NCERT Solutions · Class 12 Chemistry The d-and f-Block Elements

Cerium loses its 6s and 5d electrons first, not its 4f electron — so Ce³⁺ keeps exactly one f electron, and that single electron decides everything about its magnetism.Step $\displaystyle 1$ — write the electronic configuration of neutral Ce (Z = $\displaystyle 58$).Cerium is one of the lanthanides with an irregular configuration: instead of filling 4f before 5d, it puts one electron in each,\[\text{Ce (Z = 58)} : [\text{Xe}]\,4f^{1}\,5d^{1}\,6s^{2} \]where \(\displaystyle [\text{Xe}]\) stands for the full xenon core (\(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\,4p^6\,4d^{10}\,5s^2\,5p^6\), $\displaystyle 54$ electrons), and the three extra electrons sit in \(\displaystyle 4f^1\,5d^1\,6s^2\).Step $\displaystyle 2$ — remove three electrons to form Ce³⁺.When a lanthanide atom ionises, electrons are always lost from the outermost occupied shells first — \(\displaystyle 6s\) before \(\displaystyle 5d\) before \(\displaystyle 4f\) — because once the atom is stripped of its outer electrons, the 4f orbitals contract and drop below 5d and 6s in energy. This is the step that trips people up: it is tempting to remove the 4f electron since "f" looks more buried, but 4f is not the outermost shell here — 6s and 5d are, and Ce³⁺ removes exactly those three:\[\text{Ce}^{3+} = \text{Ce} - 3e^{-} : [\text{Xe}]\,4f^{1}\,5d^{0}\,6s^{0} = [\text{Xe}]\,4f^{1} \]So Ce³⁺ has a single electron sitting in the 4f subshell and nothing outside the xenon core besides it.Step $\displaystyle 3$ — apply Hund's rule to the \(\displaystyle 4f^1\) electron.The 4f subshell has \(\displaystyle l = 3\), so it has \(\displaystyle 2l+1 = 7\) degenerate orbitals (\(\displaystyle m_l = -3, -2, -1, 0, +1, +2, +3\)), each able to hold $\displaystyle 2$ electrons of opposite spin. Hund's rule says electrons fill degenerate orbitals singly, with parallel spin, before any pairing occurs. With only one electron to place, it simply drops into one of the seven empty 4f orbitals with its spin unpaired:\[4f:\ \underset{\uparrow}{\Box}\ \ \Box\ \ \Box\ \ \Box\ \ \Box\ \ \Box\ \ \Box \]There is no partner electron available to pair with, so this electron stays unpaired — giving\[n(\text{unpaired electrons}) = 1 \]Step $\displaystyle 4$ — calculate the magnetic moment from the spin-only formula.The spin-only formula for the magnetic moment of an ion is\[\mu = \sqrt{n(n+2)}\ \text{BM} \]where \(\displaystyle n\) is the number of unpaired electrons and BM (Bohr Magneton) is the unit of magnetic moment. This formula ignores orbital contribution — it counts only the spin of unpaired electrons, which is the level expected here.Substituting \(\displaystyle n = 1\):\[\mu = \sqrt{1(1+2)}\ \text{BM} = \sqrt{3}\ \text{BM} \]\[\mu = 1.732\ \text{BM} \]Rounding to three significant figures (the precision the spin-only model itself supports):\[\mu \approx 1.73\ \text{BM} \]Answer: Ce³⁺ has the configuration \(\displaystyle [\text{Xe}]\,4f^{1}\) with $\displaystyle 1$ unpaired electron, giving a spin-only magnetic moment of \(\displaystyle \mu = \sqrt{3} \approx 1.73\) BM.

The characteristic oxidation state of every lanthanoid is +$\displaystyle 3$ — but a few members swing to +$\displaystyle 2$ or +$\displaystyle 4$ whenever that gets them to an empty \(\displaystyle 4f^0\), half‑filled \(\displaystyle 4f^7\), or completely filled \(\displaystyle 4f^{14}\) configuration. These three arrangements have extra stability (empty and completely filled subshells minimise electron–electron repulsion; half‑filled subshells maximise exchange energy), and that stability can outweigh the energy cost of removing one extra electron or of not losing one that would normally be lost.Setting up the bookkeeping. A neutral lanthanoid atom is \(\displaystyle \text{[Xe]}\,4f^{1-14}5d^{0-1}6s^2\). To form \(\displaystyle Ln^{3+}\), an atom always loses its \(\displaystyle 6s^2\) electrons plus one \(\displaystyle 4f\) or \(\displaystyle 5d\) electron, so every \(\displaystyle Ln^{3+}\) ends up as a pure \(\displaystyle \text{[Xe]}\,4f^n\) ion with \(\displaystyle n = Z - 57\). It is the value of this \(\displaystyle n\) — how close it sits to $\displaystyle 0$, $\displaystyle 7$, or $\displaystyle 14$ — that decides whether an element can also be pushed to +$\displaystyle 4$ (one electron poorer than \(\displaystyle Ln^{3+}\)) or pulled back to +$\displaystyle 2$ (one electron richer than \(\displaystyle Ln^{3+}\)).Members showing +$\displaystyle 4$ (they want to lose one more electron than usual): Members showing +$\displaystyle 2$ (they resist losing their normal third electron): The pattern to notice: Ce and Tb are the two elements whose \(\displaystyle Ln^{3+}\) configuration sits one electron above an extra‑stable count (\(\displaystyle f^1\) above \(\displaystyle f^0\); \(\displaystyle f^8\) above \(\displaystyle f^7\)), so they are pulled up to +4. Eu and Yb are the two elements whose neutral atom sits exactly at an extra‑stable count once the \(\displaystyle 6s^2\) pair is removed (\(\displaystyle f^7\), \(\displaystyle f^{14}\)), so they are held back at +2. Pr, Nd, and Sm are one electron further away from the nearest magic number in each direction, so the same effect operates for them but more weakly, giving oxidation states that are known but less common and less stable than Ce(IV)/Tb(IV) or Eu(II)/Yb(II).Answer: +$\displaystyle 4$ oxidation state — Ce, Pr, Nd, Tb (Ce⁴⁺ = \(\displaystyle 4f^0\) and Tb⁴⁺ = \(\displaystyle 4f^7\) are the most stable, since they are one electron short of the parent \(\displaystyle Ln^{3+}\) configuration). +$\displaystyle 2$ oxidation state — Sm, Eu, Yb (Eu²⁺ = \(\displaystyle 4f^7\) and Yb²⁺ = \(\displaystyle 4f^{14}\) are the most stable, since they retain one electron more than the parent \(\displaystyle Ln^{3+}\) configuration). In every case the driving force is the extra stability of an empty, half‑filled, or completely filled 4f subshell.

