Q.The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Consider the following two statements Statement – I: The order of energies of 2s-orbitals of H, Li and Na is $E_{2s}(H) < E_{2s}(Li) < E_{2s}(Na)$ Statement – II: The photoelectric effect explains the wave nature of light The correct answer is (A) Both statements – I and II are correct (B) Statement – I is correct but Statement – II is not correct (C) Statement – I is not correct but Statement – II is correct (D) Both statements – I and II are not correct

   ›Reveal solutionSolution

   Both statements test classic conceptual traps: orbital-energy ordering across atoms (Statement I) and what the photoelectric effect actually demonstrates (Statement II) — both statements, as worded, are wrong.

   Concept and Intuition

   Statement I: Comparing "the 2s orbital" across different atoms is not like comparing orbitals within the same atom. Even though all three species have an electron described as "2s", the effective nuclear charge experienced by that electron increases sharply with atomic number ($Z_H=1$, $Z_{Li}=3$, $Z_{Na}=11$), since a higher-$Z$ nucleus attracts even a shielded 2s electron much more strongly. A more tightly bound electron has more negative (lower) energy. So going from H to Li to Na, the 2s energy should become progressively more negative — i.e. $E_{2s}(Na) < E_{2s}(Li) < E_{2s}(H)$ — exactly the reverse of the statement's claimed order.

   Statement II: The photoelectric effect (instantaneous emission with a threshold frequency, and kinetic energy depending on frequency not intensity) can only be explained by treating light as discrete photons (quanta) — it directly demonstrates the particle nature of light, not the wave nature (which explains interference/diffraction instead).

   Step-by-Step Solution

   1. Evaluate Statement I: higher $Z$ (Na) $\Rightarrow$ 2s electron more tightly bound $\Rightarrow$ more negative energy. So actual order is $E_{2s}(Na) < E_{2s}(Li) < E_{2s}(H)$, not $E_{2s}(H) < E_{2s}(Li) < E_{2s}(Na)$ as claimed — Statement I is incorrect.
   2. Evaluate Statement II: the photoelectric effect is explained by the photon (particle) model of light, so it demonstrates the particle nature, not the wave nature — Statement II is incorrect.
   3. Since both statements are false, the answer is that both I and II are not correct.

   Common Mistakes

   • Assuming "same orbital label (2s)" means "same energy trend as within one atom" — cross-atom comparisons depend on effective nuclear charge, which changes a lot.
   • Confusing the photoelectric effect (particle nature) with interference/diffraction (wave nature) — a very common mix-up.

   ✓Final answer

   The correct option is (D) — Both statements I and II are not correct.

   ANSWER: D

2. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.In which of the following, elements are not in correct order with respect to the property mentioned in brackets? (A) $S < P < N < O$ (Electronegativity) (B) $Br < Ge < Ga < Ca$ (Atomic radius) (C) $Al < Mg < S < P$ (First ionization enthalpy) (D) $Mg < Ca < K < Cs$ (Metallic nature)

   ›Reveal solutionSolution

   Checking each listed periodic trend against real values shows (A)'s electronegativity order has S and P swapped — the real order is $P<S<N<O$, not $S<P<N<O$.

   Concept and Intuition

   Periodic trends (electronegativity, atomic radius, ionization enthalpy, metallic character) generally increase or decrease smoothly across periods and down groups, but P and S sit close together with a well-known small anomaly worth double-checking against actual Pauling electronegativity values rather than relying purely on the "increases left-to-right" rule of thumb.

   Step-by-Step Solution

   1. (A) Electronegativity — Pauling values: $P\approx2.1$, $S\approx2.5$, $N\approx3.0$, $O\approx3.5$. True increasing order: $P < S < N < O$. The option states $S<P<N<O$ (S before P) — this is wrong, since $S(2.5) > P(2.1)$, not less.
   2. (B) Atomic radius — approximate values: $Br\approx114$ pm, $Ge\approx122$ pm, $Ga\approx135$ pm, $Ca\approx197$ pm. Increasing order $Br<Ge<Ga<Ca$ matches the option — correct.
   3. (C) First ionization enthalpy — approximate values: $Al\approx577$, $Mg\approx738$, $S\approx1000$, $P\approx1012\ kJ/mol$. Increasing order $Al<Mg<S<P$ matches (note P's extra stability from a half-filled 3p³ subshell makes it slightly higher than S) — correct.
   4. (D) Metallic character — increases down a group and towards the left across a period. Mg(gp 2) < Ca(gp 2, lower period) < K(gp 1, more metallic than Ca in the same period) < Cs(gp 1, lower period than K) — the stated order $Mg<Ca<K<Cs$ is correct.
   5. Only (A) fails.

