Write down the electronic configuration of: (i) Cr³⁺ (ii) Pm³⁺ (iii) Cu^+ (iv) Ce⁴⁺ (v) Co²⁺ (vi) Lu²⁺ (vii) M

Exercises · 4.1

Q. Write down the electronic configuration of:

(i)

Cr^{3+}

(ii)

Pm^{3+}

(iii)

Cu^+

(iv)

Ce^{4+}

(v)

Co^{2+}

(vi)

Lu^{2+}

(vii)

Mn^{2+}

(viii)

Th^{4+}

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams

When asked "Explain the stability of oxidation states of [element]", follow this mental checklist:

Write the electronic configuration of the atom.
Write configurations for each possible oxidation state.
Look for half-filled, fully-filled, or inert pair effects.
Check if the state can disproportionate (common for +1 states of Cu, Au, and +3 states of Mn).
Mention the medium (acidic/alkaline) if relevant.

Tip

For d-block elements, remember: $d^0$ , $d^5$ , and $d^{10}$ are especially stable. For p-block, the inert pair effect makes lower oxidation states more stable as you go down the group.

The Bottom Line

Stability of an oxidation state is a measure of how strongly an atom holds onto that oxidation number — how hard it is to push it up or down. It's determined by electronic structure, the element's position in the periodic table, and the chemical environment. Master this, and you'll predict redox behaviour without memorising every reaction.

Stability of oxidation states among transition and inner-transition elements is discussed in the NCERT/CBSE Class 12 Chemistry chapter on d- and f-Block Elements, and ‘stability of oxidation states in transition elements’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Predicting which oxidation state is most stable is a reasoning skill regularly tested in competitive-exam inorganic chemistry MCQs.

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds:

The free energy change $\Delta G^\circ$ for the reaction is negative. This happens when the intermediate oxidation state is less stable than the extremes.

Formula (for aqueous ions):

If $E^\circ_{\text{reduction}}$ for the higher state is more positive than for the lower state, disproportionation is spontaneous.

Summary Table: Why Each "Formula" Holds

Principle	Why it works	Key exam example
Inert pair effect	6s² electrons are too tightly bound	$\ce{Pb^{2+}}$ stable, $\ce{Pb^{4+}}$ oxidising
Half-filled stability	Extra exchange energy	$\ce{Mn^{2+}}$ > $\ce{Mn^{3+}}$
Hydration vs ionisation	Energy balance in solution	$\ce{Cu^{2+}}$ stable, $\ce{Cu^+}$ not
Disproportionation	$\Delta G < 0$ for intermediate state	$\ce{Cu^+}$ in water

Final Takeaway for Exams

Never memorise stability blindly. Always ask:

Is the electronic configuration special? (half-filled / inert pair)
Is the medium aqueous or solid? (hydration vs lattice)
Does the element belong to a heavier group? (inert pair effect)

The "formula" is really a balance of energies — and the reasoning is what gets you marks.

The key idea is to first write the ground-state configuration of the neutral atom, then remove electrons from the outermost shells (highest $n$ , then highest $l$ within that $n$ ) to form the cation. The final configurations are: (i) $[Ar]3d^3$ ,

(ii) $[Xe]4f^4$ ,

(iii) $[Ar]3d^{10}$ ,

(iv) $[Xe]4f^0$ ,

(v) $[Ar]3d^7$ ,

(vi) $[Xe]4f^{14}5d^1$ ,

(vii) $[Ar]3d^5$ ,

(viii) $[Rn]$ .

When writing electronic configurations for ions, the most common mistake is to remove electrons from the last filled subshell in the neutral atom. That is wrong. The correct rule: electrons are removed from the orbital with the highest principal quantum number $n$ first. If two orbitals share the same $n$ , remove from the one with the higher azimuthal quantum number $l$ (i.e., $p$ before $s$ , $d$ before $p$ , etc.). This is because orbitals with higher $n$ are farther from the nucleus and less tightly bound.