The row of orbitals being filled — 4f for one series, 5f for the other — is the single fact that drives every difference below: how well it shields the nucleus decides how many oxidation states show up and how reactive the metal is.(i) Electronic configurationLanthanoids (Ce to Lu, following La) have the general configuration \[[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^{2} \] Actinoids (Th to Lr, following Ac) have the general configuration \[[\text{Rn}]\,5f^{1-14}\,6d^{0-1}\,7s^{2} \] So structurally the two series run in parallel — one is the 4f row, the other the 5f row, each sitting one period below the last d-block element (La and Ac respectively), and each showing the same kind of "extra stability" exceptions: an electron shifts from f into d whenever that empties, half-fills, or fully fills the f subshell. That is why La is \(\displaystyle [\text{Xe}]5d^{1}6s^{2}\) (no f electron at all) and Gd is \(\displaystyle [\text{Xe}]4f^{7}5d^{1}6s^{2}\) (f half-filled); the same pattern gives Ac as \(\displaystyle [\text{Rn}]6d^{1}7s^{2}\) and Cm as \(\displaystyle [\text{Rn}]5f^{7}6d^{1}7s^{2}\).The difference is in how messy this filling gets. In the lanthanoids, \(\displaystyle 4f\), \(\displaystyle 5d\) and \(\displaystyle 6s\) are close in energy, so exceptions are limited. In the actinoids, \(\displaystyle 5f\), \(\displaystyle 6d\) and \(\displaystyle 7s\) are even closer in energy over a much wider stretch of the series — so actinoid configurations are far less regular, with more exceptions and even some configurations still disputed experimentally.The knock-on effect is size: because the \(\displaystyle 5f\) orbitals are more diffuse than \(\displaystyle 4f\) and shield the growing nuclear charge poorly, the actinoid contraction (the steady fall in atomic/ionic radius along Ac→Lr) is greater, element for element, than the lanthanoid contraction.(ii) Oxidation statesLanthanoids are overwhelmingly monotonous: +$\displaystyle 3$ is the characteristic and dominant oxidation state for every member, because losing the \(\displaystyle 6s^{2}\) and one \(\displaystyle 5d\)/\(\displaystyle 4f\) electron reaches this state easily and it is the one from which the fourth ionisation energy rises sharply. A few members also show +$\displaystyle 2$ or +$\displaystyle 4$, but only where it buys a special stability — an empty, half-filled, or fully filled \(\displaystyle 4f\) subshell: \(\displaystyle \text{Ce}^{4+}\) (\(\displaystyle f^{0}\)), \(\displaystyle \text{Tb}^{4+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \text{Eu}^{2+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \text{Yb}^{2+}\) (\(\displaystyle f^{14}\)).Actinoids, by contrast, show a genuinely wide spread of oxidation states — commonly +$\displaystyle 3$ to +$\displaystyle 6$, and up to +$\displaystyle 7$ for some (e.g. uranium shows +$\displaystyle 3$, +$\displaystyle 4$, +$\displaystyle 5$, +$\displaystyle 6$; neptunium and plutonium reach +$\displaystyle 7$ under strong oxidising conditions). This is a direct consequence of point (i): because \(\displaystyle 5f\), \(\displaystyle 6d\) and \(\displaystyle 7s\) are so close in energy, more electrons from all three sets can take part in bonding, not just the outer three. So while +$\displaystyle 3$ is common to both families, it is only one option among many for actinoids, whereas for lanthanoids it is essentially the only option.(iii) Chemical reactivityLanthanoids are electropositive metals whose reactivity falls off along the series: the early members (La, Ce, Pr) react with water almost as vigorously as calcium, while the later, smaller members (a consequence of the lanthanoid contraction) behave more like aluminium. They tarnish in air, burn in oxygen to the oxide \(\displaystyle M_2O_3\), combine with halogens to give \(\displaystyle MX_3\), and react with dilute acids liberating \(\displaystyle H_2\) gas; their oxides and hydroxides, \(\displaystyle M_2O_3\) and \(\displaystyle M(OH)_3\), are distinctly basic.Actinoids are markedly more reactive metals, particularly when finely divided. They react with boiling water to give a mixture of the oxide and the hydride, combine directly with most non-metals at only moderate temperature, and dissolve readily in mineral acids such as hydrochloric acid, though alkalis have little effect on them. This heightened reactivity traces back to the same cause as their oxidation-state variety: the diffuse, poorly-shielding \(\displaystyle 5f\) electrons are more easily drawn into bonding, and the early actinoids in particular are larger and more electropositive than their lanthanoid