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

When a metal atom loses electrons to become a cation, the electrons come out of the outermost shell first — the general order of removal is \(\displaystyle ns\) before \(\displaystyle (n-1)d\) before \(\displaystyle (n-2)f\), even though these are filled in the reverse order going up the periodic table. People get this backwards: they assume ions are built by "undoing" the Aufbau filling order, so they leave the \(\displaystyle s\) electrons in place and strip a \(\displaystyle d\) or \(\displaystyle f\) electron instead. The correct picture is that once a \(\displaystyle d\) or \(\displaystyle f\) subshell starts filling, it drops below the outer \(\displaystyle ns\) level in energy, so the loosely-held \(\displaystyle ns\) electrons are the first to go.Work out each neutral atom's configuration first, then remove electrons from the outside in.(i) \(\displaystyle \mathrm{Cr^{3+}}\) Chromium, \(\displaystyle Z = 24\), is itself an exception to the simple filling order: \(\displaystyle Cr = [Ar]\,3d^5\,4s^1\) (a half-filled \(\displaystyle 3d\) subshell is extra stable, so one \(\displaystyle 4s\) electron shifts into \(\displaystyle 3d\)). Removing $\displaystyle 3$ electrons takes the single \(\displaystyle 4s^1\) first, then two from \(\displaystyle 3d^5\): \[Cr^{3+} = [Ar]\,3d^3 \](ii) \(\displaystyle \mathrm{Pm^{3+}}\) Promethium, \(\displaystyle Z = 61\), is a lanthanide: \(\displaystyle Pm = [Xe]\,4f^5\,6s^2\). Removing $\displaystyle 3$ electrons takes both \(\displaystyle 6s\) electrons first, then one from \(\displaystyle 4f\): \[Pm^{3+} = [Xe]\,4f^4 \](iii) \(\displaystyle \mathrm{Cu^{+}}\) Copper, \(\displaystyle Z = 29\), is also an exception: \(\displaystyle Cu = [Ar]\,3d^{10}\,4s^1\) (a filled \(\displaystyle 3d^{10}\) is extra stable). Removing $\displaystyle 1$ electron takes the \(\displaystyle 4s^1\): \[Cu^{+} = [Ar]\,3d^{10} \](iv) \(\displaystyle \mathrm{Ce^{4+}}\) Cerium, \(\displaystyle Z = 58\), is another exception: \(\displaystyle Ce = [Xe]\,4f^1\,5d^1\,6s^2\). Removing $\displaystyle 4$ electrons empties \(\displaystyle 6s^2\), \(\displaystyle 5d^1\), and \(\displaystyle 4f^1\) in turn — that is all four outer electrons, leaving the closed-shell xenon core: \[Ce^{4+} = [Xe]\;\;(\text{i.e. } 4f^0) \](v) \(\displaystyle \mathrm{Co^{2+}}\) Cobalt, \(\displaystyle Z = 27\): \(\displaystyle Co = [Ar]\,3d^7\,4s^2\). Removing $\displaystyle 2$ electrons takes the \(\displaystyle 4s^2\): \[Co^{2+} = [Ar]\,3d^7 \](vi) \(\displaystyle \mathrm{Lu^{2+}}\) Lutetium, \(\displaystyle Z = 71\), closes the lanthanide series with the \(\displaystyle 4f\) shell already full: \(\displaystyle Lu = [Xe]\,4f^{14}\,5d^1\,6s^2\). Removing $\displaystyle 2$ electrons takes the \(\displaystyle 6s^2\) first — the \(\displaystyle 5d^1\) and the full \(\displaystyle 4f^{14}\) are left untouched: \[Lu^{2+} = [Xe]\,4f^{14}\,5d^1 \](vii) \(\displaystyle \mathrm{Mn^{2+}}\) Manganese, \(\displaystyle Z = 25\): \(\displaystyle Mn = [Ar]\,3d^5\,4s^2\). Removing $\displaystyle 2$ electrons takes the \(\displaystyle 4s^2\), leaving the stable half-filled \(\displaystyle 3d^5\): \[Mn^{2+} = [Ar]\,3d^5 \](viii) \(\displaystyle \mathrm{Th^{4+}}\) Thorium, \(\displaystyle Z = 90\), is an actinide exception with no \(\displaystyle 5f\) electrons in the neutral atom: \(\displaystyle Th = [Rn]\,6d^2\,7s^2\). Removing all $\displaystyle 4$ outer electrons (\(\displaystyle 7s^2\) then \(\displaystyle 6d^2\)) leaves the closed-shell radon core: \[Th^{4+} = [Rn]\;\;(\text{i.e. } 5f^0\,6d^0\,7s^0) \]Answer: (i) \(\displaystyle Cr^{3+} = [Ar]\,3d^3\) (ii) \(\displaystyle Pm^{3+} = [Xe]\,4f^4\) (iii) \(\displaystyle Cu^{+} = [Ar]\,3d^{10}\) (iv) \(\displaystyle Ce^{4+} = [Xe]\) (v) \(\displaystyle Co^{2+} = [Ar]\,3d^7\) (vi) \(\displaystyle Lu^{2+} = [Xe]\,4f^{14}\,5d^1\) (vii) \(\displaystyle Mn^{2+} = [Ar]\,3d^5\) (viii) \(\displaystyle Th^{4+} = [Rn]\)

A half-filled d-subshell carries extra stability from exchange energy, and the two oxidations move in opposite directions relative to that stable configuration.Write out the electron configurations built from each atom's ground state.Manganese, \(\displaystyle Z = 25\): \(\displaystyle \mathrm{Mn} = [Ar]\,3d^{5}4s^{2}\)\[\mathrm{Mn^{2+}} = [Ar]\,3d^{5} \qquad \xrightarrow{\text{oxidation, } -e^-} \qquad \mathrm{Mn^{3+}} = [Ar]\,3d^{4} \]Iron, \(\displaystyle Z = 26\): \(\displaystyle \mathrm{Fe} = [Ar]\,3d^{6}4s^{2}\)\[\mathrm{Fe^{2+}} = [Ar]\,3d^{6} \qquad \xrightarrow{\text{oxidation, } -e^-} \qquad \mathrm{Fe^{3+}} = [Ar]\,3d^{5} \]The step that decides everything here is which side of the \(\displaystyle d^5\) configuration each ion sits on. A half-filled subshell (\(\displaystyle d^5\), one electron in each of the five d orbitals, all spins parallel) is unusually stable. This is not about orbital energy in the ordinary sense — every one of those five orbitals is singly occupied, so electrons of like spin can swap places among degenerate orbitals without violating the Pauli exclusion principle. Each such swap lowers the energy of the ion by a fixed increment called the exchange energy, and \(\displaystyle d^5\) (with parallel spins) has the maximum possible number of these exchanging pairs for a single d-subshell. A \(\displaystyle d^5\) ion is also spherically symmetric in its charge distribution, which adds a further small stabilization. Both effects disappear once the configuration moves away from \(\displaystyle d^5\).Now compare the two oxidations directly: This is the point people skip: the comparison isn't "which ion has a lower charge is more stable" — it's which product configuration the extra exchange energy rewards. Mn is penalized for leaving \(\displaystyle d^5\); Fe is rewarded for reaching it. That is exactly the pattern seen in the measured third ionisation enthalpies across the first transition series: Mn's third ionisation enthalpy is anomalously high compared to its neighbours Cr and Fe, because it corresponds to breaking a stable \(\displaystyle d^5\) core, whereas Fe's third ionisation enthalpy dips below what the general periodic trend would predict, because it corresponds to forming one.Answer: Mn²⁺ (3d⁵) already has the extra-stable, half-filled d-subshell, so oxidizing it to Mn³⁺ (3d⁴) destroys that stability and is energetically unfavourable — Mn²⁺ resists oxidation. Fe²⁺ (3d⁶) does not have this stability, and oxidizing it to Fe³⁺ (3d⁵) creates the stable half-filled configuration, releasing extra exchange energy that favours the oxidation — so Fe²⁺ is readily oxidized while Mn²⁺ is not.

