When chemicals react, they can either lose or gain electrons. Oxidation states are numbers that show us whether an atom has lost or gained electrons in a reaction. If an atom loses electrons, its oxidation number goes up, but if it gains electrons, its oxidation number goes down. We use these numbers to see how many electrons an ion has lost or gained compared to the uncombined element. If the element lost electrons, it has a positive oxidation state, and if it gained electrons, it has a negative oxidation state.

Let's take magnesium oxide (MgO) as an example. When magnesium and oxygen react, two electrons from magnesium transfer to oxygen. This means that magnesium has lost two electrons, so it has an oxidation state of +2. Conversely, oxygen has gained two electrons, so it has an oxidation state of -2. When we write oxidation states, we put the sign before the number. So we would write +2 or -6, for instance. Understanding these oxidation states can help us better understand chemical reactions and the behavior of different elements. This is especially important for transition elements, which can have varying oxidation states.

Variable oxidation state definition

When atoms bond with other atoms, the number of electrons available for bonding determines its oxidation state. In transition metals, both the 4s and 3d electrons are available for bonding. This unique feature gives transition metals something called variable oxidation states.

Variable oxidation states are important because they determine the charge on an atom under certain conditions. To better understand variable oxidation states, let's imagine that only ionic bonding is possible. If an atom can only lose up to three electrons to make bonds with another atom, it has 3 variable oxidation states, depending on the atom it bonds to. Now, let's dive deeper into why transition metals have variable oxidation states. We will explore the redox potential of a transition metal when going from a high to a lower oxidation state, which depends on the pH and ligand. We will also discover the five oxidation states of vanadium through a reaction between vanadate(V) and zinc. Finally, we will carry out the ‘silver mirror’ test or the Tollens' test. Understanding variable oxidation states is important for understanding how transition metals behave and interact with other elements.

Variable oxidation states of transition metals

Transition metals have the unique ability to use both the 4s and 3d electrons for bonding. This is because there is not a significant energy gap between the 3d orbital and the 4s orbital. In fact, the 4s electrons are lost first. In transition metals, the ionisation energy required to remove the third electron (on the 3d sub-shell) is not much larger than the ionisation energies required to remove the first two (on the 4s sub-shell).

To better understand this concept, let's examine the graph below. It shows the first ionisation energies of vanadium (a transition metal), calcium, and scandium (non-transition metals). Notice how the first ionisation energies of vanadium are closer together than those of scandium and calcium? This small difference between the ionisation energies makes it easy for transition metals like vanadium to have variable oxidation states. Not much energy is required to remove a small number of electrons from transition metals.

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The table below shows examples of transition metals and their common oxidation states.

The atoms with lower oxidation states can exist as simple ions like Mn2+ or Mn3+. The higher oxidation states can only exist as complex ions covalently bonded to significantly electronegative elements like oxygen. For example, MnO4-.The higher oxidation states are oxidising while the lower ones are reducing. Examples include the MnO4- and the Cr2O72- ions which make excellent oxidising agents. On the other hand, Cr2+ and Fe2+ are reducing.

Transition metals are not the only elements on the periodic table with variable oxidation states. Most elements have variable oxidation states! Oxidation states can be crucial in helping a chemist balance an equation. But it can be tricky to figure out the oxidation state of an element with variable oxidation states, so chemists follow several rules to make it easier. The most straightforward rule is that the sum of the oxidation states of all the elements in a compound must always equal zero. Some other oxidation state rules that chemists follow are listed below. You may already be familiar with some of them. Metals always have a positive oxidation state. Metals cannot have an oxidation state below zero. The oxidation state of a single ion is the same as its ionic charge. The lowest oxidation state for non-metals is 8 minus the group number of the element.

Redox potential

The redox potential of a substance is a measure of its tendency to undergo reduction or oxidation reactions. It tells us how easily a species accepts or donates electrons. In the context of redox reactions, we typically only consider the transfer of electrons and write what is called a half-reaction. For example, the half-reaction below shows the reduction of zinc 2+ ion (Zn2+) to zinc (Zn) with an oxidation state of zero by accepting two electrons:

Zn2+ (aq) + 2e- ⇌ Zn (s) Eº = -76 V

The Eº value of -76 V represents the electrical potential of this half-reaction. It is the difference between the demand for electrons in the reduction half-reaction and the tendency to lose electrons in the oxidation half-reaction. This value is also referred to as the electrode potential or reduction potential, and it shows how easily a substance can be reduced.