   Common Mistakes

   • Assuming electronegativity simply increases monotonically left-to-right without checking actual values — S and P are frequently swapped by students who forget S(2.5) is a full unit-ish above P(2.1).
   • Overlooking that P's ionization enthalpy exceeds S's due to the extra stability of a half-filled p-subshell, which could make one wrongly suspect (C) instead.

   ✓Final answer

   The correct option is (A) — $S < P < N < O$ (Electronegativity) is NOT correctly ordered.

   ANSWER: A

3. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.In which of the following, elements are correctly arranged in the decreasing order of atomic radius? (A) $\mathrm{Cl} > \mathrm{Si} > \mathrm{C} > \mathrm{F}$ (B) $\mathrm{Si} > \mathrm{Cl} > \mathrm{C} > \mathrm{F}$ (C) $\mathrm{Si} > \mathrm{Cl} > \mathrm{F} > \mathrm{C}$ (D) $\mathrm{Cl} > \mathrm{Si} > \mathrm{F} > \mathrm{C}$

   ›Reveal solutionSolution

   Comparing periods and groups (Si and Cl in period 3; C and F in period 2; Si/C in group 14; Cl/F in group 17) gives the order Si $>$ Cl $>$ C $>$ F.

   Concept and Intuition

   Atomic radius increases down a group (extra electron shells) and decreases across a period left-to-right (increasing effective nuclear charge pulling the outer electrons in). Here we have two period-2 elements (C, F) and two period-3 elements (Si, Cl), so we compare within each period first, then note that period-3 elements are generally bigger than period-2 elements due to the extra shell.

   Step-by-Step Solution

   1. Period 3 comparison: Si (group 14) is to the left of Cl (group 17), so Si has lower effective nuclear charge on its valence shell and hence a larger radius: $\text{Si} > \text{Cl}$ (approx. 117 pm vs 99 pm).
   2. Period 2 comparison: C (group 14) is to the left of F (group 17), so similarly $\text{C} > \text{F}$ (approx. 77 pm vs 71 pm).
   3. Cross-period comparison: even though Cl is further right (group 17) than C (group 14), Cl is in period 3 (one shell more) while C is in period 2, and the extra shell dominates, so $\text{Cl} > \text{C}$ (99 pm vs 77 pm).
   4. Combining: $\text{Si}(117) > \text{Cl}(99) > \text{C}(77) > \text{F}(71)$.

   Common Mistakes

   • Assuming radius decreases monotonically with atomic number without accounting for the shell change between periods.
   • Comparing Cl and C purely by period position and concluding C $>$ Cl (ignoring the shell effect).

   ✓Final answer

   The correct option is (B) — $\mathrm{Si} > \mathrm{Cl} > \mathrm{C} > \mathrm{F}$.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The order of negative electron gain enthalpy of Li, Na, S, $Cl$ is (A) Na > S > $Cl$ > Li (B) $Cl$ > S > Li > Na (C) $Cl$ > Li > S > Na (D) Li > Na > S > $Cl$

   ›Reveal solutionSolution

   Tests recall/reasoning about periodic trends in electron gain enthalpy; halogens have the most negative values, and among the two alkali metals Li is slightly more negative than Na.

   Concept and Intuition

   Electron gain enthalpy becomes more negative (a bigger energy release) as atoms get smaller and have a higher effective nuclear charge pulling in the extra electron, generally increasing (more negative) across a period and less negative down a group — though the very first member of a group (like Li) sometimes releases slightly less energy than expected due to small size causing electron-electron repulsion; empirically Li's magnitude is still marginally greater than Na's.