The +$\displaystyle 2$ state gets harder and harder to push further to +$\displaystyle 3$, because the electron being removed to make M³⁺ is coming out of a 3d subshell that grows steadily more resistant to losing it.Across Sc, Ti, V, Cr, Mn (\(\displaystyle Z = 21\) to \(\displaystyle 25\)), the metal atom loses its two \(\displaystyle 4s\) electrons first, so the M²⁺ ion in every case has a pure \(\displaystyle 3d^{n}\) configuration:\(\displaystyle \text{Sc}^{2+}:[\text{Ar}]3d^{1}\qquad \text{Ti}^{2+}:[\text{Ar}]3d^{2}\qquad \text{V}^{2+}:[\text{Ar}]3d^{3}\qquad \text{Cr}^{2+}:[\text{Ar}]3d^{4}\qquad \text{Mn}^{2+}:[\text{Ar}]3d^{5}\)"How stable is M²⁺" is not asking how easily the atom formed that ion (that would be \(\displaystyle \Delta_iH_1+\Delta_iH_2\)). It is asking how hard it is to strip out one more electron and go to M³⁺ — that is the third ionisation enthalpy, \(\displaystyle \Delta_iH_3\). People often reach for the wrong ionisation step here; the +$\displaystyle 2$/+$\displaystyle 3$ question always turns on \(\displaystyle \Delta_iH_3\), the energy of the step \(\displaystyle \text{M}^{2+}\rightarrow \text{M}^{3+}+e^-\).The measured \(\displaystyle \Delta_iH_3\) values (in kJ mol⁻¹) for this stretch of the series are:\[\text{Sc}=2389,\quad \text{Ti}=2652,\quad \text{V}=2828,\quad \text{Cr}=2987,\quad \text{Mn}=3248 \]These rise almost monotonically from Sc to Mn. Two effects combine to produce that climb: Because \(\displaystyle \Delta_iH_3\) keeps rising from Sc to Mn, converting M²⁺ to M³⁺ costs more and more energy at each step — so the +$\displaystyle 3$ state becomes progressively harder to reach, and correspondingly the +$\displaystyle 2$ state becomes progressively more stable, as atomic number increases across the first half of the series. This is exactly why \(\displaystyle \text{Sc}^{2+}\) is essentially unknown (its lone, unstabilised \(\displaystyle 3d^{1}\) electron is lost with very little energy, so Sc goes straight to the far more stable \(\displaystyle \text{Sc}^{3+}\), \(\displaystyle [\text{Ar}]\)), while \(\displaystyle \text{Mn}^{2+}\) (\(\displaystyle 3d^5\)) is the most stable ion Mn forms, and \(\displaystyle \text{Mn}^{3+}\) behaves as a fairly strong oxidising agent because it "wants" to fall back to the exchange-stabilised \(\displaystyle 3d^{5}\) state.Answer: Going from Sc to Mn, the 3d electron removed to form M³⁺ becomes progressively harder to pull out — the third ionisation enthalpy climbs steadily ($\displaystyle 2389$ → $\displaystyle 2652$ → $\displaystyle 2828$ → $\displaystyle 2987$ → $\displaystyle 3248$ kJ mol⁻¹), peaking at Mn²⁺ because its half-filled \(\displaystyle 3d^5\) configuration has extra exchange-energy stability. So the +$\displaystyle 2$ oxidation state becomes more and more stable relative to +$\displaystyle 3$ as atomic number increases across the first half of the first transition series.

In the first transition series, the stability of an oxidation state is decided mainly by whether losing (or keeping) electrons leaves the metal ion with a \(\displaystyle 3d^0\), \(\displaystyle 3d^5\), or \(\displaystyle 3d^{10}\) configuration — an empty, an exactly half‑filled, or a completely filled d‑subshell is always unusually stable, because a half‑filled shell keeps every electron spin‑parallel (maximum exchange energy) and a filled or empty shell is spherically symmetric.A 3d-metal atom has the configuration \(\displaystyle (n-1)d^{x}\,ns^{2}\) (with the well‑known exceptions Cr: \(\displaystyle 3d^{5}4s^{1}\) and Cu: \(\displaystyle 3d^{10}4s^{1}\)). To form a cation \(\displaystyle M^{n+}\), the \(\displaystyle 4s\) electrons are always removed first, then \(\displaystyle 3d\) electrons. So the oxidation state you end up with fixes the resulting \(\displaystyle d\)-count, and that \(\displaystyle d\)-count is what decides how stable the state is.1. Highest oxidation states (Sc → Mn): stable because they reach \(\displaystyle d^0\). For the early members, all the valence electrons can be stripped away (or