counterparts. Unlike the lanthanoids — where only promethium is radioactive — every actinoid is radioactive, which is itself an added complication in studying their chemistry (self-heating and radiolytic decomposition of compounds).Answer: Lanthanoids fill \(\displaystyle 4f\) (general configuration \(\displaystyle [Xe]4f^{1-14}5d^{0-1}6s^2\)), show almost exclusively the +$\displaystyle 3$ state (with rare +$\displaystyle 2$/+$\displaystyle 4$ only when an empty, half-filled, or filled \(\displaystyle 4f\) results), and are moderately reactive, electropositive metals of gradually falling reactivity along the series. Actinoids fill \(\displaystyle 5f\) (general configuration \(\displaystyle [Rn]5f^{1-14}6d^{0-1}7s^2\)) with far more irregular configurations and a greater actinoid contraction (poorer \(\displaystyle 5f\) shielding); they show a much wider range of oxidation states (+$\displaystyle 3$ up to +$\displaystyle 7$, since \(\displaystyle 5f\), \(\displaystyle 6d\), \(\displaystyle 7s\) are comparable in energy); and they are more reactive than the lanthanoids, besides being uniformly radioactive.

A lanthanide or actinide's configuration is the previous noble gas core plus the leftover electrons split between the s-orbital and the f (or d) orbital that is filling — and for the early actinides you cannot assume the f-orbital fills before the d one.For any atom, write its configuration as [core]\(\displaystyle +\) remaining electrons, where the core is the noble gas with the atomic number closest below it. Beyond period $\displaystyle 4$, the filling order runs \(\displaystyle 6s < 4f < 5d\) for period $\displaystyle 6$, and \(\displaystyle 7s < 5f < 6d\) for period 7. Each s-subshell holds at most $\displaystyle 2$ electrons, each f-subshell at most $\displaystyle 14$, each d-subshell at most $\displaystyle 10$ — so once you know how many electrons are left after the core and the s-electrons, you know how many are left to place in f (and, for actinides, d).Z = $\displaystyle 61$ — Promethium (Pm), a lanthanideThe last noble gas below $\displaystyle 61$ is xenon, \(\displaystyle Z = 54 \). Electrons remaining: \[61 - 54 = 7 \] Of these, $\displaystyle 2$ fill \(\displaystyle 6s\) (its maximum), leaving \(\displaystyle 7 - 2 = 5\) for \(\displaystyle 4f\): \[\text{Pm: } [\text{Xe}]\,4f^{5}\,6s^{2} \]Z = $\displaystyle 91$ — Protactinium (Pa), an actinideThe last noble gas below $\displaystyle 91$ is radon, \(\displaystyle Z = 86 \). Electrons remaining: \[91 - 86 = 5 \] Two fill \(\displaystyle 7s\), leaving $\displaystyle 3$ electrons to be split between \(\displaystyle 5f\) and \(\displaystyle 6d\). This is the step where the actinides differ from the lanthanides: early in the actinide series \(\displaystyle 5f\) and \(\displaystyle 6d\) sit so close in energy that electrons occupy both before \(\displaystyle 5f\) is anywhere near full — filling all $\displaystyle 3$ straight into \(\displaystyle 5f\) is the mistake to watch for. For protactinium the accepted ground-state split is $\displaystyle 2$ electrons in \(\displaystyle 5f\) and $\displaystyle 1$ in \(\displaystyle 6d\): \[\text{Pa: } [\text{Rn}]\,5f^{2}\,6d^{1}\,7s^{2} \]Z = $\displaystyle 101$ — Mendelevium (Md), an actinideElectrons remaining beyond radon: \[101 - 86 = 15 \] By this point in the actinide series the stray \(\displaystyle 6d\) electron seen in elements like Pa has been reabsorbed, and filling runs regularly through \(\displaystyle 5f\). Two electrons fill \(\displaystyle 7s\), leaving \(\displaystyle 15 - 2 = 13\) for \(\displaystyle 5f\): \[\text{Md: } [\text{Rn}]\,5f^{13}\,7s^{2} \]Z = $\displaystyle 109$ — Meitnerium (Mt), a d-block elementElectrons remaining beyond radon: \[109 - 86 = 23 \] Two fill \(\displaystyle 7s\), leaving $\displaystyle 21$ electrons for \(\displaystyle 5f\) and \(\displaystyle 6d\) together. \(\displaystyle 5f\) fills completely first (its maximum is $\displaystyle 14$), leaving \(\displaystyle 21 - 14 = 7\) for \(\displaystyle 6d\): \[\text{Mt: } [\text{Rn}]\,5f^{14}\,6d^{7}\,7s^{2} \] This also matches where Mt sits on the periodic table — directly below cobalt, rhodium, and iridium in group $\displaystyle 9$ — so a filled \(\displaystyle 5f^{14}\) core plus \(\displaystyle 6d^{7}\) is exactly the configuration its position predicts.Answer: Z = $\displaystyle 61$ (Pm): \(\displaystyle [\text{Xe}]\,4f^{5}\,6s^{2}\); Z = $\displaystyle 91$ (Pa): \(\displaystyle [\text{Rn}]\,5f^{2}\,6d^{1}\,7s^{2}\); Z = $\displaystyle 101$ (Md): \(\displaystyle [\text{Rn}]\,5f^{13}\,7s^{2}\); Z = $\displaystyle 109$ (Mt): \(\displaystyle [\text{Rn}]\,5f^{14}\,6d^{7}\,7s^{2}\).

