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Two chapters, twenty questions

Every chemistry question from the test that was marked wrong or left blank, arranged by chapter and sub-topic so a whole chapter can be revised in one pass. Each question carries what was given, what was asked, the concept and formula behind it, the steps in full, and the fastest route through.

20
Questions lost
8
Attempted, wrong
12
Left blank
2
Chapters involved

p-Block Elements 8 · Chemical Kinetics 12. On NEET marking (+4 / −1) these twenty carried 80 marks; the eight wrong attempts also cost 8 more.

The pattern is unusually clean. Not one of these twenty is a hard question. The p-Block losses are all trend-direction slips — the right fact recalled and then applied the wrong way round. The Kinetics losses are mostly not losses of knowledge at all: ten of the twelve were never started.

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

The p-Block Elements

8 questions · 6 wrong · 2 blank

Q46Q47Q49Q50Q59Q63Q66Q70

Chemical Kinetics

12 questions · 2 wrong · 10 blank

Q75Q78Q79Q81Q82Q83Q85Q86Q87Q88Q89Q90

Chapter

The p-Block Elements

Groups 15 and 16  ·  8 questions · 6 wrong · 2 blank

Group 15 — nitrogen family

4 questions · 3 wrong · 1 blank
Q46 Group 15 · dπ–dπ bonding & hydride basicity Marked wrong

A group 15 element forms dπ–dπ bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form dπ–dπ bond. The atomic number of the element is

  •  7
  • KEY15
  •  33
  • MARKED51
Given
  • The element belongs to group 15.
  • It is able to form dπ–dπ bonds with transition metals.
  • Among only those group-15 members that can form dπ–dπ bonds, its hydride is the strongest base.
Asked
The atomic number (Z) of the element.
Concept to use
dπ–dπ back-bonding needs vacant d orbitals on the group-15 atom. Nitrogen sits in period 2 — its valence shell is only 2s and 2p, there is no 2d subshell — so nitrogen is disqualified immediately. For the survivors, basicity of the hydride depends on how available the lone pair is: as you go down the group the atom gets bigger, the lone pair sits in a larger, more diffuse orbital, and donation gets weaker.
Formula to use
Basicity of group-15 hydrides:  NH3 > PH3 > AsH3 > SbH3 > BiH3
Baby steps
  1. Ask which members can form dπ–dπ bonds. P, As, Sb and Bi have vacant 3d / 4d / 5d orbitals — they qualify. N has no d orbitals — it is out.
  2. So the hydrides in the competition are only PH3, AsH3, SbH3, BiH3. NH3 is not a candidate, even though it is the most basic of all five.
  3. Basicity falls down the group, so the strongest base among the survivors is the topmost one: PH3.
  4. The element is phosphorus. Z = 15.
Answer
15  (phosphorus)
Shortcut
Strike nitrogen out the moment “d orbitals” or “dπ–dπ” appears, then pick the topmost element still standing. In group 15, that two-move routine solves most of these.
Where it went wrong
51 is antimony. Sb does form dπ–dπ bonds, so it clears the first filter — but SbH3 is one of the weakest bases in the group, not the strongest. The basicity trend was applied in the wrong direction.
Q47 Nitrogen — oxides vs halides Marked wrong

Given below are two statements.
Statement I: Nitrogen exhibits oxidation states ranging from +1 to +5 in its oxides due to its ability to form pπ–pπ multiple bonds with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In light of the above statements, choose the correct answer from the options given below.

