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

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Which of the following is not the correct order of atomic radius of the elements given? (A) $Br < Ge < Ga < Ca$ (B) $Cr < V < Ti < Sc$ (C) $F < Cl < K < Cs$ (D) $O < P < K < Ge$

   ›Reveal solutionSolution

   Checking each sequence against real periodic trends shows options (A)-(C) are consistent, but (D) wrongly places the large alkali metal potassium before the much smaller metalloid germanium. Answer: (D).

   Concept and Intuition

   Atomic radius decreases across a period (left to right, as nuclear charge increases while shielding stays roughly constant) and increases down a group (as a new shell is added). When a sequence mixes elements from very different groups, it's easy to state an order that looks plausible but violates the actual measured sizes — this question is checking exactly that.

   Step-by-Step Solution

   1. (A) $Br<Ge<Ga<Ca$: all four (Ca, Ga, Ge, Br) lie in period 4, increasing atomic number left to right ($Ca\to Ga\to Ge\to Br$), so radius should decrease in that order — i.e. increase as $Br<Ge<Ga<Ca$. This matches the normal periodic trend: correct.
   2. (B) $Cr<V<Ti<Sc$: these are consecutive 3d transition metals (Sc, Ti, V, Cr), whose atomic radii decrease steadily left to right due to increasing effective nuclear charge with poor 3d shielding — so increasing order $Cr<V<Ti<Sc$ is correct.
   3. (C) $F<Cl<K<Cs$: $F$ and $Cl$ are small period-2/3 halogens; $K$ and $Cs$ are much larger alkali metals from periods 4 and 6. Numerically $F(\approx64\text{pm}) < Cl(\approx99\text{pm}) < K(\approx227\text{pm}) < Cs(\approx265\text{pm})$ — correct.
   4. (D) $O<P<K<Ge$: $O(\approx66\text{pm})<P(\approx110\text{pm})$ is fine, but then it claims $K<Ge$. Actually $K$ (Group 1, period 4, $\approx227$ pm) is one of the largest atoms in period 4, while $Ge$ (Group 14, period 4, $\approx122$ pm) is much smaller since radius falls sharply moving right across period 4. So $K>Ge$, not $K<Ge$ — this order is wrong.

   Common Mistakes

   • Assuming atomic number alone predicts radius even when comparing across very different groups — the group trend (down = bigger) can outweigh a modest atomic-number increase within the same period, especially for Group-1 metals.
   • Not recognizing potassium (an alkali metal) as unusually large for its period when it appears in a list otherwise dominated by smaller p-block elements.

   ✓Final answer

   The correct option is (D) — $O < P < K < Ge$.

   ANSWER: D

2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Identify the correct order of the given compounds against the given property (A) $\mathrm{BeCl_2 < MgCl_2 < CaCl_2}$ --- Covalent character (B) $\mathrm{CaSO_4 < SrSO_4 < BaSO_4}$ --- Solubility (C) $\mathrm{Mg(OH)_2 < Ca(OH)_2 < Ba(OH)_2}$ --- Basic Character (D) $\mathrm{BaCO_3 < SrCO_3 < MgCO_3}$ --- Thermal stability

   ›Reveal solutionSolution

   Checking each option against the real periodic trend for Group 2 compounds shows only the basic-character order of the hydroxides is stated correctly; the others reverse their true trend.

   Concept and Intuition

   Down Group 2, cation size increases and polarising power decreases, which increases ionic/basic character of hydroxides, changes solubility of different salts depending on lattice vs hydration energy balance, and increases thermal stability of oxosalts (carbonates/sulfates) because the larger cation stabilizes the larger anion in the solid lattice.

   Step-by-Step Solution

   1. (A) Covalent character $BeCl_2<MgCl_2<CaCl_2$ — WRONG: covalent character actually decreases down the group (Be²⁺, smallest & most polarising, gives the most covalent chloride).
   2. (B) Solubility $CaSO_4<SrSO_4<BaSO_4$ — WRONG: sulfate solubility decreases down the group, so BaSO₄ is the least soluble, not the most.
   3. (C) Basic character $Mg(OH)_2<Ca(OH)_2<Ba(OH)_2$ — CORRECT: basicity of the hydroxides genuinely increases down the group.
   4. (D) Thermal stability $BaCO_3<SrCO_3<MgCO_3$ — WRONG: thermal stability of carbonates increases down the group (true order is $MgCO_3<SrCO_3<BaCO_3$).

