Properties and Uses of Transition Metals (Senior 6 Chemistry): free lessons | Ganzaa

Properties and Uses of Transition Metals

Lessons

Senior 6 · Chemistry

Properties and Uses of Transition Metals

5 lessons

Lesson 1 of 5

The First-Row Transition Elements and Their Electron Configurations

25 min

Ten metals in the middle of Period 4

A hoe from the market is steel: iron with a little carbon, often toughened with manganese. The wiring in a house is copper. A roofing sheet is steel coated with zinc so that it does not rust. Iron, manganese, copper and zinc sit side by side in the periodic table, in the block of ten elements from scandium (Z = 21) to zinc (Z = 30). In S4 you learned that this is the d-block, where the 3d subshell is being filled. In this unit you will see how the 3d electrons explain what makes most of these metals special: several oxidation states, coloured compounds, complex ions and catalytic activity.

Definition

d-block element: an element whose outermost (highest-energy) electron enters a d subshell. The first row runs from Sc to Zn, filling 3d.

Transition element (the A-level definition used in this unit): a d-block element that forms at least one stable ion with a partly filled d subshell. Example: iron forms $\mathrm{Fe^{3+}}$ , [Ar] $3 d^{5}$ .

IUPAC also counts an element whose atom has an incomplete d subshell, which brings scandium in. By either definition, zinc is not a transition element.

Writing the configurations

In the atoms, 4s fills before 3d, so every configuration from Sc to Zn is argon's core, [Ar] = $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ , followed by 3d and 4s electrons. Table 1.1 lists them. Two do not follow the simple pattern: chromium and copper.

Table 1.1: Electron configurations of the first-row d-block atoms

Element	Z	Configuration	Unpaired electrons
Scandium, Sc	21	[Ar] $3d^1\,4s^2$	1
Titanium, Ti	22	[Ar] $3d^2\,4s^2$	2
Vanadium, V	23	[Ar] $3d^3\,4s^2$	3
Chromium, Cr	24	[Ar] $3d^5\,4s^1$ (not $3d^4\,4s^2$ )	6
Manganese, Mn	25	[Ar] $3d^5\,4s^2$	5
Iron, Fe	26	[Ar] $3d^6\,4s^2$	4
Cobalt, Co	27	[Ar] $3d^7\,4s^2$	3
Nickel, Ni	28	[Ar] $3d^8\,4s^2$	2
Copper, Cu	29	[Ar] $3d^{10}\,4s^1$ (not $3d^9\,4s^2$ )	1
Zinc, Zn	30	[Ar] $3d^{10}\,4s^2$	0

Figure 1.1: Orbital box diagrams for chromium and copper. Each box is an orbital; each arrow is an electron, and a pair of arrows has opposite spins.

Why chromium and copper break the pattern

The 3d and 4s subshells are very close in energy. Moving one electron from 4s to 3d costs little, and in chromium and copper it is repaid: a half-filled ( $d^{5}$ ) or completely filled ( $d^{10}$ ) subshell spreads the electrons evenly around the nucleus and lowers the repulsion between them. The same preference for $d^{5}$ and $d^{10}$ will explain several properties of ions later in this lesson.

Definition

Configuration of a transition-metal ion: when a first-row d-block atom forms a positive ion, the 4s electrons are removed first, then 3d electrons. Once 3d is occupied it lies below 4s in energy. Example: Mn is [Ar] $3d^5\,4s^2$ , so $\mathrm{Mn^{2+}}$ is [Ar] $3 d^{5}$ , not [Ar] $3d^3\,4s^2$ .

Worked example

Problem. Write the electron configurations of Fe, $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$ . How many unpaired electrons does each have? Suggest why iron(II) compounds are slowly oxidised to iron(III) compounds in air.

Solution. Fe (Z = 26): [Ar] $3d^6\,4s^2$ . Remove the two 4s electrons: $\mathrm{Fe^{2+}}$ is [Ar] $3 d^{6}$ . Remove one more, from 3d: $\mathrm{Fe^{3+}}$ is [Ar] $3 d^{5}$ .

Fill the five 3d orbitals singly before pairing (Hund's rule). $3 d^{6}$ : five singles and one pair, so 4 unpaired electrons in both Fe and $\mathrm{Fe^{2+}}$ (the two 4s electrons in Fe are paired). $3 d^{5}$ : 5 unpaired.

