Q.Metallic radii of some transition elements are given below. Which of these elements will have highest density?
Element: Fe, Co, Ni, Cu
Metallic radii/pm: Fe =126, Co =125, Ni =125, Cu =128
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 (Zeff).
Zeff=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 Zeff 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: Zeff 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 2s2 subshell. B has 2s22p1. 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.
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 2p3 subshell — each 2p orbital has one electron. This is an especially stable arrangement (exchange energy stabilisation). O has 2p4 — 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.
| Period | Group 15 (IE) | Group 16 (IE) | Which is higher? |
|--------|--------------|--------------|------------------|
| 2 | N (1402) | O (1314) | N > O |
| 3 | P (1012) | S (1000) | P > S |
| 4 | As (947) | Se (941) | As > Se |
The pattern holds for all periods.
The Big Picture: What You Must Remember
Ionization energy increases across a period (with two dips) and decreases down a group.
The dips occur at Group 13 (lower than Group 2) and Group 16 (lower than Group 15).
The underlying reason is always the same: effective nuclear charge and distance. When Zeff is high and the electron is close, IE is high. When the electron is far or repulsion helps it leave, IE is low.
A Final Check: First vs Second Ionization Energy
Removing one electron from an atom leaves a positive ion. Removing a second electron from that ion is always harder — the ion has a higher positive charge pulling on the remaining electrons.
Second IE > First IE — always. For example, Na: first IE = 496 kJ/mol, second IE = 4562 kJ/mol. That's nearly 10 times larger. This huge jump tells you that the second electron comes from a different shell (closer to the nucleus).
In exams, this jump is used to identify the group of an element — a sudden large increase in successive ionization energies indicates you've stripped off all valence electrons and are now pulling from a core shell.
"Ionization energy trends periodic table" and "periodicity class 11 chemistry important questions" are extremely common searches, both anchored in the Classification of Elements and Periodicity chapter of the NCERT/CBSE Class 11 Chemistry curriculum. The Group 13 and Group 16 exceptions in particular are a favourite trap question in board exams and JEE Main.
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=r2k⋅Zeff⋅e2
Where:
- Zeff = 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, Zeff increases — the outer electrons feel a stronger pull.
Result:
IE∝Zeff
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.
- Zeff increases only slightly (not enough to compensate for distance).
Result:
IE∝r21
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:
IE=n213.6eV⋅Z2
For multi-electron atoms, we replace Z with Zeff:
IE≈n213.6eV⋅Zeff2
Why this holds:
- Zeff accounts for shielding by inner electrons.
- n is the principal quantum number of the electron being removed.
- The 1/n2 dependence comes from the Bohr model — energy levels scale as En∝−Z2/n2.
5. Exceptions (Why the Trend Isn’t Perfect)
a) Group 13 vs Group 2 (e.g., Al vs Mg)
- Al has a p-orbital electron (higher energy, easier to remove) than Mg’s s-orbital.
- Also, p-orbitals are more shielded by s- and p-electrons.
b) Group 16 vs Group 15 (e.g., O vs N)
- N has a half-filled p-subshell (extra stability).
- O has one paired electron — electron-electron repulsion makes removal easier.
6. Summary Table
| Factor | Across Period (→) | Down Group (↓) |
|---|---|---|
| Zeff | Increases | Increases slightly |
| r (distance) | Decreases slightly | Increases |
| Shielding | Constant | Increases |
| IE | Increases | Decreases |
Final Takeaway
Ionization energy is not just a number — it’s a direct consequence of Coulomb’s law, modified by shielding and orbital shape.
The trend is driven by Zeff (across) and distance + shielding (down).
Always ask: “How strongly is this electron held?” — and the answer lies in the balance of nuclear charge, distance, and shielding.
Concept: Density depends on mass per unit volume. For metals in the same period, atomic mass increases faster than atomic volume, so density generally increases across a series.
Reasoning:
- Density ∝atomic volumeatomic mass. Atomic volume ∝(radius)3.
- Atomic masses (approx.): Fe = 55.8, Co = 58.9, Ni = 58.7, Cu = 63.5 g/mol.
- Radii: Fe = 126, Co = 125, Ni = 125, Cu = 128 pm. Volume scales as r3, so Cu has the largest volume, but its mass is significantly higher.
- Compare mass/volume ratios: Cu has the highest atomic mass with only a slightly larger radius, giving it the greatest density.