The first (3d) series looks like a preview of the second (4d) and third (5d) series, but the third series breaks the pattern everywhere — because of the lanthanoid contraction sitting just before it.(i) Electronic configurations. All three series share the general outer configuration \(\displaystyle (n-1)d^{1-10}\,ns^{0-2} \): In a vertical column the outer configuration usually matches — but the 4d and 5d elements deviate from the "expected" pattern far more often than the 3d elements, because in the heavier series the \(\displaystyle (n-1)d \) and \(\displaystyle ns \) orbitals lie closer in energy, so an electron slips from \(\displaystyle ns \) into \(\displaystyle (n-1)d \) whenever that gives a filled or half-filled d subshell. This is the step where students expect a clean, unbroken \(\displaystyle d^{n}s^{2} \) pattern down a group and get caught out: compare \(\displaystyle \mathrm{Ni}\,3d^{8}4s^{2} \) with \(\displaystyle \mathrm{Pd}\,4d^{10}5s^{0} \) and \(\displaystyle \mathrm{Pt}\,5d^{9}6s^{1} \) — three very different splits of the same ten valence electrons in one group.(ii) Oxidation states. Going down a column, transition metals show more oxidation states, and the higher ones become more stable — the opposite of the inert-pair trend in the p-block. In Group $\displaystyle 7$, \(\displaystyle \mathrm{Mn} \) is most stable as \(\displaystyle \mathrm{Mn}^{2+} \), and its \(\displaystyle +7 \) state (\(\displaystyle \mathrm{MnO_4^-} \)) is a strong, easily-reduced oxidiser; but \(\displaystyle \mathrm{Tc} \) and \(\displaystyle \mathrm{Re} \) form stable \(\displaystyle +7 \) oxides (\(\displaystyle \mathrm{Tc_2O_7}, \mathrm{Re_2O_7} \)) and the perrhenate ion is a far weaker oxidant than permanganate. In Group $\displaystyle 8$, iron rarely gets past \(\displaystyle +3 \) in ordinary chemistry (its \(\displaystyle +6 \) ferrate is a strong oxidiser and hard to isolate), while \(\displaystyle \mathrm{Ru} \) and \(\displaystyle \mathrm{Os} \) readily form the stable, isolable \(\displaystyle +8 \) oxides \(\displaystyle \mathrm{RuO_4} \) and \(\displaystyle \mathrm{OsO_4} \). The reason is that in the 4d/5d metals, the valence d and s orbitals are close enough in energy that more electrons can be pulled into bonding, so high formal charges are stabilised rather than punished as the group is descended.(iii) Ionisation enthalpies. Along any one period, ionisation enthalpy rises overall (with dips at half-filled and filled configurations). Down a column, though, d-block ionisation enthalpy increases from 3d to 4d to 5d — this is the step people misapply, carrying over the s/p-block rule that ionisation enthalpy falls down a group. It doesn't fall here because the atomic radius does not expand as much as an extra shell should allow (see point iv), so the effective nuclear charge felt by the outermost electron actually grows. Group $\displaystyle 11$ shows this clearly: first ionisation enthalpies run \(\displaystyle \mathrm{Cu}\approx 745\ \text{kJ mol}^{-1} \), \(\displaystyle \mathrm{Ag}\approx 731\ \text{kJ mol}^{-1} \), \(\displaystyle \mathrm{Au}\approx 890\ \text{kJ mol}^{-1} \) — gold's is the highest of the three, not the lowest, because the poorly-shielding 4f electrons filled in just before it leave its nucleus more exposed than expected.(iv) Atomic sizes. From the 3d to the 4d series, atomic radius increases normally, as it does anywhere a new principal shell is added. But from 4d to 5d, that expected increase is almost completely cancelled by the lanthanoid contraction: the steady, cumulative shrinkage caused by the poorly-shielding 4f electrons that fill across \(\displaystyle \mathrm{La} \) to \(\displaystyle \mathrm{Lu} \) before the 5d series even starts. The result is that 5d atoms are nearly the same size as the 4d atoms directly above them — sometimes marginally smaller. \(\displaystyle \mathrm{Zr} \) (4d, radius \(\displaystyle \approx 160\ \text{pm} \)) and \(\displaystyle \mathrm{Hf} \) (5d, radius \(\displaystyle \approx 159\ \text{pm} \)) are essentially identical in size, despite Hf carrying a whole extra filled 4f shell that "should" have made it larger. This is exactly why Zr and Hf (and similarly Nb/Ta) are so hard to separate chemically — same size, same charge density, nearly identical chemistry.Putting the four together: the first series behaves as the textbook d-block pattern, while the second and third series increasingly resemble each other rather than the first, because the lanthanoid contraction inserted between them suppresses the size increase (and so also the ionisation-enthalpy decrease) that would otherwise be expected on descending the group, while simultaneously making higher oxidation states more accessible and stable.Answer: Configurations follow \(\displaystyle (n-1)d^{1-10}ns^{0-2}\) in all three series, but 4d/5d elements deviate more often (e.g. Pd \(\displaystyle 4d^{10}5s^0\), Pt \(\displaystyle 5d^9 6s^1\)); oxidation states increase in number and the higher ones grow more stable down a group (Mn²⁺ vs. stable Re⁷⁺; Fe vs. stable Ru/Os⁸⁺); ionisation enthalpy rises from 3d→4d→5d (opposite to the s/p-block trend); and atomic radius rises from 3d→4d but then stalls from 4d→5d because of the lanthanoid contraction (Zr ≈ $\displaystyle 160$ pm ≈ Hf, $\displaystyle 159$ pm) — so the 4d and 5d elements of a group resemble each other much more closely than either resembles the 3d element above them.

