Q.Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Concept: Electron Configuration – filling order follows the Aufbau principle (n+l rule) and the (n-1)d, (n-2)f blocks fill after the ns orbital.

Reasoning steps:

1. Atomic number 61 (Promethium, Pm):

   After Xe (54 electrons), the next 7 electrons go into 6s² and then 4f.

   Configuration: $[\text{Xe}] \, 6s^2 \, 4f^5$ (since 4f fills before 5d).

2. Atomic number 91 (Protactinium, Pa):

   After Rn (86 electrons), the next 5 electrons fill 7s² and then 5f², with one electron going into 6d¹ (anomaly due to stability).

   Configuration: $[\text{Rn}] \, 7s^2 \, 5f^2 \, 6d^1$.

3. Atomic number 101 (Mendelevium, Md):

   After Rn (86), the next 15 electrons fill 7s², 5f¹³, and then 6d⁰ (no 6d electron).

   Configuration: $[\text{Rn}] \, 7s^2 \, 5f^{13}$.

4. Atomic number 109 (Meitnerium, Mt): …

Why Electron Configuration Works This Way

Electrons fill orbitals in order of increasing energy, not just increasing principal quantum number $n$. The rule is: an orbital with lower $(n + l)$ fills first; if two have the same $(n + l)$, the one with lower $n$ fills first. This is the Aufbau principle, and it explains why the 4f subshell fills after 6s, and 5f after 7s.

For elements beyond lanthanum (atomic number 57), the 4f orbitals begin to fill. Similarly, beyond actinium (89), the 5f orbitals fill. But there are exceptions — half-filled (f⁷) and fully-filled (f¹⁴) subshells are extra stable, so sometimes an electron from the s-orbital moves into the f-orbital to achieve that stability.

Let’s work through each element.

1. Atomic number 61 — Promethium (Pm)

The nearest noble gas is xenon (Xe, Z=54). That gives us a core of $[Xe]$.

Remaining electrons: $61 - 54 = 7$ electrons.

The filling order after Xe is: 6s (2 electrons), then 4f (up to 14 electrons), then 5d, then 6p.

So we put 2 electrons into 6s: $6s^2$.

That leaves $7 - 2 = 5$ electrons. These go into the 4f subshell: $4f^5$.

No special stability is reached here (f⁷ would be half-filled, but we only have 5), so no exception occurs.

Configuration: $[Xe]\,4f^5\,6s^2$

2. Atomic number 91 — Protactinium (Pa)

Nearest noble gas: radon (Rn, Z=86). Core: $[Rn]$.

Remaining electrons: $91 - 86 = 5$ electrons.

After Rn, the filling order is: 7s (2), then 5f (14), then 6d (10), then 7p.

First, 2 electrons go into 7s: $7s^2$.

That leaves $5 - 2 = 3$ electrons. According to the Aufbau order, the next orbital is 5f. So we would expect $5f^3$.

But here’s the catch: for protactinium, the 5f and 6d orbitals are very close in energy. Experimentally, the configuration is $[Rn]\,5f^2\,6d^1\,7s^2$, not $[Rn]\,5f^3\,7s^2$. Why? Because having one electron in the 6d orbital (which is slightly lower in energy for Pa) is more stable than putting all three into 5f.

Configuration: $[Rn]\,5f^2\,6d^1\,7s^2$

3. Atomic number 101 — Mendelevium (Md)

Core: $[Rn]$ (Z=86).

Remaining electrons: $101 - 86 = 15$ electrons. …

Method: Aufbau Principle with (n + ℓ) Rule

This is the standard method for writing ground-state electron configurations. It uses the order of increasing orbital energy determined by the sum $(n + \ell)$ — and for equal sums, by lower $n$ first.

Steps

1. Identify the atomic number (Z) — this equals the total number of electrons in a neutral atom.
2. Follow the Aufbau order (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p...).
3. Fill each subshell to its maximum capacity:
   • $s$: 2 electrons
   • $p$: 6 electrons
   • $d$: 10 electrons
   • $f$: 14 electrons
4. Stop when the total electrons equal Z.
5. Write in order of increasing principal quantum number $n$ (standard notation), not in filling order.

