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Error notes · Chemistry · Kinetics and p-Block

Two chapters, thirty-one questions

Every chemistry question saved from this paper, grouped by chapter and sub-topic. Each carries what was given, what was asked, the concept and formula behind it, the steps in full, the fastest route through, and a figure wherever one makes the answer visible.

31
Questions lost
17
Attempted, wrong
12
Left blank
2
Chapters involved

Chemical Kinetics 18 · The p-Block Elements 13. On NEET marking these thirty-one were worth 124 marks, and the seventeen wrong attempts cost 17 more. One question (Q175) had its option list cut off in the screenshot, and one (Q158) was answered correctly.

The attempt rate has transformed: seventeen attempted against twelve blank, where the previous chemistry paper ran two against fourteen. What the errors reveal is a single repeated weakness — trend directions and statement pairs. One anomaly alone (sulfur's electron gain enthalpy beating oxygen's) is tested three times here and was missed twice.

Chapters in this paper  —  red = wrong · amber = blank · green = correct

Chemical Kinetics

18 questions · 10 wrong · 7 blank · 1 correct

Q136Q137Q138Q140Q141Q142Q143Q144Q145Q146Q147Q149Q151Q154Q155Q156Q157Q158

The p-Block Elements

13 questions · 7 wrong · 5 blank · 1 uncaptured

Q159Q160Q162Q167Q168Q169Q171Q173Q174Q175Q176Q178Q179

Chapter

Chemical Kinetics

Rate laws, integrated equations and temperature dependence  ·  18 questions · 10 wrong · 7 blank · 1 correct

Rate law, order and molecularity

5 questions · 4 wrong · 1 blank
Q136 Rate ratio when concentrations change Marked wrong

Rate law for a reaction between A and B is given by R = k[A]n[B]m. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction (r2/r1) is

  • KEY2(n−m)
  •  (n − m)
  •  (m + n)
  • MARKED1/2m+n
Given
  • Rate law R = k[A]n[B]m.
  • [A] is doubled and [B] is halved.
Asked
The ratio r2/r1.
Concept to use
Take the ratio of the two rate expressions and everything except the concentration factors cancels — including k. Each concentration contributes its factor raised to its own order, and the two contributions multiply.
Formula to use
r2/r1 = (factor for A)n × (factor for B)m
Baby steps
  1. New rate: r2 = k(2[A])n ([B]/2)m.
  2. Divide by r1 = k[A]n[B]m: everything cancels except the numerical factors.
  3. r2/r1 = 2n × (1/2)m = 2n × 2−m.
  4. Add the exponents: 2(n−m).
  5. Check with a case: if n = m the two effects cancel exactly and the rate is unchanged, giving 20 = 1. ✓
Answer
2(n−m)
Shortcut
Handle each species separately and multiply. Doubling contributes 2n, halving contributes 2−m, and multiplying powers of the same base means adding exponents. Two options here are not even powers of 2 — (n − m) and (m + n) are bare numbers, so they cannot be rate ratios and can be struck out on sight.
Where it went wrong
1/2m+n treats both changes as halvings. Doubling and halving push in opposite directions, so the exponents must subtract, not add. Writing the two factors on separate lines — 2n for A, 2−m for B — makes the sign of each explicit before they are combined.
Q144 What makes a reaction zero order Marked wrong

2A → B + C. It would be a zero order reaction when

  •  The rate of reaction is proportional to square of concentration of A
  • KEYThe rate of reaction remains same at any concentration of A
  •  The rate remains unchanged at any concentration of B and C
  • MARKEDThe rate of reaction doubled if concentration of B is increased to double
Given
  • Reaction 2A → B + C.
Asked
The condition that makes it zero order.
Concept to use
Order is defined by how the rate responds to the reactant concentrations. Zero order means the rate is completely independent of [A] — changing how much A is present makes no difference to how fast the reaction goes. The products B and C do not appear in the rate law at all, so statements about them are irrelevant to the order.
Formula to use
Rate = k[A]0 = k  —  constant, whatever [A] may be
Baby steps
  1. “Proportional to [A]²” — that is second order in A, not zero. Wrong.
  2. “Rate remains same at any concentration of A” — exactly the definition of zero order in A. Correct.
  3. “Rate unchanged at any concentration of B and C” — B and C are products; they are normally absent from the rate law regardless of order. This tells you nothing. Wrong.
  4. “Rate doubles when [B] doubles” — this describes first order in a product, which is not what order in this reaction means. Wrong.
Answer
The rate of reaction remains same at any concentration of A
Shortcut
Scan for which species each option is about. Only options mentioning the reactant A can define the order, so the two options about B and C are eliminated without any thought about kinetics. That leaves two, and one of them says “squared”.
Where it went wrong
The chosen option describes the rate responding to a product. Order is defined by the reactants — asking how the rate changes with product concentration is a different question altogether. Before evaluating any statement about order, check that the species named is on the left-hand side of the equation.
Q147 Units of a third-order rate constant Marked wrong

The rate law of reaction is r = K[A][B]2. The unit of rate constant is

  •  mol L−1 s−1
  •  s−1
  • MARKEDmol−1 L−1 s−1
  • KEYmol−2 L2 s−1
Given
  • Rate law r = k[A][B]².
Asked
The units of k.
Concept to use
Rearrange the rate law for k and cancel units. Rate always carries mol L−1 s−1, and dividing by concentration raised to the overall order leaves the units of k. The general result is worth memorising because it appears in some form in almost every kinetics paper.
Formula to use
k = rate / [conc]n  →  units = mol(1−n) L(n−1) s−1
Baby steps
  1. Overall order: n = 1 + 2 = 3.
  2. k = rate/([A][B]²), so the units are (mol L−1 s−1) ÷ (mol L−1)3.
  3. Numerator: mol1 L−1 s−1. Denominator: mol3 L−3.
  4. Divide: mol1−3 L−1+3 s−1 = mol−2 L2 s−1.
  5. Check against the formula: n = 3 gives mol−2 L2 s−1. ✓
Answer
mol−2 L2 s−1
Shortcut
Note that the powers of mol and L are always equal and opposite, and both are fixed by the order: mol(1−n)L(n−1)s−1. Any option where the two exponents do not mirror each other is wrong on sight, which kills the chosen answer immediately.
Where it went wrong
mol−1 L−1 s−1 has both exponents negative, which cannot happen for any order. It also corresponds to no value of n at all. Checking that the mol and L powers mirror each other is a two-second filter that catches this class of error every time.
Q151 Order when reactants are in excess Not attempted

A + B(excess) + C(excess) —slow→ Products. Order of the reaction is

  • KEY1
  •  2
  •  3
  •  Zero
Given
  • Reaction A + B + C → products, with B and C both in excess.
Asked
The order of the reaction.
Concept to use
When a reactant is present in large excess, its concentration barely changes as the reaction proceeds, so it behaves like a constant and is absorbed into the rate constant. Only species whose concentration varies appear in the observed rate law. With B and C both in excess, only A is left — giving pseudo first order kinetics.
Formula to use
Rate = k[A][B][C] → with [B], [C] ≈ constant → Rate = k′[A], where k′ = k[B][C]
Baby steps
  1. Write the rate law assuming it reflects the stoichiometry: Rate = k[A][B][C].
  2. B is in excess, so [B] stays effectively constant. Same for C.
  3. Absorb both into the constant: Rate = k′[A] with k′ = k[B][C].
  4. The observed rate depends on one varying concentration.
  5. Order = 1 (pseudo first order).
Answer
1
Shortcut
Count only the species not in excess. Three reactants with two in excess leaves one, so the order is 1. The classic examples are ester hydrolysis and the inversion of cane sugar, where water is the excess reagent.
This is the same idea as statement C in Q156 of this paper — a pseudo-molecular reaction “does not depend on the concentration of reactants taken in excess”. Two questions in one paper on the same concept is a strong signal it is worth locking down.
Q155 What the rate expression represents Marked wrong

