15A.1-15A.3 - d-Block Configurations And Transition Metals

d-Block And Transition Metals

Iron can form $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$, copper can form $\ce{Cu+}$ and $\ce{Cu^{2+}}$, but zinc is usually not treated as a transition metal in this course even though it sits in the d-block. The difference is not the position of the neutral atom alone. It is the electron configuration of stable ions.

The period-4 d-block is the series:

Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn

These elements involve the $3\mathrm{d}$ subshell, even though the $4\mathrm{s}$ subshell is filled before $3\mathrm{d}$ in the simple filling order. The useful shorthand is the argon core:

So instead of rewriting 18 core electrons every time, you can focus on the electrons after $[\mathrm{Ar}]$. For example:

Fe atom: $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{6}$

The catch is that forming period-4 d-block cations is not just "take electrons away from the last symbol written". When these atoms form positive ions, the $4\mathrm{s}$ electrons are removed before the $3\mathrm{d}$ electrons. That one rule prevents most wrong ion configurations in this topic.

Common-error contrast

Deducing Period-4 d-Block Atom Configurations

For neutral atoms, begin from $[\mathrm{Ar}]$ and then add the $4\mathrm{s}$ and $3\mathrm{d}$ electrons. The simple pattern is $4\mathrm{s}^{2}$, then add one more $3\mathrm{d}$ electron as you move across the series.

Two atoms need special attention:

Cr: $[\mathrm{Ar}] 4\mathrm{s}^{1} 3\mathrm{d}^{5}$

Cu: $[\mathrm{Ar}] 4\mathrm{s}^{1} 3\mathrm{d}^{10}$

These are associated with especially stable half-filled or full $3\mathrm{d}$ subshells. Do not write chromium as $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{4}$ or copper as $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{9}$ when giving their ground-state atom configurations.

Worked example: deduce the configuration of cobalt

Cobalt has atomic number 27, so a neutral cobalt atom has 27 electrons. The argon core accounts for 18 electrons, leaving:

27 - 18 = 9 electrons after $[\mathrm{Ar}]$

Cobalt is not one of the chromium/copper exceptions. Fill $4\mathrm{s}^{2}$, then put the remaining 7 electrons in $3\mathrm{d}$:

Co: $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{7}$

The chemical interpretation is that cobalt is a period-4 d-block element because the configuration involves the $3\mathrm{d}$ subshell.

Guided practice

Deduce the atom configurations for titanium, manganese and copper.

Deducing Ion Configurations

For d-block ions, do not refill the ion from scratch using the neutral-atom filling order. Use this route:

[DIAGRAM: asset_name: Iron atom and ion orbital occupancy removing 4s first; asset_slug: edexcel_a_level_chemistry_l069_d_orbital_boxes; recommended_method: deterministic_chart; description: Iron atom and ion orbital occupancy removing 4s first.]

1. Write the neutral atom configuration.
2. Use the positive charge to decide how many electrons have been lost.
3. Remove $4\mathrm{s}$ electrons first.
4. If more electrons must be removed, remove them from $3\mathrm{d}$.
5. Check the final electron count.

Worked example 1: $\ce{V^{3+}}$

Vanadium has atomic number 23:

V atom: $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{3}$

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

$\ce{V^{3+}}$: $[\mathrm{Ar}] 3\mathrm{d}^{2}$

Electron-count check: $[\mathrm{Ar}]$ has 18 electrons and $3\mathrm{d}^{2}$ adds 2, so $\ce{V^{3+}}$ has 20 electrons. A $3+$ ion from vanadium should have $23 - 3 = 20$ electrons, so the configuration is consistent.

Worked example 2: $\ce{Cr^{2+}}$

Chromium is an exception as an atom:

Cr atom: $[\mathrm{Ar}] 4\mathrm{s}^{1} 3\mathrm{d}^{5}$

$\ce{Cr^{2+}}$ has lost 2 electrons. Remove the one $4\mathrm{s}$ electron first, then one $3\mathrm{d}$ electron:

$\ce{Cr^{2+}}$: $[\mathrm{Ar}] 3\mathrm{d}^{4}$

The common wrong answer is $[\mathrm{Ar}] 4\mathrm{s}^{1} 3\mathrm{d}^{3}$ or $[\mathrm{Ar}] 4\mathrm{s}^{2} 3\mathrm{d}^{2}$. Both keep $4\mathrm{s}$ electrons in a period-4 d-block cation, which is the key error.

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

Copper is also an exception as an atom:

Cu atom: $[\mathrm{Ar}] 4\mathrm{s}^{1} 3\mathrm{d}^{10}$

$\ce{Cu^{2+}}$ has lost 2 electrons. Remove the one $4\mathrm{s}$ electron first, then one $3\mathrm{d}$ electron:

$\ce{Cu^{2+}}$: $[\mathrm{Ar}] 3\mathrm{d}^{9}$

This is why copper can satisfy the transition-metal criterion even though the neutral copper atom has a full $3\mathrm{d}$ subshell: $\ce{Cu^{2+}}$ has an incomplete $3\mathrm{d}$ subshell.

Useful ion configurations

You do not need to write $4\mathrm{s}^{0}$. Once the $4\mathrm{s}$ electrons have been removed, the configuration is normally written with the remaining occupied subshells.