Electronic configuration of elements (Z = 1 to 30) — exceptions Cr, Cu: Exercises

Before we start, one reminder of the tools, each with its job:

Level 1 — Recognition

L1.1

Write the ground-state configuration of Scandium (Z = 21) in noble-gas shorthand.

Recall Solution

WHAT: Distribute 21 electrons. WHY the order: fill by $(n + l)$ . Argon core $[Ar]$ uses 18. Remaining $21 - 18 = 3$ .

$4 s$ has $(n + l) = 4$ , $3 d$ has $(n + l) = 5$ → $4 s$ fills first: $4 s^{2}$ (2 used).
Last 1 electron → $3 d^{1}$ . Answer: $[Ar] 3 d^{1} 4 s^{2}$ .

L1.2

How many electrons does a completely filled $d$ -subshell hold, and how many $l$ -values does $d$ have?

Recall Solution

$d$ means $l = 2$ . The magnetic quantum number $m$ runs $- 2, - 1, 0, + 1, + 2$ → 5 orbitals (see Quantum numbers (n, l, m, s)). Each orbital holds 2 (Pauli) → $5 \times 2 = 10$ electrons.

Level 2 — Application

L2.1

Give the ground-state configuration of Titanium (Z = 22) and count its unpaired electrons.

Recall Solution

$[Ar] = 18$ ; remaining $4$ . $4 s^{2}$ (cheaper), then $3 d^{2}$ . Config: $[Ar] 3 d^{2} 4 s^{2}$ . Unpaired: By Hund's rule of maximum multiplicity, the two $3 d$ electrons go into separate orbitals with parallel spin → 2 unpaired. $4 s^{2}$ is paired (0). Total = 2.

L2.2

Write the configuration of the ion Ca²⁺ (from Z = 20).

Recall Solution

Neutral Ca: $[Ar] 4 s^{2}$ . To form Ca²⁺ remove 2 electrons from the highest $n$ first → remove both $4 s$ . Answer: $[Ar]$ (i.e. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ ), 0 unpaired.

L2.3

Neutral Manganese (Z = 25) — configuration and number of unpaired electrons.

Recall Solution

$[Ar] = 18$ ; remaining $7$ . $4 s^{2}$ , then $3 d^{5}$ . Config: $[Ar] 3 d^{5} 4 s^{2}$ . By Hund, all five $3 d$ orbitals get one electron each, all parallel → 5 unpaired.

Level 3 — Analysis

L3.1

Using the exchange-pair count, compute the exchange stabilization gained when Chromium chooses $3 d^{5} 4 s^{1}$ over the naïve $3 d^{4} 4 s^{2}$ . Count only $d$ -electron parallel pairs.

Recall Solution

WHAT: count parallel-spin pairs $(2 k)$ for the $d$ -electrons in each option.

Naïve $3 d^{4}$ : $k = 4$ parallel → $(2 4) = \frac{4 \cdot 3}{2} = 6$ pairs.
Real $3 d^{5}$ : $k = 5$ parallel → $(2 5) = \frac{5 \cdot 4}{2} = 10$ pairs. Gain: $10 - 6 = 4$ extra exchange pairs. WHY it matters: those 4 extra parallel pairs lower the energy by more than the small cost of lifting one electron from $4 s$ into the nearly-degenerate $3 d$ . See the bar figure below.

Figure — Electronic configuration of elements (Z = 1 to 30) — exceptions Cr, Cu

L3.2

Repeat the exchange count for Copper: $3 d^{10} 4 s^{1}$ (real) vs $3 d^{9} 4 s^{2}$ (naïve). Careful — a full $d$ has both spin sets.

Recall Solution

A full $d^{10}$ has 5 spin-up and 5 spin-down electrons. Exchange pairs form only between electrons of the same spin, so total pairs $= (2 5) + (2 5) = 10 + 10 = 20$ .

Naïve $3 d^{9}$ : one spin set is full (5), the other has 4 → $(2 5) + (2 4) = 10 + 6 = 16$ .
Real $3 d^{10}$ : $(2 5) + (2 5) = 10 + 10 = 20$ . Gain: $20 - 16 = 4$ extra pairs, plus the special stability of a completely symmetric, fully-filled subshell. Answer: $[Ar] 3 d^{10} 4 s^{1}$ wins.

L3.3

Which is more paramagnetic, Cr or Mn²⁺? Compare unpaired-electron counts.

Recall Solution

Cr: $[Ar] 3 d^{5} 4 s^{1}$ → 5 unpaired in $3 d$ + 1 unpaired in $4 s$ = 6 unpaired.
Mn²⁺: neutral Mn $[Ar] 3 d^{5} 4 s^{2}$ ; remove $4 s^{2}$ → $[Ar] 3 d^{5}$ → 5 unpaired. Answer: Cr (6) > Mn²⁺ (5), so Cr is more paramagnetic. (Links to Magnetic properties of transition metals.)

