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The configurations of first-row transition metals and their ions: 4s electrons leave first, the $d$ count is group minus charge, and Hund's rule gives the unpaired electrons.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write the configuration of any common first-row transition-metal ion, find its $d$ count as group number minus charge, and count its unpaired electrons with Hund's rule.
You can write the ground-state electron configuration of an atom from its atomic number, using the aufbau order, the Pauli principle and Hund's rule, and you can write a noble-gas core in square brackets. You know that the fourth period of the periodic table runs from potassium to krypton and that its middle ten elements, scandium to zinc, are where the 3d orbitals fill. This lesson turns that knowledge toward the ions those elements form, which is where all of their chemistry happens.
| Term | What it means |
|---|---|
| Transition metal | An element whose atom or common ion has a partly filled d subshell; in the first row, titanium to copper. |
| d block | The ten columns, groups 3 to 12, in which the d subshell fills. |
| Argon core | The $18$ electrons of argon, $1s^2 2s^2 2p^6 3s^2 3p^6$, written $[\mathrm{Ar}]$. |
| $d$ count | The number of electrons in the metal's d orbitals, written $d^n$; for a first-row ion it is group number minus charge. |
| Hund's rule | Electrons entering orbitals of equal energy occupy them singly, with parallel spins, before any pair. |
| Unpaired electron | An electron alone in its orbital; each one makes a substance attracted into a magnetic field. |
| Free ion | A gaseous ion with no neighbors, whose five d orbitals all have the same energy. |
The first transition series is the row of ten elements from scandium ($Z = 21$) to zinc ($Z = 30$). In the neutral atoms the 4s orbital is occupied before 3d is complete, and the configurations run $[\mathrm{Ar}]\,3d^1 4s^2$ for scandium, $[\mathrm{Ar}]\,3d^2 4s^2$ for titanium, and so on across the row, with two exceptions. Chromium is $[\mathrm{Ar}]\,3d^5 4s^1$ and copper is $[\mathrm{Ar}]\,3d^{10} 4s^1$, because a half-filled or full 3d set together with one 4s electron has a slightly lower energy than the pattern would predict.
Transition metals are almost never met as neutral atoms in their compounds, though. They are met as ions, and the ions lose their 4s electrons first. Once a first-row atom has lost even one electron, its 3d orbitals drop below 4s in energy, because the 3d electrons are held more tightly as the effective nuclear charge rises. So the electrons that entered 4s early are the first to leave, and every electron an ion keeps beyond the argon core sits in 3d.
That gives one rule for the whole row. A first-row metal in group $g$ has $g$ electrons beyond argon, so its ion of charge $+q$ has
$$d\ \text{count} = g - q.$$
Iron is in group $8$, so $\mathrm{Fe^{2+}}$ is $d^6$ and $\mathrm{Fe^{3+}}$ is $d^5$. Copper is in group $11$, so $\mathrm{Cu^{2+}}$ is $d^9$, and the copper exception in the atom makes no difference to the ion. Chromium is in group $6$, so $\mathrm{Cr^{3+}}$ is $d^3$. The $d$ count is the number every later lesson starts from: it decides how many electrons are unpaired, what color the ion's compounds are, and how strongly they hold their surrounding molecules.
In a free ion the five 3d orbitals have the same energy, so they fill by Hund's rule: one electron to an orbital, spins parallel, until all five hold one, and only then does a sixth electron pair up. An ion with up to five $d$ electrons has every one of them unpaired; past five, each extra electron cancels one, so a $d^n$ ion with $n > 5$ has $10 - n$ unpaired electrons.
Another way: picture
Picture the five 3d orbitals as five chairs in a row and the $d$ electrons as people arriving one at a time. Nobody shares a chair while an empty one is left, so the first five people sit alone. The sixth has to share, and so does every person after that. Counting the people sitting alone is counting the unpaired electrons: it climbs to five and then falls back to zero at ten.