For transition metals and lanthanides/actinides, this means that the $ns$ electrons (where $n$ is the period number) are lost before the $(n-1)d$ or $(n-2)f$ electrons. Let’s apply this step by step.

1. $Cr^{3+}$

Neutral Cr (Z=24) has configuration: $[Ar] 3d^5 4s^1$ .

Why $3d^5 4s^1$ and not $3d^4 4s^2$ ? Because a half-filled $d$ subshell ( $d^5$ ) is extra stable — this is an exception you must remember.

To form $Cr^{3+}$ , remove 3 electrons. Start with the highest $n$ : the $4s$ electron goes first. That gives $[Ar] 3d^5$ . Then remove two more from the $3d$ subshell (since $n=3$ is now the highest). $3d^5$ minus 2 electrons = $3d^3$ .

Watch out

Do not remove $4s$ electrons last. Many students write $[Ar]3d^2 4s^1$ for $Cr^{3+}$ , which is incorrect. The $4s$ orbital is higher in energy than $3d$ once the atom is ionized.

Answer: $[Ar] 3d^3$

2. $Pm^{3+}$

Promethium (Pm, Z=61) is a lanthanide. Neutral configuration: $[Xe] 4f^5 6s^2$ .

Lanthanides fill the $4f$ subshell after $6s$ . For $Pm^{3+}$ , remove 3 electrons. Highest $n$ is 6: remove both $6s$ electrons first. Then remove one more from the $4f$ subshell (next highest $n$ is 4). $4f^5$ minus 1 = $4f^4$ .

Answer: $[Xe] 4f^4$

3. $Cu^+$

Copper (Cu, Z=29) neutral: $[Ar] 3d^{10} 4s^1$ (another exception — full $d$ subshell is stable).

Remove 1 electron. Highest $n$ is 4: remove the $4s$ electron. That leaves $[Ar] 3d^{10}$ .

Tip

$Cu^+$ has a completely filled $d$ subshell ( $d^{10}$ ), which is very stable. This is why copper(I) compounds are common.

Answer: $[Ar] 3d^{10}$

4. $Ce^{4+}$

Cerium (Ce, Z=58) neutral: $[Xe] 4f^1 5d^1 6s^2$ (NCERT Table 4.9's form; the alternative $4f^2 6s^2$ is sometimes quoted in the literature).

Remove 4 electrons. First, remove both $6s$ electrons, then the $5d$ electron, then the single $4f$ electron. So the configuration is just the noble gas core $[Xe]$ .

Answer: $[Xe]$

5. $Co^{2+}$

Cobalt (Co, Z=27) neutral: $[Ar] 3d^7 4s^2$ .

Remove 2 electrons. Highest $n$ is 4: remove both $4s$ electrons. That leaves $[Ar] 3d^7$ .

Answer: $[Ar] 3d^7$

6. $Lu^{2+}$

Lutetium (Lu, Z=71) neutral: $[Xe] 4f^{14} 5d^1 6s^2$ .

Remove 2 electrons. Highest $n$ is 6: remove both $6s$ electrons. That leaves $[Xe] 4f^{14} 5d^1$ .

Note

Lu is the last lanthanide; its $4f$ subshell is full ( $4f^{14}$ ). The $5d$ electron is present because after $4f^{14}$ , the next electron goes into $5d$ (not $4f$ ).

Answer: $[Xe] 4f^{14} 5d^1$

7. $Mn^{2+}$

Manganese (Mn, Z=25) neutral: $[Ar] 3d^5 4s^2$ .

Remove 2 electrons. Highest $n$ is 4: remove both $4s$ electrons. That leaves $[Ar] 3d^5$ .

Tip

$Mn^{2+}$ has a half-filled $d$ subshell ( $d^5$ ), which gives it extra stability. This is why manganese(II) is a common oxidation state.

Answer: $[Ar] 3d^5$

8. $Th^{4+}$

Thorium (Th, Z=90) is an actinide. Neutral: $[Rn] 6d^2 7s^2$ .