The element with the highest density is Cu (option (iv)).
Density depends on atomic mass and atomic volume (radius³). Among Fe, Co, Ni, and Cu, copper has the highest atomic mass and a relatively large radius, giving it the highest density.
Why density depends on radius and mass
Density is mass per unit volume. For a metallic crystal, the density of the element is proportional to:
Density∝(Metallic radius)3Atomic mass
The exact formula involves packing fraction and Avogadro’s number, but for comparing elements with the same crystal structure (all four are face-centered cubic at room temperature), the packing fraction cancels out. So we only need to compare:
r3Atomic mass
A higher atomic mass and a smaller radius both push density up. Let’s see which element wins.
Step-by-step comparison
1. List the given data
| Element | Metallic radius (pm) | Atomic mass (g/mol) |
|---|---|---|
| Fe | 126 | 55.85 |
| Co | 125 | 58.93 |
| Ni | 125 | 58.69 |
| Cu | 128 | 63.55 |
The radii are nearly equal — all within 3 pm of each other. So the atomic mass will be the deciding factor.
2. Compute M/r3 for each
We can work with relative values since the constant factor (packing fraction, Avogadro’s number) is the same for all.
For Fe:
126355.85=200037655.85≈2.79×10−5
For Co:
125358.93=195312558.93≈3.02×10−5
For Ni:
125358.69≈3.00×10−5
For Cu:
128363.55=209715263.55≈3.03×10−5
3. Compare the values
- Fe: 2.79×10−5 — lowest, because Fe has the smallest atomic mass and a mid-sized radius.
- Co: 3.02×10−5 — higher than Fe.
- Ni: 3.00×10−5 — very close to Co, slightly lower.
- Cu: 3.03×10−5 — the highest value.
Copper’s atomic mass is about 7–8% higher than cobalt’s, and its radius is only 2.4% larger. The cube in the denominator means a 2.4% radius increase raises the volume by about 7.4%, but the mass increase of ~7.8% more than compensates. So Cu edges ahead.
You don’t need to compute the exact numbers. Just compare ratios:
For Co vs Cu: 125358.93 vs 128363.55.
Notice 128/125=1.024, so (128/125)3≈1.074.
The mass ratio 63.55/58.93≈1.078. Since 1.078>1.074, Cu wins.
4. Final ranking
Cu > Co > Ni > Fe in density.
A common mistake is to pick the element with the smallest radius (Co or Ni) thinking that smaller radius always means higher density. But density also depends on atomic mass — copper is heavier enough to overcome its slightly larger radius.
The element with the highest density is (iv) Cu.
Method: Density Estimation from Metallic Radius and Atomic Mass
This problem uses the relationship between density, atomic mass, and atomic radius in a periodic table trend context.
Concept Behind the Method
Density (ρ) is mass per unit volume. For metallic elements in the same period:
- Mass depends on atomic mass (increases across a period)
- Volume depends on atomic radius (generally decreases across a period)
Since density ∝volumemass, and volume ∝r3, we can compare density using:
ρ∝r3Atomic mass
Steps to Solve
Step 1: List the given data
| Element | Metallic radius (pm) | Atomic mass (g/mol) |
|---|---|---|
| Fe | 126 | 55.85 |
| Co | 125 | 58.93 |
| Ni | 125 | 58.69 |
| Cu | 128 | 63.55 |
Step 2: Calculate r3 for each element
- Fe: 1263=2,000,376
- Co: 1253=1,953,125
- Ni: 1253=1,953,125
- Cu: 1283=2,097,152
Step 3: Compute r3Atomic mass for comparison
- Fe: 2,000,37655.85≈2.79×10−5
- Co: 1,953,12558.93≈3.02×10−5
- Ni: 1,953,12558.69≈3.00×10−5
- Cu: 2,097,15263.55≈3.03×10−5
Step 4: Compare the values
The highest value indicates the highest density.
Order: Cu > Co > Ni > Fe
Final Answer
Cu has the highest density.
(iv) Cu
Why This Works
Across the first transition series, atomic mass increases steadily while atomic radius remains nearly constant (due to poor shielding by d-electrons). Copper has the highest atomic mass among these four, with only a slightly larger radius — giving it the highest density.