Another way: steps
The aufbau order, $4s$ before $3d$, describes neutral atoms built up one proton and one electron at a time. For potassium and calcium, with no 3d electrons, the 4s orbital really is lower. By scandium the two orbitals lie very close, and which is lower depends on what else is in the atom. In every first-row cation the 3d orbitals are the lower ones: the extra positive charge pulls the compact 3d orbitals in more strongly than the diffuse 4s orbital, which spends much of its time far from the nucleus.
Two measured facts back this up. First, the configurations of the gaseous ions, found from their spectra, are $[\mathrm{Ar}]\,3d^n$ with no 4s electrons for every common $2+$ and $3+$ ion of the row. Second, the magnetism of transition metal compounds matches the unpaired count of $3d^{g - q}$, not of any configuration that keeps 4s. A $\mathrm{Mn^{2+}}$ salt behaves as five unpaired electrons, which is $3d^5$; the wrong configuration $3d^3 4s^2$ would give three.
So there is no contradiction to resolve, only two separate questions. In what order were the electrons added to the neutral atoms? has the aufbau answer. Which electrons does an atom lose most easily when it becomes an ion? has the answer 4s, and it is the second question this course keeps asking.
| Metal | Group | Atom | $2+$ ion | $3+$ ion |
|---|---|---|---|---|
| Sc | 3 | $3d^1 4s^2$ | — | $3d^0$ |
| Ti | 4 | $3d^2 4s^2$ | $3d^2$ | $3d^1$ |
| V | 5 | $3d^3 4s^2$ | $3d^3$ | $3d^2$ |
| Cr | 6 | $3d^5 4s^1$ | $3d^4$ | $3d^3$ |
| Mn | 7 | $3d^5 4s^2$ | $3d^5$ | $3d^4$ |
| Fe | 8 | $3d^6 4s^2$ | $3d^6$ | $3d^5$ |
| Co | 9 | $3d^7 4s^2$ | $3d^7$ | $3d^6$ |
| Ni | 10 | $3d^8 4s^2$ | $3d^8$ | — |
| Cu | 11 | $3d^{10} 4s^1$ | $3d^9$ | — |
| Zn | 12 | $3d^{10} 4s^2$ | $3d^{10}$ | — |
Every configuration of every ion is argon plus 3d electrons only, and every $2+$ ion has the atom's group number minus two. The dashes are ions that are rare in ordinary chemistry. Scandium's only common ion, $\mathrm{Sc^{3+}}$, has no $d$ electrons at all, and zinc's, $\mathrm{Zn^{2+}}$, has a full set of ten. Neither has a partly filled d subshell, which is why many chemists do not count scandium and zinc as transition metals in the strict sense. Their compounds are white or colorless and never paramagnetic because of $d$ electrons, and that is exactly what the $d$ count predicts.
The atoms of chromium and copper take one electron from 4s to complete a half-filled or full 3d set. It is a real effect, but a small one, and it is a fact about neutral atoms. The moment either metal forms an ion, the remaining 4s electron leaves first like any other, and the rule $d = g - q$ applies without exception: $\mathrm{Cr^{3+}}$ is $d^3$, $\mathrm{Cr^{2+}}$ is $d^4$, $\mathrm{Cu^{2+}}$ is $d^9$, $\mathrm{Cu^{+}}$ is $d^{10}$.
So when you are asked for the configuration of an ion, you never need the exceptions. Start from the group number, subtract the charge, and write the result as 3d electrons. Only a question about the neutral chromium or copper atom needs the special configuration, and none of the chemistry in the rest of this course asks it.
Ruby and emerald have almost nothing in common as minerals. Ruby is aluminum oxide, $\mathrm{Al_2O_3}$; emerald is a beryllium aluminum silicate, $\mathrm{Be_3Al_2Si_6O_{18}}$. Pure, both are colorless, because $\mathrm{Al^{3+}}$ and $\mathrm{Be^{2+}}$ have no $d$ electrons at all. Their color comes from a few chromium(III) ions standing in for aluminum: about one percent by mass in a deep ruby, and far less in emerald.