When a transition-metal atom loses electrons to form a cation, the electrons leave the 4s orbital first, never the 3d — even though 4s filled first when the atom was built up. In the neutral atom \(\displaystyle 4s\) sits slightly below \(\displaystyle 3d\) in energy, so it fills first. Once the \(\displaystyle 3d\) orbitals start holding electrons, though, they drop below \(\displaystyle 4s\) in energy for that species — so ionisation always strips the outer \(\displaystyle 4s\) electrons before touching \(\displaystyle 3d\). Getting this order backwards is the single most common mistake on this question.Start from each atom's ground-state configuration (atomic number in brackets), then remove electrons \(\displaystyle 4s\) first:\(\displaystyle \mathrm{Ti}\ (Z=22):\ [\mathrm{Ar}]3d^24s^2 \ \Rightarrow\ \mathrm{Ti}^{2+}: [\mathrm{Ar}]3d^2\) — $\displaystyle 2$ electrons in 3d\(\displaystyle \mathrm{V}\ (Z=23):\ [\mathrm{Ar}]3d^34s^2 \ \Rightarrow\ \mathrm{V}^{2+}: [\mathrm{Ar}]3d^3\) — $\displaystyle 3$ electrons in 3d\(\displaystyle \mathrm{Cr}\ (Z=24):\ [\mathrm{Ar}]3d^54s^1 \ \Rightarrow\ \mathrm{Cr}^{3+}\) loses the one \(\displaystyle 4s\) electron and then two more from \(\displaystyle 3d^5\): \(\displaystyle [\mathrm{Ar}]3d^3\) — $\displaystyle 3$ electrons in 3d (chromium's neutral atom is itself an exception, \(\displaystyle 3d^54s^1\) rather than \(\displaystyle 3d^44s^2\), for the extra stability of a half-filled \(\displaystyle d\) subshell)\(\displaystyle \mathrm{Mn}\ (Z=25):\ [\mathrm{Ar}]3d^54s^2 \ \Rightarrow\ \mathrm{Mn}^{2+}: [\mathrm{Ar}]3d^5\) — $\displaystyle 5$ electrons in 3d\(\displaystyle \mathrm{Fe}\ (Z=26):\ [\mathrm{Ar}]3d^64s^2 \ \Rightarrow\ \mathrm{Fe}^{2+}: [\mathrm{Ar}]3d^6\) — $\displaystyle 6$ electrons in 3d; \(\displaystyle \mathrm{Fe}^{3+}\) loses one more, from \(\displaystyle 3d\), giving \(\displaystyle [\mathrm{Ar}]3d^5\) — $\displaystyle 5$ electrons in 3d\(\displaystyle \mathrm{Co}\ (Z=27):\ [\mathrm{Ar}]3d^74s^2 \ \Rightarrow\ \mathrm{Co}^{2+}: [\mathrm{Ar}]3d^7\) — $\displaystyle 7$ electrons in 3d\(\displaystyle \mathrm{Ni}\ (Z=28):\ [\mathrm{Ar}]3d^84s^2 \ \Rightarrow\ \mathrm{Ni}^{2+}: [\mathrm{Ar}]3d^8\) — $\displaystyle 8$ electrons in 3d\(\displaystyle \mathrm{Cu}\ (Z=29):\ [\mathrm{Ar}]3d^{10}4s^1 \ \Rightarrow\ \mathrm{Cu}^{2+}\) loses the \(\displaystyle 4s\) electron and one from \(\displaystyle 3d^{10}\): \(\displaystyle [\mathrm{Ar}]3d^9\) — $\displaystyle 9$ electrons in 3d (copper's neutral atom is also an exception, \(\displaystyle 3d^{10}4s^1\), for the extra stability of a filled \(\displaystyle d\) subshell)Now place these electrons in the five 3d orbitals of the hydrated (octahedral, \(\displaystyle [\mathrm{M}(\mathrm{H_2O})_6]^{n+}\)) ion. In an octahedral field the five otherwise-degenerate 3d orbitals split into two sets: the lower-energy \(\displaystyle t_{2g}\) set (\(\displaystyle d_{xy}, d_{yz}, d_{zx}\); $\displaystyle 3$ orbitals) and the higher-energy \(\displaystyle e_g\) set (\(\displaystyle d_{z^2}, d_{x^2-y^2}\); $\displaystyle 2$ orbitals). Water is only a weak-to-intermediate field ligand for these first-row ions, so the crystal-field splitting energy \(\displaystyle \Delta_o\) is smaller than the electron-pairing energy — the hydrated ions are high-spin. That means electrons occupy the five orbitals singly (Hund's rule) across both sets before any pairing begins; pairing starts only once a fifth electron must be added, and it fills the lower \(\displaystyle t_{2g}\) set first.