  • KEYBoth statement I and statement II are correct.
  •  Both statement I and statement II are not correct.
  •  Statement I is correct, but statement II is incorrect.
  • MARKEDStatement I is incorrect, but statement II is correct.
Given
  • Statement I — N shows +1 to +5 in oxides, reason given: pπ–pπ bonding with O.
  • Statement II — N forms no +5 halide, reason given: no d orbitals.
Asked
Judge each statement independently as correct or incorrect.
Concept to use
This is not an assertion–reason question — the two statements are separate claims, so test each on its own and never ask whether II explains I. Statement I is about bonding capacity with oxygen (π bonds allowed, so high oxidation number reachable with only four orbitals). Statement II is about covalency — a +5 halide would need five σ bonds, and five σ bonds need five orbitals.
Formula to use
N2O (+1)  ·  NO (+2)  ·  N2O3 (+3)  ·  NO2/N2O4 (+4)  ·  N2O5 (+5)
Baby steps
  1. List nitrogen’s oxides and their oxidation numbers — the row above covers every value from +1 to +5 with no gaps. The range in Statement I is exactly right.
  2. Why is it possible? N is small, so its 2p orbitals overlap sideways with oxygen’s 2p very efficiently. pπ–pπ multiple bonding lets nitrogen reach a high oxidation number without needing extra σ bonds. Statement I is correct.
  3. Now the halides. NX5 would require nitrogen to make five single bonds — covalency 5. Nitrogen’s valence shell has only four orbitals (one 2s + three 2p), so maximum covalency is 4. No 3d is available to expand the octet.
  4. That is why NCl5 is unknown while PCl5 is a bench chemical. Statement II is correct.
  5. Both correct.
Answer
Both statement I and statement II are correct.
Shortcut
When two statements are joined by nothing more than “and”, mark each one T/F separately in the margin before you look at the options. Half of these questions are lost by testing a link that was never claimed.
Where it went wrong
Statement I was marked incorrect — almost certainly because −3 is missing from the range. But the statement says “in its oxides”. Bonded to oxygen, nitrogen can only be positive; −3 belongs to nitrides and NH3, which are not oxides. Read the qualifier before judging the range.
Q49 Maximum covalency & which species exist Not attempted

Number of species which are known to exist among the following list are
a) NF3   b) OCl3−   c) PF6−   d) PCl5   e) SF6   f) SCl7

  •  5
  •  2
  •  3
  • KEY4
Given
  • Six candidate species, each with a central atom and a halogen or oxygen ligand set.
Asked
How many of the six actually exist.
Concept to use
Two rules settle every item. Rule 1 — period-2 ceiling: N, O and F have only four valence orbitals (2s + three 2p), so their maximum covalency is 4, and for oxygen it is effectively 2. Rule 2 — the fluorine privilege: only fluorine is small enough and electronegative enough to force period-3 elements to their highest covalency. Chlorine is too bulky, so high-covalency chlorides simply do not form.
Formula to use
Max covalency: period 2 → 4  |  period 3 and below → 5, 6 (with F)
Baby steps
  1. a) NF3 — N makes 3 bonds, covalency 3, well within the limit of 4. Exists.
  2. b) OCl3− — oxygen would need three bonds. O is period 2, has no d orbitals and a covalency of 2. Does not exist. (The real ion with this formula is ClO3−, chlorate — chlorine central, not oxygen.)
  3. c) PF6− — covalency 6 on P, allowed via 3d, and six small F atoms fit comfortably. Exists.
  4. d) PCl5 — covalency 5, a standard reagent. Exists.
  5. e) SF6 — covalency 6 on S with six small F atoms. Exists.
  6. f) SCl7 — covalency 7 is beyond sulfur, and Cl is far too large. Even SCl6 is unknown. Does not exist.
  7. Count the survivors: a, c, d, e → 4.
Answer
4
Shortcut
Run a two-question filter on each species. (1) Is the central atom from period 2? Then covalency ≤ 4 (oxygen ≤ 2). (2) Is the halogen Cl, Br or I with covalency 5 or more demanded? Then it fails. Anything tripping either test is fictitious — you never need to draw a structure.
Q50 Group 15 trends — four claims Marked wrong

The correct arrangement of elements according to the property given is
i) N > P > As > Sb  — ionisation enthalpy
ii) N > P > As > Sb  — single bond energy between same atoms
iii) N > P > Sb > Bi  — stability in −3 oxidation state
iv) N > Sb > P > As  — electronegativity