   Step-by-Step Solution

   1. Chlorine, a halogen, has the strongest tendency to gain an electron (achieves a stable octet) — most negative value ($\approx-349$ kJ/mol).
   2. Sulphur, in group 16, also gains an electron readily but less strongly than a halogen ($\approx-200$ kJ/mol).
   3. Between the alkali metals, Li ($\approx-60$) has a slightly larger negative electron gain enthalpy than Na ($\approx-53$), because Na's larger size gives a smaller charge density despite Li's stronger electron-electron repulsion in the small $2p$ shell — the standard tabulated values place Li below (more negative than) Na.
   4. So the order of magnitude is Cl > S > Li > Na.

   Common Mistakes

   • Swapping Li and Na's order, assuming smaller atomic size always means simply "more negative" without checking standard tabulated values.
   • Confusing electron gain enthalpy trend with ionization enthalpy trend.

   ✓Final answer

   The correct option is (B) — Cl > S > Li > Na.

   ANSWER: B

5. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.Which of the following is not correct with the property mentioned against them? (A) $\mathrm{I^- > I > I^+}$ : radius (B) $\mathrm{Li > Be > B}$ : first ionization enthalpy (C) $\mathrm{Cl > S > P}$ : electronegativity (D) $\mathrm{Rb > K > Na}$ : screening effect

   ›Reveal solutionSolution

   This tests periodic trends; the odd one out is ionization enthalpy across Li, Be, B, because Be's stable, fully-filled 2s² subshell makes it higher than both its neighbours, not the lowest.

   Concept and Intuition

   Most periodic properties follow smooth trends, but ionization enthalpy across period 2 has a well-known anomaly: Be (with a completely filled 2s² subshell) has a higher first ionization enthalpy than B (which has to remove an electron from the higher-energy, less-penetrating 2p subshell), even though B comes after Be in the period. So the general "increases across a period" rule breaks at the s²→p¹ transition.

   Step-by-Step Solution

   1. Check (A): $I^->I>I^+$ for radius — correct, since adding electrons increases radius (more electron-electron repulsion, same nuclear charge) and removing electrons decreases it.
   2. Check (C): $Cl>S>P$ for electronegativity — correct, electronegativity generally increases left to right across a period, and Cl (rightmost of the three, closest to a stable octet) is the most electronegative.
   3. Check (D): $Rb>K>Na$ for screening effect — correct, screening (shielding) increases down a group as more inner shells are added.
   4. Check (B): actual first ionization enthalpies are approximately Li = 520, Be = 899, B = 801 kJ/mol. So the real order is $Be>B>Li$, NOT $Li>Be>B$ as the statement claims. This is the incorrect match.

   Common Mistakes

   • Assuming ionization enthalpy increases monotonically and smoothly across every period without exception; the Be > B anomaly (and similarly N > O) is a classic exception to memorize.

   ✓Final answer

   The correct option is (B) — $\mathrm{Li > Be > B}$: first ionization enthalpy (this is the incorrect statement).

   ANSWER: B

6. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Which one of the following indicates correct order of atomic size of the given elements? (A) $\mathrm{Li > B > F > N}$ (B) $\mathrm{N > F > Li > B}$ (C) $\mathrm{F > N > B > Li}$ (D) $\mathrm{Li > B > N > F}$

   ›Reveal solutionSolution

   Across period 2, atomic size steadily decreases with increasing atomic number: Li > B > N > F.

   Concept and Intuition

   Moving left to right across a period, electrons are added to the same principal shell while the nuclear charge (number of protons) increases. The added protons pull the entire electron cloud inward more strongly than the extra shielding from same-shell electrons can compensate for, so atomic radius shrinks fairly steadily across a period. Li, B, N, and F are all period-2 elements with increasing atomic number (3, 5, 7, 9), so their sizes should decrease in that same order.

   Step-by-Step Solution

   1. Identify that Li (Z=3), B (Z=5), N (Z=7), F (Z=9) are all in period 2.
   2. Apply the general periodic trend: atomic radius decreases left to right across a period.
   3. Order them by increasing atomic number (Li, B, N, F) and note radius decreases correspondingly: Li (largest) > B > N > F (smallest).
   4. This matches option (D): $\text{Li} > \text{B} > \text{N} > \text{F}$.