Half-reactions with positive Eº values go towards the right, indicating a greater tendency to undergo reduction, while half-reactions with negative Eº values move to the left, indicating a greater tendency to undergo oxidation.

The redox potential of transition metal ions when going from a high oxidation state to a lower one is influenced by two factors: the pH and the ligand. The pH can affect the redox potential by altering the concentration of H+ ions, which can interact with the metal ions and affect their oxidation state. The ligand, which is a molecule or ion that binds to the metal ion, can also affect the redox potential by affecting the stability of the metal-ligand complex and the ease with which the metal ion can be reduced or oxidized.

pH

pH is a measure of the acidity or basicity of a solution. It is a logarithmic scale that ranges from 0 (very acidic) to 14 (very basic). A neutral solution has a pH of 7. When transition metal ions in an aqueous solution go through a redox reaction, it usually involves hydrogen ions. In other words, it requires acidic conditions.

For example, the half-reaction for the reduction of manganate(VII) ion in an acidic solution is MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l)  Eº = +1.51 V. The positive Eº value indicates that the process moves towards the right, and manganese gets reduced from the +8 to +2 oxidation state. An excess of acid is used to ensure this reaction goes to completion.

On the other hand, if a neutral solution like water is used, manganese ions only get reduced to the +4 oxidation state. The Eº value of this reaction is much lower, as the manganese ions are less willing to accept electrons in a neutral solution. The half-reaction for this is MnO4- (aq) + 2H2O (l) + 3e- ⇌ MnO2 (s) + 4OH- (aq)  Eº = +0.59 V.

Ligand

How do ligands affect the redox potential of transition ions? Notice the Eº values of the half-reactions below. They separately show the reduction of the nickel(II) ion and the hexaaminenickel(II) ion. What can you conclude about the ligands by looking at the Eº values?

[Ni(H2O)6]2+ (aq) + 2e- ⇌ Ni (s) + 6H2O (l)    Eº = -0.26 V

[Ni(NH3)6]2+ (aq) + 2e- ⇌ Ni (s) + 6NH3 (aq) Eº = -0.49 V

By comparing the above two equations, we can state the following:

Eº becomes increasingly negative when ammonia replaces the water ligands. The ammonia ligands are more firmly attached to the nickel ions than the water ligands. The process with nickel(II) is more positive than with ammonia. So the equilibrium lies slightly more to the right.

Variable oxidation state of vanadium

Vanadium exhibits variable oxidation states, including vanadium (II), (III), (IV), and (V). Among these, vanadium(IV) is the most stable oxidation state. By carrying out a redox reaction between vanadate(V) ions and zinc in an acidic solution, we can form vanadium species in different oxidation states.

To carry out this reaction, we start with ammonium vanadate, which dissolves in hydrochloric acid to produce a yellow-coloured solution. When we add zinc to the solution, vanadium gets reduced from +5 to +2. This reduction is evident from the colour change of the solution, which goes from yellow to blue to green to violet, indicating the presence of different vanadium species in solution. It is worth noting that the higher oxidation states of vanadium do not exist as simple ions such as V5+(aq) or V4+(aq). Instead, they exist as complex species with ligands that stabilize the high oxidation states. The colour changes observed in the solution reflect the different vanadium species formed during the reduction of vanadate(V) ions by zinc.

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The vanadium(II) ions quickly oxidise in the air when you remove the zinc, because they are unstable. If you experiment, you may briefly observe a pale green colour as the solution goes from yellow (+5) to blue (+4). What you see is not a new oxidation state, but a mixture of the two colours as the vanadium(V) ions get reduced to vanadium(IV).