  • MARKEDi, ii, iii only
  •  i, iii, iv only
  • KEYi, iii only
  •  ii, iv only
Given
  • Four proposed orderings, each attached to a named property.
Asked
Which of the four orderings are correct.
Concept to use
Three of these are plain periodic trends and one is an anomaly. Ionisation enthalpy and electronegativity both fall smoothly down the group. Stability of the −3 state also falls, because forming M3−/MH3 gets harder for bigger atoms. But single-bond energy is not smooth: N–N is abnormally weak, because the two nitrogen atoms are so small that their lone pairs sit close together and repel each other across the bond.
Formula to use
Single bond enthalpy / kJ mol−1:   P–P 201 > N–N 159 > As–As 146 > Sb–Sb 121
Baby steps
  1. (i) Ionisation enthalpy N > P > As > Sb. Size increases down the group, the outer electron is further from the nucleus and better shielded, so IE falls. Correct.
  2. (ii) Single bond energy N > P > As > Sb. This puts N–N at the top, but N–N (159) is weaker than P–P (201) because of lone-pair–lone-pair repulsion in the tiny nitrogen atom. The true order is P–P > N–N > As–As > Sb–Sb. Incorrect.
  3. (iii) Stability of −3 state N > P > Sb > Bi. The −3 state needs the atom to hold three extra electrons; that gets steadily harder as the atom grows, and by Bi it is essentially unknown. Correct.
  4. (iv) Electronegativity N > Sb > P > As. Electronegativity decreases monotonically: N > P > As > Sb. Sb cannot outrank P. Incorrect.
  5. Correct statements: i and iii.
Answer
i, iii only
Shortcut
Attack (ii) and (iv) first, not (i). (iv) is a scrambled order — one glance kills it, and that removes two options. (ii) is the classic N–N anomaly — killing it leaves only one answer. You never have to verify (i) or (iii) at all.
Where it went wrong
Statement (ii) was accepted because it looks like a normal down-the-group decrease and is written with the same elements in the same order as (i). Same-looking sequences with different property labels are the whole point of this question — the property, not the sequence, decides.

Group 16 — oxygen family

4 questions · 3 wrong · 1 blank
Q59 Disproportionation — oxidation-state test Marked wrong

Which one of the following has no tendency to undergo a disproportionation reaction?

  •  HNO2
  •  H3PO3
  • KEYSF6
  • MARKEDS2Cl2
Given
  • Four species; one of them cannot disproportionate.
Asked
Identify the species with no tendency to disproportionate.
Concept to use
Disproportionation is one element being simultaneously oxidised and reduced. That is only possible if the element starts in an intermediate oxidation state — it needs room both above and below. An element already at its group maximum (or minimum) has nowhere to go in one direction, so it cannot disproportionate.
Formula to use
Disproportionation possible ⇔ minimum < oxidation state < maximum
Baby steps
  1. HNO2: N is +3. Range for N is −3 to +5, so +3 is intermediate. Indeed 3 HNO2 → HNO3 + 2 NO + H2O. Disproportionates.
  2. H3PO3: P is +3, intermediate between −3 and +5. On heating, 4 H3PO3 → 3 H3PO4 + PH3. Disproportionates.
  3. S2Cl2: two Cl at −1 give −2 total, so the two S share +2 → each S is +1. That is intermediate between −2 and +6. Disproportionates.
  4. SF6: six F at −1 give −6, so S is +6 — the maximum possible for sulfur (group 16). There is no higher state to be oxidised to, so half the reaction is impossible. Cannot disproportionate.
  5. Answer: SF6.
Answer
SF6
Shortcut
Write the oxidation number of the central atom above each formula — four quick numbers, +3, +3, +1, +6. The one sitting at the group maximum is the answer. No reactions need to be recalled.
Where it went wrong
S2Cl2 was picked because it looks unfamiliar, and unfamiliar formulas feel like “the odd one out”. Unfamiliarity is not a chemical argument. Assigning the oxidation number takes five seconds and shows S is at +1 — squarely intermediate, so it is one of the species that does disproportionate.
Q63 Oxidation number of oxygen — the fluorine exception Marked wrong

Which one of the oxidation states is shown by oxygen in OF2?

  •  +1
  • MARKED−2
  • KEY+2
  •  −1
Given
  • The neutral molecule OF2.
Asked
Oxidation number of oxygen.
Concept to use
Oxidation number is assigned by giving the shared electrons to the more electronegative atom. Fluorine is the most electronegative element that exists — nothing can out-pull it — so F is always −1 in every compound, with no exceptions at all. Oxygen’s usual −2 is only a default that applies when oxygen is the more electronegative partner. In OF2 it is not.
Formula to use
Sum of oxidation numbers in a neutral molecule = 0
Baby steps
  1. Assign fluorine first, because F has zero exceptions: each F = −1, and there are two of them → −2 total.
  2. Let oxygen be x. Then x + (−2) = 0.
  3. x = +2.
  4. Sanity check: F is more electronegative than O, so oxygen must come out positive. +2 fits; −2 would mean oxygen out-pulled fluorine, which is impossible.
Answer
+2
Shortcut
Memorise the four oxygen states as one line: −2 normally, −1 in peroxides, −½ in superoxides, positive with fluorine (OF2 = +2, O2F2 = +1). The instant an F appears in the formula, assign F before anything else.
Where it went wrong
−2 is oxygen’s reflex value and it was applied without checking the partner. This is a rule-exception miss, not a knowledge gap — the fix is a habit: look for F first, then assign O.
Q66 Group 16 — identify elements from ionisation enthalpy Not attempted

Five elements of group-16 represented by A, B, C, D and E have the first ionisation enthalpies as 813, 1000, 869, 941 and 1314 kJ mol−1 respectively. Which of the following correctly identifies the corresponding element based on the given physical properties?

