Concept and Intuition

Lanthanides normally exist as $Ln^{3+}$. A few elements also show +2 or +4 states when that unusual state happens to correspond to an especially stable $f^0$, $f^7$ (half-filled) or $f^{14}$ (fully-filled) configuration. But "isolable" is not the same as "thermodynamically preferred" — the normal +3 state is still the most stable overall for the element, so:

• An unusual +4 ion (higher than normal +3) tends to gain an electron and fall back to +3 — it therefore acts as an oxidising agent (itself gets reduced). Examples: $Ce^{4+}$ ($4f^0\to 4f^1$), $Tb^{4+}$ ($4f^7\to 4f^8$).
• An unusual +2 ion (lower than normal +3) tends to lose an electron and rise back to +3 — it therefore acts as a reducing agent (itself gets oxidised). Examples: $Eu^{2+}$ ($4f^7\to 4f^6$), $Yb^{2+}$ ($4f^{14}\to 4f^{13}$), and (more weakly) $Sm^{2+}$.

Step-by-Step Solution

1. Identify each ion's usual/unusual character: $Ce^{4+}$ (unusual +4, oxidiser), $Tb^{4+}$ (unusual +4, oxidiser), $Eu^{2+}$ (unusual +2, reducer), $Yb^{2+}$ (unusual +2, reducer).
2. Option (A) $Ce^{4+}, Yb^{2+}$ — mixes an oxidiser with a reducer, not a matching pair.
3. Option (B) $Ce^{4+}, Tb^{4+}$ — both are oxidising agents, not reducing.
4. Option (C) $Ce^{3+}, Tb^{2+}$ — $Ce^{3+}$ is just the normal, unremarkable state, and $Tb^{2+}$ is not a standard/stable species discussed for this behaviour.
5. Option (D) $Eu^{2+}, Yb^{2+}$ — both are unusual +2 ions stabilised by $f^7$/$f^{14}$ that readily give up an electron to reach the normal $Ln^{3+}$ state, i.e. both are genuine reducing agents. This is the correct pair.

Common Mistakes

• Confusing which direction (oxidation vs reduction) the special $f^0/f^7/f^{14}$ stability drives the ion — remember it is the product $Ln^{3+}$ state that is thermodynamically normal, so unusual +4 ions get reduced (oxidisers) and unusual +2 ions get oxidised (reducers).
• Mixing up $Ce$ (which shows +4) with $Eu/Yb/Sm$ (which show +2).

ANSWER: D

Concept and Intuition

Lanthanides are overwhelmingly found in the $+3$ oxidation state. Ions that deviate from $+3$ tend to revert to it if doing so gives a specially stable electron configuration (empty, half-filled, or fully-filled f subshell). $Eu^{2+}$ ($4f^7$, half-filled — already quite stable) still readily loses an electron to $Eu^{3+}$ ($4f^6$) — actually, more precisely, the drive is that $+2$ ions with configurations one electron short of a stable count are pushed to lose an electron toward +3, or +4 ions are pulled to gain one toward +3. The exam-relevant memorised fact: $Eu^{2+}$ and $Sm^{2+}$ (and here, $Yb^{2+}$, which is $4f^{14}$, fully filled) act as reducing agents by oxidising to $+3$, while $Ce^{4+}$ and $Tb^{4+}$ act as oxidising agents by being reduced to the stable $+3$ state ($Ce^{3+}$ is $4f^0$; $Tb^{3+}$ is $4f^8$/near half-filled +1 pattern that's more stable than $Tb^{4+}$).

Step-by-Step Solution

1. Recall the standard exceptions to $+3$ among lanthanides: $Ce^{4+}, Pr^{4+}, Tb^{4+}$ (oxidising agents, reduce to $+3$) and $Eu^{2+}, Sm^{2+}, Yb^{2+}$ (reducing agents, oxidise to $+3$).
2. Option (A) $Ce^{4+}, Tb^{4+}$ — both oxidising agents, not reducing.
3. Option (B) $Eu^{2+}, Yb^{2+}$ — both are the classic reducing-agent pair among lanthanide ions.
4. Options (C) and (D) list ions already in the common, stable $+3$ state ($Gd^{3+}$, $Lu^{3+}$, $La^{3+}$, $Pm^{3+}$), which show no special redox activity.
5. Hence (B) is correct.

