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.
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.Identify the element, which does not belong to fifth period of long form of periodic table (A) Pd (B) Pt (C) Cd (D) Ag
›Reveal solutionSolution
The period of an element in the periodic table is determined by the highest principal quantum number (n) of its electrons. Palladium (Pd), Cadmium (Cd), and Silver (Ag) are in the 5th period, while Platinum (Pt) is in the 6th period. The element that does not belong to the fifth period is (B) Pt.
The long form of the periodic table arranges elements based on their atomic number and recurring chemical properties. Elements with similar properties are placed in the same group, while elements with the same number of electron shells are placed in the same period.
The period number directly corresponds to the principal quantum number (n) of the outermost electron shell. For example, elements in the 1st period have their valence electrons in the 1st shell (n=1), elements in the 2nd period have their valence electrons in the 2nd shell (n=2), and so on. Therefore, to identify which element does not belong to the fifth period, we need to determine the period for each given element.
-
Understand the concept of a period:
The period number of an element in the periodic table is equal to the highest principal quantum number (n) occupied by its electrons. This means that for an element in the 5th period, its outermost electrons will be in the 5th shell (i.e., n=5).
-
Determine the period for each option:
We can determine the period by looking at the atomic number and the electron configuration, or by simply recalling their positions in the periodic table.
-
(A) Palladium (Pd):
- Atomic number (Z) = 46.
- Its electron configuration is [Kr]4d10. The highest principal quantum number is 4 for the d-subshell, but the valence shell is 5s (even if empty in this specific exception). The element is located in the 5th period.
-
(B) Platinum (Pt):
- Atomic number (Z) = 78.
- Its electron configuration is [Xe]4f145d96s1. The highest principal quantum number occupied by electrons is n=6 (for the 6s orbital). Therefore, Platinum belongs to the 6th period.
-
(C) Cadmium (Cd):
- Atomic number (Z) = 48.
- Its electron configuration is [Kr]4d105s2. The highest principal quantum number occupied by electrons is n=5 (for the 5s orbital). Therefore, Cadmium belongs to the 5th period.
-
(D) Silver (Ag):
- Atomic number (Z) = 47.
- Its electron configuration is [Kr]4d105s1. The highest principal quantum number occupied by electrons is n=5 (for the 5s orbital). Therefore, Silver belongs to the 5th period.
-
-
Identify the element that does not belong to the fifth period:
Comparing the periods:
- Pd: 5th period
- Pt: 6th period
- Cd: 5th period
- Ag: 5th period
Platinum (Pt) is the only element among the options that belongs to the 6th period, not the 5th.
Watch outWhile electron configurations can be complex for transition metals (with exceptions like Pd, Ag, Pt), the period number is always determined by the highest principal quantum number (n) of the valence shell, which is typically the s-orbital of that period. For example, for Pt, even though 4f and 5d orbitals are being filled, the 6s orbital marks the beginning of the 6th period.
The correct option is (B).
✓Final answerThe element that does not belong to the fifth period of the long form of the periodic table is (B) Pt.
-
- TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQQ.In which of the following compounds, the size of manganese is maximum? (A) KMnO4 (B) K2MnO4 (C) Mn2O7 (D) MnO2
›Reveal solutionSolution
The size of manganese is largest when it has the lowest oxidation state, because higher oxidation states pull electrons in more tightly. Among the given compounds, Mn in MnO₂ has the lowest oxidation state (+4), so (D) MnO₂ is the answer.
The key concept here is oxidation state and its effect on ionic radius. When an atom loses more electrons (higher oxidation state), the remaining electrons are held more tightly by the increased nuclear charge, shrinking the atom or ion. So, to find where manganese is largest, we need the compound where Mn has the lowest positive oxidation state.
Let’s determine the oxidation state of Mn in each compound:
-
KMnO₄: Potassium is +1, oxygen is –2 (four oxygens = –8). Let Mn be x.
(+1)+x+(−8)=0⇒x=+7.
Mn is in the +7 oxidation state — very high, so the ion is small.
-
K₂MnO₄: Two potassiums = +2, four oxygens = –8.
(+2)+x+(−8)=0⇒x=+6.
Mn is +6 — still high, but slightly lower than +7.
-
Mn₂O₇: Oxygen is –2, seven oxygens = –14. Let each Mn be x.