  •  The element with the lowest electronegativity is E.
  •  The element with the highest melting point is D.
  • KEYThe element with the highest boiling point is C.
  •  The element with the largest covalent radius is B.
Given
  • Group 16 elements labelled A, B, C, D, E.
  • First ionisation enthalpies in order: A = 813, B = 1000, C = 869, D = 941, E = 1314 kJ mol−1.
Asked
Which one of the four property statements is true.
Concept to use
First ionisation enthalpy decreases down group 16: O > S > Se > Te > Po. So ranking the five numbers from largest to smallest directly reveals which letter is which element. After that, every option is a one-line trend check. Melting and boiling points rise down the group up to tellurium, then dip slightly at polonium — so Te, not Po, holds the maxima.
Formula to use
IE1 (kJ mol−1): O 1314 > S 1000 > Se 941 > Te 869 > Po 813
Baby steps
  1. Sort the given values in descending order: 1314 (E), 1000 (B), 941 (D), 869 (C), 813 (A).
  2. Match to the known sequence O > S > Se > Te > Po. That gives E = O, B = S, D = Se, C = Te, A = Po. Write this down before touching the options.
  3. “Lowest electronegativity is E” — E is oxygen, which has the highest electronegativity in the group. The lowest belongs to Po = A. False.
  4. “Highest melting point is D” — D is selenium (490 K). Tellurium melts at 725 K, the highest in the group. That is C, not D. False.
  5. “Highest boiling point is C” — C is tellurium, b.p. ≈ 1261 K, above Po (≈ 1235 K), Se (958 K) and S (718 K). True.
  6. “Largest covalent radius is B” — B is sulfur, one of the smallest. The largest is Po = A. False.
Answer
The element with the highest boiling point is C.
Shortcut
Decode all five letters first, before reading a single option. Once “E=O, B=S, D=Se, C=Te, A=Po” is on the page, each option is a two-second true/false and the whole question takes under a minute. Trying to evaluate options while still decoding is what makes this feel hard.
Both extremes of melting and boiling point in group 16 sit at tellurium, not polonium — the trend reverses at the last member. That single fact decides two of the four options here.
Q70 Group 16 dioxides — redox character Marked wrong

Nature of the compounds SO2, TeO2 is respectively

  • MARKEDOxidising, Basic
  • KEYReducing, oxidizing
  •  Basic, Acidic
  •  Amphoteric, Oxidizing
Given
  • Two group-16 dioxides, both with the central atom in the +4 oxidation state.
Asked
The redox nature of each, in order.
Concept to use
Both S and Te are +4 here, yet they behave oppositely — and the reason is which way the +4 state wants to move. For sulfur, the +6 state is perfectly stable, so S(IV) readily climbs to S(VI): SO2 gives electrons away and is a reducing agent. For tellurium, the inert-pair effect makes the higher +6 state unstable, so Te(IV) refuses to climb and would rather drop to Te(0): TeO2 takes electrons and is an oxidising agent.
Formula to use
Stability of the higher (+6) state falls down group 16 → lighter dioxide reduces, heavier dioxide oxidises
Baby steps
  1. Oxidation number check: in SO2, S = +4; in TeO2, Te = +4. Identical starting point.
  2. For sulfur, +6 is comfortably reachable (H2SO4, SO3). So SO2 is easily oxidised — meaning it reduces whatever it meets: SO2 + Cl2 + 2H2O → H2SO4 + 2HCl. SO2 = reducing agent.
  3. For tellurium, the 5s pair is reluctant to participate (inert-pair effect), so Te(VI) is unstable and Te(IV) is the comfortable state. TeO2 therefore prefers to be reduced. TeO2 = oxidising agent.
  4. Answer, in order: reducing, oxidising.
Answer
Reducing, oxidizing
Shortcut
Same oxidation number, different rows: the lighter member is the reductant, the heavier member is the oxidant. That one sentence handles SO2 vs TeO2, SO2 vs SeO2, and the analogous group-14 and group-15 pairs.
Where it went wrong
“Oxidising, Basic” is wrong twice over. It reverses SO2’s redox role, and it also switches classification mid-answer — the question asked for nature of both, and the pairing that makes sense is redox for both. On acid–base grounds TeO2 is amphoteric, never basic; all group-16 dioxides are acidic or amphoteric, never basic.

