueous solution. In chromium the most common are +3 and +6. As a "free" ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one electron in the d shell. If you dissolve chromium(III)chloride in water you will get a complex formation.
$$\ce{CrCl3 + 6H2O &lt;=&gt; [Cr(H2O)6]^{3+} + 3Cl-}$$
Assuming an octahedral ligand alignment in crystal field theory, you are correct in assuming, that the $t_{2g}$ orbitals will be half filled, resulting in a more stable configuration.

For manganese the most common oxidation states are +2, +4 and +7. If you dissolve manganese(III)chloride, then you will also form a aqueous complex, which will slowly transform to manganese(III)hydroxide and precipitate.
$$\ce{MnCl3 + 6H2O &lt;=&gt; [Mn(H2O)6]^{3+} + 3Cl- &lt;=&gt; Mn(OH)3 v + 3H3+O + 3Cl-}$$
The resulting hydroxide may be oxidised by air to manganese(IV)oxide hydroxide.
$$\ce{2\overset{\color{\green}{+III}}{Mn}(OH)3 + \overset{\color{\green}{\pm0}}{O2} -&gt; 2\overset{\color{\green}{+IV}}{Mn}O(OH)2 + 2{}^{-}OH}$$
Over time this will convert into manganese dioxide. 
$$\ce{MnO(OH)2 -&gt; MnO2 + H2O}$$