The d- and f-Block Elements • Topic 2 of 3

Properties of Transition Metals

The partially filled d orbitals of transition metals give rise to a cluster of characteristic properties: colour, paramagnetism, catalytic activity, complex formation, the ability to trap small atoms (interstitial compounds) and easy alloy formation. Two important compounds — potassium dichromate and potassium permanganate — bring many of these ideas together.

Colour and d–d transitions

Most transition-metal ions are coloured. When light falls on the ion, an electron is promoted from a lower-energy d orbital to a higher-energy d orbital (a d–d transition). The energy gap corresponds to part of the visible spectrum; the colour we see is the complement of the absorbed colour. Ions with empty ($d^0$, e.g. $\text{Sc}^{3+}$, $\text{Ti}^{4+}$) or full ($d^{10}$, e.g. $\text{Zn}^{2+}$) d subshells are colourless because no d–d transition is possible.

Magnetic properties

Species with unpaired electrons are paramagnetic (attracted into a magnetic field); those with all electrons paired are diamagnetic. For most first-row complexes the orbital contribution is quenched, so the magnetic moment is given by the spin-only formula $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Paramagnetism therefore rises and falls as $n$ changes across a series.

Catalytic properties

Transition metals and their compounds are excellent catalysts. They can adopt variable oxidation states and provide surfaces that adsorb reactants, lowering the activation energy. Examples: iron in the Haber process, $\text{V}_2\text{O}_5$ in the Contact process, and nickel in hydrogenation.

Complex formation and interstitial compounds

Their small, highly charged ions with vacant d orbitals readily accept lone pairs from ligands, forming coordination complexes such as $[\text{Fe}(\text{CN})_6]^{3-}$. Small atoms (H, C, N) can also occupy the holes in their metal lattices to give hard, chemically inert interstitial compounds (e.g. steel, which is iron with interstitial carbon).

Alloy formation

Because transition metals have similar atomic sizes, atoms of one can replace another in the lattice to give alloys — brass (Cu–Zn), bronze (Cu–Sn) and stainless steel are common examples.

Potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$)

Prepared from chromite ore: it is fused with $\text{Na}_2\text{CO}_3$ in air to give sodium chromate, which is acidified to sodium dichromate and then treated with KCl. It is a powerful oxidising agent in acidic medium: $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6e^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$. The orange dichromate ion is in pH-dependent equilibrium with the yellow chromate ion: $\text{Cr}_2\text{O}_7^{2-}+2\text{OH}^-\rightleftharpoons2\text{CrO}_4^{2-}+\text{H}_2\text{O}$.

Potassium permanganate ($\text{KMnO}_4$)

Made by fusing pyrolusite ($\text{MnO}_2$) with KOH and an oxidant to give green manganate, $\text{K}_2\text{MnO}_4$, which is then oxidised to purple permanganate. In acidic medium it is a strong oxidant: $\text{MnO}_4^-+8\text{H}^++5e^-\rightarrow\text{Mn}^{2+}+4\text{H}_2\text{O}$. It oxidises ferrous ions, oxalate and iodide, and is widely used in volumetric analysis.

Spin-only magnetic moments of some 3d ions
Iond-configurationUnpaired electrons (n)μ = √[n(n+2)] BM
Sc3+d000 (diamagnetic)
Ti3+d111.73
V3+d222.84
Mn2+d555.92
Fe2+d644.90
Ni2+d822.84
Zn2+d1000 (diamagnetic)
1
Worked Example
Calculate the spin-only magnetic moment of $\text{Mn}^{2+}$.
Solution
  1. Mn is $[\text{Ar}]3d^5\,4s^2$; losing two $4s$ electrons gives $\text{Mn}^{2+}=3d^5$.
  2. Five d electrons occupy five orbitals singly, so $n=5$ unpaired electrons.
  3. $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$ BM.

Answer: $\mu=5.92$ BM.

2
Worked Example
A divalent 3d ion has a spin-only magnetic moment of 4.90 BM. Identify the ion.
Solution
  1. Set $\sqrt{n(n+2)}=4.90$, so $n(n+2)=24$, giving $n=4$.
  2. A $2+$ ion of the 3d series with 4 unpaired electrons is $d^6$.
  3. $d^6$ for an $M^{2+}$ ion corresponds to iron(II).

Answer: The ion is $\text{Fe}^{2+}$ ($3d^6$, 4 unpaired electrons).