For the reaction 2HI → H2 + I2, the expression −½ d[HI]/dt represents

  •  The rate of formation of HI
  •  The rate of disappearance of HI
  • KEYThe instantaneous rate of the reaction
  • MARKEDBoth B and C
Given
  • Reaction 2HI → H2 + I2.
  • Expression −½ d[HI]/dt.
Asked
What this expression represents.
Concept to use
The factor of ½ is the whole point. Without it, −d[HI]/dt is simply the rate at which HI disappears — a species-specific quantity. Dividing by the stoichiometric coefficient converts that into the unique rate of reaction, the single number that is the same whichever species you monitor. So the ½ changes what the expression means.
Formula to use
Rate of reaction = −(1/2) d[HI]/dt = +d[H2]/dt = +d[I2]/dt
Baby steps
  1. −d[HI]/dt on its own is the rate of disappearance of HI.
  2. Dividing by the coefficient 2 gives the rate of the reaction — the quantity that equals d[H2]/dt and d[I2]/dt as well.
  3. The expression given includes the ½, so it is the rate of reaction, not the rate of disappearance.
  4. “Formation of HI” is wrong outright — HI is being consumed, and the minus sign in front confirms it.
  5. Answer: the instantaneous rate of the reaction.
Answer
The instantaneous rate of the reaction
Shortcut
Look for the stoichiometric divisor. With the 1/coefficient it is the rate of reaction; without it, it is that species' own rate of change. One glance at whether the fraction is present answers the question.
Where it went wrong
“Both B and C” treats the two as interchangeable, but they differ by exactly the factor of 2 — they are numerically different quantities and cannot both describe the same expression. Since only one option can be correct, an option combining two mutually exclusive readings is self-defeating. The ½ is doing real work, not decoration.

Integrated rate laws and half life

5 questions · 1 wrong · 4 blank
Q142 First-order gas reaction in terms of pressure Marked wrong

For a first order reaction A(g) → 2B(g) + C(g), p0 is initial pressure of A and pt the total pressure at time t. The integrated rate equation is

track the pressure of A alone, not the totalA → 2B + Cstart: A = p0, B = C = 0let x of A reactA = p0 − x  ·  B = 2x  ·  C = xtotal = p0 + 2xx = (pt − p0)/2pA = p0 − x= (3p0 − pt)/2k = (2.303/t) log(p0/pA)= (2.303/t) log[2p0/(3p0−pt)]general rule: A → nB gives pA = (np0 − pt)/(n − 1)here the moles go 1 → 3, so n = 3 and the denominator is 2
  • MARKEDK = (2.303/t) log[p0/(p0 − pt)]
  • KEYK = (2.303/t) log[2p0/(3p0 − pt)]
  •  K = (2.303/t) log[p0/(2p0 − pt)]
  •  K = (2.303/t) log[2p0/(2p0 − pt)]
Given
  • A(g) → 2B(g) + C(g), first order.
  • Initial pressure of A = p0; total pressure at time t = pt.
Asked
The integrated rate equation.
Concept to use
The rate law needs the pressure of A alone, but the measurable quantity is the total pressure. Since 1 mole of gas becomes 3 moles, the total pressure rises as the reaction proceeds, and the excess above p0 tells you how much A has reacted. Express pA in terms of p0 and pt, then substitute into the standard first-order law.
Formula to use
k = (2.303/t) log(p0/pA), with pA found from the stoichiometry
Baby steps
  1. Let x be the pressure of A consumed at time t. Then A = p0 − x, B = 2x, C = x.
  2. Total: pt = (p0 − x) + 2x + x = p0 + 2x.
  3. So x = (pt − p0)/2.
  4. pA = p0 − x = p0 − (pt − p0)/2 = (3p0 − pt)/2.
  5. Substitute: k = (2.303/t) log[p0 ÷ (3p0−pt)/2] = (2.303/t) log[2p0/(3p0 − pt)].
Answer
K = (2.303/t) log[2p0/(3p0 − pt)]
Shortcut
Use the general result: for 1 mole of gas becoming n moles, pA = (np0 − pt)/(n − 1). Here 1 → 3 so n = 3, giving (3p0 − pt)/2 immediately. Then sanity-check at t = 0, where pt = p0 must make the log ratio 1 — only the correct option passes.
Where it went wrong
The chosen option, p0/(p0 − pt), treats pt as though it were the pressure consumed. But pt is the growing total, so p0 − pt is negative — the logarithm of a negative number, which is impossible. This is the same question shape as Q75 (A → 3B) in the earlier chemistry paper, so it is now worth learning as a formula rather than deriving each time.
Q143 Rate constant from 75% completion Not attempted

75% of first order reaction is completed in 30 min. Calculate its rate constant

  •  7.7 × 10−4 min−1
  •  7.7 × 10−4 M sec−1
  • KEY7.7 × 10−4 sec−1
  •  7.7 min−1
Given
  • First order reaction, 75% complete in 30 minutes.
Asked
The rate constant.
Concept to use
75% completion means one quarter of the reactant remains, so [A]0/[A] = 4. Note that 75% is exactly two half lives, which gives a quick alternative route. The real decision in this question is the unit: three of the four options share the same digits, so the answer hinges on whether the time is expressed in minutes or seconds.
Formula to use
k = (2.303/t) log([A]0/[A])  |  first-order k has units of time−1 only
Baby steps
  1. 75% reacted leaves 25%, so [A]0/[A] = 4.
  2. Convert the time: t = 30 min = 1800 s.
  3. k = (2.303/1800) log 4 = (2.303/1800)(0.602).
  4. k = 1.386/1800 = 7.7 × 10−4 s−1.
  5. Alternative route: 75% complete = 2 half lives, so t½ = 15 min = 900 s, and k = 0.693/900 = 7.7 × 10−4 s−1. ✓
Answer
7.7 × 10−4 sec−1
Shortcut
Recognise 75% as two half lives and 87.5% as three — then k = 0.693/t½ avoids the logarithm entirely. Here t½ = 15 min, and the only remaining decision is the time unit.
Two distractors are unit traps. M sec−1 is impossible — a first-order rate constant never carries a concentration unit, only time−1. And 7.7 × 10−4 min−1 uses the right digits with the wrong time base; in minutes the answer would be 4.6 × 10−2. Check the unit before checking the digits.
Q145 Radioactive decay through several half lives Not attempted

The half-life of a radio isotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is

  • KEY3.125 g
  •  2.084 g
  •  1.042 g
  •  4.167 g
Given
  • Half life = 4 hours.
  • Initial mass = 200 g.
  • Elapsed time = 24 hours.
Asked
Mass remaining undecayed.
Concept to use
Radioactive decay is first order, so the half life is constant regardless of how much material is left. Count how many half lives have elapsed, then halve the mass that many times. The only step that goes wrong is counting.
Formula to use
n = t/t½  |  remaining = initial / 2n
Baby steps
  1. Number of half lives: n = 24/4 = 6.
  2. Divide by 2 six times: 26 = 64.
  3. Remaining mass = 200/64.
  4. = 3.125 g.
  5. Trace it if you prefer: 200 → 100 → 50 → 25 → 12.5 → 6.25 → 3.125. Six steps. ✓
Answer
3.125 g
Shortcut
Halve the number out loud six times — it takes about five seconds and is less error-prone than computing 26 and dividing. The doubling sequence 2, 4, 8, 16, 32, 64 is worth having instantly available for exactly this purpose.
The distractors correspond to n = 6.5, 7.5 and 5.6 — all products of miscounting the half lives rather than of faulty arithmetic. Write n down explicitly on its own line before dividing anything.
Q146 Fraction reacted after two half lives Not attempted