Remove 4 electrons. Highest $n$ is 7: remove both $7s$ electrons. Then remove two from $6d$ : $6d^2$ minus 2 = $6d^0$ . So the configuration is just $[Rn]$ .

Answer: $[Rn]$

✓Final answer

[Ar]3d^3

Method: Electronic Configuration Using the Aufbau Principle + Ionization Sequence

This method uses the Aufbau order (filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p) and the rule that for ions, electrons are removed first from the outermost shell (highest n), not from the subshell that was filled last.

Steps

Write the ground-state configuration of the neutral atom using the Aufbau order.
Remove electrons equal to the positive charge, starting from the highest principal quantum number (n) shell.
Write the final configuration in order of increasing n (and within same n, increasing ℓ).

Solutions

(i) $Cr^{3+}$

Neutral Cr (Z = 24): $[Ar]\, 3d^5\, 4s^1$ (Exception: half-filled d-subshell stability)
Remove 3 electrons: first from 4s (1 e⁻), then from 3d (2 e⁻)
$Cr^{3+}$ : $[Ar]\, 3d^3$

(ii) $Pm^{3+}$

Neutral Pm (Z = 61): $[Xe]\, 4f^5\, 6s^2$
Remove 3 electrons: from 6s (2 e⁻), then from 4f (1 e⁻)
$Pm^{3+}$ : $[Xe]\, 4f^4$

(iii) $Cu^+$

Neutral Cu (Z = 29): $[Ar]\, 3d^{10}\, 4s^1$ (Exception: fully filled d-subshell)
Remove 1 electron: from 4s
$Cu^+$ : $[Ar]\, 3d^{10}$

(iv) $Ce^{4+}$

Neutral Ce (Z = 58): $[Xe]\, 4f^1\, 5d^1\, 6s^2$ (Actual ground state: $[Xe]\, 4f^1\, 5d^1\, 6s^2$ )
Remove 4 electrons: from 6s (2 e⁻), then 5d (1 e⁻), then 4f (1 e⁻)
$Ce^{4+}$ : $[Xe]$ (noble gas core)

(v) $Co^{2+}$

Neutral Co (Z = 27): $[Ar]\, 3d^7\, 4s^2$
Remove 2 electrons: from 4s
$Co^{2+}$ : $[Ar]\, 3d^7$

(vi) $Lu^{2+}$

Neutral Lu (Z = 71): $[Xe]\, 4f^{14}\, 5d^1\, 6s^2$
Remove 2 electrons: from 6s (2 e⁻)
$Lu^{2+}$ : $[Xe]\, 4f^{14}\, 5d^1$

(vii) $Mn^{2+}$

Neutral Mn (Z = 25): $[Ar]\, 3d^5\, 4s^2$
Remove 2 electrons: from 4s
$Mn^{2+}$ : $[Ar]\, 3d^5$

(viii) $Th^{4+}$

Neutral Th (Z = 90): $[Rn]\, 6d^2\, 7s^2$
Remove 4 electrons: from 7s (2 e⁻), then 6d (2 e⁻)
$Th^{4+}$ : $[Rn]$ (noble gas core)

Key Exam Insight

Stability of oxidation states is linked to half-filled ( $d^5$ , $f^7$ ) or fully filled ( $d^{10}$ , $f^{14}$ ) subshells.

For example:

$Mn^{2+}$ ( $d^5$ ) is stable → half-filled stability

$Cu^+$ ( $d^{10}$ ) is stable → fully filled stability

$Ce^{4+}$ ( $[Xe]$ ) is stable → noble gas configuration

Here are the most common mistakes students make when writing electronic configurations for ions, especially in the context of Stability of Oxidation States (d- and f-block elements), and how to avoid each.

Mistake 1: Forgetting that electrons are removed from the 4s orbital first (for d-block ions)

The Error:

For $Cr^{3+}$ , writing $[Ar] 3d^1 4s^2$ — removing all three electrons from 3d and leaving the 4s pair untouched.

The same wrong removal order turns $Co^{2+}$ into $[Ar] 3d^5 4s^2$ instead of the correct $[Ar] 3d^7$ .