Ion	3d electrons	\(\displaystyle t_{2g}\)	\(\displaystyle e_g\)
\(\displaystyle \mathrm{Ti}^{2+}\)	$\displaystyle 2$	\(\displaystyle t_{2g}^{2}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{V}^{2+}\)	$\displaystyle 3$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{Cr}^{3+}\)	$\displaystyle 3$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{Mn}^{2+}\)	$\displaystyle 5$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Fe}^{2+}\)	$\displaystyle 6$	\(\displaystyle t_{2g}^{4}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Fe}^{3+}\)	$\displaystyle 5$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Co}^{2+}\)	$\displaystyle 7$	\(\displaystyle t_{2g}^{5}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Ni}^{2+}\)	$\displaystyle 8$	\(\displaystyle t_{2g}^{6}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Cu}^{2+}\)	$\displaystyle 9$	\(\displaystyle t_{2g}^{6}\)	\(\displaystyle e_g^{3}\)

Reading the table row by row: up to \(\displaystyle d^3\) (\(\displaystyle \mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}\)) the electrons go singly into the three \(\displaystyle t_{2g}\) orbitals with none in \(\displaystyle e_g\) yet. At \(\displaystyle d^5\) (\(\displaystyle \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}\)) all five orbitals now hold one electron each — three in \(\displaystyle t_{2g}\), two in \(\displaystyle e_g\) — the maximum-multiplicity, all-unpaired arrangement. Beyond \(\displaystyle d^5\) every extra electron must pair up, and pairing goes into the lower \(\displaystyle t_{2g}\) set first: \(\displaystyle \mathrm{Fe}^{2+}\) (\(\displaystyle d^6\)) pairs one \(\displaystyle t_{2g}\) electron to give \(\displaystyle t_{2g}^{4}e_g^{2}\), \(\displaystyle \mathrm{Co}^{2+}\) (\(\displaystyle d^7\)) gives \(\displaystyle t_{2g}^{5}e_g^{2}\), \(\displaystyle \mathrm{Ni}^{2+}\) (\(\displaystyle d^8\)) gives \(\displaystyle t_{2g}^{6}e_g^{2}\) (the \(\displaystyle t_{2g}\) set now fully paired), and \(\displaystyle \mathrm{Cu}^{2+}\) (\(\displaystyle d^9\)) must finally start pairing in \(\displaystyle e_g\) too, giving \(\displaystyle t_{2g}^{6}e_g^{3}\).Answer: 3d electron counts — \(\displaystyle \mathrm{Ti}^{2+}=2,\ \mathrm{V}^{2+}=3,\ \mathrm{Cr}^{3+}=3,\ \mathrm{Mn}^{2+}=5,\ \mathrm{Fe}^{2+}=6,\ \mathrm{Fe}^{3+}=5,\ \mathrm{Co}^{2+}=7,\ \mathrm{Ni}^{2+}=8,\ \mathrm{Cu}^{2+}=9\). For the hydrated (octahedral) ions, all are high-spin, with orbital occupations \(\displaystyle t_{2g}^{2}e_g^{0},\ t_{2g}^{3}e_g^{0},\ t_{2g}^{3}e_g^{0},\ t_{2g}^{3}e_g^{2},\ t_{2g}^{4}e_g^{2},\ t_{2g}^{3}e_g^{2},\ t_{2g}^{5}e_g^{2},\ t_{2g}^{6}e_g^{2},\ t_{2g}^{6}e_g^{3}\) respectively.

The claim is true, and the reason behind almost every difference is the same: 3d orbitals are compact and poorly shielded, while 4d and 5d orbitals are large, diffuse, and — for the 5d series — squeezed in size by the lanthanoid contraction that sits just before them. Once you see that one geometric fact, each "different property" below stops looking like a separate rule and becomes a consequence of it.1. Atomic and ionic radii. Going down a group, radius normally increases (more shells). It does for Sc→Y (3d→4d), but Y→La→Hf (4d→5d) barely changes size at all. Between La (Z = $\displaystyle 57$) and Hf (Z = $\displaystyle 72$) the $\displaystyle 14$ lanthanoids are filling the inner 4f subshell, which shields the nucleus very poorly. The steady, cumulative shrinkage this causes — the lanthanoid contraction — almost exactly cancels the size increase expected on adding a whole new shell. The result: Zr and Hf, Nb and Ta, Mo and W are nearly identical in radius, and hence in chemistry (which is why separating Zr from Hf is notoriously hard), while the first-row member of each triad (Ti, Nb's partner V, Mo's partner Cr) is distinctly smaller and behaves differently from both.2. Density. Density is mass divided by volume, and the lanthanoid contraction fixes the volume of a 5d atom close to that of its 4d partner while its atomic mass is far higher. So density roughly doubles from the first series to the second, and rises only a little further into the third: iron is about \(\displaystyle 7.9\ \mathrm{g\,cm^{-3}}\), but osmium and iridium — its group-mates two rows down — are around \(\displaystyle 22.6\) and \(\displaystyle 22.4\ \mathrm{g\,cm^{-3}}\), among the densest elements known. First-series metals are comparatively light.3. Enthalpy of atomization and melting point. 