   Common Mistakes

   • Assuming "bigger cation = more of everything" without checking whether the specific property actually increases or decreases down the group.
   • Confusing the covalent-character trend (decreases down the group) with the basic-character trend (increases down the group).

   ✓Final answer

   The correct option is (C) — $Mg(OH)_2<Ca(OH)_2<Ba(OH)_2$ — basic character.

   ANSWER: C

3. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Identify the pairs of elements, which possess almost the same size as well as properties I. $Ti, Zr$ II. $Zr, Hf$ III. $Mo, W$ IV. $Nb, Ta$ V. $Cr, Mo$ The correct answer is (A) I & II only (B) II, III & IV only (C) III, IV & V only (D) I, II & V only

   ›Reveal solutionSolution

   This tests the lanthanide contraction effect, which makes specific pairs of 2nd/3rd-row transition elements (in the same group) nearly identical in size and chemistry.

   Concept and Intuition

   The lanthanide contraction is the steady decrease in atomic/ionic radius across the lanthanides (due to poor shielding by 4f electrons), which causes elements right after the lanthanides (3rd transition series) to have almost the same size as their 2nd-series counterparts one period above, despite the extra shell. This makes Zr–Hf, Nb–Ta, and Mo–W famously similar in radius and chemical behaviour — a textbook example of near-identical "twin" pairs.

   Step-by-Step Solution

   1. Zr (Z=40, 4d) and Hf (Z=72, 5d): lanthanide contraction cancels the expected radius increase → almost identical radii and properties. (Pair II ✓)
   2. Mo (Z=42, 4d) and W (Z=74, 5d): same effect → very similar radii/chemistry. (Pair III ✓)
   3. Nb (Z=41, 4d) and Ta (Z=73, 5d): same effect. (Pair IV ✓)
   4. Ti (Z=22, 3d) and Zr (Z=40, 4d): Ti is in the first transition series (no lanthanides precede it), so there's no contraction effect here — Zr is noticeably larger than Ti. (Pair I ✗)
   5. Cr (Z=24, 3d) and Mo (Z=42, 4d): again first-vs-second series without a compensating contraction — not nearly identical. (Pair V ✗)

   Common Mistakes

   • Assuming any same-group pair is automatically "nearly identical" — the lanthanide contraction effect specifically applies to 2nd-vs-3rd series pairs, not 1st-vs-2nd.
   • Forgetting that Ti and Cr are first-series (3d) elements, not compensated by the lanthanide contraction.

   ✓Final answer

   The correct option is (B) — II, III & IV only.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Which of the following orders are correct against the stated property? I) $\mathrm{NaO_2 < KO_2 < RbO_2 < CsO_2}$ - stability II) $\mathrm{Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2}$ - basic strength III) $\mathrm{MgCO_3 < CaCO_3 < SrCO_3}$ - thermal stability (A) I & III only (B) II & III only (C) I & II only (D) I, II & III

   ›Reveal solutionSolution

   This tests three parallel s-block periodic trends (superoxide stability, hydroxide basicity, carbonate thermal stability), all of which increase down their respective groups as cation size increases. All three stated orders are correct.

   Concept and Intuition

   A recurring theme in s-block chemistry is that a LARGER, less polarizing cation better stabilizes a LARGE anion (superoxide $O_2^-$, carbonate $CO_3^{2-}$) — this is Fajans'-rule reasoning: a small, highly polarizing cation distorts the anion's electron cloud, weakening it and making the compound less stable (or, for hydroxides, making the M–OH bond more covalent and hence less readily releasing $OH^-$, i.e. less basic). So down a group, as cation size grows and polarizing power falls, superoxide stability, hydroxide basicity, and carbonate thermal stability all increase together.

   Step-by-Step Solution

   1. Statement I (superoxide stability): Alkali metal superoxides ($MO_2$) are stabilized by larger cations that better match the large superoxide ion's size/charge density (a lattice-energy/radius-ratio argument). Down the group Na→K→Rb→Cs, cation size increases, so stability increases: $NaO_2<KO_2<RbO_2<CsO_2$. Correct.
   2. Statement II (basic strength of Group 2 hydroxides): Down the group Mg→Ca→Sr, the metal becomes more electropositive and the M–OH bond becomes more ionic, so the hydroxide ionises more readily and basicity increases: $Mg(OH)_2<Ca(OH)_2<Sr(OH)_2$. Correct.
   3. Statement III (thermal stability of Group 2 carbonates): Down the group Mg→Ca→Sr, the larger cation causes less polarization/distortion of the carbonate ion, so the carbonate resists decomposition (to the oxide + $CO_2$) at higher temperatures: $MgCO_3<CaCO_3<SrCO_3$. Correct.
   4. All three statements are correct.

