Q.Which of the following species will have the largest and the smallest size? $Mg,\ Mg^{2+},\ Al,\ Al^{3+}$.

Concept and Intuition

This problem tests your understanding of periodic trends: ionization enthalpy, metallic character, ionic size, and electron gain enthalpy. Each option claims a specific order; we must verify each against known periodic behavior. The key is to recall that metallic character decreases across a period and increases down a group, while ionization enthalpy generally increases across a period and decreases down a group. Ionic size decreases with increasing positive charge for isoelectronic species. Electron gain enthalpy becomes more negative (more energy released) across a period, but fluorine is an exception due to its small size.

Step-by-step verification

1. Option (A): First Ionization Enthalpies – Na < Al?

   • Ionization enthalpy generally increases across a period (left to right).
   • Na (Group 1) has a low ionization enthalpy; Al (Group 13) is to the right, so Al’s first ionization enthalpy is higher than Na’s.
   • The given order “Na < Al” is correct.
   • Conclusion: (A) is a correct match.

2. Option (B): Metallic character – Al > Mg > B > K?

   • Metallic character decreases across a period and increases down a group.
   • K (Group 1, Period 4) is highly metallic, more so than Mg (Group 2, Period 3).
   • Al (Group 13, Period 3) is less metallic than Mg.
   • B (Group 13, Period 2) is a metalloid, least metallic here.
   • The correct decreasing order: K > Mg > Al > B.
   • The given order (Al > Mg > B > K) is wrong because K should be the most metallic, not the least.
   • Conclusion: (B) is an incorrect match.

3. Option (C): Ionic size – Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺?

   • These ions are isoelectronic (all have 10 electrons: Ne configuration).
   • For isoelectronic species, ionic radius decreases as nuclear charge increases.
   • Nuclear charges: Na⁺ (11), Mg²⁺ (12), Al³⁺ (13), Si⁴⁺ (14).
   • So size order: Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺ is correct.
   • Conclusion: (C) is a correct match.

4. Option (D): Electron gain enthalpy – F > Cl > Br > I?

   • Electron gain enthalpy becomes less negative down a group (Cl has the most negative, then Br, then I).
   • Fluorine is an exception: its small size causes high electron-electron repulsion, so its electron gain enthalpy is less negative than chlorine’s.
   • The correct order (most negative to least): Cl > F > Br > I.
   • The given order “F > Cl > Br > I” is incorrect because Cl should be before F.
   • Wait — the option says “F > Br > I” but omits Cl? Actually the option reads: “Electron Gain enthalpy – F > Br > I”. That is incomplete; it likely means F > Cl > Br > I? The printed text shows “F Br > I” which is ambiguous. However, typical exam questions list “F > Cl > Br > I” as incorrect because Cl has more negative electron gain enthalpy than F. So (D) is also incorrect.
   • But the question asks for one incorrect match. Since (B) is clearly wrong, and (D) is also wrong, we must check the original wording: Option (D) says “$\quad-\mathrm{F}\mathrm{Br}>\mathrm{I}$” — likely a typo for “F > Cl > Br > I”. In standard periodic trends, that order is indeed incorrect because Cl > F. So both (B) and (D) appear incorrect.
   • However, many such questions consider (D) as correct if they mean “F has a more negative electron gain enthalpy than Br and I” (which is true: F is more negative than Br and I, but less than Cl). The phrasing “F > Br > I” is actually correct if comparing only those three: F (−328 kJ/mol), Br (−325), I (−295). So F > Br > I is correct. The missing Cl is not an error if the list only includes those three.
   • Therefore, (D) is a correct match for the given elements.
   • Conclusion: (D) is a correct match.

Final determination

Only option (B) presents an incorrect order of metallic character.

Step 1 — Hybridisation in $\mathrm{AlCl_3}$.

Aluminium: $Z = 13$, configuration $[\mathrm{Ne}]\,3s^2 3p^1$ — three valence electrons. In $\mathrm{AlCl_3}$ it forms three $\sigma$ bonds to three Cl atoms and retains no lone pair.

A steric number of 3 requires three equivalent hybrid orbitals, formed from one $s$ and two $p$ orbitals:

Crucially, this leaves Al with an empty, unhybridised $3p$ orbital perpendicular to the molecular plane. With only 6 electrons around it, Al is electron-deficient — an octet-incomplete Lewis acid.

Step 2 — What the reaction is.

This is a Lewis acid–base reaction. The chloride ion (a Lewis base, rich in lone pairs) donates a lone pair into aluminium's vacant $p$ orbital, forming a coordinate (dative) covalent bond. Aluminium thereby completes its octet.

Step 3 — Hybridisation in $\mathrm{AlCl_4^-}$.

Al now has four $\sigma$ bonds (three normal covalent + one coordinate — they are indistinguishable once formed) and still no lone pair:

Four equivalent hybrid orbitals need one $s$ and all three $p$ orbitals:

Step 4 — The change and the elimination of the other options.

• (A) "No change" — false; the steric number demonstrably rises from 3 to 4, and the geometry changes from planar to tetrahedral.
• (C) "$\mathrm{sp^3}$ to $\mathrm{sp^3}d$" — would require five bonds around Al. $\mathrm{AlCl_4^-}$ has only four, and Al (period 3, but with no accessible expansion here) does not form $\mathrm{AlCl_5^{2-}}$ in this reaction.
• (D) "$\mathrm{sp^3}$ to $\mathrm{sp^2}$" — the reverse of the truth; it would mean Al loses a bond, whereas it gains one.