4d and 5d valence orbitals are larger and more diffuse than 3d orbitals, so they overlap more extensively with neighbouring atoms, giving stronger metal–metal bonding in the bulk metal. Enthalpies of atomization and melting points therefore climb sharply down each group — tungsten (5d) has the highest melting point of any metal, far above its 3d counterpart chromium — whereas the first-series metals atomize and melt far more readily.4. Ionization enthalpy and reactivity. The poor shielding from the intervening 4f electrons leaves the nucleus of a 5d atom unusually effective at holding on to its valence electrons, so first ionization enthalpies rise irregularly but markedly from 3d to 5d. This is why the heavy end of the 5d row — platinum, gold, mercury — are noble metals, resistant to oxidation and to attack by dilute acids, while several first-series metals (Fe, Mn, Zn) dissolve in dilute acids liberating \(\displaystyle \mathrm{H_2}\) with ease.5. Oxidation states. First-series metals reach high oxidation states reluctantly, and those states are strong oxidizers once formed — \(\displaystyle \mathrm{Mn}\) in \(\displaystyle \mathrm{MnO_4^-}\) is +$\displaystyle 7$, but \(\displaystyle \mathrm{MnO_4^-}\) is a powerful, easily-reduced oxidizing agent. Heavier transition metals reach even higher oxidation states and hold them stably: \(\displaystyle \mathrm{OsO_4}\) has Os in +$\displaystyle 8$, \(\displaystyle \mathrm{ReF_7}\) has Re in +$\displaystyle 7$, and both are comparatively stable, isolable compounds rather than aggressive oxidants. The larger, more diffuse 4d/5d orbitals form stronger, more covalent M–O and M–halogen bonds that can support the higher charge.6. Metal–metal bonding and cluster chemistry. Because 4d and 5d orbitals are more diffuse, they overlap with each other far more effectively than 3d orbitals do. Second- and third-row transition metals form extensive metal–metal bonded clusters — species such as \(\displaystyle \mathrm{[Re_3Cl_{12}]^{3-}}\) or \(\displaystyle \mathrm{[Mo_6Cl_8]^{4+}}\) — a type of chemistry the first series shows only rarely (a little in Cr(II) chemistry).7. Magnetic behaviour. For most 3d complexes, orbital angular momentum is largely quenched by the surrounding ligand field, so the observed magnetic moment matches the spin-only formula \(\displaystyle \mu = \sqrt{n(n+2)}\) BM (n = number of unpaired electrons) quite closely. In 4d and 5d complexes, spin–orbit coupling is much stronger, so moments deviate substantially from this simple formula, and many are diamagnetic altogether because the crystal field splitting \(\displaystyle \Delta_o\) is large enough to force full electron pairing.8. Crystal field splitting and spin state. \(\displaystyle \Delta_o\) is roughly $\displaystyle 1.5$–$\displaystyle 2$ times larger for 4d and 5d metal ions than for their 3d counterparts, because the larger d orbitals interact more strongly with ligand orbitals. Pairing energy exceeds \(\displaystyle \Delta_o\) for many 3d ions, favouring high-spin complexes; for 4d/5d ions the reverse is usually true, so low-spin complexes (and higher coordination numbers, since the bigger ion has more room) dominate.9. Abundance. First-series elements such as Fe, Ti, and Mn are among the more abundant elements in the Earth's crust, while their 4d/5d congeners (Os, Ir, Re) are among the rarest — a purely geochemical difference, but one more way the "first ten" behave unlike the thirty that follow them.Every one of these contrasts — size, density, melting point, reactivity, favoured oxidation states, tendency to metal–metal bond, and magnetic behaviour — traces back to the same cause: 3d orbitals are small and poorly-shielding, 4d/5d orbitals are large and diffuse, and the lanthanoid contraction squeezes the 5d row down to almost the same size as the 4d row above it. That is why the statement is correct.Answer: True — first-series (3d) transition elements differ from the heavier (4d, 5d) transition elements in atomic/ionic radii, density, enthalpy of atomization, ionization enthalpy, oxidation-state stability, tendency to form metal–metal bonds, and magnetic behaviour, all traceable to the compact, poorly-shielded 3d orbitals versus the larger, more diffuse 4d/5d orbitals (whose size is further fixed by the lanthanoid contraction).