   Common Mistakes

   • Assuming smaller cations always give more stable ionic compounds — this only holds for compounds of small anions (higher lattice energy from closer approach); for LARGE anions like superoxide and carbonate, the trend actually favors bigger cations (better size match, less polarization).
   • Mixing up "thermal stability of carbonates" with "solubility of carbonates" (which decreases down the group) — these are different trends and easy to confuse.

   ✓Final answer

   The correct option is (D) — I, II & III.

   ANSWER: D

5. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.Match the following List-I (element) — List-II (atomic radius in pm) A) Al — I) 64 B) F — II) 117 C) N — III) 143 D) Si — IV) 74 The correct answer is (A) A-III, B-I, C-IV, D-II (B) A-III, B-IV, C-II, D-I (C) A-I, B-III, C-II, D-IV (D) A-I, B-IV, C-III, D-II

   ›Reveal solutionSolution

   This is a periodic-trends recall question. Matching known atomic radii values (Al=143, F=64, N=74, Si=117 pm) to the given list gives A-III, B-I, C-IV, D-II.

   Concept and Intuition

   Atomic radius generally decreases across a period (increasing effective nuclear charge) and increases down a group. Al (Period 3, Group 13) has a much larger radius than F (Period 2, Group 17, near the top-right of the periodic table where radii are smallest). N (Period 2, Group 15) is smaller than Al but bigger than F because F is further right in the same period. Si (Period 3, Group 14) is smaller than Al (further right in Period 3) but noticeably bigger than second-period N.

   Step-by-Step Solution

   1. Recall/rank the known covalent atomic radii: Al $\approx 143$ pm, Si $\approx 117$ pm, N $\approx 74$ pm, F $\approx 64$ pm.
   2. Match against List-II values: I=64, II=117, III=143, IV=74.
   3. A) Al = 143 = III.
   4. B) F = 64 = I.
   5. C) N = 74 = IV.
   6. D) Si = 117 = II.
   7. This gives A-III, B-I, C-IV, D-II — exactly option (A).

   Common Mistakes

   • Mixing up N and F radii (F is smaller since it's further right in the same period).
   • Forgetting that going down a group (N→ heavier group 15, or comparing periods) increases radius, causing wrong period-3 vs period-2 comparisons.

   ✓Final answer

   The correct option is (A) — A-III, B-I, C-IV, D-II.

   ANSWER: A

6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Among the ions $\mathrm{Mg^{2+}}$, $\mathrm{O^{2-}}$, $\mathrm{Al^{3+}}$, $\mathrm{F^{-}}$, $\mathrm{Na^{+}}$ and $\mathrm{N^{3-}}$, the ion with largest size and ion with smallest size are respectively (A) $\mathrm{N^{3-}, Mg^{2+}}$ (B) $\mathrm{O^{2-}, F^{-}}$ (C) $\mathrm{Al^{3+}, N^{3-}}$ (D) $\mathrm{N^{3-}, Al^{3+}}$

   ›Reveal solutionSolution

   All six ions have exactly 10 electrons (the neon configuration); for isoelectronic species, higher nuclear charge pulls the same electron cloud in tighter, so size decreases as atomic number increases from N to Al.

   Concept and Intuition

   Isoelectronic species have the same number of electrons and the same electron configuration (all reach the stable Ne core here), so the outer electron cloud has essentially the same shape and shell structure across the series. The only thing that differs is the nuclear charge $Z$ pulling on that identical electron cloud. A larger $Z$ means a stronger effective pull per electron, contracting the ion; a smaller $Z$ means a weaker pull, letting the ion stay larger. So within an isoelectronic series, ionic radius decreases monotonically as $Z$ increases.

