Match the properties given in Column I with the metals given in Column II. Column I (Property): (i) Element wi

NCERT Exemplar · Q66

Q. Match the properties given in Column I with the metals given in Column II. Column I (Property):

(i) Element with highest second ionisation enthalpy

(ii) Element with highest third ionisation enthalpy

(iii) M in

M(CO)_6

(iv) Element with highest heat of atomisation
Column II (Metal):

(a) Co

(b) Cr

(d) Zn

(e) Ni

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Concept understanding — Ionization Energy Trends

Ionization Energy: The First Meeting

Imagine you're holding onto something precious — say, a favourite pen. How hard would someone have to pull to take it from your hand? That's the core idea behind ionization energy. In an atom, the "something precious" is an electron, and the "pulling force" is energy.

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

Why "gaseous" and "ground state"? Because we want a fair comparison — no extra energy from neighbours or from the atom being already excited. We measure how tightly the atom holds its outermost electron when it's alone and calm.

The unit you'll see most often in exams: kJ/mol (kilojoules per mole of atoms).

The Intuition: What Controls the Grip?

Two factors decide how hard an atom holds its outermost electron:

Nuclear charge — more protons in the nucleus means a stronger pull on the electron. Simple: bigger positive charge, tighter grip.
Distance from the nucleus — the farther the electron is, the weaker the pull. Think of a magnet: it holds a paperclip strongly up close, but barely at arm's length.

But there's a subtle twist: shielding (or screening). Inner electrons partially block the nuclear charge from reaching the outer electron. The outermost electron doesn't "feel" the full nuclear charge — it feels only the effective nuclear charge ( $Z_{\text{eff}}$ ).

Note

$Z_{\text{eff}} = Z - S$ , where $Z$ is the atomic number and $S$ is the shielding constant (roughly the number of inner electrons). This is the net positive charge pulling on the outer electron.

So the real question becomes: How large is $Z_{\text{eff}}$ for the outermost electron, and how far away is it?

The Precise Trend: Across a Period

As you move left to right across a period (say, from Li to Ne in period 2):

Nuclear charge increases steadily (more protons).
Electrons are added to the same shell — no new inner layers.
Shielding stays roughly constant (same number of inner electrons).
Result: $Z_{\text{eff}}$ increases → the outer electron is pulled in tighter → ionization energy increases.

Across a period: IE increases (generally).

Example:

Li (IE = 520 kJ/mol) → Be (900) → B (801) → C (1086) → N (1402) → O (1314) → F (1681) → Ne (2081)

Wait — why does B have lower IE than Be? And O lower than N? That's the exception, not the rule. We'll come back to it.

The Precise Trend: Down a Group

As you move down a group (say, from Li to Cs in group 1):

Nuclear charge increases (more protons).
But electrons are added to new, higher shells — the outermost electron is much farther from the nucleus.
Shielding also increases significantly (more inner electrons).
Result: distance dominates → the outer electron is held more loosely → ionization energy decreases.

Down a group: IE decreases.

Example:

Li (520) → Na (496) → K (419) → Rb (403) → Cs (376) — all in kJ/mol.

The Two Exceptions (and Why They Matter)

Exception 1: Group 13 vs Group 2 (e.g., B vs Be)

Be has a full $2s^2$ subshell. B has $2s^2 2p^1$ . The $2p$ electron is slightly higher in energy and slightly better shielded by the $2s$ electrons than the $2s$ electrons shield each other. So removing the $2p$ electron from B takes less energy than removing a $2s$ electron from Be.

Watch out

Don't memorise "IE increases across a period" blindly. Group 13 always has lower IE than Group 2 in the same period.

Exception 2: Group 16 vs Group 15 (e.g., O vs N)

N has a half-filled $2p^3$ subshell — each $2p$ orbital has one electron. This is an especially stable arrangement (exchange energy stabilisation). O has $2p^4$ — one orbital gets a second electron. That extra electron experiences electron-electron repulsion, making it easier to remove. So O has lower IE than N.

Tip

Why this formula?

Ionization Energy Trends: The Why Behind the Trends

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

It is measured in kJ/mol or eV/atom.

