Electronic Configurations of d-Block Elements

The General (n−1)d ns Rule

The d-block occupies the broad central section of the periodic table, flanked by the s-block on the left and the p-block on the right. What defines this block is that the d orbitals of the penultimate (second-outermost) energy level receive the differentiating electrons, giving rise to four rows of transition metals: the 3d, 4d, 5d and 6d series. NCERT compresses this into a single outer-shell formula:

$(n-1)d^{1-10}\,ns^{1-2}$

Here $(n-1)$ refers to the inner d orbitals, which may hold anywhere from one to ten electrons, while the outermost $ns$ orbital carries one or two. The principal quantum number $n$ is that of the outermost shell — 4 for the 3d series, 5 for the 4d series, and so on. The single most-quoted exception to this generalisation is palladium, whose ground-state configuration is $4d^{10}5s^{0}$ — no outer s electron at all.

One consequence of this formula is immediately worth flagging. The elements zinc, cadmium, mercury and copernicium share the configuration $(n-1)d^{10}ns^{2}$. Their d orbitals are completely filled in both the ground state and in their common $+2$ oxidation state. Because a transition element is defined as one having an incomplete d subshell in at least one of its common ions, this group is not regarded as truly transition — a point NEET tests directly.

Why 4s Fills Before 3d

To write a 3d configuration correctly you must first understand the ordering of the 4s and 3d orbitals — and recognise that this ordering is not fixed. For the elements up to argon and into potassium, the 4s orbital lies lower in energy than 3d, so it fills first (this is the familiar $(n+l)$ rule outcome). At calcium the two are almost equal. From scandium onwards, however, the 3d orbitals drop below 4s in energy once they begin to be occupied.

The practical rule for the neutral atom is therefore: fill 4s first, then begin filling 3d. The energy difference between the $(n-1)d$ and $ns$ sets is genuinely small — small enough that it can be reversed by subtle electron–electron interactions, which is exactly why the exceptions arise. The figure below shows the schematic filling sequence across the start of the 3d series.

The 3d Series: Sc to Zn

The first transition series runs from scandium ($Z=21$) to zinc ($Z=30$). Following the rule, each successive element adds one electron to the 3d set while 4s stays as $4s^2$ — except at chromium and copper, where the 3d/4s split intervenes. The full ground-state outer configurations are tabulated below; commit the two highlighted exceptions to memory.

Notice the smooth $3d^1 \to 3d^2 \to 3d^3$ progression through vanadium, the jump to $3d^5$ at chromium, the return to $3d^5 4s^2$ at manganese, then $3d^6$ through $3d^8$, and finally the $3d^{10}$ jump at copper. Zinc closes the series with both inner d and outer s completely filled.

The Chromium and Copper Exceptions

NCERT attributes both anomalies to a single cause: the energy gap between the 3d and 4s sets is "small enough to prevent the electron entering the 3d orbitals" in the expected way, combined with the fact that exactly half-filled and completely filled sets of orbitals are relatively more stable. Two stabilising influences underlie that extra stability.

For chromium, promoting one 4s electron into 3d converts $3d^44s^2$ into $3d^54s^1$ — now both the d and s sets are half-filled. For copper, the same shift converts $3d^94s^2$ into $3d^{10}4s^1$, completing the d subshell. In each case the small energy cost of moving the electron is more than repaid by the gain in exchange and symmetry stabilisation. The orbital-box diagram makes this concrete.

Configurations of Ions: 4s Lost First

Writing the configuration of a transition-metal cation trips up more students than any other part of this topic, because of one counter-intuitive rule. Although 4s fills before 3d in the neutral atom, the 4s electrons are removed before the 3d electrons on ionisation. The reason is the energy inversion shown in Figure 1: once the 3d orbitals are occupied they lie below 4s, so the higher-energy 4s electrons depart first.

This rule explains a host of NEET favourites: $\ce{Sc^3+}$ is $[Ar]3d^0$ (the noble-gas core, hence colourless and diamagnetic), $\ce{Zn^2+}$ is $[Ar]3d^{10}$, $\ce{Cr^3+}$ is $[Ar]3d^3$, and $\ce{Ni^2+}$ is $[Ar]3d^8$. From these $d^n$ counts you read off unpaired-electron numbers directly, which feeds straight into spin-only magnetic-moment calculations.

Exceptions in the 4d and 5d Series

Beyond the 3d row, the $(n-1)d$–$ns$ energy gap shrinks further, so the second (4d) and third (5d) transition series carry more numerous and less regular exceptions. NEET does not demand that you memorise the entire 4d/5d table, but you should recognise the headline cases — especially palladium and silver, which appear in option lists.

The pattern is clear: half-filled and fully-filled d targets recur (Mo mirrors Cr; Ag and Au mirror Cu), and palladium goes one step further with $4d^{10}5s^0$. These heavier-series anomalies are also linked to relativistic stabilisation of the s orbital, but at NEET level the half/fully-filled-stability argument is the expected explanation.

Why Configuration Governs Everything Else

Electronic configuration is not an isolated piece of bookkeeping — it is the master variable for the whole chapter. Almost every characteristic property of the transition metals can be traced back to the number and arrangement of d electrons.

NCERT highlights one configuration-driven anomaly worth noting: the unusually high stability of $\ce{Mn^2+}$ ($3d^5$, half-filled) and of $\ce{Fe^3+}$ ($3d^5$) follows directly from the half-filled-d stabilisation you have just studied. Likewise the prevalence of $\ce{Cu^2+}$ over $\ce{Cu+}$ in aqueous solution, while ultimately driven by hydration energy, starts from the $3d^{10}4s^1$ configuration of copper. Once you can write a configuration confidently, the rest of the chapter becomes a set of consequences rather than a set of facts to memorise.

This is also why the d-block sits where it does in the periodic table. The progressive filling of the inner $(n-1)d$ shell — while the outer $ns$ stays nearly constant — produces the gradual, gentle variation in properties across a transition series, in contrast to the sharper changes seen across an s- or p-block period. For the broader placement logic, see the companion note on the block's position in the table.