Level 4 — Synthesis

L4.1

Predict, using the reasoning behind Cr (not memory), whether Molybdenum (Z = 42), directly below Cr, is $[Kr] 4 d^{4} 5 s^{2}$ or $[Kr] 4 d^{5} 5 s^{1}$ .

Recall Solution

WHAT: apply the exact half-filled logic to the $4 d /5 s$ pair. $[Kr] = 36$ ; remaining $6$ . $5 s$ and $4 d$ are nearly degenerate near this column (same reason as Cr). A half-filled $4 d^{5}$ gives $(2 5) = 10$ parallel pairs vs $4 d^{4}$ 's $(2 4) = 6$ — a gain of 4, outweighing the small promotion cost. Prediction: $[Kr] 4 d^{5} 5 s^{1}$ — Mo really is an exception, just like Cr. Understanding, not memory, got us there.

L4.2

Total number of unpaired electrons in neutral Copper (Z = 29), and state whether Cu is more or less magnetic than Ni ( $3 d^{8} 4 s^{2}$ ).

Recall Solution

Cu $[Ar] 3 d^{10} 4 s^{1}$ : $3 d^{10}$ fully paired (0 unpaired), $4 s^{1}$ has 1 unpaired. Total = 1. Ni $[Ar] 3 d^{8} 4 s^{2}$ : $3 d^{8}$ = five orbitals, three doubly filled + two singly filled → 2 unpaired. Answer: Cu (1) < Ni (2), so Cu is less paramagnetic than Ni.

L4.3

Build the configuration of Zn²⁺ (from Z = 30) and explain why it is diamagnetic.

Recall Solution

Neutral Zn: $[Ar] 3 d^{10} 4 s^{2}$ . Remove 2 from highest $n$ ( $4 s$ ): Zn²⁺ $= [Ar] 3 d^{10}$ . Every $3 d$ orbital holds a spin-up/spin-down pair → 0 unpaired → diamagnetic (no net spin to align with a field).

Level 5 — Mastery

L5.1

Given only the fact "an atom minimizes total energy," construct the argument for why $[Ar] 3 d^{5} 4 s^{1}$ beats $[Ar] 3 d^{4} 4 s^{2}$ in Cr. State the two competing energy terms and the sign of each change, then the verdict.

Recall Solution

Compare promotion cost vs exchange gain as we go $3 d^{4} 4 s^{2} \to 3 d^{5} 4 s^{1}$ .

Promotion cost $Δ E_{promo} > 0$ : one electron leaves $4 s$ (slightly lower) for $3 d$ (slightly higher). Because $3 d$ and $4 s$ are near-degenerate here (see Shielding and penetration effect), this cost is small.
Exchange gain $Δ E_{exch} < 0$ (stabilizing): parallel $d$ -pairs rise from $(2 4) = 6$ to $(2 5) = 10$ , i.e. $+ 4$ pairs. Plus the extra symmetry stability of a half-filled shell. Verdict: $∣Δ E_{exch} ∣ > Δ E_{promo}$ , so the total energy of $3 d^{5} 4 s^{1}$ is lower. The atom picks it. Aufbau was only an approximate ordering; total energy is the true judge. See the balance figure:

L5.2

A hypothetical atom has a $d$ -subshell containing 7 electrons. Compute its total exchange pairs and its number of unpaired electrons, splitting by spin.

Recall Solution

$d^{7}$ : fill 5 orbitals singly (spin-up) then pair 2 (spin-down).

Spin-up set: 5 electrons → $(2 5) = 10$ pairs.
Spin-down set: 2 electrons → $(2 2) = (2 2) = \frac{2 \cdot 1}{2} = 1$ pair.
Total exchange pairs $= 10 + 1 = 11$ .
Unpaired = $5 - 2 = 3$ (five up, two of them now paired by a down-electron, leaving 3 lone up-spins).

L5.3

For the first transition series (Sc→Zn), which neutral atom has the maximum number of unpaired electrons? Justify with a count.

Recall Solution

Unpaired count peaks where the $3 d$ subshell is exactly half-filled with a lone $4 s$ available.

Cr $3 d^{5} 4 s^{1}$ : $5 + 1 = 6$ unpaired.
Mn $3 d^{5} 4 s^{2}$ : $5 + 0 = 5$ .
Fe $3 d^{6} 4 s^{2}$ : $4 + 0 = 4$ . No other neutral atom in the row exceeds 6. Answer: Chromium, with 6 unpaired electrons — a direct payoff of the $3 d^{5} 4 s^{1}$ exception.

Recall One-line takeaways

Why is Cr $3 d^{5} 4 s^{1}$ and not $3 d^{4} 4 s^{2}$ ? ::: The +4 exchange pairs plus half-filled symmetry lower the total energy more than the small $4 s \to 3 d$ promotion cost. How do you count exchange pairs in a subshell? ::: Split by spin, do $(2 k)$ in each spin set, add. Which neutral 3d-series atom has the most unpaired electrons? ::: Chromium, 6 unpaired. From which orbital do electrons leave first when forming a cation? ::: The highest principal quantum number $n$ (e.g. $4 s$ before $3 d$ ).