used in covalent bonds to O/F) to leave an empty d-subshell: \[\mathrm{Sc}\;(3d^{1}4s^{2}) \to \mathrm{Sc}^{3+}\;(3d^{0}),\quad \mathrm{Ti}\;(3d^{2}4s^{2}) \to \mathrm{Ti}^{4+}\;(3d^{0}), \] \[\mathrm{V}^{5+}\text{ in }\mathrm{VO_2^{+}}\;(3d^{0}),\quad \mathrm{Cr}^{6+}\text{ in }\mathrm{CrO_4^{2-}}\;(3d^{0}),\quad \mathrm{Mn}^{7+}\text{ in }\mathrm{MnO_4^{-}}\;(3d^{0}). \] Because the group oxidation state (equal to the total number of \(\displaystyle 3d+4s\) electrons) always lands on \(\displaystyle d^0\), it is stable, and this is exactly why the maximum oxidation state rises by one at each element from Sc(+$\displaystyle 3$) to Mn(+$\displaystyle 7$).The step people get wrong here: it looks like this trend should keep climbing — Fe should give +$\displaystyle 8$, Co +$\displaystyle 9$, and so on. It doesn't. Past Mn, the effective nuclear charge felt by the 3d electrons rises sharply (3d electrons shield each other poorly), so pulling out enough electrons to reach \(\displaystyle d^0\) again becomes too costly in ionisation energy. That is why Fe, Co, Ni essentially never reach oxidation states anywhere near "electron count," and their chemistry is dominated instead by +$\displaystyle 2$/+3.2. Mid-series oxidation states: stabilised by \(\displaystyle d^5\) or \(\displaystyle d^{10}\), even against the "should climb higher" expectation. 3. \(\displaystyle d^{10}\) stability, and where the rule breaks. So configuration decides stability to a large extent — it correctly explains the rising maximum oxidation states up to Mn, and the special stability of \(\displaystyle \mathrm{Mn}^{2+}\), \(\displaystyle \mathrm{Fe}^{3+}\), \(\displaystyle \mathrm{Cr}^{3+}\), and \(\displaystyle \mathrm{Zn}^{2+}\) — but not completely, since ionisation‑energy trends past Mn and hydration/lattice energies (as in the \(\displaystyle \mathrm{Cu}^{+}/\mathrm{Cu}^{2+}\) case) can tip the balance the other way.Answer: Electronic configuration governs oxidation-state stability to a large extent — \(\displaystyle d^0\) (Sc³⁺, Ti⁴⁺, VO₂⁺, CrO₄²⁻, MnO₄⁻), half-filled \(\displaystyle d^5\) (Mn²⁺, Fe³⁺), and filled \(\displaystyle d^{10}\) (Zn²⁺) configurations are always the most stable — but it is not the sole factor: rising effective nuclear charge caps the highest oxidation states after Mn, and hydration/lattice energies can override the configurational preference, as seen when aqueous \(\displaystyle \mathrm{Cu}^{2+}\) (\(\displaystyle d^9\)) proves more stable than \(\displaystyle \mathrm{Cu}^{+}\) (\(\displaystyle d^{10}\)).

A transition metal's stable oxidation states come from how many electrons it must lose to reach a \(\displaystyle d^0 \), a half-filled \(\displaystyle d^5 \), or a full \(\displaystyle d^{10} \) shell — not simply from "which group it's in." To use that idea, first pin down which first-row transition atom actually carries each d-electron count in its ground state, remembering that chromium and copper break the simple filling order to grab the extra stability of a half-filled or fully-filled 3d subshell.The ground-state configurations of the first transition series (Sc to Zn) are\[\text{Sc: }3d^14s^2,\quad \text{Ti: }3d^24s^2,\quad \text{V: }3d^34s^2,\quad \text{Cr: }3d^54s^1,\quad \text{Mn: }3d^54s^2, \] \[\text{Fe: }3d^64s^2,\quad \text{Co: }3d^74s^2,\quad \text{Ni: }3d^84s^2,\quad \text{Cu: }3d^{10}4s^1,\quad \text{Zn: }3d^{10}4s^2 . \]\(\displaystyle 3d^3 \): This is vanadium, \(\displaystyle \text{V}=[\text{Ar}]3d^34s^2 \). Losing all five valence electrons \(\displaystyle (3d^34s^2) \) empties the d subshell completely, giving \(\displaystyle \text{V}^{5+}\,(3d^0) \). Vanadium sits early enough in the series that its successive ionization energies rise gently, so this full loss is easy to achieve — \(\displaystyle +5 \) is vanadium's most stable, most common oxidation state, seen in \(\displaystyle \text{V}_2\text{O}_5 \), vanadates \(\displaystyle (\text{VO}_3^-,\ \text{VO}_4^{3-}) \), and \(\displaystyle \text{VOCl}_3 \). States \(\displaystyle +2, +3, +4 \) also exist but are progressively easier to oxidise up to \(\displaystyle +5 \).