Common Mistakes

• Mixing up which unusual oxidation states are oxidising agents (the $+4$ ions) versus reducing agents (the $+2$ ions).
• Assuming $Gd^{3+}$ or $La^{3+}$ (already stable $+3$) have any special reducing/oxidising character.

ANSWER: B

Concept and Intuition

Lanthanides are overwhelmingly +3 because that oxidation state matches a filled 6s/5d loss while leaving a reasonably stable 4f configuration. A few members show +4 (or +2) only when losing (or gaining) one more electron gets them to, or close to, an empty ($f^0$), half-filled ($f^7$), or fully-filled ($f^{14}$) 4f sub-shell — extra stability that offsets the higher ionisation energy.

Step-by-Step Solution

1. Ce (4f¹5d¹6s²): losing one more electron beyond +3 gives Ce⁴⁺ with $f^0$ — very stable, +4 is in fact its best-known state (e.g., $\mathrm{CeO_2}$).
2. Pr and Nd: Pr⁴⁺ ($f^1$) and Nd⁴⁺ ($f^2$) are known, though less stable than Ce⁴⁺ (e.g., $\mathrm{PrO_2}$, and Nd⁴⁺ in a few solid oxides).
3. Pm: radioactive, poorly characterised; +4 not established.
4. Sm, Eu: these instead favour +2 ($f^6$ close to half-filled for Sm²⁺, and $f^7$ half-filled exactly for Eu²⁺) — the opposite direction, not +4.
5. Gd: Gd³⁺ is already $f^7$ (half-filled, maximally stable); removing a further electron destroys this stability, so Gd⁴⁺ is not favoured.
6. Tb: Tb³⁺ is $f^8$; going to Tb⁴⁺ gives $f^7$ (half-filled) — favourable, so Tb⁴⁺ is well known (e.g., $\mathrm{TbO_2}$).
7. Dy: Dy⁴⁺ ($f^8$) is known, though less common, in a few compounds.
8. Count: Ce, Pr, Nd, Tb, Dy = 5.

Common Mistakes

• Including Sm or Eu, which are famous for +2, not +4.
• Including Gd, whose f⁷ stability is at +3, not +4.

ANSWER: A

Concept and Intuition

This question tests several standard d- and f-block facts together: (i) lanthanide ions in unusual oxidation states tend to revert to the characteristic, most stable +3 state, so $\mathrm{Eu^{2+}}$/$\mathrm{Yb^{2+}}$ act as reducing agents (they get oxidized to +3) while $\mathrm{Ce^{4+}}$/$\mathrm{Tb^{4+}}$ act as oxidizing agents (they get reduced to +3);

(ii) magnetism of lanthanide ions follows from unpaired 4f electrons — a fully-filled or fully-empty 4f subshell is diamagnetic;

(iii) the acid-base character of transition-metal oxides follows their oxidation state — low oxidation states give basic oxides, high oxidation states give acidic oxides, with intermediate ones amphoteric;

(iv) common alloy compositions (brass vs bronze) are a factual recall point.

Step-by-Step Solution

1. (A) $\mathrm{Yb^{2+}}$: has configuration $4f^{14}$ (fully filled, stable), so it readily loses an electron to attain the general lanthanide-favoured $\mathrm{Yb^{3+}}$ state — meaning $\mathrm{Yb^{2+}}$ itself gets oxidized, i.e. it is a reducing agent, not an oxidant. Statement false.
2. (B) $\mathrm{Lu^{3+}}$: Lu (Z=71) is $[\mathrm{Xe}]4f^{14}5d^16s^2$; removing 3 electrons ($5d^1,6s^2$) gives $\mathrm{Lu^{3+}} = [\mathrm{Xe}]4f^{14}$ — completely filled 4f, hence diamagnetic, not paramagnetic. Statement false.
3. (C) CrO: chromium here is in the low +2 oxidation state. Per the general trend (lower oxidation state $\Rightarrow$ more ionic/basic oxide; e.g. MnO basic, Mn$_2$O$_7$ acidic), CrO is basic, Cr$_2$O$_3$ amphoteric, CrO$_3$ acidic. Statement true.
4. (D) Brass is an alloy of copper and zinc (bronze is the Cu–Sn alloy), so "Brass is an alloy of Cu, Sn" is false.
5. Only (C) survives as correct.