2x+(−14)=0⇒x=+7.
Again, Mn is +7 — same as in KMnO₄.
-
MnO₂: Oxygen is –2, two oxygens = –4.
x+(−4)=0⇒x=+4.
Mn is +4 — the lowest oxidation state among the options.
TipA common pitfall is to think that Mn₂O₇ has a different oxidation state than KMnO₄ because the compound looks different. But both have Mn in +7 — the oxidation state depends only on the bonding partners, not the overall formula’s complexity.
Since ionic size decreases as oxidation state increases, the Mn with the lowest oxidation state (+4 in MnO₂) will be the largest.
Watch outDon’t confuse “size of manganese” with the size of the whole compound or the anion. The question specifically asks about the manganese atom/ion itself.
✓Final answerThe correct option is (D).
ANSWER: D
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- TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQQ.In acidic medium one mole each of MnO4− and Cr2O72− is reduced by x and y moles of ferrous ions. The sum of x and y is (A) 14 (B) 12 (C) 10 (D) 11
›Reveal solutionSolution
The key is to balance the half‑reactions in acidic medium: each mole of MnO4− accepts 5 electrons, each mole of Cr2O72− accepts 6 electrons, and each ferrous ion donates 1 electron. Thus x=5, y=6, and x+y=11.
Concept & Intuition
The problem is about redox stoichiometry. In acidic solution, permanganate and dichromate are strong oxidizers; ferrous ion (Fe2+) is a common reducing agent that gets oxidized to ferric ion (Fe3+). The number of moles of Fe2+ needed to reduce one mole of oxidizer equals the number of electrons that oxidizer gains. So we just need the electron‑change per mole for each oxidizer, then add them.
Step‑by‑step reasoning
- Determine the electron change for MnO4− in acidic medium The half‑reaction is:
MnO4−+8H++5e−→Mn2++4H2O
Manganese goes from oxidation state +7 to +2, a gain of 5 electrons.
Therefore, 1 mole of MnO4− requires 5 moles of electrons.
- Relate electrons to ferrous ions Each Fe2+ loses one electron when oxidized to Fe3+:
Fe2+→Fe3++e−
So 1 mole of Fe2+ supplies exactly 1 mole of electrons.
Hence, to supply 5 moles of electrons, we need 5 moles of Fe2+.
Thus x=5.
- Determine the electron change for Cr2O72− in acidic medium The half‑reaction is:
Cr2O72−+14H++6e−→2Cr3++7H2O
Each chromium goes from +6 to +3, so two chromium atoms gain a total of 6 electrons.
Therefore, 1 mole of Cr2O72− requires 6 moles of electrons.
-
Convert to moles of ferrous ions
Again, each Fe2+ gives 1 electron, so 6 moles of electrons need 6 moles of Fe2+.
Thus y=6.
-
Sum x and y
x+y=5+6=11
Watch outA common mistake is to forget that dichromate contains two chromium atoms; its electron gain is 6, not 3. Also, ensure you use the acidic‑medium half‑reactions — the electron counts differ in basic medium.
TipYou can shortcut: MnO4− changes oxidation state by 5 (from +7 to +2), Cr2O72− changes by 6 (each Cr from +6 to +3, times 2). Since Fe2+ changes by 1, the mole ratios are directly those numbers.
✓Final answerThe correct option is (D).
ANSWER: D
- TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQQ.Work functions of four metals M1, M2, M3 and M4 are 4.8, 4.3, 4.75 and 3.75 eV respectively. The metals which do not show photoelectric effect when light of wavelength 310 nm falls on the metals are (A) M1, M2 only (B) M1, M3 only (C) M1, M2, M3 only (D) M1, M2, M4 only
›Reveal solutionSolution
The photoelectric effect occurs only if the incident photon energy exceeds the metal’s work function. For λ = 310 nm, the photon energy is 4.0 eV; metals with work function > 4.0 eV (M₁, M₂, M₃) do not emit electrons. The correct option is (C).
Concept & Intuition
The photoelectric effect requires that the energy of a single photon be at least as large as the work function (the minimum energy needed to eject an electron). If the photon energy is too small, no electrons are emitted, no matter how intense the light. So we compute the photon energy for the given wavelength and compare it to each metal’s work function. Metals with work function greater than the photon energy will not show the effect.