Common Mistakes & How to Avoid Them
Mistake 1: Assuming the smallest radius always gives the highest density
Why students make it:
They see Co and Ni have the smallest radii (125 pm) and assume "smaller atom = more tightly packed = denser," picking (iii) Co or (ii) Ni without checking atomic mass.
Why it's wrong:
Density depends on mass per unit volume, not radius alone. A smaller radius does increase density for a fixed mass, but here the atomic masses are also different — and that difference decides the outcome.
How to avoid:
Always write the formula:
Density∝Atomic volumeAtomic mass
For a spherical atom, volume ∝r3. So:
Density∝r3Atomic mass
Check the numbers:
| Element | Atomic mass (g/mol) | Radius (pm) | r3 (× 106 pm³) | Mass/r3 (relative) |
|---|---|---|---|---|
| Fe | 55.85 | 126 | 2.00 | 27.9 |
| Co | 58.93 | 125 | 1.95 | 30.2 |
| Ni | 58.69 | 125 | 1.95 | 30.1 |
| Cu | 63.55 | 128 | 2.10 | 30.3 |
Result: Cu has the highest mass per unit volume → highest density, even though its radius is the largest of the four — its atomic mass is high enough to overcome the larger volume.
Correct answer: (iv) Cu
Mistake 2: Forgetting that density depends on atomic mass, not just radius
Why students make it:
They focus only on the given radii and ignore the atomic masses (which are not directly given but must be recalled from periodic table knowledge).
How to avoid:
Always recall or note the atomic masses of the elements in the series. For 3d transition series:
- Fe ≈ 56
- Co ≈ 59
- Ni ≈ 58.7
- Cu ≈ 63.5
Cu has both the largest radius and the largest mass among these four — so the ratio, not either number alone, decides the answer.
Mistake 3: Thinking density tracks radius alone across the series
Why students make it:
They remember that atomic radii change only slightly across Fe–Co–Ni–Cu, so they assume the element with the smallest radius must be densest.
Why it's wrong:
Across the 3d series, density is set by the combination of mass and volume. Cu's radius is about 2.4% larger than Co's/Ni's, which raises its volume by about 7.4% — but Cu's atomic mass is about 7.8% higher than Co's, which more than compensates.
How to avoid:
Compare ratios directly instead of eyeballing radius or mass alone:
- Fe → Co: mass ↑ noticeably, radius ↓ slightly → density increases
- Co → Ni: mass and radius both nearly unchanged → density nearly the same
- Ni → Cu: mass ↑ more than radius ↑ → density increases again, reaching its highest value at Cu
So the peak is at Cu, not Co.
Mistake 4: Confusing metallic radius with density formula
Why students make it:
They try to use the metallic radius directly in density without converting to volume.
How to avoid:
Remember: density uses volume, which scales as r3. A 2% change in radius causes ~6-7% change in volume — significant enough to matter here.
Quick Summary — Do This Instead
| Step | Action |
|---|---|
| 1 | Note atomic masses (from memory or periodic table) |
| 2 | Compute r3 for each element |
| 3 | Compute mass/r3 (relative comparison is enough) |
| 4 | Pick the largest ratio |
Final answer: (iv) Cu
Showing the 12 most recent of 38 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.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. ✓
- Electronegativity (III): F(3.98) > C(2.55) far exceed Li(0.98) and K(0.82). Between Li and K (same group), electronegativity DECREASES down the group, so Li > K. The correct order is F > C > Li > K, not the stated F > C > K > Li. So III is incorrect. ✗
- Therefore only I and II are correctly matched.
Common Mistakes
- Assuming all group-1 trends are identical in magnitude to period-2 trends and mixing up Li/K ordering for electronegativity.
- Confusing ionization enthalpy trend (which decreases down a group) with electronegativity trend — both decrease down a group and increase across, so students often get III "by pattern" without checking Li vs K specifically.
✓Final answerThe correct option is (A) — I, II only.
ANSWER: A
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.Which of the following statement is not correct? (A) The increasing order of first ionization enthalpy of Be, B, C is B < Be < C (B) The IUPAC name of element Livermorium is ununhexium (C) The element Tennessene belongs to group 16 in periodic table (D) Al2O3, As2O3 are amphoteric oxides
›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 2s2 configuration (extra stability), while B's outermost electron is in 2p1 (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 → "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.