Chapter

Chemical Kinetics

Rate laws, order and temperature dependence  ·  12 questions · 2 wrong · 10 blank

Rate of reaction and order

5 questions · 1 wrong · 4 blank
Q81 Rate of reaction vs rate of species Not attempted

For 3A → xB, d[B]/dt is found to be 2/3rd of −d[A]/dt. Then, the value of x is

  •  1.5
  •  3
  •  1/2
  • KEY2
Given
  • Balanced reaction 3A → xB.
  • d[B]/dt = (2/3) × (−d[A]/dt).
Asked
The stoichiometric coefficient x.
Concept to use
A reaction has one rate, but each species changes at its own speed. To get the single unique rate, divide each species’ rate of change by its stoichiometric coefficient — reactants with a minus sign, products with a plus.
Formula to use
Rate = −(1/3) d[A]/dt = +(1/x) d[B]/dt
Baby steps
  1. Write the equality from the balanced equation: −(1/3) d[A]/dt = (1/x) d[B]/dt.
  2. Substitute the given relation d[B]/dt = (2/3)(−d[A]/dt):   −(1/3) d[A]/dt = (1/x) × (2/3) × (−d[A]/dt).
  3. The factor (−d[A]/dt) appears on both sides — cancel it:   1/3 = 2 / (3x).
  4. Cross-multiply: 3x = 6 → x = 2.
  5. Check: with 3A → 2B, d[B]/dt = (2/3)(−d[A]/dt). Matches the given statement.
Answer
2
Shortcut
If d[B]/dt = f × (−d[A]/dt) for a reaction aA → xB, then x = a × f directly. Here x = 3 × 2/3 = 2, in one step, no algebra.
Dropping the stoichiometric coefficients is the single most common mechanical error in this topic. Write the −(1/a) and +(1/x) prefixes before substituting anything.
Q86 Order from concentration–rate ratios Not attempted

For A + B → C + D, when [A] alone is doubled, rate gets doubled, but when [B] alone is increased by 9 times, rate gets tripled. Then order of reaction is

  •  3/4
  • KEY3/2
  •  4/9
  •  2
Given
  • Doubling [A] at constant [B] doubles the rate.
  • Increasing [B] nine-fold at constant [A] triples the rate.
Asked
Overall order of the reaction.
Concept to use
Order is found experimentally, never from the balanced equation. When one concentration is changed and the other held fixed, the rate ratio equals the concentration ratio raised to that species’ order. Orders need not be whole numbers.
Formula to use
Rate = k[A]a[B]b  →  (rate ratio) = (conc. ratio)order  →  overall order = a + b
Baby steps
  1. For A: 2a = 2 → a = 1.
  2. For B: 9b = 3. Rewrite both sides to base 3: (32)b = 31, so 32b = 31.
  3. Equate the exponents: 2b = 1 → b = ½.
  4. Overall order = a + b = 1 + ½ = 3/2.
Answer
3/2
Shortcut
Convert both numbers to a common base rather than reaching for logarithms. 9 = 32 and 3 = 31, so the exponent is ½ on sight. In general, a nine-fold change producing a three-fold effect always means order ½.
The distractors 3/4 and 4/9 are built from mis-pairing the numbers (3 and 4, or 4 and 9). Write a and b separately with their labels before adding.
Q87 Rate law from an initial-rate data table Not attempted

The following results have been obtained during the kinetic studies of the reaction 2P + Q → R + S.

