Oxidation states, colour and magnetism

Many oxidation states

Sodium has one oxidation state, $+1$. Aluminium has one, $+3$. Manganese has at least six. Why can a transition element give up different numbers of electrons?

The reason is the energy of the orbitals. The $4s$ and $3d$ electrons have almost the same energy. Once the $4s$ electrons have gone, a $3d$ electron can leave with only a small extra cost, and then another. Each loss gives a new oxidation state.

The chart shows the states. There are three patterns to look for.

• Steps of one. Manganese shows $+2$, $+3$, $+4$, $+5$, $+6$ and $+7$. In the p block the states usually differ by two, as in $\text{Sn}^{2+}$ and $\text{Sn}^{4+}$.
• The highest state rises from Sc to Mn. It is $+3$, $+4$, $+5$, $+6$ and $+7$, which is the total number of $4s$ and $3d$ electrons. Manganese uses all seven.
• After manganese the highest state falls. Iron reaches $+6$ (in the ferrate ion, $\text{FeO}_4^{2-}$, which is rare). Cobalt and nickel reach $+4$. Copper reaches $+3$ only in rare compounds, and zinc has $+2$ alone. The nuclear charge is now so high that the $3d$ electrons are held tightly.

The program below finds the number of outer electrons for each element and puts it next to the highest state known.

Which oxidation state is stable?

Three ideas help you to judge.

Higher states are held by oxygen and fluorine. Both atoms are small and very electronegative. Oxygen can also form double bonds. Manganese reaches $+7$ in the oxoanion $\text{MnO}_4^-$ and in the oxide $\text{Mn}_2\text{O}_7$. Its highest fluoride is only $\text{MnF}_4$. Fluorine can only form single bonds, while oxygen can double bond to the metal, so oxygen holds a higher state.

Higher states become more stable down a group. In group 6, chromium(VI) is a strong oxidising agent. Molybdenum(VI) and tungsten(VI) are stable and are not oxidising. The $4d$ and $5d$ electrons are further from the nucleus and take part in bonding more easily.

An ion with a half-filled or full d subshell is specially stable. Look at the standard potentials of the $\text{M}^{3+}/\text{M}^{2+}$ couples:

• Ti $-0.37\ \text{V}$, V $-0.26\ \text{V}$, Cr $-0.41\ \text{V}$
• Mn $+1.57\ \text{V}$, Fe $+0.77\ \text{V}$, Co $+1.97\ \text{V}$

A negative value means that $\text{M}^{3+}$ is the more stable ion, and $\text{M}^{2+}$ is a reducing agent. A large positive value means $\text{M}^{3+}$ is hard to keep and is a strong oxidising agent. Link each value to the electrons.

• Mn, $+1.57\ \text{V}$. $\text{Mn}^{2+}$ is $3d^5$, specially stable. $\text{Mn}^{3+}$ gladly takes an electron to return to it. This agrees with the high third ionisation enthalpy of manganese in the last lesson.
• Fe, $+0.77\ \text{V}$. $\text{Fe}^{3+}$ is $3d^5$, so $\text{Fe}^{2+}$ loses an electron fairly easily to reach it. That is why iron(II) solutions slowly turn to iron(III) in air. In the last lesson, iron had the lowest ratio of the energy for the third electron to the energy for the first two.
• Co, $+1.97\ \text{V}$. $\text{Co}^{3+}$ is so strong an oxidising agent that it oxidises water. Cobalt(III) lasts only in complexes.
• Cr, $-0.41\ \text{V}$. $\text{Cr}^{3+}$ is $3d^3$, which is a stable arrangement in a field of ligands. So $\text{Cr}^{2+}$ is a good reducing agent and chromium(III) is the common state.

Copper(I) falls apart in water

Copper has two states, $+1$ and $+2$. The $\text{Cu}^+$ ion has the full $3d^{10}$ subshell, so you might expect it to be the more stable. In water it is not. It disproportionates: the same ion is oxidised and reduced at once.