Answer: losing the one paired 3d electron turns $3 d^{6}$ into the stable half-filled $3 d^{5}$ , so $\mathrm{Fe^{2+}}$ is easily oxidised to $\mathrm{Fe^{3+}}$ .

Worked example

Problem. An ion $\mathrm{M^{2+}}$ of a first-row d-block element has the configuration [Ar] $3 d^{9}$ . Identify M, state whether M is a transition element, and give the configuration of $\mathrm{M^{+}}$ .

Solution. $\mathrm{M^{2+}}$ has $18 + 9 = 27$ electrons, so the atom has $27 + 2 = 29$ electrons: M is copper. $\mathrm{Cu^{2+}}$ has a partly filled 3d subshell ( $3 d^{9}$ ), so copper is a transition element. The atom is [Ar] $3d^{10}\,4s^1$ ; removing the single 4s electron gives $\mathrm{Cu^{+}}$ = [Ar] $3d^{10}$ .

Scandium and zinc: d-block, but not typical

Table 1.2: The only ions of scandium and zinc

Element and atom	Only common ion	Its 3d subshell	Consequences
Sc, [Ar] $3d^1\,4s^2$	$\mathrm{Sc^{3+}}$ : [Ar]	empty ( $3 d^{0}$ )	one oxidation state (+3); white compounds; colourless solutions
Zn, [Ar] $3d^{10}\,4s^2$	$\mathrm{Zn^{2+}}$ : [Ar] $3d^{10}$	full ( $3d^{10}$ )	one oxidation state (+2); white compounds; colourless solutions

Neither ion has a partly filled d subshell, so neither shows the typical transition-metal behaviour of variable oxidation states and coloured ions (Lesson 4 explains why colour needs a partly filled d subshell). Both still form complex ions, such as $\mathrm{[Zn(NH_3)_4]^{2+}}$ , because that needs only empty orbitals to accept lone pairs.

Physical properties: hard, dense, high-melting metals

Table 1.3: Physical data for the first-row d-block metals, with calcium for comparison

Element	Melting point / °C	Density / g cm⁻³	First ionisation energy / kJ mol⁻¹
Ca (s-block)	842	1.55	590
Sc	1541	2.99	633
Ti	1668	4.51	659
V	1910	6.11	651
Cr	1907	7.19	653
Mn	1246	7.21	717
Fe	1538	7.87	762
Co	1495	8.90	760
Ni	1455	8.91	737
Cu	1085	8.96	745
Zn	420	7.14	906

Three patterns stand out.

High melting points and hardness. Both 4s and 3d electrons can be delocalised into the metallic bond, so the bonding is stronger than in calcium, which gives only two. Zinc, with a full 3d subshell, melts at only 420 °C.
High density. Across the row the nuclear charge rises but the atomic radius barely changes, so more mass is packed into a similar volume.
Similar first ionisation energies. Each extra electron goes into the inner 3d subshell, where it shields the outer 4s electrons from most of the extra nuclear charge. The first ionisation energy therefore rises only slowly, from 633 to 745 kJ mol⁻¹ between Sc and Cu, far less than the rise across Period 3.

Definition

Paramagnetism: weak attraction of a substance into a magnetic field, caused by unpaired electrons. A substance with all its electrons paired is diamagnetic and is very slightly repelled. Example: $\mathrm{Fe^{3+}}$ (5 unpaired electrons) is strongly paramagnetic; $\mathrm{Zn^{2+}}$ ( $3d^{10}$ ) is diamagnetic.

Warning

4s out first. The most common exam error is to remove 3d electrons first, for example writing $\mathrm{Fe^{2+}}$ as [Ar] $3d^4\,4s^2$ . The 4s subshell fills first in the atom but also empties first in the ion.

Tip

For an ion $\mathrm{M^{n+}}$ of a first-row metal, the number of 3d electrons is simply group number − n, counting groups 3 to 12. For $\mathrm{Co^{3+}}$ : $9 - 3 = 6$ , so [Ar] $3 d^{6}$ . It works for chromium and copper too, because an ion has lost all its 4s electrons anyway: $\mathrm{Cr^{3+}}$ is $6 - 3 = 3$ , $\mathrm{Cu^{2+}}$ is $11 - 2 = 9$ .