50% completion of a first order reaction takes place in 16 minutes. The fraction that would react in 32 minutes from the beginning

  •  1/2
  •  1/4
  •  1/8
  • KEY3/4
Given
  • First order reaction, 50% complete in 16 minutes.
  • Total elapsed time 32 minutes.
Asked
The fraction that has reacted.
Concept to use
50% completion is the half life, so t½ = 16 min and 32 min is exactly two half lives. After two half lives one quarter remains — so three quarters have reacted. The question asks for the fraction that reacted, which is the complement of what the halving gives you directly.
Formula to use
Remaining after n half lives = 1/2n  |  reacted = 1 − 1/2n
Baby steps
  1. t½ = 16 min, so 32 min = 2 half lives.
  2. After the first: 1/2 remains. After the second: 1/4 remains.
  3. The question asks what has reacted: 1 − 1/4 = 3/4.
Answer
3/4
Shortcut
Halving always gives you the fraction left. Read the final word of the question — “react” or “remain” — and subtract from 1 if needed. That one habit covers every question of this type.
1/4 is on the option list because it is the fraction remaining — the right calculation answering the wrong question. Both 1/4 and 3/4 being offered is the paper checking whether you read to the end of the sentence.
Q154 Rate constant from a concentration ratio Not attempted

In a first order reaction the ratio [A0]/[At] was found to be 8 after 10 minute. The rate constant is

  • KEY(2.303 × 3 log 2)/10
  •  (2.303 × 2 log 3)/10
  •  10 × 2.303 × 2 log 3
  •  10 × 2.303 × 3 log 2
Given
  • First order reaction.
  • [A0]/[At] = 8 after t = 10 min.
Asked
The rate constant, as an expression.
Concept to use
Substitute directly into the integrated first-order law. The only manipulation needed is rewriting log 8 as 3 log 2, since 8 = 2³ — the options are written in that form to see whether you make the conversion.
Formula to use
k = (2.303/t) log([A0]/[At])
Baby steps
  1. Substitute: k = (2.303/10) log 8.
  2. Rewrite 8 as 2³: log 8 = log 2³ = 3 log 2.
  3. So k = (2.303 × 3 log 2)/10.
  4. Check the shape: t = 10 belongs in the denominator, because k has units of time−1. Two options place it in the numerator and are dimensionally impossible.
Answer
(2.303 × 3 log 2)/10
Shortcut
Use the position of t as a filter before looking at the logarithms. A first-order k carries units of time−1, so t must divide — that eliminates half the options at once. Then 8 = 2³ settles the rest.
A ratio of 8 means three half lives have passed, so an equivalent route is k = 0.693/t½ with t½ = 10/3 min. That gives the same number and is a useful cross-check when the answer is required numerically rather than as an expression.

Temperature dependence and Arrhenius

6 questions · 3 wrong · 2 blank · 1 correct
Q137 Effect of temperature on reaction rate Marked wrong

Statement I: With increase in temperature rate of both exo and endothermic reactions increases.
Statement II: Increasing the temperature of the substance decreases the fraction of molecules which collide with energies greater than Ea.

  •  Both statements I & II are incorrect
  • KEYStatement I is correct, but statement II is incorrect
  •  Statement II is correct but statement I is incorrect.
  • MARKEDBoth the statements I & II are correct
Given
  • Two statements about how temperature affects rate.
Asked
Which statements are correct.
Concept to use
Raising the temperature shifts the Maxwell–Boltzmann distribution to higher energies, so the fraction of molecules exceeding Ea increases — which is why the rate rises. This happens regardless of whether the reaction is exothermic or endothermic, because Ea is a barrier that must be climbed in either case. Statement II asserts the exact opposite of the mechanism.
Formula to use
k = A e−Ea/RT  →  T increases ⇒ k increases, for every reaction
Baby steps
  1. Statement I. The Arrhenius equation contains no reference to ΔH, so k rises with T for exothermic and endothermic reactions alike. Correct.
  2. Statement II. Heating moves the distribution curve to the right, so more molecules exceed Ea, not fewer. Incorrect.
  3. Answer: Statement I correct, Statement II incorrect.
Answer
Statement I is correct, but statement II is incorrect
Shortcut
Statement II contradicts Statement I. If a smaller fraction of molecules had enough energy, the rate would fall — so the two cannot both be true. Spotting that they conflict removes “both correct” and “both incorrect” together, leaving only which one to keep.
Where it went wrong
Marking both correct means Statement II was read for its shape (“temperature, molecules, Ea”) rather than for the word decreases. This is the same single-word reversal that appeared in Q105 of the physics paper in this same set. Underline the direction word in any statement about a trend before judging it.
Q140 Four statements about rate and order Marked wrong

Identify correct statements from the following.
(A) Activation energy of reaction varies with temperature
(B) Rate constant of an exothermic reaction decreases with increase in temperature
(C) For a zero order reaction, concentration of reactants remains constant with time.
(D) For a first order reaction, [A]t = [A]0e−kt

  •  A, B and C only
  •  A, B, C and D
  • KEYD only
  • MARKEDA and C only
Given
  • Four statements about kinetics.
Asked
Which are correct.
Concept to use
Three of these are near-misses of true statements, each altered by one idea. Activation energy is treated as a constant for a given reaction. The rate constant always rises with temperature, exothermic or not. And in a zero-order reaction it is the rate that is constant, not the concentration — concentration falls steadily in a straight line.
Formula to use
Zero order: [A]t = [A]0 − kt (linear fall)  |  First order: [A]t = [A]0e−kt (exponential fall)
Baby steps
  1. (A) Ea is a property of the reaction pathway and is taken as independent of temperature. Incorrect.
  2. (B) From k = A e−Ea/RT, k increases with T for every reaction. Exothermicity affects the equilibrium constant, not the direction in which k moves. Incorrect.
  3. (C) In a zero-order reaction the rate is constant, so the concentration falls linearly with time. It does not stay constant. Incorrect.
  4. (D) This is the standard integrated first-order law. Correct.
  5. Answer: D only.
Answer
D only
Shortcut
Statement D is a formula you either know or do not — check it first, because it is the only one that can be verified with certainty. Once D is confirmed correct, every option that excludes D falls away, and here only one option contains D.
Where it went wrong
Statement C is the trap that was taken. “Zero order” means the rate does not depend on concentration — but the reactant is still being consumed at a steady pace, so its concentration drops in a straight line to zero. Confusing “rate is constant” with “concentration is constant” is the classic zero-order misreading. A zero-order reaction that never consumed anything would never finish.
Q141 Temperature at which two rate constants are equal Not attempted

For gaseous reaction, the following data is given: A → B, K1 = 1015e−2000/T; C → D, K2 = 1014e−1000/T. The temperature at which K1 = K2 is

