Oxides and potassium dichromate

Rust, paint and a green pigment

Think of three things you have seen. Rust on an iron gate. White paint on a wall. The green pigment in some paints. All three are oxides of d-block metals. Rust is mainly iron(III) oxide. White paint holds titanium dioxide. The green pigment is chromium(III) oxide.

Oxygen reacts with almost every metal, so these oxides are everywhere. First we see how the nature of an oxide depends on the oxidation state of its metal. Then we study one compound of chromium that every laboratory keeps: potassium dichromate.

The oxides of the first series

A metal can form several oxides, because it has several oxidation states. Here are the common ones, with the oxidation state of the metal in brackets.

• Scandium and titanium: $\text{Sc}_2\text{O}_3$ (+3). $\text{TiO}$ (+2), $\text{Ti}_2\text{O}_3$ (+3), $\text{TiO}_2$ (+4).
• Vanadium: $\text{VO}$ (+2), $\text{V}_2\text{O}_3$ (+3), $\text{VO}_2$ (+4), $\text{V}_2\text{O}_5$ (+5).
• Chromium: $\text{CrO}$ (+2), $\text{Cr}_2\text{O}_3$ (+3), $\text{CrO}_3$ (+6).
• Manganese: $\text{MnO}$ (+2), $\text{Mn}_2\text{O}_3$ (+3), $\text{MnO}_2$ (+4), $\text{Mn}_2\text{O}_7$ (+7).
• Iron: $\text{FeO}$ (+2), $\text{Fe}_2\text{O}_3$ (+3), $\text{Fe}_3\text{O}_4$ (+2 and +3).
• Cobalt, nickel, copper, zinc: $\text{CoO}$, $\text{NiO}$, $\text{Cu}_2\text{O}$ (+1), $\text{CuO}$, $\text{ZnO}$.

Look at the highest state in each oxide from scandium to manganese. It is +3, +4, +5, +6 and +7: it equals the group number. After manganese the d electrons are held more tightly, and the highest state in an oxide stops following the group number.

Oxygen is very good at bringing out the highest states. Manganese reaches +7 in $\text{Mn}_2\text{O}_7$ and in the permanganate ion. Its fluoride stops at $\text{MnF}_4$, which is +4. Oxygen can form a double bond with the metal, because its p orbitals overlap with the d orbitals of the metal. Fluorine can form only single bonds.

Basic, amphoteric or acidic?

Ask what an oxide does with an acid and with an alkali. A basic oxide reacts with an acid only. An acidic oxide reacts with an alkali only. An amphoteric oxide reacts with both.

Chromium shows all three in one family:

• $\text{CrO}$ is basic: $\text{CrO} + 2\text{H}^+ \rightarrow \text{Cr}^{2+} + \text{H}_2\text{O}$.
• $\text{Cr}_2\text{O}_3$ is amphoteric. It dissolves in acid: $\text{Cr}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{H}_2\text{O}$. It also dissolves in alkali: $\text{Cr}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Cr(OH)}_4]^-$.
• $\text{CrO}_3$ is acidic. It dissolves in alkali: $\text{CrO}_3 + 2\text{OH}^- \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O}$. With water it gives chromic acid: $\text{CrO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{CrO}_4$.

Manganese does the same. $\text{MnO}$ is basic and $\text{MnO}_2$ is weakly amphoteric. $\text{Mn}_2\text{O}_7$ is acidic, and it gives permanganic acid with water: $\text{Mn}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{HMnO}_4$. Vanadium(V) oxide is amphoteric: $\text{V}_2\text{O}_5 + 6\text{OH}^- \rightarrow 2\text{VO}_4^{3-} + 3\text{H}_2\text{O}$, and $\text{V}_2\text{O}_5 + 2\text{H}^+ \rightarrow 2\text{VO}_2^+ + \text{H}_2\text{O}$. Zinc oxide is amphoteric too: $\text{ZnO} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2\text{O}$, and $\text{ZnO} + 2\text{OH}^- + \text{H}_2\text{O} \rightarrow [\text{Zn(OH)}_4]^{2-}$.

The rule: for one metal, the higher the oxidation state, the more acidic the oxide. Low states are basic, middle states are amphoteric, high states are acidic.

Why? A metal ion with a small positive charge holds the electrons of oxygen weakly. In water the bond between the metal and oxygen breaks, and the hydroxide ion goes free. The oxide is basic. A metal in a high oxidation state pulls the electrons of oxygen strongly. Now the O-H bond breaks instead, and $\text{H}^+$ goes free. The oxide is acidic. A middle state can do either.

Chromate and dichromate

Add a little dilute acid to a yellow solution of potassium chromate. The solution turns orange. Add alkali, and the yellow comes back. Is this a redox change? Check the oxidation numbers. In $\text{CrO}_4^{2-}$, $x + 4(-2) = -2$, so chromium is $+6$. In $\text{Cr}_2\text{O}_7^{2-}$, $2x + 7(-2) = -2$, so chromium is again $+6$. Nothing is oxidised and nothing is reduced.

The change is an acid and base equilibrium:

Acid adds $\text{H}^+$ and pushes the equilibrium to the right. Alkali removes $\text{H}^+$ and pushes it to the left, so $\text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O}$.