The key trend is:

IE increases across a period (left → right) and decreases down a group (top → bottom).

But why? Let’s break down the reasoning step-by-step.

1. The Core Formula: Coulomb’s Law

The energy needed to remove an electron is fundamentally governed by the electrostatic attraction between the electron and the nucleus.

F = \frac{k \cdot Z_{\text{eff}} \cdot e^2}{r^2}

Where:

$Z_{\text{eff}}$ = effective nuclear charge (net positive charge felt by the electron)
$r$ = distance of the electron from the nucleus
$e$ = charge of electron
$k$ = Coulomb constant

Key insight: The stronger the attraction, the higher the ionization energy.

2. Why IE Increases Across a Period

Reasoning:

As you move left → right, protons increase in the nucleus.
Electrons are added to the same principal energy level (same shell).
Shielding by inner electrons remains roughly constant (same number of inner shells).
Therefore, $Z_{\text{eff}}$ increases — the outer electrons feel a stronger pull.

Result:

\text{IE} \propto Z_{\text{eff}}

So IE increases across a period.

Example:

Na (Z=11): IE = 496 kJ/mol
Mg (Z=12): IE = 738 kJ/mol
Al (Z=13): IE = 578 kJ/mol (slight dip due to p-orbital shielding — see exception below)

3. Why IE Decreases Down a Group

Reasoning:

As you move down a group, principal quantum number $n$ increases.
The outermost electron is farther from the nucleus ( $r$ increases).
Shielding increases because more inner electron shells are present.
$Z_{\text{eff}}$ increases only slightly (not enough to compensate for distance).

Result:

\text{IE} \propto \frac{1}{r^2}

So IE decreases down a group.

Example:

Li (n=2): IE = 520 kJ/mol
Na (n=3): IE = 496 kJ/mol
K (n=4): IE = 419 kJ/mol

4. The Mathematical Expression (Approximation)

For a hydrogen-like atom (single electron), the ionization energy is given by:

\text{IE} = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2}

For multi-electron atoms, we replace $Z$ with $Z_{\text{eff}}$ :

\text{IE} \approx \frac{13.6 \, \text{eV} \cdot Z_{\text{eff}}^2}{n^2}

Why this holds:

$Z_{\text{eff}}$ accounts for shielding by inner electrons.
$n$ is the principal quantum number of the electron being removed. …

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Correct matches: (i) → (c) Cu, (ii) → (d) Zn, (iii) → (b) Cr, (iv) → (e) Ni.

Highest second ionisation enthalpy → Cu. $IE_2$ removes an electron from $M^{+}$ . For copper, $Cu^{+} = [Ar]3d^{10}$ — a stable, fully-filled $d$ subshell — so removing the next electron is exceptionally difficult. Cu has the highest $IE_2$ of the 3d series.
Highest third ionisation enthalpy → Zn. $IE_3$ removes an electron from $M^{2+}$ . For zinc, $Zn^{2+} = [Ar]3d^{10}$ (stable filled $d$ subshell), so its $IE_3$ is the highest of the series. (iii) M in $M(CO)_6$ → Cr. By the 18-electron rule, six CO ligands donate $6\times2 = 12$ electrons, so the metal must supply $18-12 = 6$ . Chromium ( $[Ar]3d^5\,4s^1$ , 6 valence electrons) meets this, giving the stable $Cr(CO)_6$ . …

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Method: Electronic Configuration & Periodic Trend Analysis

This method uses electronic configurations and periodic trends (ionization enthalpy, stability of half-filled/d orbitals, and metallic bonding strength) to match properties with metals.