  •  2000 K
  •  868.8 K
  • KEY434.2 K
  •  1000 K
Given
  • K1 = 1015e−2000/T, K2 = 1014e−1000/T.
Asked
The temperature at which the two rate constants are equal.
Concept to use
Set the two expressions equal and take natural logarithms. The exponentials collapse into a single term in 1/T, and the pre-exponential factors collapse into a single logarithm — leaving one linear equation in 1/T.
Formula to use
Set K1 = K2, take ln of both sides, solve for T
Baby steps
  1. 1015e−2000/T = 1014e−1000/T.
  2. Divide both sides by 1014e−2000/T: 10 = e−1000/T + 2000/T = e1000/T.
  3. Take natural logs: ln 10 = 1000/T.
  4. T = 1000 / ln 10 = 1000 / 2.303.
  5. T ≈ 434.2 K.
Answer
434.2 K
Shortcut
Group the like parts before doing anything else — powers of ten on one side, exponentials on the other. Here that gives 10 = e1000/T in one move, and the answer is 1000/2.303. Recognising 2.303 as ln 10 turns the last step into a single division.
868.8 K is exactly twice the answer, which is what you get by writing 2000/T instead of 1000/T after subtracting the exponents. Subtracting negative exponents is where the sign slips, so write the step out: −1000/T − (−2000/T) = +1000/T.
Q149 Activation energy from a rate doubling Marked wrong

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314 J mol−1 K−1)

  •  342 kJ mol−1
  •  269 kJ mol−1
  • KEY34.7 kJ mol−1
  • MARKED15.1 kJ mol−1
Given
  • Rate doubles: k2/k1 = 2.
  • T1 = 20°C = 293 K, T2 = 35°C = 308 K.
  • R = 8.314 J mol−1 K−1.
Asked
The activation energy Ea.
Concept to use
The two-temperature form of the Arrhenius equation relates the ratio of rate constants to Ea. The two things that must be right are the conversion to kelvin — the formula needs absolute temperatures and their product — and remembering to convert the final answer from joules to kilojoules.
Formula to use
log(k2/k1) = Ea(T2 − T1) / (2.303 R T1T2)
Baby steps
  1. Convert temperatures: T1 = 293 K, T2 = 308 K, so T2 − T1 = 15 K.
  2. Rearrange for Ea: Ea = 2.303 R T1T2 log(2) / (T2 − T1).
  3. Numerator: 2.303 × 8.314 × 293 × 308 × 0.301.
  4. 2.303 × 8.314 = 19.15; 293 × 308 = 90 244; 19.15 × 90 244 × 0.301 ≈ 520 000.
  5. Divide by 15: Ea ≈ 34 700 J mol−1 = 34.7 kJ mol−1.
Answer
34.7 kJ mol−1
Shortcut
Sanity-check the magnitude. Typical activation energies sit between about 20 and 200 kJ mol−1, and a rate that merely doubles over 15 degrees implies a modest barrier. 342 and 269 kJ mol−1 are implausibly large for such a small effect, so the answer is one of the two small options before any calculation.
Where it went wrong
15.1 is close to half the correct value, which is the signature of using the temperature difference somewhere the product T1T2 belongs, or of dropping the 2.303. Write the formula out fully before substituting — both T1T2 in the numerator and (T2−T1) in the denominator must appear.
Q157 Slope of an Arrhenius plot Not attempted

If we plot a graph between log k and 1/T of Arrhenius equation, the slope is:

take logs of the Arrhenius equation and read off the slopelog k1/Trunrisenegative slopek = A e−Ea/RTtake log10 of both sideslog k = log A − Ea/(2.303 RT)compare with y = c + m xslope = −Ea/2.303Rthe 2.303 appears only because the axis is log10 — a plot of ln k would give −Ea/R
  •  −Ea/R
  •  +Ea/R
  • KEY−Ea/2.303R
  •  +Ea/2.303R
Given
  • Arrhenius equation k = A e−Ea/RT.
  • Graph plotted as log k against 1/T.
Asked
The slope of the line.
Concept to use
Two decisions. The sign is negative because Ea appears with a minus in the exponent — raising T raises k, and since the x-axis is 1/T the line falls. The 2.303 appears only because the axis uses log10 rather than ln; converting between them introduces that factor.
Formula to use
log k = log A − Ea/(2.303 R T)  →  compare with y = c + mx
Baby steps
  1. Start with k = A e−Ea/RT and take natural logs: ln k = ln A − Ea/RT.
  2. Convert to base 10 by dividing through by 2.303: log k = log A − Ea/(2.303 R T).
  3. Compare with y = c + m x, where y = log k and x = 1/T.
  4. The coefficient of 1/T is the slope: m = −Ea/2.303R.
  5. The intercept is log A.
Answer
−Ea/2.303R
Shortcut
Two quick filters. The slope must be negative, since k rises with T while 1/T falls — that removes two options. And log10 always brings a 2.303, whereas ln does not — that removes one more. No derivation needed.
The companion result is worth storing alongside: a plot of ln k against 1/T has slope −Ea/R, with no 2.303. Which axis is used decides whether the factor appears, so read the axis label before answering.
Q158 Rate increase over a large temperature rise Answered correctly

The rate of reaction becomes 2 times for every 10°C rise in temperature. How many times the rate of reaction will increase when temperature is increased from 30°C to 80°C

  •  16
  • KEY32
  •  64
  •  28
Given
  • Temperature coefficient = 2 per 10°C rise.
  • Temperature raised from 30°C to 80°C.
Asked
The factor by which the rate increases.
Concept to use
Count how many 10-degree blocks fit into the temperature rise, then raise the coefficient to that power. The growth is multiplicative, not additive — each 10-degree step multiplies the rate again by 2.
Formula to use
factor = μn, where n = ΔT/10
Baby steps
  1. Temperature rise: ΔT = 80 − 30 = 50°C.
  2. Number of 10-degree steps: n = 50/10 = 5.
  3. Factor = 25 = 32.
Answer
32
Shortcut
Count the steps on your fingers — 30→40→50→60→70→80 is five doublings, giving 2, 4, 8, 16, 32. Counting the arrows rather than the temperatures avoids the off-by-one that produces 16 or 64.
16 is 24 and 64 is 26 — both come from miscounting the steps by one in either direction. The number of steps is the number of intervals, not the number of temperatures listed. This was answered correctly.

Graphs and matching

2 questions · 2 wrong
Q138 Half life against initial concentration Marked wrong

Match List – I (Order) with List – II (Graph of t½ vs [R]0): A = order 0, B = order 1, C = order 2. Graph I is a horizontal line, graph II is a falling curve, graph III is a rising straight line.

how t½ depends on [R]0 is different for every orderZero ordert½ = [R]0 / 2krises in a straight lineFirst ordert½ = 0.693 / kflat — no dependence at allSecond ordert½ = 1 / k[R]0falls as a hyperbolaA(0) → rising line  ·  B(1) → flat  ·  C(2) → falling curve
  •  A – I, B – II, C – III
  • MARKEDA – I, B – III, C – II
  • KEYA – III, B – I, C – II
  •  A – III, B – II, C – I
Given
  • Three orders: 0, 1 and 2.
  • Three graph shapes of t½ against initial concentration.
Asked
The correct matching.
Concept to use
Each order has its own half-life expression, and the shape follows directly from where [R]0 sits in it. Zero order puts [R]0 in the numerator, giving a straight line through the origin. First order has no [R]0 at all, giving a horizontal line — this independence is the defining feature of first-order kinetics. Second order puts [R]0 in the denominator, giving a falling hyperbola.
Formula to use
Zero: t½ = [R]0/2k  |  First: t½ = 0.693/k  |  Second: t½ = 1/k[R]0
Baby steps
  1. A, zero order. t½ ∝ [R]0 — directly proportional, so a rising straight line → graph III.
  2. B, first order. t½ = 0.693/k contains no concentration, so the graph is flat → graph I.
  3. C, second order. t½ ∝ 1/[R]0 — inversely proportional, so a falling curve → graph II.
  4. Matching: A–III, B–I, C–II.
Answer
A – III, B – I, C – II
Shortcut
Anchor on first order, the one you know best: its half life is famously independent of concentration, so it must be the flat graph. Fixing B–I alone eliminates three of the four options here.
Where it went wrong
The chosen answer swaps the zero-order and first-order graphs, giving first order the rising line. But “constant half life whatever the starting amount” is the single most quoted property of first-order kinetics — it is why radioactive decay has a fixed half life. The general rule worth carrying: t½ ∝ [R]0(1−n), so n = 0 gives a line, n = 1 gives a constant, n = 2 gives an inverse.
Q156 Matching kinetics terms to descriptions Marked wrong

Match column I with column II.