Experiment[P] mol/lit[Q] mol/litRate (mol/lit/sec)
I0.10.16 × 10−3
II0.40.12.4 × 10−2
III0.40.29.6 × 10−2
IV0.80.476.8 × 10−2
The correct option for the rate equation for above reaction is

  •  Rate = k[P]2[Q]
  •  Rate = k[P]2[Q]0
  •  Rate = k[P][Q]
  • KEYRate = k[P]1[Q]2
Given
  • Four experiments with initial concentrations of P and Q and the corresponding initial rates.
Asked
The experimentally correct rate equation.
Concept to use
Pick pairs of experiments in which only one concentration changes. The ratio of rates then isolates the order in that species. The coefficients 2 and 1 in the balanced equation are irrelevant — a rate law is measured, not read off the equation.
Formula to use
Rate = k[P]a[Q]b  →  a = ln(rate ratio) / ln([P] ratio), same for b
Baby steps
  1. Find a using I → II ([Q] is fixed at 0.1). [P] goes 0.1 → 0.4, a factor of 4. Rate goes 6 × 10−3 → 2.4 × 10−2, also a factor of 4.
  2. So 4a = 4 → a = 1. Note this is not 2, despite the 2P in the equation.
  3. Find b using II → III ([P] is fixed at 0.4). [Q] goes 0.1 → 0.2, a factor of 2. Rate goes 2.4 × 10−2 → 9.6 × 10−2, a factor of 4.
  4. So 2b = 4 → b = 2.
  5. Verify with IV. From III to IV, [P] doubles and [Q] doubles, so the rate should rise by 21 × 22 = 8×. Check: 9.6 × 10−2 × 8 = 76.8 × 10−2. ✓ Matches exactly.
  6. Rate = k[P]1[Q]2.
Answer
Rate = k[P][Q]2
Shortcut
Scan the table for the two rows where a single column changes — those two pairs give both orders and nothing else is needed. Row IV exists only as a check, so use it to confirm rather than to derive. Beware the option Rate = k[P]2[Q], which is the balanced equation dressed up as a rate law: it is on the list precisely to catch that reflex.
Overall order here is 1 + 2 = 3, so k carries units of mol−2 L2 s−1 — worth checking if a variant of this question asks for units.
Q89 Order vs molecularity Marked wrong

Given below are two statements:
Statement I: Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions.
Statement II: Reactions with the molecularity three (or) more are very rare as the probability that more than three molecules can collide and react simultaneously is very small.
In light of the above statements, choose the correct answer from the options given below.

  • KEYBoth statements I and II are true
  •  Both statements I and II are false
  •  Statement I is true but statement II is false
  • MARKEDStatement I is false but statement II is true
Given
  • Two statements comparing order and molecularity.
Asked
The truth value of each statement.
Concept to use
Order is an experimental quantity read off the measured rate law. Every reaction has a rate law, however complicated its mechanism, so order is defined for complex reactions too — and it may be zero, fractional or even negative. Molecularity counts the species colliding in a single elementary step. A multi-step reaction has no single collision to count, so molecularity is undefined for it; where it is defined, it is always a small whole number (1, 2, rarely 3).
Formula to use
Order → experimental, any real value, any reaction  |  Molecularity → theoretical, whole number, elementary steps only
Baby steps
  1. Statement I. Order comes from the experimentally determined rate law — that exists for elementary and complex reactions alike. Molecularity requires a single identifiable collision, so it is restricted to elementary steps. True.
  2. Statement II. For a termolecular step, three particles must meet at the same instant, with enough energy and the right orientation. The probability of that is already small; for four it is effectively zero. Hence molecularity three is rare and above three unknown. True.
  3. Both statements are true.
Answer
Both statements I and II are true
Shortcut
Hold one sentence in memory: ‘Order is measured and can be anything; molecularity is counted and only exists for a single step.’ Nearly every order-vs-molecularity question in NEET is a rephrasing of that line.
Where it went wrong
Statement I was marked false — most likely by reading ‘order applies to complex reactions’ as the error. It is not: order is the one of the pair that does extend to complex reactions. The restriction belongs to molecularity, and the statement says exactly that. When two properties are contrasted in one sentence, check which one carries the restriction before judging.
Q90 Match the reaction to its order Not attempted

Match the following.

