Q.Write down the electronic configuration of:

Concept: Stability of Oxidation States — ions with half-filled, fully-filled, or empty $d$/$f$ subshells are particularly stable.

Reasoning:

1. Write the ground-state configuration of the neutral atom.
2. Remove electrons from the outermost orbitals ($ns$ before $(n-1)d$ for transition metals; $6s$ before $4f$ for lanthanides/actinides).
3. Adjust for stability: $Cr^{3+}$ loses the $4s^1$ electron first, then two $3d$ electrons.

1. $Cr$: $[Ar]\,3d^5\,4s^1$ → remove $4s^1$ + two $3d$ → $[Ar]\,3d^3$
2. $Pm$: $[Xe]\,4f^5\,6s^2$ → remove $6s^2$ + one $4f$ → $[Xe]\,4f^4$
3. $Cu$: $[Ar]\,3d^{10}\,4s^1$ → remove $4s^1$ → $[Ar]\,3d^{10}$
4. $Ce$: $[Xe]\,4f^1\,5d^1\,6s^2$ → remove $6s^2$ + $5d^1$ + one $4f$ → $[Xe]$ (empty $4f$)
5. $Co$: $[Ar]\,3d^7\,4s^2$ → remove $4s^2$ → $[Ar]\,3d^7$
6. $Lu$: $[Xe]\,4f^{14}\,5d^1\,6s^2$ → remove $6s^2$ (2 e⁻, matching the +2 charge) → $[Xe]\,4f^{14}\,5d^1$
7. $Mn$: $[Ar]\,3d^5\,4s^2$ → remove $4s^2$ → $[Ar]\,3d^5$
8. $Th$: $[Rn]\,6d^2\,7s^2$ → remove $7s^2$ + $6d^2$ → $[Rn]$
   ✓Final answer

   (i) $[Ar]\,3d^3$ (ii) $[Xe]\,4f^4$ (iii) $[Ar]\,3d^{10}$ (iv) $[Xe]$ (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 = $3d^3$.

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 $[Ar] 3d^{10}$.

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

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 $[Ar] 3d^5$.

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: $[Rn]$

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

1. Write the ground-state configuration of the neutral atom using the Aufbau order.
2. Remove electrons equal to the positive charge, starting from the highest principal quantum number (n) shell.
3. 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 $Cu^+$:
  • Neutral Cu: $[Ar] 3d^{10} 4s^1$
  • Remove 1 electron (from 4s) → $[Ar] 3d^{10}$ (fully filled) — this is the correct configuration.
• For $Cr^{3+}$:
  • 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

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.