Column IColumn II
A. Mathematical representation of rate of reactionI. Order and molecularity are equal.
B. Catalyst of a reactionII. Does not depend on the concentration of reactants taken in excess.
C. The rate of the pseudo-molecular reactionIII. Rate law
D. Elementary reactionIV. Lowers the activation energy of reaction
V. Is independent of the concentration of all reactants.

  • KEYA–III, B–IV, C–II, D–I
  • MARKEDA–III, B–IV, C–V, D–I
  •  A–III, B–II, C–IV, D–V
  •  A–I, B–IV, C–II, D–III
Given
  • Four kinetics terms and five descriptions — one description is a decoy.
Asked
The correct matching.
Concept to use
The question hinges on separating pseudo-order from zero order. A pseudo-molecular reaction is independent of the reactants held in excess — but it still depends on the one reactant that varies. A reaction independent of the concentration of all reactants is zero order, which is a different thing entirely. Descriptions II and V are placed together to test exactly that distinction.
Formula to use
Pseudo first order: Rate = k′[A], independent of the excess species only  |  Zero order: Rate = k, independent of everything
Baby steps
  1. A. The mathematical representation of the rate of reaction is the rate law → III.
  2. B. A catalyst provides an alternative path with a lower activation energy → IV.
  3. C. A pseudo-molecular reaction is independent of the reactants in excess — but not of all reactants → II.
  4. D. For an elementary reaction there is a single step, so order equals molecularity → I.
  5. V is the decoy: it describes zero order, which matches nothing in column I.
  6. Matching: A–III, B–IV, C–II, D–I.
Answer
A–III, B–IV, C–II, D–I
Shortcut
Count first: five descriptions for four terms means one is a decoy. Identifying it early — here V, which describes zero order — removes the option that uses it, and this question differs between the top two choices in exactly that one position.
Where it went wrong
C was matched to V instead of II, and the two options differ only at that position. The word doing the work is “in excess” in description II: a pseudo-first-order reaction still depends on the reactant that is not in excess, so “independent of all reactants” overstates it. When two options differ in one slot, that slot is where the question is decided — check it first.

Chapter

The p-Block Elements

Groups 15, 16 and 18 — trends, anomalies and oxidation states  ·  13 questions · 7 wrong · 5 blank · 1 uncaptured

Electron gain enthalpy and the period-2 anomaly

3 questions · 2 wrong · 1 blank
Q159 Matching electron gain enthalpies in group 16 Marked wrong

Match the element with first electron gain enthalpy.

ElementElectron affinity
1) OA. −200 kJ mol−1
2) SB. −195 kJ mol−1
3) SeC. −190 kJ mol−1
4) TeD. −142 kJ mol−1

oxygen breaks the trend — sulfur is the most negativeO−142anomalyS−200Se−195Te−190taller bar = more negative electron gain enthalpy (kJ mol−1)why oxygen is the odd oneit is so small that the addedelectron is crowded and repelledfrom S onwardsthe normal trend resumes:S > Se > Te in magnitude
  • MARKEDi–A, ii–B, iii–C, iv–D
  •  i–D, ii–C, iii–B, iv–A
  •  i–A, ii–D, iii–C, iv–B
  • KEYi–D, ii–A, iii–B, iv–C
Given
  • Group 16 elements O, S, Se, Te.
  • Four electron gain enthalpy values: −200, −195, −190, −142 kJ mol−1.
Asked
The correct matching.
Concept to use
Oxygen is the anomaly of group 16. Its atom is so small that the incoming electron enters a compact 2p shell already crowded with electrons, and the repulsion it meets outweighs the nuclear attraction. So oxygen's electron gain enthalpy is the least negative in the group, not the most. From sulfur downwards the normal trend takes over and the values become progressively less negative as the atoms grow.
Formula to use
ΔegH: S (−200) > Se (−195) > Te (−190) in magnitude, with O (−142) out of line
Baby steps
  1. Sulfur has the most negative value in the group: −200 → A.
  2. Selenium, one place below, is slightly less negative: −195 → B.
  3. Tellurium, lower still: −190 → C.
  4. Oxygen, despite being at the top, has the least negative value because of electron–electron repulsion in its small 2p shell: −142 → D.
  5. Matching: O–D, S–A, Se–B, Te–C.
Answer
i–D, ii–A, iii–B, iv–C
Shortcut
Place oxygen first, because it is the exception. Once O takes the odd value (−142), the remaining three follow the ordinary down-the-group order and fall into place automatically. Dealing with the anomaly before the trend is faster than working down the list.
Where it went wrong
The chosen answer pairs the elements with the values straight down the list — first with first, second with second. That works only if both columns happen to be in matching order, and here they are not. The same straight-down error appeared in Q70 of the previous zoology paper. Anchor on the one item you are sure of and build outwards from it, rather than pairing by position.
Q168 Electron gain enthalpy of oxygen versus sulfur Marked wrong

Statement I: Among Chalcogens, oxygen has the less negative electron gain enthalpy, while sulphur has more negative electron gain enthalpy value.
Statement II: Oxygen possesses smallest size among Chalcogens.

oxygen breaks the trend — sulfur is the most negativeO−142anomalyS−200Se−195Te−190taller bar = more negative electron gain enthalpy (kJ mol−1)why oxygen is the odd oneit is so small that the addedelectron is crowded and repelledfrom S onwardsthe normal trend resumes:S > Se > Te in magnitude
  • KEYBoth statement I and statement II are correct
  •  Both statement I and statement II are incorrect
  •  Statement I is correct but statement II is incorrect
  • MARKEDStatement I is incorrect while statement II is correct
Given
  • Two statements about the chalcogens (group 16).
Asked
Which statements are correct.
Concept to use
Both statements are true, and the second is the reason for the first. Oxygen is the smallest chalcogen, and that small size means the incoming electron enters a compact shell where electron–electron repulsion is severe. The repulsion partly cancels the nuclear attraction, so oxygen's electron gain enthalpy is less negative than sulfur's despite oxygen being higher in the group.
Formula to use
O: −142 kJ mol−1  |  S: −200 kJ mol−1 — sulfur is the more negative
Baby steps
  1. Statement I. Oxygen −142, sulfur −200. Sulfur is indeed the more negative, so oxygen has the less negative value. Correct.
  2. Statement II. Atomic size increases down the group, so oxygen, the first member, is the smallest chalcogen. Correct.
  3. Both are correct — and II is the underlying cause of I, even though this question only asks for truth values.
  4. Answer: both correct.
Answer
Both statement I and statement II are correct
Shortcut
Statement II is a plain size fact with no subtlety — oxygen is the first member, therefore the smallest. Confirming it takes a second and eliminates half the options straight away.
Where it went wrong
Statement I was marked incorrect, most likely because the general expectation is that the topmost element in a group has the most negative electron gain enthalpy. Oxygen is the standing exception to that, and this paper tests it twice — here and in Q159, where the same numbers had to be matched. Both were missed, so the anomaly is worth committing to memory as a single fact: S is more negative than O.
Q171 Most negative electron gain enthalpy in two groups Not attempted

Statement – I: Chlorine has most negative electron gain enthalpy in its group.
Statement – II: Sulphur has most negative electron gain enthalpy in its group.

