Identify the correct trend given below: (Atomic No. = Ti: 22, Cr: 24 and Mo: 42)

$\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}>$ $\displaystyle \left[\mathrm{Mo}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}>\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

$\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\displaystyle >\left[\mathrm{Mo}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}<\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

$\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\displaystyle <\left[\mathrm{Mo}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}>\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

$\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\displaystyle <\left[\mathrm{Mo}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\displaystyle \Delta \mathrm{o}$ of $\displaystyle \left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}<\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$