- (D) Al2O3 is a textbook amphoteric oxide (reacts with both acids and bases). As2O3, the oxide of a heavier p-block/metalloid-adjacent element, is likewise classified as amphoteric in standard periodic-trends tables (basic/amphoteric character increases down a group for p-block oxides). So (D) is a correct statement.
- Only (C) contains a factual error, making it the statement that is "not correct."
Common Mistakes
- Second-guessing (B) as "wrong" because Livermorium's final name isn't Ununhexium — but the statement is about the historical/systematic IUPAC name, which is indeed accurate, unlike (C)'s definite group-number error.
- Mixing up group 16 (chalcogens, ending at Livermorium) with group 17 (halogens, ending at Tennessine).
✓Final answerThe correct option is (C) — Tennessine actually belongs to group 17 (the halogens), not group 16.
ANSWER: C
- AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQQ.The spin only magnetic moment of the element having highest third ionization enthalpy among Ti,V,Cr,Mn,Fe in its +3 state (in BM) is (A) 3.87 (B) 4.90 (C) 5.92 (D) 2.84
›Reveal solutionSolution
The element with the anomalously highest third ionization enthalpy here is Mn (removing an electron from its stable d5 M2+ ion); Mn3+ (d4) 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). Mn2+ has the very stable 3d5 configuration, so removing a third electron from Mn (i.e. converting Mn2+→Mn3+) 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: Mn2+ is [Ar]3d5, a half-filled and hence very stable configuration.
- Removing the next electron (Mn2+→Mn3+) 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 Mn3+.
- Mn3+: [Ar]3d4. In the free ion/high-spin case, this is 4 unpaired electrons (each of the 4 d-orbitals singly occupied before pairing).
- Spin-only formula: μ=n(n+2)BM=4×6=24≈4.90BM.
Common Mistakes
- Assuming Cr has the highest third IE (Cr's anomaly is actually in the second IE, since Cr+ is 3d5 stable) — the third-IE anomaly belongs to Mn.
- Miscounting unpaired electrons in d4 as 2 instead of 4 (forgetting Hund's rule fills all 5 orbitals singly before pairing).
✓Final answerThe correct option is (B) — 4.90 BM.
ANSWER: B
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.Which of the following statements is not correct regarding third period elements? (A) Cl has the highest electron gain enthalpy (B) Ar has the highest first ionization enthalpy (C) Mg has higher ionization enthalpy than Al (D) P has a lower first ionization enthalpy than S
›Reveal solutionSolution
This tests the period-3 periodic trends, especially the P/S ionization-enthalpy anomaly caused by the half-filled 3p3 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 (p3) or fully-filled (p6) subshells creates local humps. Between P and S, phosphorus's half-filled 3p3 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 3p5) gains an electron most exothermically among period-3 elements (ΔegH≈−349 kJ/mol) — correct, so not the answer.
- (B) Noble gases have the highest IE1 in their period due to stable filled shells — Ar has the highest IE1 in period 3 — correct, so not the answer.
- (C) Mg (3s2, filled subshell, stable) has a higher IE1 than Al (3s23p1, the lone 3p electron is easier to remove due to shielding by 3s2 and slightly higher energy of the 3p orbital) — correct, so not the answer.
- (D) P (3s23p3, half-filled, extra-stable) actually has a HIGHER IE1 (1012 kJ/mol) than S (3s23p4, 1000 kJ/mol, where the fourth p-electron must pair up causing repulsion that makes removal easier). The statement claims P is lower than S — this is the reverse of reality.
Common Mistakes
- Assuming ionization enthalpy increases monotonically across a period without accounting for half-filled/fully-filled subshell stability dips.
- Confusing the Mg–Al dip (Mg higher) with the P–S dip (P higher) — both are "anomalies" in the same direction (the earlier element being higher).
✓Final answerThe correct option is (D) — P has a lower first ionization enthalpy than S is the statement that is NOT correct (in reality IE1(P)>IE1(S)).
ANSWER: D
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.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 (ns2) or exactly half-filled (np3) 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 IE1 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 ≈2.5, P ≈2.1. So C > P. Statement II is TRUE.
- Statement III: Group 14, period 3 ⇒ Silicon (Si), atomic number 14. A neutral atom's electron count equals its atomic number, so it has 14 electrons. Statement III is TRUE.
- All three are correct.
Common Mistakes
- Assuming the "exception" only occurs once (Mg–Al) and forgetting the P–S anomaly, so under-crediting statement I.