oxygen breaks the trend — sulfur is the most negativeO−142anomalyS−200Se−195Te−190taller bar = more negative electron gain enthalpy (kJ mol−1)why oxygen is the odd oneit is so small that the addedelectron is crowded and repelledfrom S onwardsthe normal trend resumes:S > Se > Te in magnitude
  • KEYBoth Statement-I and Statement II are true
  •  Statement -I is true but Statement-II is false
  •  Both Statement-I and Statement-II are false
  •  Statement I is false but Statement II is true
Given
  • Statement I about the halogens, Statement II about the chalcogens.
Asked
The truth value of each.
Concept to use
The second-period anomaly applies to both groups, for the same reason. F and O are so small that adding an electron creates severe repulsion in the compact 2p shell, which offsets the strong nuclear attraction. So in each group it is the second member — Cl and S — that has the most negative electron gain enthalpy, not the first.
Formula to use
Group 17: Cl (−349) > F (−328)  |  Group 16: S (−200) > O (−142), in magnitude
Baby steps
  1. Statement I. Fluorine's value is −328 kJ mol−1, chlorine's is −349. Chlorine is the more negative. True.
  2. Statement II. Oxygen is −142, sulfur is −200. Sulfur is the more negative. True.
  3. Both follow from the same cause: the small size of the period-2 element makes the added electron unusually crowded.
  4. Answer: both true.
Answer
Both Statement-I and Statement II are true
Shortcut
One rule covers both statements: in groups 16 and 17, the second member wins on electron gain enthalpy, not the first. Recognising that they are two instances of a single anomaly means one decision instead of two.
This is the third question in this paper resting on the same fact — Q159 required the numbers, Q168 tested the O-versus-S comparison, and this one extends it to F versus Cl. Three questions, one anomaly. Learning it properly is worth 12 marks in this paper alone.

Group 16 — properties and oxidation states

5 questions · 3 wrong · 2 blank
Q162 Weakest acid among group 16 hydrides Marked wrong

Which of the following is a weakest acid in its aqueous solution?

  •  H2Te
  •  H2Se
  • KEYH2S
  • MARKEDH2Po
Given
  • Four group 16 hydrides: H2Te, H2Se, H2S, H2Po.
Asked
The weakest acid.
Concept to use
Acid strength in a group of hydrides is governed by bond strength, not electronegativity. Going down the group the central atom gets larger, the H–E bond gets longer and weaker, and the proton is released more easily — so acidity increases down the group. The weakest acid is therefore the highest member present in the list.
Formula to use
Acid strength: H2O < H2S < H2Se < H2Te < H2Po
Baby steps
  1. Down group 16 the atomic size increases, so the H–E bond lengthens and weakens.
  2. A weaker bond releases H+ more readily, so acidity increases down the group.
  3. Ranking the four given: H2S < H2Se < H2Te < H2Po.
  4. The weakest is therefore the topmost of those listed: H2S.
  5. (H2O would be weaker still, but it is not among the options.)
Answer
H2S
Shortcut
For hydrides of a group, acidity simply increases down the column — so “weakest acid” means “highest in the group” and “strongest acid” means “lowest”. Locate the elements in the periodic table and read off the answer without any reasoning about bonds.
Where it went wrong
H2Po was chosen, which is the strongest acid of the four — the trend was applied in reverse. The likely source is reasoning from electronegativity: oxygen is the most electronegative, so one expects H2O to be the strongest acid. It is not. This is the same point Q179 in this paper makes explicitly, and the two questions are worth reading together.
Q167 Melting points of oxygen and sulfur Not attempted

Assertion (A): The melting point of oxygen is very much lesser than that of sulfur.
Reason (R): Oxygen is diatomic whereas sulfur octaatomic.

  • KEYBoth assertion and reason are correct statements, but reason is the correct explanation of the assertion
  •  Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion
  •  Assertion is correct, but the reason wrong statement.
  •  Assertion is wrong, but the reason is correct statement
Given
  • Oxygen exists as O2, sulfur as S8.
Asked
Judge each statement and whether R explains A.
Concept to use
Both are simple molecular solids held together by van der Waals forces, and the strength of those forces grows with the size and electron count of the molecule. O2 is tiny and diatomic, so its dispersion forces are very weak and it melts at about 55 K. S8 is a large eight-atom ring with far more electrons, so its dispersion forces are much stronger and it melts near 390 K. The molecular size named in R is precisely the cause of the difference in A.
Formula to use
Larger molecule → more electrons → stronger dispersion forces → higher melting point
Baby steps
  1. Is A true? Yes — oxygen melts at about −218°C, sulfur at about 119°C. An enormous gap. True.
  2. Is R true? Yes — oxygen is O2 and sulfur is S8 under ordinary conditions. True.
  3. Does R explain A? Apply the removal test: if sulfur were also diatomic, its dispersion forces would be far weaker and its melting point far lower. The molecular size is the reason.
  4. Both true, and R is the correct explanation.
Answer
Both assertion and reason are correct statements, but reason is the correct explanation of the assertion
Shortcut
For simple molecular substances, melting point tracks molecular size almost directly. Once you note O2 against S8, the assertion follows without recalling a single temperature — and that is exactly what makes R the explanation rather than an incidental fact.
Assertion–reason questions have been the most consistent weakness across these papers. The removal test settles nearly all of them: if R were false, would A still hold? If the answer is no, R is the explanation.
Q169 Oxidation states in group 16 Not attempted

Statement (I): Oxygen being the first member of group 16 exhibits only −2 oxidation state.
Statement (II): Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases.

down the group the higher oxidation state gets less stableSSeTePo+6+4down the group+6 stability decreasesthe inert pair effect+4 stability increasesthe s electrons stop participatingso “+6 increases down the group” is always false
  •  Statement I is correct but Statement II is incorrect
  •  Both Statement I and Statement II are correct
  • KEYBoth Statement I and Statement II are incorrect
  •  Statement I is incorrect but Statement II is correct
Given
  • Two statements about oxidation states in group 16.
Asked
Which statements are correct.
Concept to use
Both statements are wrong, each in a different way. Oxygen shows more than just −2: −1 in peroxides, −½ in superoxides and positive values with fluorine (+2 in OF2). And the group 16 trend runs the opposite way to statement II: because of the inert pair effect, the higher +6 state becomes less stable down the group while +4 becomes more stable.
Formula to use
Inert pair effect: down the group, lower oxidation state (+4) gains stability, higher (+6) loses it
Baby steps
  1. Statement I. The word “only” makes it false. Oxygen shows −1 in H2O2, −½ in superoxides, and +2 in OF2. Incorrect.
  2. Statement II. The real trend is the reverse: +6 stability decreases down the group (SO3 is stable, PoO3 is not) and +4 stability increases. Incorrect.
  3. Both statements are incorrect.
Answer
Both Statement I and Statement II are incorrect
Shortcut
Two flags settle this without any chemistry recall. Statement I contains the absolute word “only”, which is nearly always the planted error. Statement II asserts that a higher oxidation state gains stability down a group, which contradicts the inert pair effect that governs every p-block group.
This paper tests the same inert pair trend twice — here and in Q178, where statement IV makes the identical false claim. Learning it once in the correct direction (lower state more stable down the group) covers both, and it applies equally to group 13, 14 and 15.
Q178 Oxidation states of group 16 elements Marked wrong

Which is/are true statements w.r.t oxidation states of group 16 elements?
(I) Bonding in +4, +6 oxidation states is primarily covalent.
(II) S, Se and Te show +6 oxidation state with F.
(III) Stability of −2 oxidation state decreases down the group.
(IV) Stability of +6 oxidation state increases down the group.