Now that you know what is happening between the molecules, consider the half-reactions that show the stages of the reaction between vanadate(V) and zinc:Stage 12VO2+ (aq) + 4H+ (aq) + Zn (s)  ⟶ 2VO2+ (aq) + 2H2O (l) + Zn2+ (aq) The half-equations for the above reaction are below. You get the ionic equation above by putting the two half-reactions for vanadate(V) and zinc together and balancing them out. Notice how the half-equation for zinc moves towards the left because of its negative Eº value. VO2+ (aq) + 2H+ (aq) + e- ⇌ VO2+ (aq) + H2O (l)                                          Eº = +1.00 VZn (s) ⇌ Zn2+ (aq) + 2e-                                                                              Eº = -0.76 V                       2VO2+ (aq) + 4H+ (aq) + Zn (s)  ⟶ 2VO2+ (aq) + 2H2O (l) + Zn2+ (aq) Find out how to write an ionic equation in Balancing Equations! Stage 2 VO2+ (aq) + 2H+ (aq) + e- ⇌ V3+ (aq) + H2O (l)  Eº = +0.34 VStage 3V3+ (aq) + e- ⇌ V2+ (aq)                                         Eº = -0.26 VWe can also use tin as the reducing agent instead of zinc to achieve the same results. As long as vanadium has the more positive Eº value, the reaction will proceed in the direction where the vanadium(V) ion gets reduced to the vanadium(II) ion.

The 'silver mirror' test is another reaction that uses the variable oxidation states of a complex ion. Read on to discover why we call it that!

Variable oxidation state of dichromate ions

As mentioned previously, transition metal ions can have a variety of oxidation states. We have gone through this for vanadium, so now we will be exploring dichromate ions.

Firstly, let's explore how the dichromate(VI) ion, Cr2O72− can be reduced to Cr3+ and Cr2+ ions. This can be done using zinc and a dilute acid such as sulphuric acid or hydrochloric acid. Cr2O72−  is orange and when it is reduced using zinc and a dilute acid, it can form Cr3+ (green), which can be further reduced to Cr2+ (blue). This is represented with the following equations.

This equation shows the reduction from +6 to +3.

Cr2O72- + 14H+ + 3Zn → 2Cr3+ + 7H2O + 3 Zn2+

This shows the reduction from +3 to +2.

2Cr3+ + Zn → 2Cr2+ + Zn2+

Now we shall explore how dichromate ions can be produced from the oxidation of Cr3+. This is done using hydrogen peroxide in alkaline conditions which is then followed by acidification.  When a transition metal in a low oxidation state is in an alkaline solution, it is more easily oxidised than when it is in an acidic solution. This can be seen in the following equation:

[Cr(H2O)6]3+ (aq) → [Cr(OH)6]3- (aq) in excess sodium hydroxide (NaOH)

The reduction can be seen here:H2O2 + 2e- → 2OH-The oxidation can be seen here: [Cr(OH)6]3- + 2OH- → CrO42-+ 3e- + 4H2O

This then leads to the following equation:

2[Cr(OH)6]3- + 3H2O2 → 2CrO42- + 2OH- + 8H2O

Finally, let us explore how dichromate(VI) ion Cr2O72−, can be converted into chromate(VI). Chromate can be converted to dichromate using this equilibrium equation:

2CrO42- + 2H+ ⇌ Cr2O72- + H2O

It is important to note that this reaction is not a redox reaction. This is because the oxidation number is alway +6. is instead an acid base reaction.

CrO42- is a yellow solution and can be turned into Cr2O72-, an orange solution by adding dilute sulphuric acid. To change from the orange solution to the yellow solution, we need the addition of sodium hydroxide.

Reduction of Tollens' reagent

We carry out a test to distinguish between an aldehyde or ketone by using the complex ion diamminesilver(I). This test, also called Tollens' test, is one of the ways we use to identify the functional group in an unknown organic compound. Diamminesilver(I) [Ag(NH3)2]+ is also known as Tollens' reagent after the German chemist Bernard Tollens.

Learn more about ketones and aldehydes in Aldehydes and Ketones.

The test involves reducing Tollens' reagent, which contains silver(I) nitrate, to metallic silver. In order to carry out the test, you must first prepare Tollens' reagent. We prepare it for each test since Tollens' reagent is unstable in solution.