Step 1: Write electronic configurations of all metals

Metal	Atomic No.	Configuration
Co	27	$[Ar]\,3d^7\,4s^2$
Cr	24	$[Ar]\,3d^5\,4s^1$
Cu	29	$[Ar]\,3d^{10}\,4s^1$
Zn	30	$[Ar]\,3d^{10}\,4s^2$
Ni	28	$[Ar]\,3d^8\,4s^2$

Step 2: Match (i) — Highest second ionisation enthalpy

Second IE = energy to remove one electron from $M^+$ ion.
After losing one electron, Cu becomes $[Ar]\,3d^{10}$ — fully filled, very stable.
Removing a second electron from this filled shell requires very high energy, giving Cu the highest second IE among the given metals.
Result: (i) → (c) Cu

Step 3: Match (ii) — Highest third ionisation enthalpy

After losing two electrons, Zn becomes $[Ar]\,3d^{10}$ — fully filled, very stable.
Removing a third electron from this stable $d^{10}$ core needs extremely high energy.
Result: (ii) → (d) Zn

Step 4: Match (iii) — M in $M(CO)_6$

Metal carbonyls follow the 18-electron rule.
For $M(CO)_6$ , each CO donates 2 electrons → 12 from CO.
M must contribute 6 electrons to reach 18.
Cr has configuration $3d^5\,4s^1$ — total 6 valence electrons.
Result: (iii) → (b) Cr

Step 5: Match (iv) — Highest heat of atomisation

Heat of atomisation depends on metallic bond strength; the actual experimental trend does not simply track the half-filled/fully-filled stability rule used for ionisation enthalpy. …

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Here are the common mistakes students make when solving this specific question on ionization enthalpy trends, along with how to avoid each.

Mistake 1: Confusing "Highest Second IE" with "Highest First IE"

The Error:

Students often pick Zn for (i) because Zn has a high first ionization enthalpy due to its stable $3d^{10}4s^2$ configuration. However, the question asks for second ionization enthalpy.

Why It’s Wrong:

Zn’s second IE is low because after losing one electron, it becomes $3d^{10}4s^1$ — losing the second electron gives a stable $3d^{10}$ configuration, which is easy.
The element with the highest second IE is Cu.
- Cu: $[Ar] 3d^{10}4s^1$ → after losing one electron → $3d^{10}$ (stable). Removing a second electron from a filled d-subshell requires a huge amount of energy.

How to Avoid:

Always write the electronic configuration of the atom and the ion after the first removal.
Look for the stability of the resulting configuration — a filled or half-filled d-subshell makes the next removal very hard.

Correct match: (i) → (c) Cu

Mistake 2: Forgetting that Third IE depends on Core Stability

The Error:

Students sometimes pick Cu again for (iii) or guess Ni without checking the configuration after two removals.

Why It’s Wrong:

After losing two electrons, Cu becomes $3d^9$ — not particularly stable.
The element with the highest third IE is Zn.
- Zn: $[Ar] 3d^{10}4s^2$ → after losing two electrons → $3d^{10}$ (stable). Removing a third electron from a filled d-subshell is extremely difficult.

How to Avoid:

Track the ion after each removal.
For third IE, check which element reaches a noble gas core or a filled d-subshell after two removals.

Correct match: (ii) → (d) Zn

Mistake 3: Misidentifying the Metal in $M(CO)_6$

The Error:

Students often pick Co or Ni because they are common in carbonyl complexes, but they forget the 18-electron rule.

Why It’s Wrong:

$M(CO)_6$ means the metal is bonded to 6 CO ligands. Each CO donates 2 electrons → total 12 electrons from ligands.
For the complex to be stable, the metal must contribute 6 electrons to reach 18.
Cr has atomic number 24: $[Ar] 3d^5 4s^1$ → it contributes 6 electrons (5 from 3d + 1 from 4s).
Co and Ni would contribute 9 and 10 electrons respectively, leading to electron counts >18, which is unstable for this geometry.

How to Avoid:

Memorize the 18-electron rule for carbonyls.
For $M(CO)_6$ , the metal must be in zero oxidation state and have 6 valence electrons.

Correct match: (iii) → (b) Cr

Mistake 4: Assuming "Highest Heat of Atomisation" means "Highest Melting Point"

The Error:

Students pick Cr because it has a very high melting point, but they don't check the actual trend in atomisation enthalpy.