down the group the higher oxidation state gets less stableSSeTePo+6+4down the group+6 stability decreasesthe inert pair effect+4 stability increasesthe s electrons stop participatingso “+6 increases down the group” is always false
  •  I, II, III, IV
  • MARKEDI, III, IV only
  •  I, II only
  • KEYI, II, III only
Given
  • Four statements about oxidation states in group 16.
Asked
Which are true.
Concept to use
The single discriminator is the inert pair effect. Down any p-block group the higher oxidation state becomes less stable and the lower one more stable, because the ns electrons become increasingly reluctant to participate in bonding. Statement IV claims the opposite, which makes it the false one.
Formula to use
Down the group: higher oxidation state (+6) less stable, lower (+4) more stable
Baby steps
  1. (I) In the +4 and +6 states these elements bond covalently — SO2, SF6, SO3 are all covalent. True.
  2. (II) Fluorine, being small and the most electronegative, forces the highest covalency: SF6, SeF6, TeF6 all exist. True.
  3. (III) The −2 state requires the atom to hold two extra electrons, which becomes harder as the atom grows. So its stability falls down the group. True.
  4. (IV) The inert pair effect makes +6 less stable down the group, not more. False.
  5. Answer: I, II, III only.
Answer
I, II, III only
Shortcut
Test statement IV first, because it is the only one making a claim about a trend direction — and direction claims are where papers plant errors. Rejecting IV leaves only two options, and one of them omits II, which is easily confirmed by SF6.
Where it went wrong
Statement III was rejected and IV accepted — exactly the wrong way round on both. Note that III and IV are not independent: both describe stability trends down the group, and the inert pair effect makes the lower state gain and the higher state lose. Getting the effect the right way round settles both at once. This is the same trend as statement II of Q169 in this paper, which was also missed.
Q179 Why H2S is more acidic than H2O Marked wrong

Statement 1: Oxygen is more electronegative than sulphur, yet H2S is more acidic than H2O
Statement 2: H–S bond is weaker than O–H bond.

  • KEYStatement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1
  • MARKEDStatement 1 is True, Statement 2 is true; Statement 2 is NOT a correct explanation for Statement 1
  •  Statement 1 is True, Statement 2 is False
  •  Statement 1 is False, Statement 2 is True
Given
  • Statement 1: O is more electronegative than S, yet H2S is the stronger acid.
  • Statement 2: the H–S bond is weaker than the O–H bond.
Asked
Judge each statement and whether 2 explains 1.
Concept to use
This question exists to correct a common misconception. Electronegativity would suggest H2O should release H+ more easily — but for hydrides down a group, bond strength dominates. Sulfur is larger, so the H–S bond is longer and weaker, and the proton leaves more readily. Statement 2 supplies exactly the mechanism that resolves the apparent paradox in Statement 1, so it is the correct explanation.
Formula to use
Acidity of hydrides ∝ ease of H–E bond breaking, not electronegativity
Baby steps
  1. Is Statement 1 true? Yes on both counts — oxygen is more electronegative, and H2S (Ka ≈ 10−7) is far more acidic than water (Ka ≈ 10−16). True.
  2. Is Statement 2 true? Yes — sulfur's larger size gives a longer, weaker H–S bond. True.
  3. Does 2 explain 1? Apply the removal test: if the H–S bond were not weaker, H2S would not be the stronger acid and the paradox would not resolve. Statement 2 is the mechanism.
  4. Both true, and 2 is the correct explanation.
Answer
Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1
Shortcut
When Statement 1 poses a puzzle (“yet”, “despite”, “although”) and Statement 2 resolves it, Statement 2 is the explanation almost by construction. The word yet in Statement 1 is signalling that an explanation is wanted.
Where it went wrong
Both statements were accepted as true but the link was denied — the same shape as Q149 in the zoology paper and Q128 in the botany paper. The recurring lesson: after marking both true, do not stop. Ask what would happen if the reason were removed. Note also that this question directly answers Q162 in the same paper, where the acidity trend was applied backwards.

Group 15 — properties and configuration

4 questions · 1 wrong · 2 blank · 1 uncaptured
Q173 Disproportionation in nitrogen and phosphorus Marked wrong

Statement-I: In the case of Nitrogen all the oxidation states from +1 to +4 tend to undergo disproportionation in acidic solution
Statement-II: In the case of phosphorous nearly all the oxidation states readily undergoes disproportionation in acidic and alkali medium

  •  Statement I and II are correct
  • KEYStatement-I correct and statement-II incorrect
  • MARKEDStatement-I incorrect and statement-II correct
  •  Both statements are incorrect
Given
  • Two statements about disproportionation in group 15.
Asked
Which statements are correct.
Concept to use
Statement I is a direct NCERT line: nitrogen's intermediate oxidation states from +1 to +4 all disproportionate in acid. Statement II overstates the phosphorus case. Phosphorus does disproportionate readily — but characteristically in alkaline medium, not in both media indiscriminately. The word “and” joining acidic to alkali is what makes it false.
Formula to use
N: intermediate states +1 to +4 disproportionate in acid  |  P: disproportionates chiefly in alkali
Baby steps
  1. Statement I. Nitrogen's intermediate oxidation states are unstable with respect to disproportionation in acidic solution — for example 3HNO2 → HNO3 + 2NO + H2O. Correct.
  2. Statement II. The claim of “nearly all oxidation states, in acidic and alkali medium” is too broad. Phosphorus disproportionation is characteristically an alkaline-medium phenomenon. Incorrect.
  3. Answer: I correct, II incorrect.
Answer
Statement-I correct and statement-II incorrect
Shortcut
Look at the qualifiers rather than the chemistry. Statement I is bounded and specific — a named range of oxidation states, one named medium. Statement II sweeps: “nearly all”, “readily”, “acidic and alkali”. In statement questions the over-general option is usually the false one.
Where it went wrong
The two statements were swapped, with I marked false and II true. That is the third statement-pair in this paper answered by reversing the two halves — the same pattern as Q137 and Q168. The habit to build: judge each statement on its own and write T or F beside it before reading a single option, so the two cannot be transposed.
Q174 Maximum covalency of a group 15 non-metal Not attempted

The maximum covalency of a non-metallic group 15 element 'E' with weakest E–E bond is:

  •  5
  •  3
  •  6
  • KEY4
Given
  • A non-metallic element of group 15 with the weakest E–E single bond.
Asked
Its maximum covalency.
Concept to use
Two steps. First identify the element: among the group 15 non-metals (N and P), the N–N single bond is anomalously weak — the two small nitrogen atoms bring their lone pairs close together and repel. So E is nitrogen. Second, nitrogen sits in period 2 with only four valence orbitals (one 2s and three 2p) and no d subshell, so it cannot form more than four bonds.
Formula to use
Single bond enthalpy: P–P (201) > N–N (159) kJ mol−1  |  period 2 → max covalency 4
Baby steps
  1. Restrict to non-metals of group 15: nitrogen and phosphorus.
  2. Compare their single bond strengths: N–N is 159 kJ mol−1, P–P is 201. The weaker is N–N, because lone-pair repulsion is severe in the small nitrogen atom.
  3. So E = nitrogen.
  4. Nitrogen has four valence orbitals and no accessible d orbitals, so it can form at most four bonds.
  5. Maximum covalency = 4 (as in NH4+).
Answer
4
Shortcut
Two anomalies decide this and both are worth storing: N–N is weaker than P–P (lone pair repulsion), and period 2 elements cap out at covalency 4 (no d orbitals). The second rules out 5 and 6 straight away, whichever element you settle on.
The N–N anomaly appeared as statement (ii) of Q50 in the very first chemistry paper of this series, where it was also missed. It is one of the highest-yield facts in the p-block: the smooth-looking trend N > P > As > Sb is wrong for single bond energy.
Q175 Matching elements to valence configurations Options not captured

Match column I (element) with column II (valence electronic configuration).