- Find the photon energy The energy of a photon is
E=λhc
where h=4.135667×10−15 eV⋅s (Planck’s constant in eV·s), c=3.00×108 m/s, and λ=310 nm=310×10−9 m.
Compute:
E=310×10−9(4.1357×10−15)(3.00×108) eV
First, hc=1240 eV⋅nm (a handy constant to remember).
So
E=310 nm1240 eV⋅nm=4.0 eV.
TipMemorize hc≈1240 eV⋅nm — it turns wavelength-in-nm to energy-in-eV instantly.
-
Compare with each work function
- M1: work function = 4.8 eV → 4.8 > 4.0 → no photoelectric effect.
- M2: work function = 4.3 eV → 4.3 > 4.0 → no photoelectric effect.
- M3: work function = 4.75 eV → 4.75 > 4.0 → no photoelectric effect.
- M4: work function = 3.75 eV → 3.75 < 4.0 → yes, photoelectric effect occurs.
-
Identify the metals that do NOT show the effect
These are M1, M2, and M3 — three metals.
Watch outA common mistake is to think that a work function equal to the photon energy still allows emission. In reality, the threshold is strict: the photon energy must be greater than or equal to the work function. Here, all three are strictly greater, so they fail.
- Match with the options The set {M1,M2,M3} corresponds to option (C).
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQQ.Identify the correct orders regarding atomic radii i. Cl > F > Li ii. P > C > N iii. Tm > Sm > Eu iv. Sr > Ca > Mg (A) i, ii, iii only (B) ii, iv only (C) ii, iii, iv only (D) iii, iv only
›Reveal solutionSolution
Atomic radii generally increase down a group and decrease across a period, but the lanthanide contraction (with the Eu half-filled anomaly) causes exceptions among the f-block elements. After checking each sequence, only orders ii and iv are correct, so the answer is (B).
Concept & Intuition
Atomic radius trends are governed by two main forces:
- Down a group: Adding electron shells increases size.
- Across a period: Increasing nuclear charge pulls electrons inward, shrinking the atom. For the f-block (lanthanides), the lanthanide contraction — a gradual decrease in radius across the series due to poor shielding of 4f electrons — makes later elements smaller than earlier ones. We must check each pair/sequence against these principles.
Step-by-step reasoning
-
Order i: Cl > F > Li
- Chlorine (Cl) is below fluorine (F) in Group 17, so Cl is larger than F. That part is correct.
- But fluorine (F) is in Period 2, lithium (Li) is also in Period 2, and across a period atomic radius decreases from left to right. Li (Group 1) is much larger than F (Group 17). So F > Li is false; actually Li > F.
- Therefore the whole order Cl > F > Li is incorrect.
-
Order ii: P > C > N
- Phosphorus (P) is in Period 3, carbon (C) and nitrogen (N) are in Period 2. P has an extra shell, so P is larger than both C and N. That part is fine.
- Between C and N: both in Period 2, but N has a higher nuclear charge (7 vs. 6), so N is smaller than C. Hence C > N is correct.
- So P > C > N is correct.
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Order iii: Tm > Sm > Eu
- These are lanthanides: Tm (thulium, atomic number 69), Sm (samarium, 62), Eu (europium, 63).
- Across the lanthanide series, atomic radius decreases (lanthanide contraction). So later elements (higher Z) are smaller.
- Tm (Z=69) comes after Sm (Z=62) and Eu (Z=63), so Tm is smaller than both. The order Tm > Sm would mean Tm is larger — that’s false.
- The given order is Tm > Sm > Eu. Tm (Z=69) comes later in the series, so by the lanthanide contraction it is the smallest, not the largest. Moreover, Eu is anomalously larger than Sm because of its stable half-filled 4f⁷ shell (radii ≈ Eu 185 pm, Sm 185 pm but Eu slightly larger, Tm ≈ 176 pm). The true order is therefore Eu > Sm > Tm — the reverse of what is given. So order iii is incorrect.
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Order iv: Sr > Ca > Mg
- Strontium (Sr) is below calcium (Ca) in Group 2, so Sr > Ca.
- Calcium (Ca) is below magnesium (Mg), so Ca > Mg.
- This follows the group trend perfectly. So correct.