\(\displaystyle 3d^5 \): Two elements share this d-count, because it is exactly what makes each of them anomalous. \(\displaystyle \text{Cr}=[\text{Ar}]3d^54s^1 \) (one 4s electron has jumped into the d subshell purely to make it half-filled), and \(\displaystyle \text{Mn}=[\text{Ar}]3d^54s^2 \). Half-filled and fully-filled d subshells are unusually stable — that is the entire reason Cr and Cu skip the "obvious" configuration, and it is exactly why \(\displaystyle \text{Mn}^{2+}\,(3d^5) \) is so hard to disturb.\(\displaystyle 3d^8 \): This is nickel, \(\displaystyle \text{Ni}=[\text{Ar}]3d^84s^2 \). Losing just the two 4s electrons gives \(\displaystyle \text{Ni}^{2+}\,(3d^8) \), and that is overwhelmingly nickel's stable oxidation state — almost all nickel chemistry is \(\displaystyle \text{Ni(II)} \). Reaching \(\displaystyle \text{Ni}^{4+}\,(3d^6) \) would mean pulling electrons out of the d subshell itself, which costs far more ionization energy than nickel compounds can recover, so higher states are rare and strongly oxidising whenever they do appear.\(\displaystyle 3d^4 \): No neutral atom of the first transition series actually has this ground-state configuration. Simple filling would predict chromium to be \(\displaystyle 3d^44s^2 \), but chromium instead promotes one 4s electron into the d subshell to reach the more stable half-filled \(\displaystyle 3d^54s^1 \) — the same exchange stabilisation invoked above. So \(\displaystyle 3d^4 \) shows up only as an ion, and precisely because it sits one electron away from a more stable arrangement on either side, it is an unstable oxidation state: \(\displaystyle \text{Cr}^{2+}\,(3d^4) \) readily loses one more electron to become \(\displaystyle \text{Cr}^{3+}\,(3d^3) \), making \(\displaystyle \text{Cr}^{2+} \) a strong reducing agent, and \(\displaystyle \text{Mn}^{3+}\,(3d^4) \) is likewise unstable, disproportionating into \(\displaystyle \text{Mn}^{2+}\,(3d^5,\text{ half-filled}) \) and \(\displaystyle \text{Mn}^{4+} \) (as \(\displaystyle \text{MnO}_2 \)).Answer: \(\displaystyle 3d^3\to \text{V} \), most stable state \(\displaystyle +5\,(3d^0) \); \(\displaystyle 3d^5\to \text{Cr} \) (most stable \(\displaystyle +3,\,3d^3 \); group state \(\displaystyle +6,\,3d^0 \)) and \(\displaystyle \text{Mn} \) (most stable \(\displaystyle +2,\,3d^5 \); group state \(\displaystyle +7,\,3d^0 \)); \(\displaystyle 3d^8\to \text{Ni} \), most stable state \(\displaystyle +2\,(3d^8) \); \(\displaystyle 3d^4 \) does not occur in any neutral ground-state atom of the series, and as an oxidation state (\(\displaystyle \text{Cr}^{2+} \) or \(\displaystyle \text{Mn}^{3+} \)) it is unstable, converting toward the more stable \(\displaystyle 3d^3 \) or \(\displaystyle 3d^5 \) states.

The oxidation state you can pull out of a first-row transition metal by putting it in an oxoanion only climbs as high as the group number up to manganese — after that, holding onto every valence electron costs more energy than the metal is willing to pay.The first transition series runs \(\displaystyle \text{Sc} \to \text{Zn} \), and the group numbers (old IUPAC style, counting all electrons outside the last noble-gas core) are\[\text{Sc}(3),\ \text{Ti}(4),\ \text{V}(5),\ \text{Cr}(6),\ \text{Mn}(7),\ \text{Fe}(8),\ \text{Co}(9),\ \text{Ni}(10),\ \text{Cu}(11),\ \text{Zn}(12) \]To find where the metal's oxidation state in an oxoanion actually matches its group number, use the rule that a neutral oxide-oxygen contributes \(\displaystyle -2\), and the sum of oxidation states equals the ion's charge.Vanadium, group 5. In the vanadate ion \(\displaystyle \text{VO}_4^{3-} \) (or the metavanadate \(\displaystyle \text{VO}_3^{-} \)), let \(\displaystyle x\) be V's oxidation number: \[x + 4(-2) = -3 \implies x = +5 \] That is +$\displaystyle 5$, matching V's group number, 5.Chromium, group 6. In the chromate ion \(\displaystyle \text{CrO}_4^{2-} \): \[x + 4(-2) = -2 \implies x = +6 \] +$\displaystyle 