Column IColumn II
A. PI. [Kr]4d105s25p3
B. AsII. [Ar]3d104s24p3
C. McIII. [Rn]5f146d107s27p3
D. SbIV. [Ra]5f146d107s27p3
V. [Ne]3s23p3
The option list was cut off in the screenshot, so the derived matching is given below.

  • KEYA–V, B–II, C–III, D–I  (derived — options not captured)
Given
  • Four group 15 elements: P, As, Mc (moscovium, Z = 115), Sb.
  • Five candidate configurations — one is a decoy.
Asked
The correct matching.
Concept to use
Every group 15 element ends in ns2np3, so the only thing that distinguishes them is the period — that is, which noble gas core precedes them. Match by the core, and the whole question becomes a check on which noble gas sits at the end of each period.
Formula to use
Group 15 valence: ns2np3; the core identifies the period
Baby steps
  1. A. Phosphorus is in period 3, so the core is [Ne] → [Ne]3s23p3 = V.
  2. B. Arsenic, period 4, core [Ar], and the 3d shell is filled first → [Ar]3d104s24p3 = II.
  3. C. Moscovium (Z = 115), period 7, core [Rn] → [Rn]5f146d107s27p3 = III.
  4. D. Antimony, period 5, core [Kr] → [Kr]4d105s25p3 = I.
  5. Entry IV is the decoy: it uses [Ra] as a core, but radium is not a noble gas and never serves as a core.
  6. Matching: A–V, B–II, C–III, D–I.
Answer
A–V, B–II, C–III, D–I
Shortcut
Read only the core symbol in each configuration — [Ne], [Ar], [Kr], [Rn] — and match it to the period. The rest of each configuration is identical in pattern, so it carries no information. Spotting the fake core [Ra] eliminates the decoy instantly.
The options were not captured in the screenshot, so this matching is derived from the periodic table rather than read from a key. The decoy is instructive in itself: a core must always be a noble gas, so [Ra] is not a legitimate notation.
Q176 Atomic numbers belonging to group 15 Not attempted

The following set of atomic numbers of elements belongs to group 15 in periodic table

  •  34, 52
  • KEY33, 51
  •  32, 50
  •  31, 49
Given
  • Four pairs of atomic numbers.
Asked
Which pair belongs to group 15.
Concept to use
Group 15 runs N (7), P (15), As (33), Sb (51), Bi (83). Rather than memorising the whole list, note that the options differ by one unit each — 31/49 is group 13, 32/50 is group 14, 33/51 is group 15, 34/52 is group 16 — so anchoring on any one of them fixes the rest by counting.
Formula to use
Group 15: N 7, P 15, As 33, Sb 51, Bi 83
Baby steps
  1. Recall the group 15 sequence: N (7), P (15), As (33), Sb (51), Bi (83).
  2. The pair 33, 51 is arsenic and antimony — both group 15. ✓
  3. Check the others: 32, 50 is Ge and Sn (group 14); 34, 52 is Se and Te (group 16); 31, 49 is Ga and In (group 13).
  4. Answer: 33, 51.
Answer
33, 51
Shortcut
Anchor on a number you are sure of and count sideways. If you know Se is 34 (group 16), then group 15 must be 33 — one place to the left. The four options are deliberately consecutive, so a single anchor plus counting answers the question.
Arsenic (33) and antimony (51) also appear in Q175 of this paper, matched to their configurations. The two questions reinforce each other, and knowing the group 15 atomic numbers answers both.

Group 18 — the noble gases

1 question · 1 wrong
Q160 Properties of the noble gases Marked wrong

Consider following properties of the noble gases.
I: They readily react and form compounds which are colourless.
II: They generally do not form ionic compounds.
III: Xenon has variable oxidation states in its compounds.
IV: The larger Xe form compounds due to low ionisation energy.
Select correct properties.

  •  I, II, III
  • KEYII, III, IV
  •  I, III, IV
  • MARKEDAll
Given
  • Four statements about the noble gases.
Asked
Which are correct.
Concept to use
The defining feature of the noble gases is that they are inert — the word “readily” in statement I contradicts the entire basis of the group. Only the heavier members react at all, and only with the most electronegative elements. The reason they do so is that ionisation energy falls down the group, so xenon's outer electrons are the easiest to disturb.
Formula to use
Ionisation energy falls down group 18 → only Xe (and to a small extent Kr, Rn) form compounds, all covalent
Baby steps
  1. I. Noble gases do not react readily — that is precisely what makes them noble. Only Xe forms a reasonable set of compounds, and only under forcing conditions. Incorrect.
  2. II. Their compounds (XeF2, XeF4, XeO3) are covalent, formed with F and O. Ionic compounds would require losing or gaining electrons outright, which they resist. Correct.
  3. III. Xenon shows +2, +4, +6 and +8 in its various compounds. Correct.
  4. IV. Ionisation energy decreases down the group, so the larger Xe is the most reactive of the practically available noble gases. Correct.
  5. Answer: II, III, IV.
Answer
II, III, IV
Shortcut
Statement I contains “readily react”, which contradicts the name of the group. Rejecting it at once removes both “All” and any option containing I — here that is three of the four options, leaving the answer without examining II, III or IV at all.
Where it went wrong
Choosing “All” means every statement was accepted without testing the one that conflicts with the group's basic character. “All of the above” is correct far less often than it looks, and it should raise suspicion rather than confidence — especially when one statement contradicts the defining property of the family.

What the thirty-one have in common

Two chapters, seventeen attempted and missed, twelve blank, one correct and one whose options were cut off. The attempt rate has risen sharply — and the errors that emerged are strikingly repetitive.

1 · The same anomaly was tested three times and missed twice — Q159, Q168, Q171
All three rest on one fact: sulfur has a more negative electron gain enthalpy than oxygen, because oxygen is so small that the incoming electron meets severe repulsion. Q159 asked for the numbers and got a straight-down-the-list match. Q168 asked the O-versus-S comparison directly and was marked backwards. Q171 extends the same anomaly to F versus Cl and was left blank. Fix: one line covers all three — in groups 16 and 17 the second member wins, not the first. Worth 12 marks in this paper alone.

2 · Trend directions reversed — Q137, Q162, Q169, Q178
Q137 accepted that heating decreases the fraction of molecules above Ea. Q162 picked the strongest acid when asked for the weakest. Q178 accepted that +6 stability increases down group 16, and Q169 tests the identical claim. Each is one word flipped inside an otherwise familiar sentence. Fix: underline the direction word — increases/decreases, strongest/weakest, more/less — before judging any trend statement. And for the p-block, hold one master rule: down a group the lower oxidation state gains stability and the higher one loses it.

3 · Statement pairs answered by transposing the two halves — Q137, Q168, Q173, Q179
In each case the truth values of statements I and II were swapped, or both were accepted true and the causal link denied. Q179 is the clearest: both statements are true and the second explains the first, since the “yet” in statement 1 poses a puzzle that statement 2 resolves. Fix: write T or F beside each statement before reading any option, then apply the removal test for the link. This is now the most persistent weakness across every paper in this set.

4 · Questions that answer each other were treated separately
Q179 explains exactly why H2S is more acidic than H2O — which is the fact Q162 needed and applied backwards. Q169 and Q178 test the same inert pair trend. Q151 and Q156 both test pseudo-order. Reading a paper for these pairings is itself a technique: getting one right often hands you another.

Seven results that cover both chapters

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. All figures have been drawn fresh for these notes, and every numerical answer here was checked computationally before being written in.