Exercises — General electronic configuration (n−1)d¹⁻¹⁰ ns⁰⁻²

Period 4 means the outermost principal quantum number is ==$n=4$==.

• $(n-1)=3$, so $(n-1)d$ is the $3d$ subshell.
• The general recipe is $3d^{1-10}4s^{0-2}$. The $3d$ is the penultimate (one-inside) shell; the $4s$ is the outer valence shell.

Recall $\ell$: $s\to 0$, $d\to 2$.

• (a) $4s$: $n+\ell =4+0=4$. $3d$: $3+2=5$. Smaller wins → $4s$ fills first.
• (b) $5s$: $5+0=5$. $4d$: $4+2=6$. Smaller wins → $5s$ fills first. Same pattern one row down: the $s$ always beats the $d$ of the row below it on the $(n+\ell )$ score.

• Argon accounts for 18 electrons.
• $22-18=4$ electrons to place. By $(n+\ell )$: fill $4s$ first ($4s^2$, uses 2), then $3d$ ($3d^2$, uses the last 2). Answer: $[\text{Ar}]3d^{2}4s^2$.

• V: $23-18=5$ → $4s^2$ then $3d^3$ → $[\text{Ar}]3d^{3}4s^2$.
• Ni: $28-18=10$ → $4s^2$ then $3d^8$ → $[\text{Ar}]3d^{8}4s^2$. Ni has 8 in $d$; it is not an exception ($d^8$ is neither half- nor fully-filled).

Naïve Madelung fill: place $4s^2$, then $3d^4$ → $[\text{Ar}]3d^{4}4s^1{}^{,}4s^1$? No — $[\text{Ar}]3d^{4}4s^2$. The atom instead promotes one $4s$ electron into $3d$, reaching $3d^{5}4s^1$. Why does this pay off?

• A half-filled $3d^5$ has one electron in each of the five $d$-orbitals, all spins parallel.
• Parallel-spin electrons that can "swap places" earn exchange stabilisation — a purely quantum bonus (see Exchange Energy and Hund's Rule).
• The exchange energy released by making the set half-filled exceeds the small energy cost of moving an electron $4s\to 3d$. Net energy drops, so $3d^{5}4s^1$ is the true ground state.

• Cu naïve: $3d^{9}4s^2$. Promote one $4s\to 3d$ → $3d^{10}4s^1$: a completely filled $d$ set, maximally symmetric spherical charge cloud → large stabilisation. Promotion pays off.
• Ni is $3d^8$. Promoting one $4s\to 3d$ would give $3d^{9}4s^1$ — that is neither half- nor fully-filled, so it earns no special symmetry/exchange bonus to offset the promotion cost. Hence Ni stays $3d^{8}4s^2$. The rule of thumb: promotion happens only when it lands you exactly on $d^5$ or $d^{10}$.

• Atom Fe: $26-18=8$ → $4s^2$, $3d^6$ → $[\text{Ar}]3d^{6}4s^2$.
• ${\text{Fe}}^{2+}$: remove the two $4s$ (removal is $ns$ first) → $[\text{Ar}]3d^6$. In $d^6$: five orbitals, first five electrons go in singly (all parallel), the 6th pairs up → $N=4$ unpaired. $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90{\mu }_B$.
• ${\text{Fe}}^{3+}$: remove two $4s$ and one $3d$ → $[\text{Ar}]3d^5$. In $d^5$: one electron per orbital → $N=5$ unpaired. $\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92{\mu }_B$. ${\text{Fe}}^{3+}$ ($d^5$) is more magnetic than ${\text{Fe}}^{2+}$ ($d^6$).

• Mn (Z=25): $[\text{Ar}]3d^{5}4s^2$. Remove all 7 valence electrons ($4s^2$ then $3d^5$) → ${\text{Mn}}^{7+}=[\text{Ar}]$, i.e. $d^0$.
• With $d^0$ there are no $d$-electrons to jump between $d$-levels, so colour cannot come from $d$–$d$ transitions (see Colour and d-d Transitions).
• The deep purple comes instead from charge transfer: an electron jumps from an oxygen ligand into the metal's empty $d$-orbitals. This absorbs visible light strongly, so the colour is intense.
• Typical hydrated ions (e.g. ${\text{Cu}}^{2+}$, $d^9$) are coloured by weaker $d$–$d$ transitions, giving paler hues.

• Mo is a period-5 d-block element, so the recipe is $(n-1)d=4d$, outer $5s$. Core is $[\text{Kr}]$ (Z=36).
• $42-36=6$ electrons. Naïve: $5s^{2}4d^4$. Like Cr, promote one $5s\to 4d$ to reach the half-filled, exchange-stabilised $4d^5$: Mo $=[\text{Kr}]4d^{5}5s^1$.
• ${\text{Mo}}^{3+}$: remove $5s^1$ first, then two $4d$ → $[\text{Kr}]4d^3$.

• Core $[\text{Kr}]$ (36). $46-36=10$ electrons.
• Naïve Madelung: $5s^{2}4d^8$. But Pd achieves the fully-filled $4d^{10}$ by pulling both $5s$ electrons into $4d$: Pd $=[\text{Kr}]4d^{10}5s^0$ — the only element with a completely filled $(n-1)d$ and empty $ns$ in its ground state.
• $4d^{10}$ is fully paired → $N=0$ unpaired electrons → diamagnetic ($\mu =\sqrt{0(0+2)}=0{\mu }_B$).

• Sc (Z=21): $[\text{Ar}]3d^{1}4s^2$. The atom has a partially filled $d$ ($3d^1$) — that alone is enough for IUPAC.
• But its common ion ${\text{Sc}}^{3+}$: remove $4s^2$ then $3d^1$ → $[\text{Ar}]$, i.e. $d^0$ — empty $d$.
• IUPAC calls an element a transition element if the atom OR a stable ion has a partially filled $d$. Sc's atom qualifies, so Sc is in the family — but its main $+3$ ion, being $d^0$, is colourless and non-magnetic, so it behaves like a borderline member (much as Zn/Cd/Hg sit at the other, $d^{10}$, edge). See Variable Oxidation States of Transition Metals.