ANSWER: B

Check each:

• (A) $\mathrm{Mg^{2+}<Na^{+}<F^{-}<O^{2-}}$ (increasing ionic size): all isoelectronic (10 e$^-$); size increases as nuclear charge falls (Mg$^{2+}$:12 → Na$^+$:11 → F$^-$:9 → O$^{2-}$:8). Correct.
• (B) $\mathrm{I<Br<F<Cl}$ (increasing $-\Delta_{eg}H$): magnitudes are Cl (349) > F (328) > Br (325) > I (295), so increasing order I < Br < F < Cl. Correct.
• (C) $\mathrm{Na<K<Rb<Cs}$ (increasing metallic radii): radius increases down group 1. Correct.
• (D) $\mathrm{B<C<N<O}$ (increasing first ionisation enthalpy): actual values are B (801) < C (1086) < O (1314) < N (1402). Nitrogen's half-filled 2p$^3$ is extra-stable, so $IE_1(\text{N}) > IE_1(\text{O})$. The listed order B < C < N < O is therefore incorrect.

Step 1 — Definition

An amphoteric oxide reacts with an acid and with a base, forming a salt and water in each case. It sits at the boundary between metallic (basic) and non-metallic (acidic) character — typically the oxides of Be, Al, Zn, Sn, Pb.

Step 2 — Examine each candidate oxide

BeO — amphoteric ✓

Beryllium is the classic diagonal-relationship element (with Al); its small, highly polarising $\mathrm{Be^{2+}}$ gives its oxide covalent character.

ZnO — amphoteric ✓

$\mathrm{Li_2O}$ — basic ✗

An alkali-metal oxide. It dissolves in water to give the strong base LiOH and reacts only with acids, never with alkali.

MgO — basic ✗

A typical alkaline-earth oxide, ionic and basic; it neutralises acids (hence its use as an antacid) but does not dissolve in NaOH. (Beryllium is the only group-2 element whose oxide is amphoteric — the anomalous first member.)

$\mathrm{BO_3}$ — not a valid species ✗

Boron's oxide is $\mathrm{B_2O_3}$, and it is acidic (it gives boric acid with water), not amphoteric. "$\mathrm{BO_3}$" is not a real formula.

Step 3 — Screen the options

ANSWER: D

Concept: Atomic-radius trend in group 13 - the d-block (and later f-block) contraction breaks the simple 'increases down the group' rule.

Radii (pm): B 85, Al 143, Ga 135, In 167, Tl 170.

Gallium is SMALLER than aluminium because Ga comes immediately after the first transition series: the ten added 3d electrons shield the nuclear charge poorly, so the effective nuclear charge felt by Ga's outer electrons is unusually high and the atom contracts.

Hence increasing atomic radius: Ga < Al < In < Tl.

ANSWER: B

[!TLDR]

Because of poor 3d shielding, gallium is smaller than aluminium, so the increasing order of atomic radii is Ga < Al < In < Tl.

Concept

In group 13 (B, Al, Ga, In, Tl) atomic radius normally rises down the group as new shells are added. Gallium breaks this trend: after Al we cross the first transition series, and the poorly-shielding 3d electrons let the nucleus pull the valence electrons in more tightly, so Ga is slightly smaller than Al (a standard NCERT Class 11 p-block point).

Solution

Approximate NCERT atomic radii (pm):

Arranging in increasing order:

The only anomaly to remember is Ga < Al; from In onward the normal increase resumes.

[!ANSWER]

(B) $\mathrm{Ga} < \mathrm{Al} < \mathrm{In} < \mathrm{Tl}$

Step 1 — The concept: amphoterism.

Most metals above hydrogen in the reactivity series displace $H_2$ from dilute acids. But dissolving in a strong base as well is a special property: it requires the metal (and its oxide/hydroxide) to be amphoteric — able to react with both acids and alkalis.

The common amphoteric metals are Zn, Al, Sn, Pb (and Be). Only these liberate hydrogen with NaOH, by forming a soluble complex/oxyanion (zincate, aluminate, stannate, plumbate).

Step 2 — Test zinc against the acid.

Zn lies above hydrogen in the reactivity series, so it displaces $H_2$ from dilute HCl:

✓ Hydrogen evolved.

Step 3 — Test zinc against the alkali.

Because Zn is amphoteric, it also dissolves in hot aqueous NaOH, forming sodium zincate and liberating hydrogen:

or, written with the hydroxo complex (the more modern representation):

✓ Hydrogen evolved. Zn satisfies both conditions.

Step 4 — Why the others fail the second test.

All of Fe, Mg and Ca are above hydrogen in the reactivity series, so each does give $H_2$ with dil. HCl:

But none of them is amphoteric:

• (A) Fe — iron and its oxides are basic (Fe is a transition metal, not amphoteric). No reaction with NaOH(aq). ✗
• (C) Mg — an alkaline earth metal; $\mathrm{MgO}$/$\mathrm{Mg(OH)_2}$ are basic. No reaction with NaOH(aq) — a base does not react with a base. ✗
• (D) Ca — likewise a strongly basic alkaline earth metal; $\mathrm{Ca(OH)_2}$ is itself a base. No reaction with NaOH(aq). ✗

(Note the trap: Be, uniquely among the alkaline earth metals, is amphoteric — but Be is not offered here; Mg and Ca are not.)

Step 5 — Conclusion.

Only Zn passes both tests, because only Zn is amphoteric.

ANSWER: B