The number of unpaired electrons sets the spin-only magnetic moment; when the observed value runs higher than that, the excess comes from unquenched orbital motion, not from extra unpaired spins.Step $\displaystyle 1$ — get the oxidation state and \(\displaystyle d\)-electron count of the metal in each complex.For \(\displaystyle K_4[Mn(CN)_6]\): four \(\displaystyle K^+\) balance a \(\displaystyle 4-\) charge on the complex ion, and each \(\displaystyle CN^-\) carries \(\displaystyle -1\): \[x + 6(-1) = -4 \implies x = +2 \] So the metal is \(\displaystyle Mn^{2+}\). Neutral Mn is \(\displaystyle [Ar]3d^54s^2\); the \(\displaystyle 4s\) electrons leave first on ionisation, so \(\displaystyle \mathrm{Mn^{2+}}\) is \(\displaystyle [Ar]3d^5\) — a \(\displaystyle d^5\) ion.For \(\displaystyle [Fe(H_2O)_6]^{2+}\): water is neutral, so the \(\displaystyle +2\) charge sits entirely on the metal: \(\displaystyle \mathrm{Fe^{2+}}\) is \(\displaystyle [Ar]3d^6\), a \(\displaystyle d^6\) ion.For \(\displaystyle K_2[MnCl_4]\): two \(\displaystyle K^+\) balance a \(\displaystyle 2-\) charge on the complex, and four \(\displaystyle Cl^-\) contribute \(\displaystyle -4\): \[x + 4(-1) = -2 \implies x = +2 \] Again \(\displaystyle \mathrm{Mn^{2+}}\), \(\displaystyle d^5\).Aside — get the oxidation state from the counter-ions and ligand charges first; reading it straight off the metal's "usual" valence is exactly where this kind of problem goes wrong.Step $\displaystyle 2$ — assign the spin state from the ligand and the geometry.\(\displaystyle CN^-\) is a strong-field ligand, so \(\displaystyle [Mn(CN)_6]^{4-}\) is low spin: the five \(\displaystyle d\)-electrons crowd into the lower \(\displaystyle t_{2g}\) set, \(\displaystyle t_{2g}^5e_g^0\), leaving \(\displaystyle n = 1\) unpaired electron.\(\displaystyle H_2O\) is a weak-field ligand, so \(\displaystyle [Fe(H_2O)_6]^{2+}\) is high spin: \(\displaystyle t_{2g}^4e_g^2\), giving \(\displaystyle n = 4\) unpaired electrons.\(\displaystyle [MnCl_4]^{2-}\) has only four ligands, so it is tetrahedral. The tetrahedral splitting \(\displaystyle \Delta_t\) is always too small to force pairing, so tetrahedral complexes are high spin regardless of the ligand: the configuration is \(\displaystyle e^2t_2^3\), and all \(\displaystyle n = 5\) electrons stay unpaired.Step $\displaystyle 3$ — compute the spin-only moment and compare with the given value.Formula (spin-only magnetic moment): \(\displaystyle \mu = \sqrt{n(n+2)}\) BM, where \(\displaystyle n\) is the number of unpaired electrons and BM (Bohr magneton) is the unit of magnetic moment.For \(\displaystyle [Mn(CN)_6]^{4-}\), \(\displaystyle n = 1\): \[\mu_{calc} = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \text{ BM} \] Given (observed): \(\displaystyle 2.2\) BM — noticeably above \(\displaystyle 1.73\) BM.For \(\displaystyle [Fe(H_2O)_6]^{2+}\), \(\displaystyle n = 4\): \[\mu_{calc} = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \text{ BM} \] Given (observed): \(\displaystyle 5.3\) BM — again above the calculated value.For \(\displaystyle [MnCl_4]^{2-}\), \(\displaystyle n = 5\): \[\mu_{calc} = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ BM} \] Given (observed): \(\displaystyle 5.9\) BM — matches the calculated value almost exactly.Step $\displaystyle 4$ — read the pattern.Aside — the spin-only formula assumes the electron's orbital motion contributes nothing to the magnetic moment, which is true only when the ground term has no orbital degeneracy. When that assumption fails, the observed moment sits above the spin-only number (never below), and a small excess does not mean extra unpaired electrons — you have to check whether it is a small bump (orbital contribution) or a jump all the way to the value for one more unpaired electron.In the two octahedral complexes, the odd or asymmetrically-placed electrons sit in the triply degenerate \(\displaystyle t_{2g}\) set, giving orbitally degenerate ground terms — \(\displaystyle {}^2T_{2g}\) for \(\displaystyle [Mn(CN)_6]^{4-}\) and \(\displaystyle {}^5T_{2g}\) for \(\displaystyle [Fe(H_2O)_6]^{2+}\). That residual orbital angular momentum adds to the spin moment, so both observed values (\(\displaystyle 2.2\) BM and \(\displaystyle 5.3\) BM) land above their spin-only predictions without signalling any extra unpaired electron.In \(\displaystyle [MnCl_4]^{2-}\), the high-spin \(\displaystyle d^5\) tetrahedral configuration \(\displaystyle e^2t_2^3\) has ground term \(\displaystyle {}^6A_1\) — an orbitally non-degenerate (A-type) term, so there is no orbital contribution to add. Its observed moment (\(\displaystyle 5.9\) BM) therefore lands right on the spin-only value (\(\displaystyle 5.92\) BM) for \(\displaystyle 5\) unpaired electrons.Answer: The values confirm $\displaystyle 1$, $\displaystyle 4$, and $\displaystyle 5$ unpaired electrons respectively — \(\displaystyle [Mn(CN)_6]^{4-}\) is low-spin \(\displaystyle d^5\) (\(\displaystyle t_{2g}^5\), calc. $\displaystyle 1.73$ BM, obs. $\displaystyle 2.2$ BM), \(\displaystyle [Fe(H_2O)_6]^{2+}\) is high-spin \(\displaystyle d^6\) (\(\displaystyle t_{2g}^4e_g^2\), calc. $\displaystyle 4.90$ BM, obs. $\displaystyle 5.3$ BM), and \(\displaystyle [MnCl_4]^{2-}\) is high-spin tetrahedral \(\displaystyle d^5\) (\(\displaystyle e^2t_2^3\), calc. $\displaystyle 5.92$ BM, obs. $\displaystyle 5.9$ BM); the first two exceed their spin-only values because their T-term ground states carry orbital contribution, while the third's \(\displaystyle {}^6A_1\) ground term has none, so its observed and calculated moments coincide.