Solutions

Atomic number 61 — Promethium (Pm)

• Z = 61
• Fill: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f⁵
• Final configuration (by n): $\mathbf{1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10} \ 4f^5 \ 5s^2 \ 5p^6 \ 6s^2}$

Atomic number 91 — Protactinium (Pa)

• Z = 91
• Fill: ... up to 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹
• Final configuration: $\mathbf{1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10} \ 4f^{14} \ 5s^2 \ 5p^6 \ 5d^{10} \ 5f^2 \ 6s^2 \ 6p^6 \ 6d^1 \ 7s^2}$

Note: Pa is an anomaly — the 5f and 6d are very close in energy. The above is the accepted ground state.

Atomic number 101 — Mendelevium (Md)

• Z = 101
• Fill: ... up to 7s² 5f¹³
• Final configuration: $\mathbf{1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^{10} \ 4s^2 \ 4p^6 \ 4d^{10} \ 4f^{14} \ 5s^2 \ 5p^6 \ 5d^{10} \ 5f^{13} \ 6s^2 \ 6p^6 \ 7s^2}$

Atomic number 109 — Meitnerium (Mt)

• Z = 109
• Fill: ... up to 7s² 5f¹⁴ 6d⁷
• Final configuration: …

🧠 The Core Idea First

Electronic configuration follows the Aufbau principle (fill lowest energy orbitals first), Hund’s rule (maximize unpaired spins), and the Pauli exclusion principle. But for elements beyond atomic number 57 (La), the energy ordering of orbitals changes due to nuclear charge and shielding effects.

The correct filling order is:

✗ Common Mistake #1: Forgetting the f-block (lanthanides & actinides)

The error:

Students write configurations for atomic numbers 61, 91, 101, 109 as if they are normal d-block elements, skipping the f-subshell entirely.

Example of wrong answer for Z=61:

$[\text{Xe}] 6s^2 4f^1$ ✗ (a 57-electron count — and not even real La, whose actual configuration is $[\text{Xe}]\,5d^1\,6s^2$)

Why it happens:

They memorise the order but forget that after La (Z=57), the 4f subshell starts filling — not 5d.

How to avoid:

• Remember: Lanthanides (Z=58 to 71) fill the 4f subshell.
• Actinides (Z=90 to 103) fill the 5f subshell.
• Use the n + ℓ rule to confirm: 4f has n+ℓ = 4+3 = 7, 5d has 5+2 = 7, but 4f is lower in energy because of lower n.

Correct for Z=61 (Promethium, Pm):

$[\text{Xe}] 6s^2 4f^5$ ✓

✗ Common Mistake #2: Misplacing the 5f and 6d orbitals for Z=91 and 101

The error:

For Z=91 (Protactinium), students write $[\text{Rn}] 7s^2 5f^3$ ✗ — but the actual configuration has a 5f² 6d¹ arrangement.

Why it happens:

They assume the 5f subshell fills strictly after 7s, ignoring that 6d can be slightly lower in energy for early actinides.

How to avoid:

• For early actinides (Th, Pa, U, Np), the 6d orbital may get one electron before 5f fills completely.
• Memorise the exceptions for Pa (Z=91): $[\text{Rn}] 7s^2 5f^2 6d^1$ ✓
• For later actinides (Am onwards), 5f fills normally.

Correct for Z=91 (Protactinium):

$[\text{Rn}] 7s^2 5f^2 6d^1$ ✓

Correct for Z=101 (Mendelevium, Md):

$[\text{Rn}] 7s^2 5f^{13}$ ✓ (no 6d electron here)

✗ Common Mistake #3: Forgetting the d-block exception for Z=109

The error:

For Z=109 (Meitnerium, Mt), students write $[\text{Rn}] 7s^2 5f^{14} 6d^7$ — the same electrons, but listed in filling order. This is a presentation slip rather than wrong chemistry: the convention is to list subshells in order of increasing $n$, so present it as $[\text{Rn}] 5f^{14} 6d^7 7s^2$.

Why it happens:

They write orbitals in filling order (7s before 6d) but forget that in the periodic table, we write by increasing n (principal quantum number), not filling order.

How to avoid: …