   Step-by-Step Solution

   1. Confirm all six are isoelectronic (10 electrons each): $\mathrm{N}(Z=7)+3e^-=10$; $\mathrm{O}(Z=8)+2e^-=10$; $\mathrm{F}(Z=9)+1e^-=10$; $\mathrm{Na}(Z=11)-1e^-=10$; $\mathrm{Mg}(Z=12)-2e^-=10$; $\mathrm{Al}(Z=13)-3e^-=10$.
   2. List by increasing atomic number (increasing nuclear charge): $\mathrm{N}(7) < \mathrm{O}(8) < \mathrm{F}(9) < \mathrm{Na}(11) < \mathrm{Mg}(12) < \mathrm{Al}(13)$.
   3. Ionic radius decreases as $Z$ increases along this isoelectronic series, so the order of size (largest to smallest) is $\mathrm{N^{3-}} > \mathrm{O^{2-}} > \mathrm{F^-} > \mathrm{Na^+} > \mathrm{Mg^{2+}} > \mathrm{Al^{3+}}$.
   4. Largest = $\mathrm{N^{3-}}$ (smallest $Z=7$); smallest = $\mathrm{Al^{3+}}$ (largest $Z=13$).

   Common Mistakes

   • Comparing charge magnitude alone instead of recognizing the isoelectronic-series rule (all have the same electron count, so it's purely a nuclear-charge argument).
   • Mixing up which end of the series is largest — a common slip is picking the highest-charge cation as "largest" by mistakenly reasoning more charge means bigger, when the opposite is true for isoelectronic cations (more positive charge = smaller ion).

   ✓Final answer

   The correct option is (D) — $\mathrm{N^{3-}, Al^{3+}}$.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.Observe the following data.

   Ion	$\mathrm{Q^{a+}}$	$\mathrm{X^{b+}}$	$\mathrm{Y^{c+}}$	$\mathrm{Z^{d+}}$
   Radius (pm)	53	66	40	100

   $\mathrm{Q^{a+}, X^{b+}, Y^{c+}, Z^{d+}}$ are respectively (A) $\mathrm{Mg^{2+}, Al^{3+}, Na^{+}, Si^{4+}}$ (B) $\mathrm{Al^{3+}, Si^{4+}, Mg^{2+}, Na^{+}}$ (C) $\mathrm{Mg^{2+}, Si^{4+}, Al^{3+}, Na^{+}}$ (D) $\mathrm{Al^{3+}, Mg^{2+}, Si^{4+}, Na^{+}}$

   ›Reveal solutionSolution

   This tests isoelectronic ionic radius trends; matching radii to Al³⁺, Mg²⁺, Si⁴⁺, Na⁺ gives option D.

   Concept and Intuition

   Na$^+$, Mg$^{2+}$, Al$^{3+}$, and Si$^{4+}$ are all isoelectronic — each has lost electrons down to the neon configuration (10 electrons). For an isoelectronic series, all ions have the same number of electrons but different nuclear charges. A higher nuclear charge pulls the same electron cloud in more tightly, so radius decreases as the ionic charge (and atomic number) increases: $\mathrm{Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}}$.

   Step-by-Step Solution

   1. List the known approximate ionic radii: $\mathrm{Na^+} \approx 100\,\mathrm{pm}$, $\mathrm{Mg^{2+}} \approx 66\,\mathrm{pm}$, $\mathrm{Al^{3+}} \approx 53\,\mathrm{pm}$, $\mathrm{Si^{4+}} \approx 40\,\mathrm{pm}$ — decreasing in that order.
   2. Match to the table: $Q = 53\,\mathrm{pm} \Rightarrow \mathrm{Al^{3+}}$; $X = 66\,\mathrm{pm} \Rightarrow \mathrm{Mg^{2+}}$; $Y = 40\,\mathrm{pm} \Rightarrow \mathrm{Si^{4+}}$; $Z = 100\,\mathrm{pm} \Rightarrow \mathrm{Na^+}$.
   3. So $Q,X,Y,Z = \mathrm{Al^{3+}, Mg^{2+}, Si^{4+}, Na^+}$.

   Common Mistakes

   • Assuming radius increases with charge (it's the opposite for isoelectronic cations — more protons pull electrons in tighter).
   • Mixing up which radius value belongs to which ion when several are close together.

   ✓Final answer

   The correct option is (D) — $\mathrm{Al^{3+}, Mg^{2+}, Si^{4+}, Na^{+}}$.