- Confusing electronegativity trend direction and thinking P > C.
- Second-guessing statement III as a trick (e.g., thinking of an ion) when it plainly describes the neutral atom.
✓Final answerThe correct option is (A) — I, II, III.
ANSWER: A
- AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQQ.Which of the following orders are not correct regarding first ionisation enthalpy of the given elements? I) B>Al II) Al>Ga III) Ga>In IV) In>Tl Correct answer is (A) I, III only (B) II, III only (C) II, IV only (D) I, IV only
›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 3d10 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 4f14 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≈801, Al≈577, Ga≈579, In≈558, Tl≈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 3d10 shielding).
- Test statement III: Ga>In gives 579>558, true.
- Test statement IV: In>Tl gives 558>589, which is false (Tl is higher due to the 4f14 lanthanide contraction on top of poor d-shielding).
- The question asks which orders are not correct: II and IV.
Common Mistakes
- Assuming ionisation enthalpy falls monotonically down every group -- Group 13 (and 14, similarly) has this well-documented anomaly.
- Forgetting that BOTH the 3d (Ga, Tl) and 4f (Tl only) poor-shielding effects stack for thallium, making it even higher than gallium, not just close to indium.
✓Final answerThe correct option is (C) -- II, IV only.
ANSWER: C
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.Which of the following orders is not correct about the property shown against it? (A) N>O>P>S - First ionisation enthalphy (B) F>Cl>O>S - Negative electron gain enthlapy (C) Fe3+<Fe2+<Fe - Size (D) O>N>S>P - Non-metallic character
›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 2p3 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 Fe3+<Fe2+<Fe: Removing electrons from a fixed nuclear charge increases effective nuclear pull per remaining electron, shrinking the ion. Going from neutral Fe to Fe2+ to Fe3+, size decreases, so increasing order is Fe3+<Fe2+<Fe. Order is correct.
- (D) Non-metallic character O>N>S>P: Non-metallic character tracks electronegativity/oxidizing power: O(3.44)>N(3.04)>S(2.58)>P(2.19), consistent with the stated order. Order is correct.
- Only (B) is wrong.
Common Mistakes
- Assuming electronegativity order and electron-gain-enthalpy magnitude order are always identical — they diverge exactly at F vs Cl and O vs S due to the small-atom repulsion anomaly.
- Forgetting that first ionisation enthalpy also has an N>O anomaly (many default to assuming enthalpy strictly increases left-to-right, missing the half-filled stability boost).
✓Final answerThe correct option is (B) — the order F>Cl>O>S for negative electron gain enthalpy is NOT correct.
ANSWER: B
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.Electronic configurations of four elements A, B, C, D are given below A) 1s22s22p63s1 B) 1s22s22p63s23p1 C) 1s22s22p63s2 D) 1s22s22p63s23p2 The correct order of first ionization enthalpy of these elements is (A) D>B>C>A (B) C>D>B>A (C) C>A>B>D (D) D>C>B>A
›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 = 1s22s22p63s1 = Na, B = 1s22s22p63s23p1 = Al, C = 1s22s22p63s2 = Mg, D = 1s22s22p63s23p2 = 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.
- Across the period ignoring the anomaly, IE would rise Na < Mg < Al < Si, but the Mg/Al anomaly flips the middle pair to give: Na < Al < Mg < Si.
- In descending order: Si > Mg > Al > Na, i.e. D>C>B>A.
Common Mistakes
- Ranking purely by atomic number/period position and missing the Mg > Al anomaly.
- Confusing this with the analogous N > O anomaly (half-filled 2p³ in N).
✓Final answerThe correct option is (D) — D>C>B>A.
ANSWER: D
- AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQQ.Which of the following orders is not correct for the given property ? (A) Li<Na<K - metallic radius (B) Br<F<Cl - electron gain enthalpy (C) C<N<O - first ionization enthalpy (D) Mg2+<Na+<F− - ionic radius
›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 (p3) or fully-filled (p6) subshell is extra stable, so removing an electron from it needs more energy than the simple increasing-across-the-period trend predicts. Nitrogen (1s22s22p3) has exactly this half-filled 2p3 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: Li<Na<K — correct order.
- (B) Electron gain enthalpy (magnitude) for halogens: 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: Br<F<Cl — correct order.