Why It’s Wrong:

Heat of atomisation depends on metallic bond strength, which is influenced by the number of unpaired electrons in the d-subshell. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 38 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$Q. Which of the following sets are correctly matched? Order | Property I. K > Li > C > F | Atomic radius II. F > C > Li > K | First ionization enthalpy III. F > C > K > Li | Electronegativity The correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III$
›Reveal solutionSolution
This tests periodic trends across the same four elements (K, Li, C, F) for three properties; I and II match the real trends, but III swaps K and Li incorrectly, so only I, II are correct.

Concept and Intuition

Across a period, effective nuclear charge rises with roughly constant shielding, so atoms shrink, ionization enthalpy rises, and electronegativity rises. Down a group, an added shell dominates, so atoms grow, ionization enthalpy falls, and electronegativity falls. K and Li are both alkali metals (group 1), C and F are both period-2 nonmetals, so comparing K vs Li is a group trend and C vs F (and Li vs C, Li vs F) mixes period and group trends — this is exactly where a student must be careful.

Step-by-Step Solution

Atomic radius (I): K (period 4, group 1) is the largest of the four; within period 2, Li > C > F (radius shrinks left to right, though Li is a different period so it's simply larger than both C and F too). Real order: K(227 pm) > Li(152 pm) > C(77 pm) > F(72 pm) — matches I. ✓

First ionization enthalpy (II): F has the highest IE1 among these (small, high effective nuclear charge), K the lowest (large, well-shielded, easy to remove the outer 4s electron). Real values (kJ/mol): F(1681) > C(1086) > Li(520) > K(419) — matches II exactly. ✓ …
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AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\mathrm{Al_2O_3}$
›Reveal solutionSolution
This is a periodic-table-trivia elimination question; three statements check out against real data, but Tennessine (Z=117) is a halogen in group 17, not group 16 — making statement (C) the incorrect one.

Concept and Intuition

The superheavy elements 113–118 were assigned to the periodic table strictly by their group continuation: element 116 (Livermorium) continues the chalcogen family (group 16, below Po), element 117 (Tennessine) continues the halogen family (group 17, below At), and element 118 (Oganesson) continues the noble gases (group 18, below Rn). Getting the group assignment of Tennessine wrong (calling it group 16 instead of 17) is the error to catch here. The other three statements are standard, verifiable periodic-trends/nomenclature facts.

Step-by-Step Solution

(A) First ionization enthalpy of Be, B, C: Be has a stable filled $2s^2$ configuration (extra stability), while B's outermost electron is in $2p^1$ (easier to remove, lower IE than Be) despite B having one more proton. So actual order is B (~801) $<$ Be (~899) $<$ C (~1086) kJ/mol — matches the statement, so (A) is a correct statement.

(B) Before a permanent name is ratified, IUPAC assigns superheavy elements a systematic placeholder name built from digits of the atomic number: element 116 $\to$ "un-un-hex-ium" = Ununhexium. This was indeed Livermorium's provisional systematic name, so (B) is a correct (historically accurate) statement.

(C) Tennessine (Ts), Z=117, sits directly below Astatine in the periodic table, in the halogen family — group 17, not group 16. Group 16 (the chalcogens: O, S, Se, Te, Po) is completed at Z=116 by Livermorium. So this statement's group assignment is wrong. …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Ti, V, Cr, Mn, Fe$
›Reveal solutionSolution
The element with the anomalously highest third ionization enthalpy here is Mn (removing an electron from its stable $d^5$ $M^{2+}$ ion); $Mn^{3+}$ ( $d^4$ ) has a spin-only moment of 4.90 BM.

Concept and Intuition

Ionization enthalpy trends among the 3d series show irregular jumps at half-filled and fully-filled configurations because those configurations are extra stable (exchange energy). $Mn^{2+}$ has the very stable $3d^5$ configuration, so removing a third electron from Mn (i.e. converting $Mn^{2+}\to Mn^{3+}$ ) costs anomalously more energy than for its neighbours — making Mn's third ionization enthalpy the highest in this series.

Step-by-Step Solution

Identify which M(II) has extra stability: $Mn^{2+}$ is $[Ar]3d^5$ , a half-filled and hence very stable configuration.

Removing the next electron ( $Mn^{2+}\to Mn^{3+}$ ) breaks this stability, so the third IE of Mn is anomalously high — the highest among Ti, V, Cr, Mn, Fe.