   Common Mistakes

   • Forgetting that all four elements belong to the same period, and instead trying to apply a group trend (which would give the opposite, increasing-down-the-group behavior).
   • Mixing up the exact order of B and N (B is metalloid-like and slightly larger than N, consistent with the general decreasing trend, not an exception here).

   ✓Final answer

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

   ANSWER: D

7. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.The correct order of atomic radii of B, Be, N and C is (A) Be < B < C < N (B) N < B < C < Be (C) N < C < B < Be (D) Be < C < B < N

   ›Reveal solutionSolution

   Moving across Period 2 (Be, B, C, N), increasing nuclear charge with the same principal shell pulls electrons in tighter, so atomic radius steadily decreases; the correct increasing-radius order is N < C < B < Be.

   Concept and Intuition

   Within a period, all atoms add electrons to the same outermost shell while the nuclear charge (number of protons) increases. Electrons in the same shell do not shield each other very effectively, so the effective nuclear charge felt by outer electrons rises across the period, pulling the electron cloud inward and steadily shrinking atomic radius from left to right.

   Step-by-Step Solution

   1. List the elements in period order: Be (Z=4) < B (Z=5) < C (Z=6) < N (Z=7).
   2. Since they are in the same period (same principal shell, n=2), radius should decrease as Z increases (left to right).
   3. So Be has the largest radius, then B, then C, then N has the smallest.
   4. Writing this as an increasing-radius sequence: N < C < B < Be.

   Common Mistakes

   • Confusing the period trend (radius decreases left to right) with the group trend (radius increases down a group) and applying the wrong one.
   • Forgetting that this simple monotonic decrease across a period is a reliable trend (unlike some ionization-energy anomalies), so no special-case correction is needed here.

   ✓Final answer

   The correct option is (C) — N < C < B < Be.

   ANSWER: C

8. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Assertion (A): Fluorine has smaller negative electron gain enthalpy than chlorine Reason (R): The electron – electron repulsion is higher in chlorine than in fluorine (A) Both (A) and (R) are correct and (R) is the correct explanation of (A). (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (C) (A) is correct but (R) is incorrect. (D) (A) is incorrect but (R) is correct.

   ›Reveal solutionSolution

   Fluorine's smaller (less negative) electron gain enthalpy compared to chlorine is real, but the stated reason has the repulsion backwards — it is fluorine's small size that causes higher electron-electron repulsion, not chlorine's.

   Concept and Intuition

   Electron gain enthalpy generally becomes more negative going down a group as atoms increase in size and add electrons to progressively larger, less repulsive orbitals — but the very first member of a group is often an anomaly. Fluorine is exceptionally small, so its outermost $2p$ subshell is already tightly packed with electron density; adding one more electron to this small, compact shell causes unusually large electron-electron repulsion, which offsets much of the favourable nuclear attraction. Chlorine, being noticeably larger (its valence electrons occupy the more diffuse $3p$ subshell), can accommodate the incoming electron with much less repulsion, so more energy is released — a more negative electron gain enthalpy.

   Step-by-Step Solution

   1. Compare experimental values: $\Delta_{eg}H(\mathrm{F}) \approx -328\ \text{kJ/mol}$, $\Delta_{eg}H(\mathrm{Cl}) \approx -349\ \text{kJ/mol}$. Fluorine's value is indeed the smaller (less negative) one — Assertion (A) is true.
   2. The reason given for this ordering is "electron-electron repulsion is higher in chlorine than in fluorine." Physically, the opposite is true: it is fluorine's small atomic/orbital size that packs electrons closer together and produces greater repulsion, which is exactly why fluorine's electron gain enthalpy is less negative than expected.
   3. So the Reason (R), as literally stated, is false — the repulsion effect it describes is reversed.
   4. Hence (A) correct, (R) incorrect — option (C).

   Common Mistakes

   • Assuming electron gain enthalpy becomes steadily more negative up a group (ignoring the fluorine anomaly).
   • Not checking the direction of the repulsion argument carefully — it is easy to skim past and assume (R) explains (A) correctly.

   ✓Final answer

   The correct option is (C) — (A) is correct but (R) is incorrect.