- (D) These are isoelectronic species with 10 electrons each: Mg2+(Z=12),Na+(Z=11),F−(Z=9). Higher nuclear charge pulls the same electron cloud in tighter, so radius decreases as Z increases: Mg2+<Na+<F− — correct order.
- (C) Naively across Period 2 ionization enthalpy should rise smoothly, giving C<N<O. But actual first IE values are approximately C:1086, N:1402, O:1314 kJ/mol. Because N's half-filled 2p3 is extra-stable, removing an electron from O (which has to break a paired 2p4 configuration, releasing some electron-electron repulsion) is actually easier than removing one from N. So the true order is C<O<N, not C<N<O.
- Hence option (C) states an order that is not correct — this is the answer.
Common Mistakes
- Assuming ionization enthalpy always increases monotonically across a period, ignoring the half-filled/fully-filled subshell stability exception.
- Confusing electron gain enthalpy trends (which do increase smoothly down halogens, except F's anomaly) with ionization enthalpy trends.
✓Final answerThe correct option is (C) — C < N < O for first ionization enthalpy is NOT correct (actual order is C < O < N).
ANSWER: C
- AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQQ.Match the following List-I (Element): A - Be, B - O, C - N, D - B List-II (First Ionization enthalpy, in kJmol−1): I - 801, II - 899, III - 1314, IV - 1402 (A) A - I, B - IV, C - III, D - II (B) A - II, B - IV, C - III, D - I (C) A - II, B - III, C - IV, D - I (D) A - I, B - III, C - IV, D - II
›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 2p1 electron shielded by the filled 2s2) 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 2p3 configuration at N). This gives the order B<Be<O<N, i.e. 801<899<1314<1402.
Step-by-Step Solution
- Recall/assign known values: Be=899, B=801, N=1402, O=1314 kJmol−1.
- Match List-I to List-II: A (Be) → II (899); B (O) → III (1314); C (N) → IV (1402); D (B) → I (801).
- This exactly matches option (C): A-II, B-III, C-IV, D-I.
Common Mistakes
- Assuming ionization enthalpy increases monotonically across the period and missing the B and O anomalies.
- Confusing which anomaly (half-filled stability at N) makes O's value lower than N's, not higher.
✓Final answerThe correct option is (C) — A - II, B - III, C - IV, D - I.
ANSWER: C
- AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQQ.The first ionization enthalpy (IE1) and second ionization enthalpy (IE2) of Mg(g) are 178 and 348 k cal mol−1 respectively. The energy required for the reaction Mg(g)→Mg2+(g)+2e− (in k. cal mol−1) is (A) +170 (B) +526 (C) -170 (D) -526
›Reveal solutionSolution
Converting Mg(g) to Mg2+(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 IE1 (to form Mg+) plus IE2 (to form Mg2+ from Mg+).
Step-by-Step Solution
- Mg(g)→Mg+(g)+e−: energy =IE1=+178 kcal/mol.
- Mg+(g)→Mg2+(g)+e−: energy =IE2=+348 kcal/mol.
- Adding (Hess's law): total energy =178+348=+526 kcal/mol.
Common Mistakes
- Giving the answer a negative sign — ionization is never exothermic.
- Subtracting instead of adding the two ionization energies.
✓Final answerThe correct option is (B) — +526.
ANSWER: B
- AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQQ.If the first ionisation enthalpy of Li, Be and C respectively are 520, 899, 1086 kJmol−1, the first ionisation enthalpy (in kJmol−1) of B will be (A) 487 (B) 950 (C) 801 (D) 1402
›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 2s2 configuration, which is harder to disturb, while Boron's outermost electron is in a 2p1 orbital that is higher in energy and more shielded from the nucleus by the 2s2 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 2s2 subshell in Be, and the higher-energy, less-penetrating 2p orbital in B, IE1(B)<IE1(Be).
- The known literature values are Li = 520, Be = 899, B = 801, C = 1086 kJ/mol -- B's value sits below Be's, consistent with the anomaly, and then C resumes the rising trend.
- Among the options, only 801 kJ/mol matches this real value and is consistent with IE1(B)<IE1(Be)<IE1(C).
Common Mistakes
- Assuming ionisation enthalpy increases monotonically across every period and picking a value between 899 and 1086 that follows a naive linear trend (e.g., option (B) 950) instead of recognizing the well-known Be-B anomaly.
✓Final answerThe correct option is (C) — 801 kJ/mol.
ANSWER: C
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