6$ matches Cr's group number, 6. (The dichromate \(\displaystyle \text{Cr}_2\text{O}_7^{2-} \) gives the same +$\displaystyle 6$ for Cr.)Manganese, group 7. In the permanganate ion \(\displaystyle \text{MnO}_4^{-} \): \[x + 4(-2) = -1 \implies x = +7 \] +$\displaystyle 7$ matches Mn's group number, 7. This is the highest oxidation state any element in the whole first transition series ever shows.Why the pattern stops at manganese. The one place this trick trips people up is expecting it to continue — an \(\displaystyle \text{FeO}_4^{4-}\)-type anion with Fe in +$\displaystyle 8$ does not exist. As you move past Mn (group $\displaystyle 7$) to Fe, Co, Ni (groups $\displaystyle 8$–$\displaystyle 10$), the nuclear charge keeps rising while d-electrons shield each other poorly, so each successive d-electron is held more tightly. Stripping out enough electrons to reach an oxidation state equal to the group number would cost far more energy than is recovered by bond formation, so these later metals settle instead for lower, more stable states (mainly +$\displaystyle 2$ and +$\displaystyle 3$) and never form oxoanions with the metal at +$\displaystyle 8$, +$\displaystyle 9$, or +10.So the oxometal anions where the metal's oxidation state equals its group number are: Answer: Vanadate \(\displaystyle \text{VO}_4^{3-} \) (V, +$\displaystyle 5$), chromate \(\displaystyle \text{CrO}_4^{2-} \) (Cr, +$\displaystyle 6$), and permanganate \(\displaystyle \text{MnO}_4^{-} \) (Mn, +$\displaystyle 7$) — the metal's oxidation state matches its group number in each.

Lanthanoid contraction is the steady, cumulative shrinkage in the size of the Ln³⁺ ions (and atoms) as you move from La to Lu, caused by poor shielding by the 4f electrons.Why it happensAcross the lanthanoid series (Z = $\displaystyle 57$ to $\displaystyle 71$), each successive element adds one proton to the nucleus and one electron to the inner 4f subshell, while the outer \(\displaystyle 5s^2 5p^6 6s^2\) configuration stays essentially unchanged. This matters because of where the added electron goes.The 4f orbitals are not simple, single-lobed shapes tucked neatly inside the atom — they have a complex nodal structure, but crucially they shield each other very poorly. One 4f electron screens another 4f electron from the nuclear charge far less effectively than, say, one d electron screens another d electron, or one s electron screens another s electron. This is called imperfect (poor) shielding of 4f electrons.Because the shielding is poor, the increase of one unit of nuclear charge at each step is not fully cancelled by the extra 4f electron. So the effective nuclear charge felt by the outer \(\displaystyle 6s\) electrons keeps rising steadily through the series. A rising effective nuclear charge pulls the outer electron cloud inward more tightly at every step, so the atomic and ionic radii fall steadily and cumulatively from La to Lu — the same imperfect-shielding logic that shrinks atoms across any period, just repeated fourteen times over because the 4f orbitals are unusually bad shields.Aside: this is a contraction, not a single big drop — the important word is "cumulative." Each individual step is small (the radii change by only a few picometers per element), but adding these small decreases over fourteen elements produces a substantial overall fall from La³⁺ (~$\displaystyle 106$ pm) to Lu³⁺ (~$\displaystyle 85$ pm).Consequences of lanthanoid contractionAnswer: Lanthanoid contraction is the steady, cumulative decrease in atomic/ionic radii across La to Lu, caused by poor (imperfect) shielding of the nucleus by 4f electrons, which lets the effective nuclear charge rise steadily through the series. Its consequences are: (i) basic strength of Ln(OH)₃ decreases from La to Lu, (ii) the lanthanoids are so similar to each other that they are difficult to separate, and (iii) it produces the near-identical atomic/ionic radii — and hence very similar properties — of corresponding 4d and 5d transition elements, e.g. Zr–Hf and Nb–Ta.