Summary of correctness:
- i: false
- ii: true
- iii: false
- iv: true
Thus only orders ii and iv are correct, which matches option (B).
Watch outA common mistake is to assume all lanthanide radii decrease monotonically — but Eu is an exception because its half-filled f-shell gives it a slightly larger radius than its neighbors. Always check f-block anomalies.
TipFor quick checks: In the lanthanide series, the two largest atoms are La and Eu; the smallest is Lu. So any order claiming a later lanthanide (like Tm) is larger than an earlier one (like Sm or Eu) is almost certainly wrong.
✓Final answerThe correct option is (B).
ANSWER: B
- TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQQ.Among Li, Na, O, S, the elements having the most negative electron gain enthalpy, the least negative electron gain enthalpy respectively are (A) O, Li (B) O, Na (C) S, Na (D) S, Li
›Reveal solutionSolution
Electron gain enthalpy becomes more negative across a period and less negative down a group. Among Li, Na, O, S, the most negative is S and the least negative is Na, so the correct option is (C).
The concept here is electron gain enthalpy — the energy change when an isolated gaseous atom gains an electron. A more negative value means the atom releases more energy, so it has a stronger tendency to accept an electron. Two trends govern this: across a period, it becomes more negative (nuclear charge increases, pulling in the electron); down a group, it becomes less negative (atomic size increases, shielding weakens the pull). But there's a twist: oxygen is an exception because of its small size and high electron-electron repulsion in the 2p subshell, making its electron gain enthalpy less negative than sulfur's.
Let's place the elements and reason step by step.
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Identify the groups and periods.
Li and Na are in Group 1 (alkali metals). Li is in period 2, Na in period 3.
O and S are in Group 16 (chalcogens). O is in period 2, S in period 3.
-
Compare within Group 16: O vs S.
Down a group, electron gain enthalpy becomes less negative (i.e., more positive). But O is an exception: its small size causes strong repulsion when adding an electron to the already half-filled 2p⁴ subshell. So O actually has a less negative electron gain enthalpy than S.
Watch outA common mistake is to assume O is more negative because it's smaller. The repulsion in the compact 2p subshell makes O's electron gain enthalpy only about −141 kJ/mol, while S's is −200 kJ/mol. So S is more negative.
-
Compare within Group 1: Li vs Na.
Alkali metals have low (less negative) electron gain enthalpies because adding an electron to a filled s-orbital is energetically unfavorable. Down the group, the value becomes less negative. Na (larger) has a less negative value than Li.
Li: about −60 kJ/mol; Na: about −53 kJ/mol. So Na is the least negative among all four.
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Rank all four from most negative to least negative.
Most negative: S (−200 kJ/mol)
Then O (−141 kJ/mol)
Then Li (−60 kJ/mol)
Least negative: Na (−53 kJ/mol)
-
Match with the options.
The most negative is S, the least negative is Na. That corresponds to option (C) S, Na.
✓Final answerThe correct option is (C).
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- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.The following orbital energies (E) are compared. Identify the correct sets (I) E2s(H)=E2p(H) (II) E2s(H)=E2s(He) (III) E2s(H)<E2s(He) (IV) E3s(He)<E3s(H) (A) I, II only (B) I, II, III, IV (C) III, IV only (D) I, IV only
›Reveal solutionSolution
The key idea is that in hydrogen, orbitals of the same principal quantum number are degenerate (same energy), but in multi-electron atoms like helium, orbital energies depend on both n and l due to shielding and penetration. The correct sets are I and IV only.
The question asks you to compare orbital energies across different atoms and subshells. This is a classic test of how well you understand the difference between a hydrogen-like atom (single electron, no electron-electron repulsion) and a multi-electron atom (where electrons shield each other from the nucleus).
Let’s break it down concept by concept.
Why does hydrogen have degenerate orbitals? In a hydrogen atom, there is only one electron. The energy of an orbital depends only on the principal quantum number n, not on the azimuthal quantum number l. So 2s and 2p have exactly the same energy. That’s statement I.
Why does helium break that degeneracy? Helium has two electrons. They repel each other. An electron in a 2s orbital spends more time closer to the nucleus (it has a higher penetration) than one in a 2p orbital. So a 2s electron in helium feels a stronger effective nuclear charge and is more tightly bound — lower (more negative) energy — than a 2p electron. But here we are comparing across atoms, not within the same atom.