Common Mistakes

• Assuming any unusual +2/+4 lanthanide ion is automatically an "oxidant" without checking which direction (toward or away from +3) it actually shifts.
• Mixing up brass (Cu-Zn) and bronze (Cu-Sn) — a very common recall error.

ANSWER: C

Concept and Intuition

Going down Group 13 (B, Al, Ga, In, Tl), the ns² electron pair becomes increasingly reluctant to participate in bonding due to poor shielding by intervening d/f electrons and relativistic contraction of the ns orbital — this is the "inert pair effect". Consequently, the +1 oxidation state (where the ns² pair stays un-ionized) becomes progressively more stable relative to +3 as we move down the group. For Al and Ga, the +3 state is the dominant, stable state. For Tl, the inert pair effect is so pronounced that +1 is actually the more stable oxidation state, which means Tl(III) compounds are strong oxidizing agents — they are readily reduced to Tl(I), releasing energy in the process.

Step-by-Step Solution

1. Rank the inert pair effect across Al, Ga, In, Tl — it strengthens down the group, being negligible for Al and dominant for Tl.
2. For Al, +3 is essentially the only common, stable oxidation state (Al³⁺ is not oxidizing).
3. For Ga and In, +3 is still the more stable state, with +1 being a minor, less common state.
4. For Tl, +1 becomes more stable than +3, so $\mathrm{Tl^{3+}}$ species act as oxidizing agents, spontaneously accepting electrons to become $\mathrm{Tl^+}$.
5. Hence the element whose +3 state is highly oxidizing is Tl.

Common Mistakes

• Applying the inert pair effect uniformly to all Group 13 elements instead of recognizing it strengthens down the group.
• Confusing "oxidizing +3 state" with "the +3 state being unstable/non-existent" — Tl(III) compounds do exist, they are just strongly oxidizing.

ANSWER: D

Concept and Intuition

Down group 14, the ns² electron pair becomes increasingly reluctant to participate in bonding (the inert-pair effect), so the lower oxidation state ($+2$) becomes progressively more stable relative to the group oxidation state ($+4$) as you go from C to Pb. This means $Pb^{4+}$ compounds readily oxidise other species while being reduced to the more stable $Pb^{2+}$.

Step-by-Step Solution

1. $SiO_2$, $GeO_2$: silicon and germanium show little inert-pair effect; $+4$ is their stable, common state, so these oxides are not strong oxidisers.
2. $SnO_2$: tin shows a mild inert-pair effect; $Sn^{4+}$ is reasonably stable, only a weak oxidiser.
3. $PbO_2$: lead shows the strongest inert-pair effect in the group; $Pb^{4+}$ is markedly less stable than $Pb^{2+}$, so $PbO_2$ readily gets reduced (e.g. to $PbO$ or $Pb^{2+}$), acting as a powerful oxidising agent (used in lead-acid batteries, oxidimetric analysis).
4. Hence $PbO_2$ is the strongest oxidising agent of the four.

Common Mistakes

• Assuming all group-14 dioxides behave similarly just because they share the same formula type $MO_2$.

ANSWER: D

Concept and Intuition

Going down Group 14, the heavier elements' $ns^2$ electron pair becomes increasingly reluctant to participate in bonding (the inert-pair effect, due to poor shielding by intervening $d$/$f$ electrons and relativistic contraction of the $ns$ orbital). This makes the lower oxidation state progressively more stable for heavier members: $Sn$ shows both $+2$ and $+4$ fairly readily, but $Pb$ strongly favours $+2$ ($PbO$, $PbCl_2$, $Pb(NO_3)_2$ are common; $Pb^{4+}$ compounds like $PbO_2$ are comparatively strong oxidizers/less stable).

Step-by-Step Solution

1. (A) $NCl_5$ does not exist (N has no accessible $d$ orbitals to expand its octet beyond 4 bonds) while $PCl_5$ does (P can use $3d$ orbitals) — this is a correct/true statement.
2. (B) "$Pb$ prefers to form tetravalent compounds" — false; due to the inert-pair effect $Pb$ actually prefers divalent compounds. This is the incorrect statement.
3. (C) In $CO_3^{2-}$, resonance delocalizes the double-bond character equally over all three C–O bonds, making them identical in length — true.
4. (D) $O_2^+$ (one unpaired electron in $\pi^*$) and $NO$ (one unpaired electron) are both paramagnetic — true.
5. Hence the incorrect statement is (B).