• $2\text{Cu}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Cu}(s)$

The standard potentials are $E^\circ(\text{Cu}^+/\text{Cu}) = +0.52\ \text{V}$ and $E^\circ(\text{Cu}^{2+}/\text{Cu}^+) = +0.16\ \text{V}$. One $\text{Cu}^+$ is reduced, so that couple is the cathode. The other is oxidised, so the second couple is the anode:

• $E^\circ_{cell} = 0.52 - 0.16 = +0.36\ \text{V}$.

It is positive, so the change happens. The deeper reason is energy. The small and doubly charged $\text{Cu}^{2+}$ ion is hydrated much more strongly than $\text{Cu}^+$. That hydration enthalpy repays the high second ionisation enthalpy of copper, $1958\ \text{kJ/mol}$. So copper(I) lasts only in insoluble solids such as $\text{Cu}_2\text{O}$ and $\text{CuI}$, or in complexes.

The same test applied to iron shows the opposite. For $\text{Fe}^{2+}$, $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\ \text{V}$ and $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$. Then $E^\circ_{cell} = -0.44 - 0.77 = -1.21\ \text{V}$, which is negative. So $\text{Fe}^{2+}$ does not disproportionate.

Magnetic properties

Every electron behaves like a tiny magnet, because it spins. Two electrons in the same orbital have opposite spins, and their magnets cancel. An unpaired electron is not cancelled.

• A substance with unpaired electrons is paramagnetic. It is pulled into a magnetic field.
• A substance with all its electrons paired is diamagnetic. It is pushed out very weakly.
• In iron, cobalt and nickel the magnets of many atoms line up together in the solid, and the metal is strongly magnetic. This is ferromagnetism.

The strength of the magnet is the magnetic moment. For an ion with $n$ unpaired electrons, the spin-only value is

Here BM is the Bohr magneton, the unit of the magnetic moment. This formula works well for the first series.

• 1 unpaired electron: $\mu = 1.73\ \text{BM}$
• 2 unpaired electrons: $2.83\ \text{BM}$
• 3 unpaired electrons: $3.87\ \text{BM}$
• 4 unpaired electrons: $4.90\ \text{BM}$
• 5 unpaired electrons: $5.92\ \text{BM}$

Use the unpaired electrons that Hund's rule gives for the ion. $\text{Ti}^{3+}$ ($3d^1$) has $\mu = 1.73\ \text{BM}$, $\text{Cr}^{3+}$ ($3d^3$) has $3.87\ \text{BM}$, and $\text{Mn}^{2+}$ and $\text{Fe}^{3+}$ ($3d^5$) have $5.92\ \text{BM}$. $\text{Zn}^{2+}$ has none, so it is diamagnetic. You can also go the other way. If a measurement gives $\mu = 3.9\ \text{BM}$, solve $n(n+2) \approx 15$, so $n = 3$.

The points rise to a maximum at five d electrons and fall again, because beyond five the electrons start to pair up.

Practise with new numbers

The colour of ions

Most compounds of the s and p blocks are white, but many transition metal ions are coloured. The sulphate of copper is blue, and nickel salts are green. Why?

In an ion that is surrounded by water molecules or other ligands, the five d orbitals no longer have the same energy. They split into two sets, a lower set of three and an upper set of two. The reason for the split belongs to the study of coordination compounds. For now, accept it as a fact.

If the lower set is not full, a d electron can jump to the upper set. It does so by taking in a photon of the right energy, and that energy falls in the visible range. This is called a d-d transition. The ion takes the colour that matches the gap out of white light, and you see the colours that remain. The right-hand panel shows the pairs. An ion that absorbs green light looks purple. An ion that absorbs orange light looks blue.

The energy per mole of ions is $E = \dfrac{N_A hc}{\lambda}$. With the wavelength in nm and the energy in kJ/mol, this becomes

The $\text{Ti}^{3+}$ ion in water has a gap of about $243\ \text{kJ/mol}$. Then $\lambda = 119\,627 / 243 = 492\ \text{nm}$, in the blue-green, and the solution looks purple.

Two rules follow from this picture.

• An ion with no d electron ($d^0$) cannot make a jump, because it has nothing to promote. $\text{Sc}^{3+}$ and $\text{Ti}^{4+}$ are colourless.
• An ion with a full d subshell ($d^{10}$) cannot make a jump, because it has no empty place. $\text{Zn}^{2+}$ and $\text{Cu}^+$ are colourless.