A transition element is defined by an incompletely filled d subshell — in the atom or in any of its common ions. That single definition is what produces every "characteristic" the exam wants you to list, so learn the definition first and the properties fall out of it.Characteristics of transition elementsWhy they are called "transition" elementsThey sit, in every period, between the strongly electropositive s-block metals (Groups $\displaystyle 1$–$\displaystyle 2$) on one side and the electronegative p-block elements on the other. Their properties are "transitional" between these two blocks — for instance, their oxides change smoothly from basic (early in the series, e.g. \(\displaystyle \mathrm{Sc_2O_3} \)) to amphoteric to acidic (late in the series, e.g. \(\displaystyle \mathrm{Mn_2O_7} \)) — so they represent the changeover, or transition, in chemical behaviour from one block to the other. Equivalently, in terms of electronic configuration, the \(\displaystyle (n-1)d \) orbitals are in the process of being filled — the elements represent a transition from the filling of the \(\displaystyle ns \) orbital to the filling of the \(\displaystyle np \) orbital.Which d-block elements are excludedGroup $\displaystyle 12$ elements — zinc, cadmium, and mercury ( \(\displaystyle \mathrm{Zn, Cd, Hg} \) ) — have the electronic configuration \(\displaystyle (n-1)d^{10}\,ns^2 \), and in their common oxidation state ( \(\displaystyle +2\) ) they lose only the \(\displaystyle ns \) electrons, leaving a completely filled \(\displaystyle (n-1)d^{10} \) shell in both the free atom and the ion. Since the defining criterion is an incompletely filled d subshell in the atom or its ions, and \(\displaystyle \mathrm{Zn, Cd, Hg} \) never satisfy this (their d subshell is always full, never partially filled), they are not regarded as transition elements, even though they belong to the d-block.Answer: Transition elements are those with a partly filled \(\displaystyle d \) subshell in the atom or common ion; this partial filling gives them metallic character, high melting points, variable oxidation states, coloured ions, catalytic activity, complex-ion formation, paramagnetism, and alloy/interstitial-compound formation. They are called "transition" elements because their properties bridge the electropositive s-block and the electronegative p-block. \(\displaystyle \mathrm{Zn, Cd,} \) and \(\displaystyle \mathrm{Hg} \) (Group $\displaystyle 12$), having a fully filled \(\displaystyle d^{10} \) configuration in both atom and ion, are not regarded as transition elements.

The electron that goes in last decides everything: for transition elements it fills a \(\displaystyle (n-1)d\) orbital; for non-transition (representative) elements it fills an \(\displaystyle ns\) or \(\displaystyle np\) orbital, never a \(\displaystyle d\) orbital.General configurationsNon-transition elements (s-block and p-block) build up their electrons only in the outermost \(\displaystyle ns\) and \(\displaystyle np\) subshells. Any inner \(\displaystyle d\) subshell present is either completely empty or completely full — it never sits half-built. \[\text{Non-transition elements: } [\text{Noble gas}]\, ns^{1-2}\ \text{or}\ [\text{Noble gas}]\, ns^{2}np^{1-6} \]Transition elements are defined by the opposite feature: the differentiating electron enters the penultimate \(\displaystyle (n-1)d\) subshell while the outer \(\displaystyle ns\) subshell stays at \(\displaystyle 0\), \(\displaystyle 1\), or \(\displaystyle 2\) electrons. \[\text{Transition elements: } [\text{Noble gas}]\, (n-1)d^{1-10}\, ns^{0-2} \]Why this matters — the "partially filled d" testA transition element is defined (IUPAC) as one whose atom, or one of whose common ions, has an incompletely filled \(\displaystyle d\) subshell. That partial filling is exactly what non-transition elements never show. The aside people miss: having a \(\displaystyle d\) subshell in the configuration is not the same as being a transition element — the \(\displaystyle d\) subshell must be partially filled. Zinc, \(\displaystyle [\text{Ar}]3d^{10}4s^{2}\), has a \(\displaystyle d\) subshell, but it is completely full (\(\displaystyle 10\) of \(\displaystyle 10\)) in the atom and in its only oxidation state \(\displaystyle \text{Zn}^{2+}\) (\(\displaystyle 3d^{10}\)), so zinc is classed with the non-transition elements, not the transition series, even though it sits in the d-block of the periodic table.**Answer: Non-transition elements have their differentiating (last-entering) electron in an outer \(\displaystyle ns\) or \(\displaystyle np\) orbital, with any inner \(\displaystyle d\) subshell either completely empty or completely full — e.g., Na \(\displaystyle [\text{Ne}]3s^{1}\), Cl \(\displaystyle [\text{Ne}]3s^{2}3p^{5}\). Transition elements have their differentiating electron entering the penultimate \(\displaystyle (n-1)d\) orbital, which is partially filled (\(\displaystyle d^{1}\) to \(\displaystyle d^{9}\)) in the atom or in at least one common ion — e.g., Sc \(\displaystyle [\text{Ar}]3d^{1}4s^{2}\), Fe \(\displaystyle [\text{Ar}]3d^{6}4s^{2}\); an element like Zn, \(\displaystyle [\text{Ar}]3d^{10}4s^{2}\), with a completely filled \(\displaystyle d^{10}\) subshell in all its states, counts as non-transition despite lying in the d-block.