The chromate ion is a tetrahedron with chromium in the middle. The dichromate ion is two such tetrahedra that share one oxygen atom. The Cr-O-Cr angle is $126^\circ$.

Both ions have chromium(VI), which has no d electrons. So the colour cannot come from a d-d jump. An electron jumps from an oxygen atom to the chromium atom. This charge transfer absorbs light strongly, which is why even a dilute solution looks bright.

Making potassium dichromate

The ore is chromite, $\text{FeCr}_2\text{O}_4$. There are three steps.

Step 1. Roast the ore with sodium carbonate in plenty of air.

Check the electrons. Four iron atoms go from $+2$ to $+3$ and lose 4 electrons. Eight chromium atoms go from $+3$ to $+6$ and lose 24. That is 28 electrons. Seven $\text{O}_2$ molecules take $7 \times 4 = 28$.

Step 2. Dissolve in water and acidify. Yellow sodium chromate dissolves. Brown iron(III) oxide does not, so it is filtered off. Sulfuric acid turns the chromate into dichromate:

Step 3. Add potassium chloride.

Orange crystals of potassium dichromate separate, because this salt is much less soluble in cold water than the others. Why make the potassium salt? Sodium dichromate takes up water from the air, so it cannot be weighed well. Potassium dichromate does not, and it can be made very pure. That is why it is a primary standard.

Dichromate as an oxidising agent

In acid, the orange dichromate ion takes six electrons. Two chromium atoms each fall from $+6$ to $+3$:

The product, $\text{Cr}^{3+}$, is green. So the colour goes from orange to green while the reducing agent is used up. In Class 11 you balanced this half-reaction. Here the reducing agent supplies the six electrons, and the balanced equations follow from matching them.

• With iodide, brown iodine forms: $\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$.
• With hydrogen sulfide, yellow sulfur forms: $\text{Cr}_2\text{O}_7^{2-} + 3\text{H}_2\text{S} + 8\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{S} + 7\text{H}_2\text{O}$.
• With ethanol, ethanoic acid forms: $2\text{Cr}_2\text{O}_7^{2-} + 3\text{C}_2\text{H}_5\text{OH} + 16\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 3\text{CH}_3\text{COOH} + 11\text{H}_2\text{O}$.

Look at the ethanol reaction. The carbon that carries the OH group goes from $-1$ to $+3$ in the acid. So each ethanol molecule loses 4 electrons. Three ethanol molecules lose 12, and two dichromate ions take $2 \times 6 = 12$. The old breath-test tube worked on this: orange crystals turned green when alcohol was present.

Dichromate is a weaker oxidising agent than permanganate. Its potential, $+1.33\ \text{V}$, is a little below that of chlorine, $+1.36\ \text{V}$. So cold, dilute hydrochloric acid can be used in a dichromate titration. The chloride is left alone.

Dichromate can also oxidise inside one compound. Orange ammonium dichromate burns once it is lit, and leaves a green fluffy powder:

Each chromium takes 3 electrons, so two take 6. The two nitrogen atoms go from $-3$ to $0$ and give 6. The same compound holds the oxidising agent and the reducing agent.

A warning. Chromium(VI) compounds are poisonous and cause cancer. A teacher handles them with care. Waste is collected and never poured down the sink.

Practise with new numbers

Try it in Python

The program matches the electrons of four reducing agents with the six electrons of one dichromate ion. It prints how many of each react, and then the dichromate needed for $0.0120\ \text{mol}$ of each.

Common mistakes

• Calling chromate to dichromate a redox change. Chromium is $+6$ on both sides. Only $\text{H}^+$ and water move.
• Mixing up the two directions. Acid gives orange dichromate. Alkali gives yellow chromate.
• Taking orange as the product. In a reaction with a reducing agent, orange dichromate is used up and green $\text{Cr}^{3+}$ is the product.
• Counting one chromium atom. Two chromium atoms take part, so one dichromate ion takes 6 electrons and not 3.
• Calling $\text{Cr}_2\text{O}_3$ acidic. It is amphoteric. $\text{CrO}_3$ is the acidic oxide.

Remember these

• For one metal, a higher oxidation state gives a more acidic oxide: basic, then amphoteric, then acidic.
• $2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$. Acid gives orange, alkali gives yellow, and chromium stays $+6$.
• Chromite is roasted with sodium carbonate and air, acidified, and treated with potassium chloride.
• In acid, dichromate takes 6 electrons and turns green: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$.
• Dichromate is a primary standard, because the salt can be made pure and does not take up water.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Chromium(III) oxide dissolves in hydrochloric acid and in sodium hydroxide solution. What kind of oxide is it?

2. Which is the more acidic oxide of manganese, $\text{MnO}$ or $\text{Mn}_2\text{O}_7$, and why?

3. A yellow solution turns orange when a few drops of acid are added. Does the oxidation number of chromium change?

4. How many electrons are lost in all when 4 formula units of $\text{FeCr}_2\text{O}_4$ are roasted with air, and where do they go?

5. Why is potassium dichromate preferred to sodium dichromate for making a standard solution?