1
$\begingroup$

Why is titanium more stable with a lower oxidation state, but vanadium more stable with a larger oxidation state?

More specifically, why is $\ce{Ti^{+}}$ more stable than $\ce{Ti^{3+}}$, but $\ce{V^{2+}}$ is less stable than $\ce{V^{4+}}$ in aqueous solution?

Both $\ce{Ti^{+}}$ ($\mathrm{3d^{2}4s^{1}}$) and $\ce{V^{2+}}$ ($\mathrm{3d^{3}4s^{0}}$) have a half filled $t^3_{\mathrm{2g}}$ orbital configuration which imparts stability during the formation of aqua complexes, assuming the $\mathrm{4s}$ electron of $\ce{Ti^{+}}$ can go to the $\mathrm{3d}$ sub-shell.

Is this discrepancy related to the charge/size ratio, which is larger in $\ce{Ti^{3+}}$ than in $\ce{Ti^+}$?

$\endgroup$
  • 2
    $\begingroup$ Ti+ is not a thing at all. $\endgroup$ – Ivan Neretin Jan 29 '18 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.