Now let’s go through each statement one by one.
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Statement I: E2s(H)=E2p(H)
In hydrogen, the energy depends only on n. Since both 2s and 2p have n=2, their energies are identical. This is true.
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Statement II: E2s(H)=E2s(He)
Hydrogen has nuclear charge Z=1; helium has Z=2. For a single electron in a 2s orbital, the energy scales as −Z2/n2. But helium’s 2s electron is not alone — the other electron (in 1s) shields it. The effective nuclear charge felt by the 2s electron in helium is much less than 2, but still greater than 1. So the 2s orbital in helium is lower in energy (more negative) than in hydrogen. They are not equal. This is false.
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Statement III: E2s(H)<E2s(He)
“Less than” here means more negative (more stable). As argued above, the 2s electron in helium feels a larger effective nuclear charge than the 2s electron in hydrogen, so its energy is lower (more negative). So E2s(H) is actually higher (less negative) than E2s(He). The inequality is reversed: E2s(H)>E2s(He). So statement III is false.
Watch outA common mistake is to think that because helium has a higher nuclear charge, its outer electrons automatically have lower energy. But you must account for shielding. The 2s electron in helium is still significantly shielded by the 1s electron, but the net effect is a higher Zeff than in hydrogen, so the energy is indeed lower (more negative). The direction of the inequality in III is wrong.
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Statement IV: E3s(He)<E3s(H)
Now compare 3s orbitals. In hydrogen, E3s=−13.6/9 eV. In helium, the 3s electron is even more shielded (by both 1s electrons), and it is in a higher n, so it feels a Zeff very close to 1. However, the 3s orbital in helium is still slightly more tightly bound than in hydrogen because the nuclear charge is 2 and the shielding is not perfect. The energy of 3s in helium is lower (more negative) than in hydrogen. So E3s(He)<E3s(H) is true.
TipFor any n and l, the orbital energy in a multi-electron atom is always lower (more negative) than in hydrogen for the same n and l, because Zeff>1 always. So Enl(He)<Enl(H) for all n,l where the orbital exists in both atoms.
So the true statements are I and IV only.
✓Final answerThe correct option is (D) I, IV only.
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- TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQQ.Identify the pair of elements in which the difference in atomic radii is maximum (A) C, N (B) O, F (C) P, S (D) Li, Be
›Reveal solutionSolution
The key idea is that atomic radii decrease across a period, and the largest drop occurs between the first two elements of a period. Among the given pairs, Li and Be (period 2) show the maximum difference in atomic radii.
The question asks for the pair with the maximum difference in atomic radii. Atomic radii generally decrease from left to right across a period due to increasing nuclear charge pulling electrons closer. The decrease is not uniform—it is steepest at the start of a period, where the effective nuclear charge jumps significantly.
Let’s examine each pair:
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Pair (A): C, N
Both are in period 2, but they are adjacent elements near the middle of the period. The decrease in radius from C to N is modest because the added proton’s pull is partially offset by electron-electron repulsion in the same shell.
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Pair (B): O, F
Also in period 2, but near the end. The radii are already small, and the difference between O and F is even smaller than between C and N—the contraction tapers off.
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Pair (C): P, S
These are in period 3. While the absolute radii are larger, the difference between adjacent elements in period 3 is similar to that in period 2, but not larger than the biggest drop in period 2.
-
Pair (D): Li, Be
These are the first two elements of period 2. The jump from Li (group 1) to Be (group 2) involves a large increase in nuclear charge (from +3 to +4) with no additional shell. This causes a relatively large contraction—the largest among all adjacent pairs in the same period. In fact, the difference in atomic radii between Li and Be is about 30–40 pm, while for C–N it’s about 10–15 pm, O–F about 5–10 pm, and P–S about 10–15 pm.
TipA common pitfall is to think that heavier elements (period 3) have larger absolute radii, so their differences must be larger. But the difference between adjacent elements is actually greatest at the start of a period, not later.
Thus, the pair with the maximum difference is Li and Be.
✓Final answerThe correct option is (D).