Why it happens:

Students memorise "4s is filled before 3d" but forget that when forming cations, electrons are removed from the 4s orbital first (because 4s is higher in energy once occupied).

How to avoid:

For any d-block ion, always remove from 4s before 3d.
Write the neutral atom configuration first, then strip the outermost s-electrons.
Example:
- Neutral Cr: $[Ar] 3d^5 4s^1$
- $Cr^{3+}$ : remove 1 from 4s, then 2 from 3d → $[Ar] 3d^3$
- Neutral Co: $[Ar] 3d^7 4s^2$
- $Co^{2+}$ : remove 2 from 4s → $[Ar] 3d^7$

Mistake 2: Ignoring the stability of half-filled and fully-filled d-subshells

The Error:

For $Cu^+$ , students write $[Ar] 3d^9 4s^0$ (which is technically possible) but miss that $Cu^+$ is actually $[Ar] 3d^{10}$ (fully filled d — very stable).

Similarly, for $Cr^{3+}$ , they might write $[Ar] 3d^2 4s^1$ instead of $[Ar] 3d^3$ .

Why it happens:

They don't check if a half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) configuration is possible after ionisation.

How to avoid:

After removing electrons, check if the d-subshell becomes $d^5$ or $d^{10}$ — these are extra stable.
For $Cu^+$ $C u^{+}$ :
- Neutral Cu: $[Ar] 3d^{10} 4s^1$
- Remove 1 electron (from 4s) → $[Ar] 3d^{10}$ (fully filled) — this is the correct configuration.
For $Cr^{3+}$ $C r^{3 +}$ :
- Neutral Cr: $[Ar] 3d^5 4s^1$
- Remove 3 electrons (1 from 4s, 2 from 3d) → $[Ar] 3d^3$ (not half-filled, but correct).

Mistake 3: Writing f-block configurations with the wrong removal order (6s must go first)

The Error:

For $Pm^{3+}$ , students remove all three electrons from 4f, writing $[Xe] 4f^2 6s^2$ instead of the correct $[Xe] 4f^4$ (remove the 6s pair first, then one 4f).

For $Ce^{4+}$ , they write $[Xe] 4f^1 5d^1$ or $[Xe] 4f^2$ instead of $[Xe]$ (empty f — $f^0$ ).

For $Lu^{2+}$ , they remove the 5d electron first, writing $[Xe] 4f^{14} 6s^1$ instead of the correct $[Xe] 4f^{14} 5d^1$ (the 6s pair goes first).

Why it happens:

f-block elements have complex filling order (4f, 5d, 6s). Students forget that for lanthanides, the 4f is filled before 5d and 6s, and that stable oxidation states often correspond to f⁰, f⁷, f¹⁴.

How to avoid:

For lanthanides ( $Ce$ to $Lu$ ), the neutral configuration is $[Xe] 4f^n 5d^0$ or $[Xe] 4f^{n-1} 5d^1$ (exceptions: La, Gd, Lu).
When forming ions, remove 6s electrons first, then 4f (if needed).
Examples:
- $Ce^{4+}$ : Neutral Ce = $[Xe] 4f^1 5d^1 6s^2$ (or $[Xe] 4f^2 6s^2$ ). Remove 4 electrons → $[Xe]$ (f⁰ — very stable).
- $Pm^{3+}$ : Neutral Pm = $[Xe] 4f^5 6s^2$ . Remove 3 electrons (2 from 6s, 1 from 4f) → $[Xe] 4f^4$ .
- $Lu^{2+}$ : Neutral Lu = $[Xe] 4f^{14} 5d^1 6s^2$ . Remove 2 electrons (from 6s) → $[Xe] 4f^{14} 5d^1$ (but note: $Lu^{2+}$ is unstable; the stable ion is $Lu^{3+}$ = $[Xe] 4f^{14}$ ).