So the element in question is Mn, and we need the magnetic moment of $Mn^{3+}$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$Cl$
›Reveal solutionSolution
This tests the period-3 periodic trends, especially the P/S ionization-enthalpy anomaly caused by the half-filled $3p^3$ stability of phosphorus.

Concept and Intuition

Ionization enthalpy generally rises across a period as effective nuclear charge increases, but extra stability from exactly half-filled ( $p^3$ ) or fully-filled ( $p^6$ ) subshells creates local humps. Between P and S, phosphorus's half-filled $3p^3$ is unusually stable (symmetric electron distribution, minimal inter-electron repulsion), so removing an electron from P actually needs more energy than from S, even though S has one more proton.

Step-by-Step Solution

(A) Electron gain enthalpy across period 3: Cl (with configuration $3p^5$ ) gains an electron most exothermically among period-3 elements ( $\Delta_{eg}H \approx -349\ kJ/mol$ ) — correct, so not the answer.

(B) Noble gases have the highest $IE_1$ in their period due to stable filled shells — Ar has the highest $IE_1$ in period 3 — correct, so not the answer.

(C) Mg ( $3s^2$ , filled subshell, stable) has a higher $IE_1$ than Al ( $3s^23p^1$ , the lone $3p$ electron is easier to remove due to shielding by $3s^2$ and slightly higher energy of the $3p$ orbital) — correct, so not the answer. …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$Q. Which of the following statements are correct? I. In third period, two elements have higher ionization enthalpy than the element immediately following them in the same period II. Electronegativity of carbon is higher than that of phosphorus III. An element X belongs to group 14 and period 3. The number of electrons present in it is 14 The correct answer is (A) I, II, III (B) I, II only (C) II, III only (D) I, III only$
›Reveal solutionSolution
All three statements are factually correct: the Mg/P ionization-enthalpy anomalies, C vs P electronegativity, and Si having 14 electrons.

Concept and Intuition

Ionization enthalpy generally rises across a period as effective nuclear charge increases, but extra stability of exactly filled ( $ns^2$ ) or exactly half-filled ( $np^3$ ) sub-shells creates two well-known exceptions in every period. Electronegativity decreases down a group and increases across a period, so carbon (top of group 14) exceeds phosphorus (below-and-left in period 3, group 15 though lower period). Locating an element by group and period simply reads off its position in the periodic table — its electron count is just its atomic number.

Step-by-Step Solution

Statement I: Period-3 IE $_1$ trend (kJ/mol): Na 496, Mg 738, Al 578, Si 786, P 1012, S 1000, Cl 1251, Ar 1521. Compare each element to the one right after it: Mg(738) $>$ Al(578) — anomaly; P(1012) $>$ S(1000) — anomaly. Exactly two elements (Mg, P) exceed the one immediately following. Statement I is TRUE.

Statement II: Electronegativity (Pauling): C $\approx 2.5$ , P $\approx 2.1$ . So C $>$ P. Statement II is TRUE. …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$B > Al$
›Reveal solutionSolution
This tests the well-known d-block/f-block contraction anomaly in Group 13 ionisation enthalpies. The real order is $B > Tl > Ga > Al > In$ , so statements II ( $Al>Ga$ ) and IV ( $In>Tl$ ) are the incorrect ones.

Concept and Intuition

Going down a group, ionisation enthalpy is normally expected to fall steadily because the outermost electron sits farther from the nucleus and is better shielded. Group 13 is a textbook exception. Between Al and Ga, a full $3d^{10}$ subshell has been filled in Ga's core; $d$ -electrons shield the nuclear charge poorly, so Ga's valence electron feels an unexpectedly strong effective nuclear charge -- enough to make Ga's IE1 edge out Al's despite being one period lower. The same thing happens even more strongly between In and Tl: Tl's core additionally has a filled $4f^{14}$ subshell (the lanthanide contraction), whose shielding is even weaker, so Tl's IE1 exceeds In's.

Step-by-Step Solution

Recall/derive the actual first ionisation enthalpies (kJ/mol): $B \approx 801$ , $Al \approx 577$ , $Ga \approx 579$ , $In \approx 558$ , $Tl \approx 589$ .