ANSWER: D
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- TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQQ.In group 14 elements, the element with highest melting point is Y and element with lowest melting point is X. X and Y respectively are (A) Si, C (B) Si, Pb (C) C, Sn (D) Sn, C
›Reveal solutionSolution
Melting points fall down Group 14 from carbon to lead, but tin is the one true exception — its melting point (232°C) is actually lower than lead's (327°C) because of tin's weaker metallic bonding. So the lowest-melting element is Sn (X) and the highest-melting is C (Y), matching option (D).
Concept & Intuition
Group 14 (C, Si, Ge, Sn, Pb) spans a non-metal (carbon), two metalloids (Si, Ge), and two metals (Sn, Pb). Carbon exists as a giant covalent solid (diamond) with very strong C–C bonds, giving it an exceptionally high melting point. Moving down the group, the bonding weakens overall, but the fall isn't perfectly smooth — tin's melting point actually dips below lead's because of how their metallic lattices pack.
Step-by-step reasoning
-
Recall the melting points across the group
Carbon (diamond) ≈ 3550°C, silicon ≈ 1414°C, germanium ≈ 938°C, tin ≈ 232°C, lead ≈ 327°C.
-
Identify the highest melting point
Carbon is far ahead of the rest, so Y (highest) = C.
-
Identify the lowest melting point
Among the group, tin (232°C) melts at a lower temperature than lead (327°C), so X (lowest) = Sn.
-
Match with the options
- (A) Si, C — X = Si is not the lowest.
- (B) Si, Pb — neither matches.
- (C) C, Sn — X and Y are reversed.
- (D) Sn, C — X = Sn (lowest), Y = C (highest) — correct.
Watch outIt's tempting to assume lead, being lower in the group, must have the lowest melting point. In reality tin melts at a lower temperature than lead — always check the actual values rather than assuming a strictly monotonic trend down a group.
TipOrder to remember: C ≫ Si > Ge > Pb > Sn — carbon towers over the rest, and tin (not lead) is the softest-melting member of the group.
✓Final answerThe correct option is (D).
ANSWER: D
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- TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQQ.The set containing the elements with positive electron gain enthalpies is (A) S, Se, Te (B) Kr, Xe, Rn (C) Cl, Br, I (D) K, Rb, Cs
›Reveal solutionSolution
Electron gain enthalpy is usually negative (energy released) for most non‑metals, but noble gases and certain heavy elements have positive values because adding an electron requires energy. Among the given sets, only the noble gases Kr, Xe, Rn have positive electron gain enthalpies, so the correct option is (B).
Concept & Intuition
Electron gain enthalpy (or electron affinity) is the energy change when an atom in the gas phase gains an electron. A negative value means energy is released (favourable), while a positive value means energy must be supplied (unfavourable). Most elements have negative electron gain enthalpies, but noble gases have completely filled shells, so adding an electron forces it into a higher-energy orbital, requiring energy input. Similarly, some heavy elements with stable half‑filled or fully‑filled subshells can also show positive values, but the most clear‑cut case is the noble gases.
Step‑by‑Step Reasoning
-
Recall the trend for electron gain enthalpy
- Across a period, electron gain enthalpy becomes more negative (more energy released) as we move toward the halogens.
- Down a group, it generally becomes less negative (or more positive) because the added electron goes into a larger orbital, experiencing less effective nuclear charge.
- Exception: Noble gases have positive electron gain enthalpies because their octet is complete; adding an electron requires overcoming repulsion and entering a new shell.
-
Examine each option
- (A) S, Se, Te – These are chalcogens (Group 16). They all have negative electron gain enthalpies (e.g., S: –200 kJ/mol, Se: –195 kJ/mol, Te: –190 kJ/mol). So not positive.
- (B) Kr, Xe, Rn – These are noble gases (Group 18). Their electron gain enthalpies are positive: Kr ≈ +60 kJ/mol, Xe ≈ +77 kJ/mol, Rn ≈ +68 kJ/mol. This matches the condition.
- (C) Cl, Br, I – These are halogens (Group 17). They have the most negative electron gain enthalpies (Cl: –349 kJ/mol, Br: –325 kJ/mol, I: –295 kJ/mol). Definitely not positive.
- (D) K, Rb, Cs – These are alkali metals (Group 1). They have negative but relatively small electron gain enthalpies (e.g., K: –48 kJ/mol, Rb: –47 kJ/mol, Cs: –46 kJ/mol). Not positive.
