5.3.1(a)-(c) - Transition element properties

Period 4 d-block configurations

The Period 4 d-block elements run from scandium to zinc:

Sc Ti V Cr Mn Fe $\ce{Co}$ $\ce{Ni}$ $\ce{Cu}$ $\ce{Zn}$

They are called d-block elements because the $\mathrm{3d}$ sub-shell is being filled across this part of the periodic table. For neutral atoms in this range, use the same filling idea you learned earlier: the $\mathrm{4s}$ sub-shell is filled before the $\mathrm{3d}$ sub-shell.

The normal neutral-atom pattern from Sc to $\ce{Zn}$ is:

$[Ar] \mathrm{4s}^{2} \mathrm{3d}^{1}$, $[Ar] \mathrm{4s}^{2} \mathrm{3d}^{2}$, $[Ar] \mathrm{4s}^{2} \mathrm{3d}^{3}$ ...

The two OCR-relevant neutral-atom exceptions are chromium and copper:

Cr: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{5} \mathrm{4s}^{1}$

$\ce{Cu}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{10} \mathrm{4s}^{1}$

Chromium has a half-filled $\mathrm{3d}^{5}$ sub-shell and copper has a full $\mathrm{3d}^{10}$ sub-shell. You do not need a deeper explanation for the stability here; you do need to remember the configurations when asked to write them.

For quick checking, the neutral atom endings after the argon core are:

In a full answer, write the whole configuration unless the question explicitly accepts a shortened argon-core form.

Worked example: iron atom

Iron has atomic number 26, so a neutral iron atom has 26 electrons. Fill up to argon, then place the remaining 8 electrons in $\mathrm{4s}$ and $\mathrm{3d}$:

Fe: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{6} \mathrm{4s}^{2}$

Some working methods write $\mathrm{4s}^{2}$ before $\mathrm{3d}^{6}$; OCR mark schemes may allow either order for the neutral atom. The safest final format for this topic is to group the $\mathrm{3d}$ and $\mathrm{4s}$ terms clearly and keep every superscript count correct.

Writing d-block ion configurations

For d-block ions, the important rule changes:

This is the error OCR repeatedly catches. A neutral atom may be built by filling $\mathrm{4s}$ before $\mathrm{3d}$, but a positive d-block ion loses its $\mathrm{4s}$ electrons first.

[DIAGRAM: asset_name: Lesson 5.3.1a-c: Transition element properties - diagram 01; asset_slug: 05_03_01a_transition_element_properties__diagram_01; recommended_method: drawn_chem; description: 16:9 summary showing Period 4 d-block Sc-Zn with Ti-Cu highlighted as transition elements, Sc3+ d0 and Zn2+ d10 exclusions, and Fe atom to Fe2+/Fe3+ electron configurations with 4s electrons removed before 3d.]

Worked example 1: $\ce{Fe^2+}$

Start with the neutral iron atom:

Fe: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{6} \mathrm{4s}^{2}$

$\ce{Fe^2+}$ has lost 2 electrons. Remove the two $\mathrm{4s}$ electrons first:

$\ce{Fe^2+}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{6}$

Do not write $\ce{Fe^2+}$ as $\mathrm{3d}^{4} \mathrm{4s}^{2}$. That keeps the wrong electrons.

Worked example 2: $\ce{Fe^3+}$

$\ce{Fe^3+}$ has lost 3 electrons. Remove both $\mathrm{4s}$ electrons, then one $\mathrm{3d}$ electron:

$\ce{Fe^3+}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{5}$

Worked example 3: $\ce{V^3+}$

Vanadium atom:

V: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{3} \mathrm{4s}^{2}$

$\ce{V^3+}$ has lost 3 electrons: first the two $\mathrm{4s}$ electrons, then one $\mathrm{3d}$ electron.

$\ce{V^3+}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{2}$

Worked example: identify the metal from its ion

An unknown Period 4 metal forms $\ce{M^2+}$ with configuration $[\mathrm{Ar}]3d^7$. The ion has $18+7=25$ electrons. Because it has lost two electrons, the neutral atom has 27, so the metal is cobalt. Its atom is $[\mathrm{Ar}]3d^7 4s^2$, and a further electron removed from $\ce{Co^2+}$ gives $\ce{Co^3+}$, $[\mathrm{Ar}]3d^6$. Check the direction: a positive ion has fewer electrons than its neutral atom; forming the ion removes 4s before 3d.

Which d-block elements are transition elements

OCR does not treat every Period 4 d-block element as a transition element. The definition depends on ions, not just atoms.

An incomplete d sub-shell means $d^{1}$ to $d^{9}$. Empty $d^{0}$ and full $d^{10}$ do not count as incomplete.

At A level, the transition elements studied are:

Ti V Cr Mn Fe $\ce{Co}$ $\ce{Ni}$ $\ce{Cu}$

Scandium and zinc are d-block elements, but they are not transition elements in this definition.

Scandium commonly forms $\ce{Sc^3+}$:

Sc: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{1} \mathrm{4s}^{2}$

$\ce{Sc^3+}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6}$