Question

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M₃O₄ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO∙M₂O₃, to permitestimation of the metal’s two oxidation states.)

(a) Sc₂O₃

(b) TiO₂

(c) V₂O₅

(d) CrO₃

(e) MnO₂

(f) Fe₃O₄

(g) Co₃O₄

(h) NiO

(i) Cu₂O

Short answer

The oxidation state are as follow:

1. Sc =+3
2. Ti =+4
3. V =+5
4. Cr =+6
5. Mn =+4
6. Fe =+2 and +3
7. Co =+2 and +3
8. Ni =+2
9. Cu =+1

Step-by-step solution

Oxidation State

The oxidation stateis the imaginary charge of an atom if all of its bonds to different atoms were fully ionic. It describes the degree of oxidation of an atom in a chemical compound.

Finding the oxidation states of various metals  

a) We have Sc₂O₃ and we can use x for the oxidation state of Sc. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Sc like this,

\(\begin{aligned}2.x + ( - 2).3 &= 0\\2.x - 6 &= 0\\2.x &= 6\\x &= \pm 3\end{aligned}\)

Which means the oxidation state of scandium is +3.

b) We have TiO₂ and we can use x for the oxidation state of Sc. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Ti like this,

\(\begin{aligned}x + ( - 2).2 &= 0\\x - 4 &= 0\\x &= + 4\end{aligned}\)

Which means the oxidation state of titanium is +3.

c) We have V₂O₅ and we can use x for the oxidation state of V. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of V like this,

\(\begin{aligned}x + ( - 2).5 &= 0\\2.x - 10 &= 0\\2.x &= 10\\x &= + 5\end{aligned}\)

Which means the oxidation state of vanadium is +5.

d) We have CrO₃ and we can use x for the oxidation state of Cr. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Cr like this,

\(\begin{aligned}x + ( - 2).3 &= 0\\x - 6 &= 0\\x &= + 6\end{aligned}\)

Which means the oxidation state of chromium is +6.

e) We have MnO₂ and we can use x for the oxidation state of Mn. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Mn like this,

\(\begin{aligned}x + ( - 2).2 &= 0\\x - 4 &= 0\\x &= + 4\end{aligned}\)

Which means the oxidation state of maganese is +4.

f) We have Fe₃O₄ is the mixed oxide which consists of FeO and Fe₂O₃.

First we can determine the oxidation state of Fe in FeO. We can use x for the oxidation state of Fe. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Fe like this,

\(\begin{aligned}x + ( - 2) &= 0\\x &= + 2\end{aligned}\)

Then we have Fe₂O₃ and we can use x for the oxidation state of Fe. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Fe like this,

\(\begin{aligned}2.x + ( - 2).3 &= 0\\2.x - 6 &= 0\\x &= + 3\end{aligned}\)

Which means the oxidation state of iron is +2 and +3.

g) We have \(C{o_3}{O_4}\) is mixed oxide which consists of CoO and Co₂O₃.

First we can determine the oxidation state of Co in CoO. We can use x for the oxidation state of Co. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Co like this,

\(\begin{aligned}x + ( - 2) &= 0\\x &= + 2\end{aligned}\)

Then we have Co₂O₃ and we can use x for the oxidation state of Co. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Co like this,

\(\begin{aligned}2.x + ( - 2).3 &= 0\\2.x - 6 &= 0\\x &= + 3\end{aligned}\)

Which means the oxidation state of cobalt is +2 and +3.

h) We have NiO and we can use x for the oxidation state of Ni. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Ni like this,

\(\begin{aligned}x + ( - 2) &= 0\\x &= + 2\end{aligned}\)

Which means the oxidation state of nickel is +2.

i) We have Cu₂O and we can use x for the oxidation state of Cu. We know that the general oxidation state of oxygen is -2, so we can determine the oxidation state of Cu like this,

\(\begin{aligned}2x + ( - 2) &= 0\\x &= + 1\end{aligned}\)

Which means the oxidation state of copper is +1.

Hence, the final answer is;

The oxidation state are as follow:

1. Sc =+3
2. Ti =+4
3. V =+5
4. Cr =+6
5. Mn =+4
6. Fe =+2 and +3
7. Co =+2 and +3
8. Ni =+2
9. Cu =+1