-
Confirm the only set with all positive values
Only the noble gases in option (B) have positive electron gain enthalpies. All other groups have negative values.
Watch outA common mistake is to think that all non‑metals have negative electron gain enthalpies. While true for most, noble gases are the clear exception — their positive values are a direct consequence of their stable electronic configuration.
TipRemember: “Noble gases are noble — they don’t like gaining electrons.” If a question asks for positive electron gain enthalpy, immediately think of Group 18 elements.
✓Final answerThe correct option is (B).
ANSWER: B
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- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.The wavenumber of first spectral line of Lyman series of He+ ion is x m−1. What is the wavenumber (in m−1) of second spectral line of Balmer series of Li2+ ion? (A) 169x (B) 916x (C) 278x (D) 827x
›Reveal solutionSolution
For a hydrogen-like ion the wavenumber is ν~=RZ2(nf21−ni21). Fixing R from the He+ Lyman first line and evaluating the Li2+ Balmer second line gives 169x, option (A).
For any one-electron (hydrogen-like) ion the Rydberg formula applies with the nuclear charge Z:
ν~=RZ2(nf21−ni21)
The Rydberg constant R is common to all such ions, so it cancels when we form the ratio of the two lines.
Step 1 - The given line (He+, Z=2).
The first line of the Lyman series is the transition ni=2→nf=1:
x=R(2)2(121−221)=4R⋅43=3R⇒R=3x
Step 2 - The required line (Li2+, Z=3).
The second line of the Balmer series is the transition ni=4→nf=2:
ν~=R(3)2(221−421)=9R(41−161)=9R⋅163=1627R
Step 3 - Substitute R=3x.
ν~=1627⋅3x=4827x=169x
Sanity check: the Z2 factor scales the line up by (3/2)2=2.25, while the transition bracket falls from 43 to 163 (a factor 41). Net 2.25×0.25=0.5625=169, confirming the result.
✓Final answerWavenumber =169x m−1. The correct option is (A).
ANSWER: A
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.In which of the following ionic pairs, second ion is smaller in size than the first ion? (A) Al3+, Mg2+ (B) F−, Na+ (C) O2−, N3− (D) Mg2+, Na+
›Reveal solutionSolution
The key idea is that ionic size depends on nuclear charge and number of electrons. For isoelectronic species, higher nuclear charge pulls electrons in tighter, making the ion smaller. The pair where the second ion is smaller is (B).
Concept & Intuition
When comparing ionic sizes, two main factors matter:
- Number of electron shells – more shells = larger radius.
- Effective nuclear charge – for ions with the same number of electrons (isoelectronic), the one with more protons pulls the electron cloud inward more strongly, making it smaller.
Here, all pairs are either isoelectronic or have different electron counts. We check each pair systematically.
-
Pair (A): Al3+ and Mg2+
- Both have 10 electrons (isoelectronic with neon).
- Al3+ has 13 protons, Mg2+ has 12 protons.
- More protons → stronger pull → smaller radius.
- So Al3+ is smaller than Mg2+.
- Here the first ion is smaller, not the second. → Not correct.
-
Pair (B): F− and Na+
- Both have 10 electrons (isoelectronic with neon).
- F− has 9 protons, Na+ has 11 protons.
- More protons → smaller radius.
- So Na+ is smaller than F−.
- The second ion is smaller. → This fits.
-
Pair (C): O2− and N3−
- Both have 10 electrons (isoelectronic).
- O2− has 8 protons, N3− has 7 protons.
- More protons → smaller radius.
- So O2− is smaller than N3−.
- Here the first ion is smaller, not the second. → Not correct.
-
Pair (D): Mg2+ and Na+
- Both have 10 electrons (isoelectronic).
- Mg2+ has 12 protons, Na+ has 11 protons.
- More protons → smaller radius.
- So Mg2+ is smaller than Na+.
- Again the first ion is smaller. → Not correct.
Watch outA common mistake is to think that more negative charge always means a larger ion. While that’s true for the same element, for different elements with the same electron count, the number of protons is the deciding factor.
TipFor isoelectronic series, just compare atomic numbers: higher Z = smaller radius. No need to memorize numbers — just count protons.
✓Final answerThe correct option is (B).
ANSWER: B
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