Common Mistakes

• Confusing $Sn$'s and $Pb$'s preferred oxidation states — it's $Sn$ that more readily forms tetravalent ($+4$) compounds, while $Pb$ is the one favouring $+2$.

ANSWER: B

Concept and Intuition

The inert-pair effect describes the reluctance of the outermost $ns^2$ electron pair to participate in bonding as atomic number increases down a group, due to poor shielding by intervening $d$/$f$ electrons and relativistic effects for heavier elements. In group 13, this makes the $+1$ oxidation state (retaining the $ns^2$ pair, only losing the single $p$ electron) increasingly favoured relative to $+3$ as you go down the group.

Step-by-Step Solution

1. Al: +3 is overwhelmingly the stable/common oxidation state; +1 compounds are rare and unstable.
2. Ga: +1 exists but is less stable than +3; +3 still dominant.
3. In: +1 and +3 are both reasonably common, with +1 gaining stability.
4. Tl: +1 is actually the more stable oxidation state (Tl$^+$ compounds are more stable than Tl$^{3+}$), the culmination of the inert-pair effect.
5. So the stability of +1 increases in the order $Al<Ga<In<Tl$.

Common Mistakes

• Reversing the trend (thinking +1 is more stable for lighter elements) — inert-pair effect strengthens down the group, not up.

ANSWER: B

Concept and Intuition

The number of oxidation states a transition metal can adopt tends to be maximised near the middle of the 3d series, where there are enough d-electrons to support both low and (through loss of many/all valence electrons) very high oxidation states, while still having partially-filled d-orbitals available for a range of intermediate states. Manganese, with configuration $3d^54s^2$, is the textbook example, spanning $+2$ (as $Mn^{2+}$) all the way to $+7$ (as $MnO_4^-$).

Step-by-Step Solution

1. $V$ (Z=23): common oxidation states $+2, +3, +4, +5$ — 4 common states.
2. $Cr$ (Z=24): common oxidation states $+2, +3, +6$ (and $+4, +5$ less commonly) — fewer well-established states than Mn.
3. $Mn$ (Z=25): common oxidation states $+2, +3, +4, +5, +6, +7$ — the widest, most well-established spread.
4. $Fe$ (Z=26): mainly $+2, +3$ (occasionally $+4, +6$) — narrower range.
5. Manganese clearly shows the greatest number of oxidation states.

Common Mistakes

• Assuming Cr (known for +6, as in dichromate/chromate) has the widest range — Cr's well-established states are actually fewer than Mn's.

ANSWER: B

Concept and Intuition

The chemistry of lanthanoids is governed by the filling of the inner $4f$ subshell, which is well shielded from the surrounding chemical environment by the outer $5s^2 5p^6$ shells. Because the $4f$ electrons are so shielded, they don't participate readily in bonding, and it is always the outermost $6s^2$ electrons (plus the occasional $5d^1$) that ionise first.

Step-by-Step Solution

1. General configuration: $[Xe]4f^{1-14}5d^{0,1}6s^2$.
2. First two ionisations remove $6s^2$ readily; where a $5d^1$ electron is present it is the third to go — giving the $Ln^{3+}$ ion with configuration $[Xe]4f^n$.
3. The 4th ionisation energy (removing an electron from $4f^n$) is very large, because the $4f$ subshell is already at a comparatively stable, contracted low-energy configuration once the outer electrons are gone.
4. Some lanthanoids additionally show $+2$ or $+4$ states, but only when it gives them an especially stable $f^0$, $f^7$, or $f^{14}$ configuration (e.g., $Ce^{4+}$ is $f^0$, $Eu^{2+}$ is $f^7$) — these are exceptions, not the norm.
5. Therefore the characteristic, most common oxidation state across the whole lanthanoid series is $+3$.

Common Mistakes

• Thinking $+4$ or $+2$ (seen for a few specific elements like Ce, Eu) is the general rule — these are special stability exceptions.
• Confusing lanthanoid oxidation-state behaviour with that of the d-block transition metals, which show many oxidation states.

ANSWER: B