Order them: $B > Tl > Ga > Al > In$ .

Test statement I: $B > Al$ gives $801 > 577$ , true.

Test statement II: $Al > Ga$ gives $577 > 579$ , which is false (Ga is marginally higher due to poor $3d^{10}$ shielding).

Test statement III: $Ga > In$ gives $579 > 558$ , true. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{N > O > P > S}$
›Reveal solutionSolution
This tests four classic periodic-trend anomalies. Three of the four stated orders (A, C, D) are textbook-correct; option (B)'s claimed order for electron gain enthalpy is the one that is wrong.

Concept and Intuition

Periodic trends are not perfectly monotonic — the second-period elements (especially F, O, N) show anomalies because they are unusually small, so adding an electron to them causes extra electron-electron repulsion in the compact $2p$ subshell. This is why F's electron gain enthalpy (magnitude) is LESS negative than Cl's, even though F is more electronegative; and O's is less negative than S's. Remembering the anomaly is key to spotting the one wrong order.

Step-by-Step Solution

(A) First ionisation enthalpy $N>O>P>S$ : N has a stable, half-filled $2p^3$ configuration, giving it an anomalously HIGH first IE — even higher than O (which has one paired $2p$ electron, easier to remove). So $N>O$ holds, and across periods IE falls going down, so $N>O>P>S$ is the correct, well-known order. Order is correct.

(B) Negative electron gain enthalpy $F>Cl>O>S$ : The real magnitudes (kJ/mol, more negative = more favourable) are approximately: F ≈ −328, Cl ≈ −349, O ≈ −141, S ≈ −200. So actually $Cl > F$ (not $F>Cl$ ) and $S > O$ (not $O>S$ ). Both parts of this stated order are backwards. Order is NOT correct — this is the answer.

(C) Size $Fe^{3+}<Fe^{2+}<Fe$ : Removing electrons from a fixed nuclear charge increases effective nuclear pull per remaining electron, shrinking the ion. Going from neutral Fe to $Fe^{2+}$ to $Fe^{3+}$ , size decreases, so increasing order is $Fe^{3+}<Fe^{2+}<Fe$ . Order is correct. …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$1s^2\,2s^2\,2p^6\,3s^1$
›Reveal solutionSolution
This tests the exceptions to the general periodic trend in ionization enthalpy caused by stable filled/half-filled subshells; the order is D > C > B > A, option (D).

Concept and Intuition

Ionization enthalpy generally increases across a period, but atoms with a fully filled or exactly half-filled subshell (extra stability) show an anomalously HIGH ionization enthalpy compared to their immediate neighbour with one extra electron in a new subshell. Here the four configurations identify: A = $1s^2 2s^2 2p^6 3s^1$ = Na, B = $1s^2 2s^2 2p^6 3s^2 3p^1$ = Al, C = $1s^2 2s^2 2p^6 3s^2$ = Mg, D = $1s^2 2s^2 2p^6 3s^2 3p^2$ = Si.

Step-by-Step Solution

Identify elements: A = Na (Z=11), B = Al (Z=13), C = Mg (Z=12), D = Si (Z=14).

Recall the anomaly: Mg (filled 3s²) has a HIGHER first IE than Al (3s²3p¹) even though Al has higher Z, because removing Al's lone 3p electron is easier than breaking Mg's stable filled 3s subshell. …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{Li} < \mathrm{Na} < \mathrm{K}$
›Reveal solutionSolution
This tests periodic trends across four different properties simultaneously; the odd one out is ionization enthalpy of C, N, O, where nitrogen's half-filled stability breaks the naive trend.

Concept and Intuition

Most periodic properties vary smoothly, but ionization enthalpy along Period 2 has a well-known anomaly: a half-filled ( $p^3$ ) or fully-filled ( $p^6$ ) subshell is extra stable, so removing an electron from it needs more energy than the simple increasing-across-the-period trend predicts. Nitrogen ( $1s^2 2s^2 2p^3$ ) has exactly this half-filled $2p^3$ configuration, so its first ionization enthalpy is anomalously high — even higher than oxygen's.

