Question

Find the oxidation state of metals in oxides, oxo-cation, oxo-anion?

$Sc_2O_3$, $TiO_2$, $VO_2^+$, $CrO_4^{2-}$, $MnO_4^-$

Answer 1

Sc: +3, Ti: +4, V: +5, Cr: +6, Mn: +7

Answer 2

To find the oxidation state of metals in the given compounds, we use the following rules:

1. The oxidation state of oxygen is typically -2.
2. The sum of oxidation states of all atoms in a neutral compound is zero.
3. The sum of oxidation states of all atoms in a polyatomic ion is equal to the charge of the ion.

Let 'x' be the oxidation state of the metal in each case.

1. $Sc_2O_3$ (Scandium oxide)

   • $2(x) + 3(-2) = 0$
   • $2x - 6 = 0$
   • $2x = 6$
   • $x = +3$
   • The oxidation state of Sc is +3.

2. $TiO_2$ (Titanium dioxide)

   • $x + 2(-2) = 0$
   • $x - 4 = 0$
   • $x = +4$
   • The oxidation state of Ti is +4.

3. $VO_2^+$ (Vanadyl cation)

   • $x + 2(-2) = +1$
   • $x - 4 = +1$
   • $x = +1 + 4$
   • $x = +5$
   • The oxidation state of V is +5.

4. $CrO_4^{2-}$ (Chromate ion)

   • $x + 4(-2) = -2$
   • $x - 8 = -2$
   • $x = -2 + 8$
   • $x = +6$
   • The oxidation state of Cr is +6.

5. $MnO_4^-$ (Permanganate ion)

   • $x + 4(-2) = -1$
   • $x - 8 = -1$
   • $x = -1 + 8$
   • $x = +7$
   • The oxidation state of Mn is +7.

Oxidation states are determined by setting the sum of oxidation states of all atoms in a neutral compound to zero, or to the charge of a polyatomic ion. Oxygen is assigned an oxidation state of -2.