   ANSWER: D

8. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.Identify the correct order of polarising power of given cations (A) $\mathrm{Be^{2+} < K^+ < Mg^{2+} < Ca^{2+}}$ (B) $\mathrm{K^+ < Ca^{2+} < Mg^{2+} < Be^{2+}}$ (C) $\mathrm{Ca^{2+} < Mg^{2+} < K^+ < Be^{2+}}$ (D) $\mathrm{K^+ < Mg^{2+} < Ca^{2+} < Be^{2+}}$

   ›Reveal solutionSolution

   Polarising power (Fajans' rules) tracks charge density; ranking these cations from lowest to highest charge density gives $\mathrm{K^+ < Ca^{2+} < Mg^{2+} < Be^{2+}}$.

   Concept and Intuition

   According to Fajans' rules, the polarising power of a cation increases with its charge and decreases with its size — i.e., it scales with charge density (charge/ionic radius). Among these four cations:

   • $\mathrm{K^+}$: charge +1, largest radius → lowest charge density.
   • $\mathrm{Ca^{2+}}$: charge +2, moderately large radius.
   • $\mathrm{Mg^{2+}}$: charge +2, smaller radius than Ca$^{2+}$ (same group, higher up).
   • $\mathrm{Be^{2+}}$: charge +2, smallest radius of all (top of the group) → highest charge density.

   Step-by-Step Solution

   1. Compare $\mathrm{K^+}$ (singly charged, large) to the divalent ions: despite its larger size, its lower charge (+1) makes its polarising power the smallest of the four.
   2. Among the +2 ions, radius decreases going up Group 2: $\mathrm{Ca^{2+}} > \mathrm{Mg^{2+}} > \mathrm{Be^{2+}}$ in size, so polarising power increases in the reverse order: $\mathrm{Ca^{2+}} < \mathrm{Mg^{2+}} < \mathrm{Be^{2+}}$.
   3. Combining: $\mathrm{K^+ < Ca^{2+} < Mg^{2+} < Be^{2+}}$.
   4. This matches option (B).

   Common Mistakes

   • Assuming $\mathrm{K^+}$'s larger size alone determines its polarising power, ignoring that its lower charge (+1 vs +2) is the dominant factor making it the weakest polariser here.
   • Reversing the Group 2 size trend (forgetting $\mathrm{Be^{2+}}$ is the smallest, hence the strongest polariser).

   ✓Final answer

   The correct option is (B) — $\mathrm{K^+ < Ca^{2+} < Mg^{2+} < Be^{2+}}$.

   ANSWER: B

9. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.The atomic radius (in pm) of Mg, P, Si, Al respectively is (A) $160, 143, 117, 110$ (B) $110, 117, 143, 160$ (C) $160, 110, 117, 143$ (D) $110, 160, 143, 117$

   ›Reveal solutionSolution

   Using the standard period-3 atomic radii (Mg 160 pm, P 110 pm, Si 117 pm, Al 143 pm), the values in the asked order (Mg, P, Si, Al) are $160, 110, 117, 143$ pm.

   Concept and Intuition

   Across a period, atomic radius generally decreases from left to right because each successive element adds a proton (increasing effective nuclear charge) while the valence electrons stay in the same shell, pulling them inward. This is why Mg (group 2) has a noticeably larger radius than the elements to its right (Al, Si, P) in period 3.

   Step-by-Step Solution

   1. Recall the standard period-3 atomic radii (pm): Na 186, Mg 160, Al 143, Si 117, P 110, S 104, Cl 99.
   2. The question asks for the radii in the specific order Mg, P, Si, Al.
   3. Substituting: Mg = 160, P = 110, Si = 117, Al = 143.
   4. So the ordered set is $160, 110, 117, 143$, matching option (C).

   Common Mistakes

   • Listing the values in strictly decreasing order of atomic number (Mg>Al>Si>P) instead of in the exact order the question asks for (Mg, P, Si, Al) — the question deliberately scrambles the periodic order.
   • Confusing atomic radius trend with ionic radius trend (which can differ for cations/anions).

   ✓Final answer

   The correct option is (C) — $160, 110, 117, 143$.

   ANSWER: C

10. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The correct order of decomposition temperature of $\mathrm{MgCO_3}$ (X), $\mathrm{BaCO_3}$ (Y), $\mathrm{CaCO_3}$ (Z) is (A) $Y > Z > X$ (B) $X > Y > Z$ (C) $Y > X > Z$ (D) $X > Z > Y$

    ›Reveal solutionSolution

    This tests the periodic trend in thermal stability of Group 2 metal carbonates, governed by cation size and lattice energy considerations. The answer is (A).