   ANSWER: C

9. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.Observe the following statements. Statement A: - In general, the ionization potential value decreases on moving down in the group. Statement B:- The $1^{st}$ ionization potential of sodium is greater from that of potassium. Correct answer is (A) Both A and B are wrong (B) Both A and B are correct (C) A is correct but B is wrong (D) A is wrong but B is correct

   ›Reveal solutionSolution

   Ionization potential decreasing down a group (A) and sodium having a higher first ionization potential than potassium (B) are both true, and B is simply a specific instance of the general trend stated in A.

   Concept and Intuition

   Moving down a group, each successive element adds a new outermost shell, increasing atomic radius and adding inner shielding electrons. The valence electron therefore experiences a weaker effective nuclear pull and is easier to remove, so ionization potential generally decreases going down a group (with some well-known exceptions elsewhere in the periodic table, but not between Na and K).

   Step-by-Step Solution

   1. Statement A: As we move down any group, atomic size increases and shielding increases, so the outermost electron is held less strongly — ionization potential generally decreases down the group. This is the standard periodic trend — true.
   2. Statement B: Sodium (period 3) lies directly above potassium (period 4) in Group 1. Applying trend A, Na should have a higher ionization potential than K.
   3. Checking known values: IE$_1$(Na) $\approx 496\ \text{kJ/mol}$ ($\approx 5.14\ \text{eV}$); IE$_1$(K) $\approx 419\ \text{kJ/mol}$ ($\approx 4.34\ \text{eV}$). Indeed Na $>$ K — true.
   4. Since both statements agree with each other and with known data, both A and B are correct.

   Common Mistakes

   • Assuming ionization potential trends have no simple pattern down Group 1 (it is, in fact, a very clean, standard decreasing trend down this group).
   • Confusing ionization potential trend with electron affinity or electronegativity trends, which can behave differently.

   ✓Final answer

   The correct option is (B) — Both A and B are correct.

   ANSWER: B

10. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.The correct order of atomic radii of the elements O, N, S and P is (A) $N < P < S < O$ (B) $N < O < P < S$ (C) $O < N < P < S$ (D) $O < N < S < P$

    ›Reveal solutionSolution

    Using periodic trends (radius decreases across a period, increases down a group), the atomic radii order is $O < N < S < P$.

    Concept and Intuition

    Atomic radius trends are governed by two competing effects: across a period, increasing nuclear charge pulls the same-shell electrons in tighter, so radius decreases left to right; down a group, an additional electron shell is added, so radius increases. Here we're comparing two period-2 elements (N, O — adjacent, Group 15 and 16) and two period-3 elements (P, S — directly below N and O). Since period-3 atoms have an extra electron shell compared to period-2, even period-3 S (further right, smaller within its period) is still noticeably larger than either period-2 element.

    Step-by-Step Solution

    1. Within period 2: N (Z=7) is to the left of O (Z=8), so N has a larger radius than O — i.e., $O < N$.
    2. Within period 3: P (Z=15) is to the left of S (Z=16), so P has a larger radius than S — i.e., $S < P$.
    3. Across periods (down the group): both P and S (period 3) are larger than both N and O (period 2), because of the extra electron shell.
    4. Combining all constraints: $O < N < S < P$.

    Common Mistakes

    • Assuming atomic number alone determines radius (ignoring the period vs. group distinction) — leads to wrongly ordering N and O relative to S and P.
    • Forgetting that within a period, radius decreases with increasing Z, so incorrectly placing O above N.

    ✓Final answer

    The correct option is (D) — $O < N < S < P$.

    ANSWER: D

11. AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
    Q.Match the following: Column-I:

    (a) F

    (b) Cl

    (c) He

    (d) Cs. Column-II:

    (i) Maximum ionization enthalpy

    (ii) Maximum atomic radius

    (iii) Maximum electro negativity

    (iv) Maximum negative electron gain enthalpy (A) (a - iv), (b - iii), (c - i), (d - ii) (B) (a - iii), (b - i), (c - iv), (d - ii) (C) (a - iii), (b - iv), (c - i), (d - ii) (D) (a - i), (b - iv), (c - iii), (d - ii)

    ›Reveal solutionSolution

    This tests four separate periodic-trend facts simultaneously: electronegativity (F is highest), electron gain enthalpy (Cl is most negative, not F), ionization enthalpy (He is highest of all elements), and atomic radius (Cs is largest among common elements). The answer is (a-iii),(b-iv),(c-i),(d-ii), option (C).