3
Worked Example
Why is the $\text{Sc}^{3+}$ ion colourless while $\text{Ti}^{3+}$ is coloured?
Solution
  1. $\text{Sc}^{3+}$ has configuration $3d^0$ — no d electrons.
  2. With an empty d subshell, no d–d transition is possible, so no visible light is absorbed.
  3. $\text{Ti}^{3+}$ is $3d^1$; the single electron can be promoted between split d orbitals, absorbing visible light.

Answer: $\text{Sc}^{3+}$ ($d^0$) cannot undergo d–d transitions and is colourless; $\text{Ti}^{3+}$ ($d^1$) can, so it is coloured (violet).

4
Worked Example
Write the balanced ionic equation for the reaction of acidified $\text{KMnO}_4$ with oxalic acid.
Solution
  1. $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$ (gain of 5 electrons each).
  2. $\text{C}_2\text{O}_4^{2-}$ is oxidised to $\text{CO}_2$ (loss of 2 electrons each).
  3. Balance electrons: 2 permanganate to 5 oxalate.

Answer: $2\text{MnO}_4^-+5\text{C}_2\text{O}_4^{2-}+16\text{H}^+\rightarrow2\text{Mn}^{2+}+10\text{CO}_2+8\text{H}_2\text{O}$.

5
Worked Example
What happens to the colour of a dichromate solution when it is made alkaline, and write the equation?
Solution
  1. Dichromate ($\text{Cr}_2\text{O}_7^{2-}$) is orange.
  2. Adding base supplies $\text{OH}^-$, shifting the equilibrium to chromate ($\text{CrO}_4^{2-}$), which is yellow.
  3. $\text{Cr}_2\text{O}_7^{2-}+2\text{OH}^-\rightleftharpoons2\text{CrO}_4^{2-}+\text{H}_2\text{O}$.

Answer: The solution turns from orange to yellow as dichromate converts to chromate.

6
Worked Example
Explain, with an example for each, why transition metals act as good catalysts.
Solution
  1. They show variable oxidation states, allowing them to form intermediates and provide alternative reaction paths.
  2. Their surfaces adsorb reactants, increasing their concentration and weakening their bonds.
  3. Examples: Fe in the Haber synthesis of $\text{NH}_3$; $\text{V}_2\text{O}_5$ in the Contact process for $\text{H}_2\text{SO}_4$.

Answer: Variable oxidation states and adsorptive surfaces lower the activation energy; e.g. Fe (Haber) and $\text{V}_2\text{O}_5$ (Contact process).

Key Points

  • Colour arises from d–d transitions; $d^0$ and $d^{10}$ ions are colourless. The observed colour is complementary to the absorbed colour.
  • Spin-only magnetic moment $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons; more unpaired electrons mean stronger paramagnetism.
  • Transition metals are good catalysts (variable oxidation states + adsorptive surfaces), form complexes and interstitial compounds, and make alloys.
  • $\text{K}_2\text{Cr}_2\text{O}_7$ is a strong oxidant: $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6e^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$; orange dichromate ⇌ yellow chromate with pH.
  • $\text{KMnO}_4$ in acid: $\text{MnO}_4^-+8\text{H}^++5e^-\rightarrow\text{Mn}^{2+}+4\text{H}_2\text{O}$; oxidises $\text{Fe}^{2+}$, oxalate and iodide.
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Q1.The spin-only magnetic moment of $\text{Ni}^{2+}$ ($3d^8$) is closest to:
Explanation: $\text{Ni}^{2+}$ has 2 unpaired electrons; $\mu=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BM.
Q2.Which ion is colourless?
Explanation: $\text{Zn}^{2+}$ is $3d^{10}$ (full d subshell) and cannot undergo d-d transitions, so it is colourless.
Q3.In acidic medium $\text{MnO}_4^-$ is reduced to:
Explanation: In acid, $\text{MnO}_4^-+8\text{H}^++5e^-\rightarrow\text{Mn}^{2+}+4\text{H}_2\text{O}$.
Q4.On adding alkali, an orange dichromate solution turns yellow because it forms:
Explanation: Base shifts the equilibrium to the yellow chromate ion: $\text{Cr}_2\text{O}_7^{2-}+2\text{OH}^-\rightleftharpoons2\text{CrO}_4^{2-}+\text{H}_2\text{O}$.