$\mathrm{Cr^{3+}}$ is $d^3$, with three unpaired electrons, and a partly filled 3d set is what lets an ion absorb visible light. In ruby the oxide ions press closely on the chromium and push its absorption toward green and blue, so ruby looks red; in emerald the chromium sits in a roomier, more weakly bonding site, the absorption shifts, and the stone looks green. The same ion, the same $d^3$, gives two different colors, and the lessons on crystal field splitting explain exactly how. The first ruby laser, built in 1960, used those same three electrons: its red beam at $694$ nm comes from a chromium(III) excited state.
Stainless steel contains at least $10.5\%$ chromium, and the common kitchen grade, 304, adds about $8\%$ nickel. Whether a steel sticks to a magnet depends on its crystal structure and on how the $d$ electrons of iron, chromium and nickel are shared in the metal, but the ions formed when steel corrodes behave exactly as this lesson predicts. Rust contains $\mathrm{Fe^{3+}}$, $d^5$ with five unpaired electrons, which is why iron(III) oxide is strongly attracted to a magnet in a strong field.
Laboratories use that difference. The magnetic susceptibility of a dissolved sample measures how many unpaired electrons its ions carry: $5$ for iron(III), $4$ for iron(II), $3$ for chromium(III), $2$ for nickel(II). A reading of five unpaired electrons per iron atom says the iron has been fully oxidized to $\mathrm{Fe^{3+}}$; four says some is still $\mathrm{Fe^{2+}}$. The arithmetic behind each of those numbers is the one in this lesson, group minus charge and then Hund's rule, and a later lesson turns the measured moment back into an electron count.
Because the aufbau order fills 4s before 3d, it is tempting to empty them in the reverse order and remove 3d electrons first. That gives $\mathrm{Fe^{3+}}$ as $[\mathrm{Ar}]\,3d^{3}4s^{2}$, with three unpaired electrons, when every iron(III) compound measured behaves as five. The filling order is a statement about neutral atoms; in an ion the 3d orbitals lie lower, and 4s electrons are lost first.
A second error is to carry the chromium or copper exception into the ion and write $\mathrm{Cu^{2+}}$ as $[\mathrm{Ar}]\,3d^{8}4s^{1}$. The exception concerns only the neutral atom; the ion is $3d^{9}$ like any other group-11 $2+$ ion. The check that catches both errors is the rule $d = g - q$ with no 4s electrons at all.
Write the neutral iron atom.
$\mathrm{Fe}\ (Z = 26): [\mathrm{Ar}]\,3d^{6}4s^{2}$
Twenty-six electrons are argon's $18$ plus eight more: two in 4s and six in 3d.
Remove the two 4s electrons.
$\mathrm{Fe^{2+}}: [\mathrm{Ar}]\,3d^{6}$
The 4s electrons are the first to leave a transition-metal atom.
Remove one 3d electron for the third unit of charge.
$\mathrm{Fe^{3+}}: [\mathrm{Ar}]\,3d^{5}$
With 4s empty, the next electron comes from 3d.
Check with the group-number rule.
$8 - 3 = 5$
Iron is in group $8$, and the charge is $+3$.
Count the unpaired electrons.
$3d^{5}: \uparrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow \Rightarrow 5\ \text{unpaired}$
Five electrons fill five orbitals singly, so none pair.
Find the $d$ count from the group number.
$10 - 2 = 8$
Nickel is in group $10$ and the ion has lost two electrons.
Write the configuration.
$\mathrm{Ni^{2+}}: [\mathrm{Ar}]\,3d^{8}$
Both electrons lost were the 4s pair.
Place the first five electrons.
$\uparrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow$
Hund's rule: one to each of the five orbitals first.
Place the remaining three.
$\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow$
Each extra electron must share an orbital, with its spin opposite.
Count the unpaired electrons.
$10 - 8 = 2$
Past five, the unpaired count is $10 - n$.
Compare with the wrong configuration.
$[\mathrm{Ar}]\,3d^{6}4s^{2} \Rightarrow 4\ \text{unpaired} \ne 2$
Removing 3d electrons first predicts the wrong magnetism for every nickel(II) salt.
Write the neutral chromium atom.
$\mathrm{Cr}\ (Z = 24): [\mathrm{Ar}]\,3d^{5}4s^{1}$
Chromium moves one 4s electron to complete a half-filled 3d set.