    Concept and Intuition

    Several periodic properties peak at different elements due to subtle competing effects, and this question tests whether those exceptions are known precisely:

    • Electronegativity increases across a period and decreases down a group, so fluorine — top-right of the periodic table (excluding noble gases) — has the single highest electronegativity of any element.
    • Electron gain enthalpy (energy released on adding an electron) would naively also peak at fluorine, but fluorine's very small atomic size causes significant electron-electron repulsion when an extra electron is forced into its already-compact 2p subshell. This makes chlorine's electron gain enthalpy more negative than fluorine's — a classic periodic-trend exception.
    • Ionization enthalpy is highest for noble gases, since their filled shell configuration is exceptionally stable; helium, being the smallest and having only a 1s2 shell, has the highest ionization enthalpy of all elements.
    • Atomic radius increases down a group and decreases across a period; caesium, at the bottom-left of the practically-occurring elements, has one of the largest atomic radii.

    Step-by-Step Solution

    1. (a) F → maximum electronegativity → matches (iii).
    2. (b) Cl → maximum (most exothermic/negative) electron gain enthalpy, due to the F-vs-Cl anomaly → matches (iv).
    3. (c) He → maximum ionization enthalpy of any element → matches (i).
    4. (d) Cs → maximum atomic radius among common elements → matches (ii).
    5. Combined: (a-iii), (b-iv), (c-i), (d-ii).

    Common Mistakes

    • Assuming fluorine has the maximum electron gain enthalpy just because it has the maximum electronegativity — these are different properties, and chlorine's larger 3p subshell accepts the extra electron with less repulsion, giving it the more negative value.
    • Forgetting that ionization enthalpy trends are dominated by noble gases, not halogens, when matching "maximum ionization enthalpy."

    ✓Final answer

    The correct option is (C) — (a - iii), (b - iv), (c - i), (d - ii).

    ANSWER: C

12. AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQ
    Q.Which order among the following is incorrect? (A) $NH_3 < PH_3 < AsH_3$ : (Acidic nature) (B) $Li < Be < B < C$ : $IE_1\ (\Delta_iH_1)$ (C) $Al_2O_3 < MgO < Na_2O < K_2O$ : (Basic nature) (D) $Li^+ < Na^+ < K^+ < Cs^+$ : (Ionic radius)

    ›Reveal solutionSolution

    This tests knowledge of periodic-trend exceptions, specifically the anomalous first ionisation energy order among Li, Be, B, C; the answer is (B).

    Concept and Intuition

    While ionisation energy generally increases across a period, there are two well-known anomalies in Period 2: Be (fully-filled $2s^2$, extra stable) has a higher $IE_1$ than B (whose lone $2p^1$ electron is easier to remove, being in a higher-energy, less-penetrating orbital), and similarly N (half-filled $2p^3$) has a higher $IE_1$ than O. So the actual trend across Li→Be→B→C is NOT monotonic if we naively expect increase; specifically Be > B.

    Step-by-Step Solution

    1. Recall/reconstruct actual $IE_1$ values (kJ/mol): Li ≈ 520, Be ≈ 899, B ≈ 801, C ≈ 1086.
    2. Ordering these: Li (520) < B (801) < Be (899) < C (1086).
    3. Option (B) claims Li < Be < B < C, i.e., Be < B — this contradicts the actual values (Be > B), so option (B) is the incorrect order.
    4. Check the others quickly: acidic strength of NH3/PH3/AsH3 increases down the group (A correct); basic character of oxides Al2O3<MgO<Na2O<K2O increases going both across (right to left) and down (C correct); ionic radius of alkali cations increases down the group (D correct).

    Common Mistakes

    • Assuming ionisation energy always increases smoothly across a period, missing the Be>B and N>O anomalies caused by stable filled/half-filled subshells.
    • Second-guessing the acidic-hydride trend (A), which is correctly increasing down the group.

    ✓Final answer

    The correct option is (B) — $Li < Be < B < C$: $IE_1\ (\Delta_iH_1)$.

    ANSWER: B