Q.Why is the $E^\circ$ value for the $Mn^{3+}/Mn^{2+}$ couple much more positive than that for $Cr^{3+}/Cr^{2+}$ or $Fe^{3+}/Fe^{2+}$? Explain.

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. …

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. …

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. …

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. …

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. …

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. …

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. …

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. …

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. …

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. …