Remove the single 4s electron.
$\mathrm{Cr^{+}}: [\mathrm{Ar}]\,3d^{5}$
4s still leaves first.
Remove two 3d electrons for the $3+$ ion.
$\mathrm{Cr^{3+}}: [\mathrm{Ar}]\,3d^{3}$
Three electrons have gone in all: one 4s and two 3d.
Check with the group-number rule.
$6 - 3 = 3$
The rule needs no exception for the ion.
Count the unpaired electrons.
$3d^{3} \Rightarrow 3\ \text{unpaired}$
Three electrons in five orbitals sit alone.
Say what the count predicts.
$\mathrm{Cr^{3+}}\ \text{compounds are paramagnetic and colored}$
A partly filled 3d set with unpaired electrons, as ruby and chrome green both show.
Find the $d$ count.
$9 - 2 = 7$
Cobalt is in group $9$.
Write the configuration.
$\mathrm{Co^{2+}}: [\mathrm{Ar}]\,3d^{7}$
Both electrons lost were 4s electrons.
Count the unpaired electrons.
Which is the ground-state electron configuration of the gaseous $\mathrm{Fe}^{3+}$ ion?
Complete the worked solution: how many 3d electrons does the $\mathrm{Fe}^{3+}$ ion have, given atomic number $26$?
Subtract the argon core from the atomic number.
$26 - 18 =$ v
These are the 4s and 3d electrons of the neutral atom.
Take away the charge, starting with the two 4s electrons.
$d\ \text{count} =$ d
Every electron the ion keeps beyond argon is now in 3d.
Check the count by the group number.
$d\ \text{count} = (\text{group number}) - (\text{charge})$
The two routes agree for a transition-metal ion, because it loses its 4s electrons first.
Match each ion to the number of electrons in its 3d orbitals.
| $d^{1}$ | $d^{3}$ | $d^{5}$ | $d^{9}$ | |
|---|---|---|---|---|
| $\mathrm{Ti}^{3+}$ | ||||
| $\mathrm{V}^{2+}$ | ||||
| $\mathrm{Fe}^{3+}$ | ||||
| $\mathrm{Cu}^{2+}$ |
manganese (atomic number $25$, group $7$) forms $2+$ and $3+$ ions. Fill in the table for the gaseous atom and both ions.
| 3d electrons | 4s electrons | unpaired electrons | |
|---|---|---|---|
| the atom | |||
| the 2+ ion | |||
| the 3+ ion |
How many 3d electrons does the vanadium(II) ion, $\mathrm{V}^{2+}$, have? The metal is in group $5$.
Answer: electrons in the 3d orbitals
A gaseous $\mathrm{Ni}^{2+}$ ion, nickel(II), has atomic number $28$. How many unpaired electrons does it have?
Answer: unpaired electrons in the free ion
Three materials owe their color or their chemistry to a transition-metal ion: ruby ($\mathrm{Cr}^{3+}$), blue cobalt glass ($\mathrm{Co}^{2+}$) and the chrome green pigment chromium(III) oxide ($\mathrm{Cr}^{3+}$). Fill in the $d$ count and the unpaired electrons of each free ion, in that order.
| 3d electrons | unpaired electrons | |
|---|---|---|
| the first material | ||
| the second material | ||
| the third material |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
manganese (atomic number $25$, group $7$) forms $2+$ and $3+$ ions. Fill in the table for the gaseous atom and both ions.
| 3d electrons | 4s electrons | unpaired electrons | |
|---|---|---|---|
| the atom | |||
| the 2+ ion | |||
| the 3+ ion |
You can find the $d$ count and the unpaired electrons of a transition-metal ion. Explain to someone why $\mathrm{Fe^{3+}}$ is $d^5$ and not $3d^3 4s^2$, using the magnetism of iron(III) compounds as your evidence.
14. Your turn: how many unpaired electrons does a gaseous $\mathrm{Co^{2+}}$ ion have?, step 3
$10 - 7 = 3$
Seven electrons in five orbitals leave two paired and three alone.