Mistake 4: Confusing actinide configurations with lanthanides

The Error:

For $Th^{4+}$ , students write $[Rn] 5f^0 6d^0 7s^0$ (which is correct) but they might write $[Rn] 5f^2$ or $[Rn] 6d^2$ because they misremember the neutral configuration.

Why it happens:

Actinides have 5f, 6d, 7s filling that is less regular than lanthanides. Thorium (Th) is an exception: neutral Th = $[Rn] 6d^2 7s^2$ , not $[Rn] 5f^2$ .

How to avoid:

Memorise key exceptions:
- Th (Z=90): $[Rn] 6d^2 7s^2$
- Pa (Z=91): $[Rn] 5f^2 6d^1 7s^2$
- U (Z=92): $[Rn] 5f^3 6d^1 7s^2$
For $Th^{4+}$ : remove 4 electrons (2 from 7s, 2 from 6d) → $[Rn]$ (noble gas core, f⁰ — very stable).

Mistake 5: Not checking for exceptions in neutral configurations before ionising

The Error:

For $Cr^{3+}$ , students start from $[Ar] 3d^4 4s^2$ (wrong neutral Cr) and then remove 3 electrons → $[Ar] 3d^3 4s^0$ (correct final, but wrong reasoning).

For $Cu^+$ , they start from $[Ar] 3d^9 4s^2$ (wrong neutral Cu) → $[Ar] 3d^9$ (wrong).

Why it happens:

They don't memorise the anomalous configurations of Cr and Cu in the neutral state.

How to avoid:

Memorise these neutral exceptions:
- Cr: $[Ar] 3d^5 4s^1$ (not $3d^4 4s^2$ )
- Cu: $[Ar] 3d^{10} 4s^1$ (not $3d^9 4s^2$ )
- Also: Nb, Mo, Ru, Rh, Pd, Ag, Pt, Au have similar anomalies.
Always write the correct neutral configuration first, then remove electrons.

Quick Reference Table for the Given Ions

Ion	Correct Configuration	Common Mistake
$Cr^{3+}$	$[Ar] 3d^3$	$[Ar] 3d^1 4s^2$ or $[Ar] 3d^2 4s^1$
$Pm^{3+}$	$[Xe] 4f^4$	$[Xe] 4f^5$ or $[Xe] 4f^3 5d^1$
$Cu^+$	$[Ar] 3d^{10}$	$[Ar] 3d^9 4s^0$
$Ce^{4+}$	$[Xe]$ (f⁰)	$[Xe] 4f^2$ or $[Xe] 4f^1 5d^1$
$Co^{2+}$	$[Ar] 3d^7$	$[Ar] 3d^5 4s^2$ (wrong removal order)
$Lu^{2+}$	$[Xe] 4f^{14} 5d^1$ (unstable)	$[Xe] 4f^{13} 5d^2$ or $[Xe] 4f^{14} 6s^2$
$Mn^{2+}$	$[Ar] 3d^5$	$[Ar] 3d^3 4s^2$
$Th^{4+}$	$[Rn]$ (f⁰)	$[Rn] 5f^2$ or $[Rn] 6d^2$

Final Exam Tip

Always write the neutral configuration first, then remove electrons from the outermost s-orbital (and then d or f if needed). Check for half-filled/full-filled stability. For f-block, remember f⁰, f⁷, f¹⁴ are especially stable.

←Previous section · summarySummary →Next question4.2 — Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to thei…

Q.Write down the electronic configuration of:

1. Cr3+Cr^{3+}Cr3+

2. Pm3+Pm^{3+}Pm3+

3. Cu+Cu^+Cu+

4. Ce4+Ce^{4+}Ce4+

5. Co2+Co^{2+}Co2+

6. Lu2+Lu^{2+}Lu2+

7. Mn2+Mn^{2+}Mn2+

8. Th4+Th^{4+}Th4+

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1. $Cr^{3+}$

2. $Pm^{3+}$

3. $Cu^+$

4. $Ce^{4+}$

5. $Co^{2+}$

6. $Lu^{2+}$

7. $Mn^{2+}$

8. $Th^{4+}$

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