Everything in between is coloured. In water:

• $\text{Ti}^{3+}$ ($d^1$) is purple, $\text{V}^{3+}$ ($d^2$) is green, $\text{Cr}^{3+}$ ($d^3$) is violet;
• $\text{Mn}^{2+}$ ($d^5$) is pale pink, $\text{Fe}^{3+}$ ($d^5$) is yellow, $\text{Fe}^{2+}$ ($d^6$) is pale green;
• $\text{Co}^{2+}$ ($d^7$) is pink, $\text{Ni}^{2+}$ ($d^8$) is green, $\text{Cu}^{2+}$ ($d^9$) is blue.

The colour also depends on the ligands, because they change the gap. Copper sulphate solution is pale blue. Add ammonia and the colour deepens, as the water molecules are replaced by ammonia. One element can show many colours: vanadium is violet as $\text{V}^{2+}$, green as $\text{V}^{3+}$, blue as $\text{VO}^{2+}$ and yellow as $\text{VO}_2^+$.

One more case. The permanganate ion, $\text{MnO}_4^-$, is deep purple, but manganese is $d^0$ in it. Here the colour comes from a different jump, in which an electron moves from an oxygen atom to the metal. This is called charge transfer.

Practise with new numbers

Complex formation

Transition metal ions are small, carry a high charge and have empty d orbitals. These are the best conditions for accepting lone pairs from other molecules or ions. So they form many complexes, such as $[\text{Fe}(\text{CN})_6]^{4-}$ and $[\text{Cu}(\text{NH}_3)_4]^{2+}$. Coordination compounds have a topic of their own.

Try it in Python

The program lists the outer electrons and the highest oxidation state of each element. Then it works out the 3d electrons, unpaired electrons, spin-only moment and colour of nine ions.

Common mistakes

• Counting all the d electrons instead of the unpaired ones. $\mu$ depends only on the unpaired electrons.
• Saying that an ion shows the colour it absorbs. It shows the colour that is left, the complementary one.
• Calling $\text{Zn}^{2+}$ or $\text{Sc}^{3+}$ coloured. There is no d-d jump, because $d^{10}$ is full and $d^0$ is empty.
• Saying that copper(I) is stable in water because it is $3d^{10}$. It disproportionates in water, as hydration favours $\text{Cu}^{2+}$.
• Using the highest oxidation state as the usual one. Chromium reaches $+6$, but chromium(III) is the common state.

Remember these

• Transition elements show many oxidation states, in steps of one, because the $4s$ and $3d$ electrons have close energies. The highest state is the total of $4s$ and $3d$ electrons, up to $+7$ for Mn.
• Higher states are held by O and F. Higher states are more stable in the heavier members of a group.
• $\text{Mn}^{2+}$ and $\text{Fe}^{3+}$ are $3d^5$ and stable. $\text{Cu}^+$ disproportionates in water: $E^\circ_{cell} = 0.52 - 0.16 = +0.36\ \text{V}$.
• $\mu = \sqrt{n(n+2)}$ BM, with $n$ unpaired electrons: $1.73\ \text{BM}$, $2.83\ \text{BM}$, $3.87\ \text{BM}$, $4.90\ \text{BM}$ and $5.92\ \text{BM}$ for $n = 1$ to $5$.
• Colour needs a d-d jump, so $d^0$ and $d^{10}$ ions are colourless. You see the complementary colour. $E = 119\,627 / \lambda$ in kJ/mol, with $\lambda$ in nm.

Check yourself

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

1. Why do the transition elements show so many oxidation states, while sodium shows only one?

2. Calculate the spin-only magnetic moment of $\text{Ni}^{2+}$.

3. A solution of an ion is colourless. Does it have to be $d^0$ or $d^{10}$?

4. $\text{Mn}^{3+}$ is a strong oxidising agent, with $E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.57\ \text{V}$. Explain this using the configuration of $\text{Mn}^{2+}$.

5. A complex absorbs light of wavelength $500\ \text{nm}$ and looks purple. What energy gap does this match?