Step-by-Step Solution

(A) Metallic radius increases down Group 1 as principal quantum number increases: $\mathrm{Li < Na < K}$ — correct order.

(B) Electron gain enthalpy (magnitude) for halogens: $\mathrm{Cl(-349) > F(-328) > Br(-325) > I(-295)}$ kJ/mol (F is anomalously low due to small size/electron-electron repulsion in the small 2p orbital). So increasing magnitude: $\mathrm{Br < F < Cl}$ — correct order.

(D) These are isoelectronic species with 10 electrons each: $\mathrm{Mg^{2+}(Z=12), Na^{+}(Z=11), F^{-}(Z=9)}$ . Higher nuclear charge pulls the same electron cloud in tighter, so radius decreases as $Z$ increases: $\mathrm{Mg^{2+} < Na^{+} < F^{-}}$ — correct order. …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\mathrm{kJ\,mol^{-1}}$
›Reveal solutionSolution
This tests the anomalous first-ionization-enthalpy trend across Period 2 (Be, B, C, N, O) — the answer is (C).

Concept and Intuition

Ionization enthalpy generally rises across a period, but Period 2 has two dips: at Boron (due to its single, easily-removed $2p^1$ electron shielded by the filled $2s^2$ ) and at Oxygen (due to inter-electron repulsion when a fourth electron pairs up in a $2p$ orbital that already has one electron, in a half-filled $2p^3$ configuration at N). This gives the order $\mathrm{B < Be < O < N}$ , i.e. $\mathrm{801 < 899 < 1314 < 1402}$ .

Step-by-Step Solution

Recall/assign known values: $\mathrm{Be = 899}$ , $\mathrm{B = 801}$ , $\mathrm{N = 1402}$ , $\mathrm{O = 1314}\ \mathrm{kJ\,mol^{-1}}$ .

Match List-I to List-II: A (Be) → II (899); B (O) → III (1314); C (N) → IV (1402); D (B) → I (801). …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{IE_1}$
›Reveal solutionSolution
Converting $\mathrm{Mg(g)}$ to $\mathrm{Mg^{2+}(g)}$ requires both the first and second ionization energies, added together, and ionization is always endothermic (positive).

Concept and Intuition

Ionization enthalpy is the energy needed to remove an electron from a gaseous atom/ion; it is always absorbed (positive) because you are pulling a negatively charged electron away from a positively charged/neutral species against electrostatic attraction. Removing two electrons in sequence needs $IE_1$ (to form $\mathrm{Mg^+}$ ) plus $IE_2$ (to form $\mathrm{Mg^{2+}}$ from $\mathrm{Mg^+}$ ).

Step-by-Step Solution

$\mathrm{Mg(g)\to Mg^+(g)+e^-}$ : energy $=IE_1=+178$ kcal/mol. …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$\mathrm{kJ\,mol^{-1}}$
›Reveal solutionSolution
This tests the well-known irregularity in the first-ionisation-enthalpy trend across Period 2 -- Boron's IE1 dips below Beryllium's. The answer is (C) 801 kJ/mol.

Concept and Intuition

Ionisation enthalpy generally increases left to right across a period because of increasing effective nuclear charge. But there are two well-documented exceptions in Period 2: Be > B (this question) and N > O. Beryllium has a stable, fully-filled $2s^2$ configuration, which is harder to disturb, while Boron's outermost electron is in a $2p^1$ orbital that is higher in energy and more shielded from the nucleus by the $2s^2$ core -- making it easier to remove despite Boron having one more proton.

Step-by-Step Solution

List the expected qualitative order: the naive trend (ignoring the anomaly) would predict B > Be, since nuclear charge increases from Be to B.

Recall the actual anomaly: because of the extra stability of the filled $2s^2$ subshell in Be, and the higher-energy, less-penetrating $2p$ orbital in B, $IE_1(B) < IE_1(Be)$ . …
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Q.Match the properties given in Column I with the metals given in Column II. Column I (Property):

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Q.Match the properties given in Column I with the metals given in Column II.
Column I (Property):

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