To prepare Tollens' reagent:

Add some sodium hydroxide to silver nitrate to produce silver(I) oxide, a brown precipitate. Add concentrated ammonia solution to redissolve the silver(I) oxide back into diamminesilver(I).

Now the Tollens' reagent is ready for Tollens' test. To carry out the test:

Add a few drops of the unknown organic compound to Tollens' reagent. Then gently warm in a water bath. If the unknown compound is a ketone, you will observe no change in the colourless solution. If the unknown compound is an aldehyde, you will get a grey silver precipitate. The Ag+ ions have reduced to Ag while the aldehyde oxidised to carboxylic acid.

We can observe the silver precipitate when the substance contains an aldehyde because aldehydes are reducing agents and reduce the silver(I) nitrate to metallic silver. The Tollens' test is also called the 'silver mirror' test because of the silver coating formed inside the test tube.

Tollens' test can also be used to detect sugars like glucose. Previously, we have used this reaction to coat mirrors with silver.

You can find the half-equations for the reduction of Tollens' reagent by an aldehyde below, as well as the net ionic equation.

Reduction of Tollens' reagent

Ag(NH3)2+ + e- ⟶ Ag + 2NH3

Oxidation of aldehyde

RCHO + 3OH- ⟶ RCOO- + 2H2O + 2e-

Net ionic equation

2Ag(NH3)2+ + RCHO + 3OH- ⟶ 2Ag + 4NH3 + RCOO- + 2H2O

Redox titrations with manganese

Sure, I can explain the procedure for these titrations.

To carry out a redox titration using potassium permanganate, we start by preparing a solution of the reducing agent we want to analyze in a flask. We then add a few drops of dilute sulfuric acid to the flask to make the solution acidic. This is important because potassium permanganate is only a strong oxidizing agent in acidic conditions.

Next, we fill a burette with a standard solution of potassium permanganate. We slowly add the potassium permanganate solution to the flask containing the reducing agent solution, swirling the flask gently to ensure complete mixing. As we add the potassium permanganate solution, we observe the colour change of the solution. At the beginning of the titration, the solution will be the colour of the reducing agent. As we add the potassium permanganate, the solution will gradually change colour until it reaches a pale pink colour. This pale pink colour indicates that all the reducing agent has been oxidized by the potassium permanganate. We stop adding the potassium permanganate solution at the point where the pale pink colour persists for at least 30 seconds. This point is known as the endpoint of the titration. To calculate the concentration of the reducing agent, we use the balanced chemical equation for the reaction between the reducing agent and potassium permanganate, and the volume and concentration of the potassium permanganate solution used in the titration.

Additionally, the redox potential of a substance can be used to predict spontaneous reactions. A spontaneous reaction occurs when the redox potential of the oxidizing agent is greater than the redox potential of the reducing agent. Transition metal ions can also form complex ions with ligands, which can affect their oxidation state and reactivity. The coordination number of a complex ion refers to the number of ligands attached to the central metal ion. The colour of a complex ion is determined by the energy difference between the d-orbitals of the metal ion when different ligands are attached. This is known as crystal field theory. Overall, understanding the variable oxidation states of transition elements is important in various fields such as chemistry, biochemistry, and materials science.

Variable Oxidation State of Transition Elements

Why do transition metals have variable oxidation states?

Transition metals have variable oxidation states, because their 3d and 4s electrons are available for bonding. The small difference between the ionisation energies makes it easy for transition metals like manganese to have variable oxidation states. Not a lot of energy is required to remove a small number of electrons from transition metals.

Which element has variable oxidation states?

Transition metals show variable oxidation states. However, transition elements are not the only elements in the periodic table that have variable oxidation states. In fact, most elements have variable oxidation states.

What is a variable oxidation state?

A variable oxidation state is a number that determines the charge on an atom depending on certain conditions.

What are the oxidation states of transition metals?

The oxidation states of transition metals vary. Vanadium, for example, has four oxidation states: vanadium (II), (III), (IV) and (V). The +5 oxidation state is the most stable.

What are the 7 oxidation states?

The 7 oxidation states are +1, +2, +3, +4, +5, +6, +7.

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