Chemistry with Dr Steph: Electronic Configuration - Transition Metals (4 min). Watch for the chromium and copper exceptions and for the rule that 4s electrons are lost first when an ion forms.

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Practice

Define a transition element. Explain why zinc is in the d-block but is not a transition element.
Write the full electron configurations (using [Ar]) of V, Cr, $\mathrm{Mn^{2+}}$ , $\mathrm{Co^{3+}}$ and $\mathrm{Cu^{+}}$ .
Explain why the configuration of chromium is [Ar] $3d^5\,4s^1$ and not [Ar] $3d^4\,4s^2$ .
How many unpaired electrons are there in $\mathrm{Fe^{2+}}$ , $\mathrm{Ni^{2+}}$ , $\mathrm{Cr^{3+}}$ and $\mathrm{Zn^{2+}}$ ? Which of these ions is diamagnetic?
An ion $\mathrm{X^{3+}}$ has the configuration [Ar] $3 d^{5}$ . Identify X.
The third ionisation energies are Mn 3248 kJ mol⁻¹ and Fe 2957 kJ mol⁻¹. Use electron configurations to explain why the value for manganese is higher, and what this suggests about $\mathrm{Mn^{2+}}$ compared with $\mathrm{Fe^{2+}}$ .
Use Table 1.3 to explain why iron melts about 700 °C higher than calcium, although both are metals in Period 4.

Show answersHide answers

A d-block element that forms at least one stable ion with a partly filled d subshell. Zinc's outermost electrons fill 3d, but its only ion, $\mathrm{Zn^{2+}}$ , is $3d^{10}$ : the d subshell is full, not partly filled.
V [Ar] $3d^3\,4s^2$ ; Cr [Ar] $3d^5\,4s^1$ ; $\mathrm{Mn^{2+}}$ [Ar] $3 d^{5}$ ; $\mathrm{Co^{3+}}$ [Ar] $3 d^{6}$ ; $\mathrm{Cu^{+}}$ [Ar] $3d^{10}$ .
3d and 4s are close in energy. Promoting one 4s electron gives a half-filled 3d subshell with the electrons spread evenly, which lowers the repulsion between them; the energy gained outweighs the small promotion cost.
$\mathrm{Fe^{2+}}$ ( $3 d^{6}$ ) 4; $\mathrm{Ni^{2+}}$ ( $3 d^{8}$ ) 2; $\mathrm{Cr^{3+}}$ ( $3 d^{3}$ ) 3; $\mathrm{Zn^{2+}}$ ( $3d^{10}$ ) 0. $\mathrm{Zn^{2+}}$ is diamagnetic.
X has $23 + 3 = 26$ electrons: iron.
The third electron removed from Mn comes from $\mathrm{Mn^{2+}}$ = $3 d^{5}$ , breaking a stable half-filled subshell. From Fe it comes from $\mathrm{Fe^{2+}}$ = $3 d^{6}$ , the one paired electron, which is repelled by its partner and leaves a stable $3 d^{5}$ . So $\mathrm{Mn^{2+}}$ is harder to oxidise to +3 than $\mathrm{Fe^{2+}}$ is.
Iron can delocalise both 4s and some 3d electrons into its metallic bond, and its ions are small and highly charged, so the attraction between the cations and the sea of electrons is much stronger. Calcium delocalises only its two 4s electrons.

Sources and credits

Written by Ganzaa. Melting points, densities and ionisation energies are standard reference values, as listed in the CRC Handbook of Chemistry and Physics.
Video: Electronic Configuration - Transition Metals, Chemistry with Dr Steph, embedded from YouTube.
The orbital box diagram, tables, worked examples and Practice items were written and drawn by Ganzaa.

1Ten metals in the middle of Period 4

2Writing the configurations

3Why chromium and copper break the pattern

4Scandium and zinc: d-block, but not typical

5Physical properties: hard, dense, high-melting metals

Ten metals in the middle of Period 4

Writing the configurations

Why chromium and copper break the pattern

Scandium and zinc: d-block, but not typical

Physical properties: hard, dense, high-melting metals