The common oxidation state is +$\displaystyle 3$ for every lanthanoid, but a few metals also show +$\displaystyle 2$ or +$\displaystyle 4$ when doing so lets them reach an empty, half-filled, or completely filled 4f subshell.A lanthanoid atom has the ground-state configuration \[[\text{Xe}]\,4f^{n}\,5d^{0-1}\,6s^{2} \] Removing the two \(\displaystyle 6s\) electrons and one more electron (either the single \(\displaystyle 5d\) electron, or — where there is no \(\displaystyle 5d\) electron — one \(\displaystyle 4f\) electron) leaves the \(\displaystyle \text{Ln}^{3+}\) ion. This is why +$\displaystyle 3$ is the characteristic and by far the most stable oxidation state across the whole series, from La(III) to Lu(III), and it is the state in which lanthanoids are found in nearly all their compounds and in solution.A smaller number of lanthanoids also show +$\displaystyle 2$ or +4. These are not random exceptions — they occur exactly where losing a different number of electrons leaves the ion with one of the three specially stable 4f arrangements: +$\displaystyle 4$ oxidation state (losing one electron more than the usual three): Cerium, \(\displaystyle \text{Ce}\;([\text{Xe}]4f^{1}5d^{1}6s^{2})\), loses all four outer electrons to give \(\displaystyle \text{Ce}^{4+}\) with configuration \(\displaystyle 4f^{0}\) — an empty, xenon-like shell, so \(\displaystyle \text{Ce}^{4+}\) is markedly stable and is a good oxidising agent used in quantitative analysis. Terbium, \(\displaystyle \text{Tb}\), similarly gives \(\displaystyle \text{Tb}^{4+}\) with configuration \(\displaystyle 4f^{7}\), a half-filled shell, so it too is more accessible than +$\displaystyle 4$ states for other members of the series.+$\displaystyle 2$ oxidation state (losing one electron fewer than the usual three, i.e., keeping one more electron in the 4f subshell): Europium, \(\displaystyle \text{Eu}\;([\text{Xe}]4f^{7}6s^{2})\), gives \(\displaystyle \text{Eu}^{2+}\) with configuration \(\displaystyle 4f^{7}\), a half-filled subshell, making \(\displaystyle \text{Eu}^{2+}\) almost as stable as \(\displaystyle \text{Eu}^{3+}\). Ytterbium, \(\displaystyle \text{Yb}\;([\text{Xe}]4f^{14}6s^{2})\), gives \(\displaystyle \text{Yb}^{2+}\) with configuration \(\displaystyle 4f^{14}\), a completely filled subshell, so \(\displaystyle \text{Yb}^{2+}\) is also well known. Samarium can likewise give \(\displaystyle \text{Sm}^{2+}\) \(\displaystyle (4f^{6})\), though this is a weaker, less-stable case since \(\displaystyle 4f^{6}\) is not one of the three specially stable arrangements — it is formed only under strongly reducing conditions.The aside worth remembering here: the +$\displaystyle 3$ state is fixed by the general pattern of removing the two \(\displaystyle 6s\) electrons plus one more, while the +$\displaystyle 2$ and +$\displaystyle 4$ states are the exceptions, and they appear only for those specific elements whose resulting ion happens to land on \(\displaystyle 4f^{0}\), \(\displaystyle 4f^{7}\), or \(\displaystyle 4f^{14}\) — it is extra subshell stability, not any general tendency of lanthanoids, that drives them.Answer: All lanthanoids show a characteristic +$\displaystyle 3$ oxidation state (from losing \(\displaystyle 6s^2\) plus one \(\displaystyle 5d\)/\(\displaystyle 4f\) electron). In addition, some members show +$\displaystyle 2$ or +$\displaystyle 4$ when it gives a more stable \(\displaystyle 4f^{0}\), \(\displaystyle 4f^{7}\), or \(\displaystyle 4f^{14}\) configuration — notably \(\displaystyle \text{Ce}^{4+}\) \(\displaystyle (4f^{0})\) and \(\displaystyle \text{Tb}^{4+}\) \(\displaystyle (4f^{7})\) for +$\displaystyle 4$, and \(\displaystyle \text{Eu}^{2+}\) \(\displaystyle (4f^{7})\), \(\displaystyle \text{Yb}^{2+}\) \(\displaystyle (4f^{14})\), and \(\displaystyle \text{Sm}^{2+}\) \(\displaystyle (4f^{6})\) for +2.