    Concept and Intuition

    The thermal decomposition of a metal carbonate ($MCO_3 \rightarrow MO + CO_2$) is favoured when the resulting oxide lattice is more stable relative to the carbonate lattice. Down a group, as the cation gets larger, the difference in lattice energy between the carbonate and the oxide becomes smaller -- practically, this manifests as an increase in thermal stability of the carbonate as you go down the group, so higher temperatures are needed to decompose the carbonates of larger cations.

    Step-by-Step Solution

    1. Identify the three carbonates given: X = MgCO3, Y = BaCO3, Z = CaCO3.
    2. Recall the trend: thermal stability of Group 2 carbonates increases down the group: $MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.
    3. Higher thermal stability means a higher decomposition temperature is required, so the decomposition temperature order follows the same direction: $BaCO_3 > CaCO_3 > MgCO_3$.
    4. Substituting the labels: $Y > Z > X$.

    Common Mistakes

    • Reversing the trend and assuming smaller cations (like Mg2+, which is more reactive in some contexts) give more thermally stable carbonates -- it is actually the opposite for this specific lattice-energy-based trend.

    ✓Final answer

    The correct option is (A) — $Y > Z > X$.

    ANSWER: A

11. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.The correct order of atomic radii of N, F, Al, Si is (A) $\mathrm{F > N > Si > Al}$ (B) $\mathrm{F > N > Al > Si}$ (C) $\mathrm{Al > Si > F > N}$ (D) $\mathrm{Al > Si > N > F}$

    ›Reveal solutionSolution

    Compare atomic radii using periodic trends: radius increases down a group/across periods (new shell) and decreases across a period (increasing nuclear charge). Answer: $Al > Si > N > F$.

    Concept and Intuition

    Atomic radius decreases across a period (left to right) because the increasing nuclear charge pulls the same-shell electrons in tighter, and it increases down a group because a new electron shell is added. N and F are period-2 elements; Al and Si are period-3 elements — being one period lower, they have an extra electron shell, making them substantially larger than any period-2 element.

    Step-by-Step Solution

    1. Within period 2: N (group 15) is to the left of F (group 17), so $N > F$ (74 pm vs 64 pm, approximately).
    2. Within period 3: Al (group 13) is to the left of Si (group 14), so $Al > Si$ (143 pm vs 117 pm, approximately).
    3. Since period 3 atoms have an extra shell compared to period 2, both $Al$ and $Si$ are larger than both $N$ and $F$.
    4. Combining: $Al > Si > N > F$.

    Common Mistakes

    • Comparing only within-period trends and forgetting that period-3 elements are categorically larger than period-2 elements due to the extra shell.
    • Mixing up the direction of the periodic trend (assuming radius increases across a period).

    ✓Final answer

    The correct option is (D) — $\mathrm{Al > Si > N > F}$.

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.The correct order of atomic radii of given elements is (A) B < Be < Mg (B) Mg < Be < B (C) Be < B < Mg (D) B < Mg < Be

    ›Reveal solutionSolution

    This tests the periodic trend of atomic radius across a period and down a group; the answer is B < Be < Mg.

    Concept and Intuition

    Atomic radius is governed by two competing effects: effective nuclear charge ($Z_{eff}$) and the number of shells (principal quantum number, $n$). Moving across a period, electrons are added to the same shell while the nuclear charge increases, so $Z_{eff}$ increases and the atom shrinks. Moving down a group, a new shell is added, which dominates over the increase in nuclear charge, so the atom expands.

    Step-by-Step Solution

    1. Locate the elements: B (Z=5, period 2, group 13), Be (Z=4, period 2, group 2), Mg (Z=12, period 3, group 2).
    2. Compare B and Be: both are in period 2, but B lies to the right of Be, so B has higher $Z_{eff}$ and a smaller radius: $r(B) < r(Be)$ (approx. 85 pm vs 112 pm).
    3. Compare Be and Mg: both are group 2, but Mg is one period below Be, adding an extra shell, so $r(Mg) > r(Be)$ (approx. 160 pm vs 112 pm).
    4. Combine the two comparisons: $r(B) < r(Be) < r(Mg)$.

    Common Mistakes

    • Assuming that being in a later group always means larger — the shell number (period) matters more than the group number when comparing across a period vs down a group.
    • Confusing atomic radius trends with ionic radius trends, which can differ when charges are involved.

    ✓Final answer

    The correct option is (A) — B < Be < Mg.

    ANSWER: A