List – I (Reaction)List – II (Order)
A.  2HI  Au, Δ→  H2 + I2I.  1
B.  2N2O5 → 4NO2 + O2II.  2
C.  2H2O2  I−, alkali→  2H2O + O2III.  1.5
D.  CHCl3 + Cl2 → CCl4 + HClIV.  0

  •  A-I, B-IV, C-II, D-III
  • KEYA-IV, B-I, C-II, D-III
  •  A-III, B-I, C-II, D-IV
  •  A-IV, B-I, C-III, D-II
Given
  • Four reactions in List I and four order values in List II.
Asked
The correct matching.
Concept to use
These four are the standard NCERT worked examples for each kind of order, and they are best learned as named anchors rather than derived. A surface-catalysed decomposition becomes zero order once the metal surface is saturated — adding more gas cannot speed it up because there is no free surface left. A catalysed reaction includes the catalyst in the rate law, which raises the overall order. A chain reaction involving a dissociating halogen typically gives a half-power term.
Formula to use
Overall order = sum of the exponents in the experimental rate law, catalyst included
Baby steps
  1. A. 2HI on a hot gold surface. Rate = k[HI]0 — the gold surface is saturated, so the rate is independent of concentration. Order 0 → IV. This is the textbook zero-order example.
  2. B. Decomposition of N2O5. Rate = k[N2O5] — the classic first-order gas-phase decomposition. Order 1 → I.
  3. C. H2O2 with iodide in alkali. Rate = k[H2O2][I−] — first order in each, and the catalyst does appear in the rate law. Overall order 1 + 1 = 2 → II.
  4. D. CHCl3 + Cl2. Rate = k[CHCl3][Cl2]½ — the half power comes from Cl2 dissociating into chlorine radicals in a fast pre-step. Overall order 1.5 → III.
  5. Matching: A-IV, B-I, C-II, D-III.
Answer
A-IV, B-I, C-II, D-III
Shortcut
Lock two anchors and stop. HI on gold = zero order and CHCl3 + Cl2 = 1.5 are the two most-asked facts in this topic. Fixing A–IV alone removes two options; adding D–III leaves exactly one. You never have to think about B or C.
Do not confuse the catalysed H2O2 decomposition (order 2, because I− is in the rate law) with the uncatalysed one, which is first order. The catalyst appearing in the rate law is the whole point of the item.

Integrated rate laws and half life

3 questions · 3 blank
Q75 First order in the gas phase — pressure form Not attempted

For a first order homogeneous gaseous reaction A → 3B, if pressure after time t was PT and initial pressure was P0 then the rate constant of the reaction is

  •  k = (1/t) ln[ P0 / 3(P0 − PT) ]
  •  k = (1/t) ln[ 2P0 / 3(P0 − PT) ]
  •  k = (1/t) ln( 3P0 / (2P0 − PT) )
  • KEYk = (1/t) ln( 2P0 / (3P0 − PT) )
Given
  • Reaction A → 3B, all gases, homogeneous, first order.
  • Initial total pressure = P0 (only A present at t = 0).
  • Total pressure at time t = PT.
Asked
An expression for the rate constant k in terms of P0, PT and t.
Concept to use
At constant volume and temperature, pressure is directly proportional to moles, so partial pressures can replace concentrations in the rate law without any conversion. The catch is that the measured quantity is total pressure, but the rate law needs the pressure of A alone. So the whole problem is one substitution: express PA in terms of P0 and PT, then drop it into the standard integrated first-order equation.
Formula to use
k = (1/t) · ln( PA(initial) / PA(at time t) )
Baby steps
  1. Let x = the pressure of A that has been consumed at time t. At t = 0: A = P0, B = 0.
  2. At time t: A = P0 − x, and B = 3x — the stoichiometric 3 must be carried, since every 1 unit of A makes 3 units of B.
  3. Total pressure PT = (P0 − x) + 3x = P0 + 2x.
  4. Rearranging: x = (PT − P0) / 2.
  5. Substitute back: PA = P0 − (PT − P0)/2 = (2P0 − PT + P0)/2 = (3P0 − PT) / 2.
  6. Put into the integrated law: k = (1/t) ln[ P0 ÷ (3P0 − PT)/2 ] = (1/t) ln[ 2P0 / (3P0 − PT) ].
Answer
k = (1/t) ln( 2P0 / (3P0 − PT) )
Shortcut
Learn the general result once: for A → nB,  PA = (nP0 − PT)/(n − 1)  and  k = (1/t) ln[ (n−1)P0 / (nP0 − PT) ]. Put n = 3 and the answer appears in one line. Then sanity-check at t = 0, where PT = P0 makes the ratio 2P0/2P0 = 1 and ln 1 = 0 — correct, no reaction yet. Two options fail that test immediately.
Q78 First order — rate constant from initial rate Not attempted

For a first order reaction, the initial rate is 0.6932 × 10−2 mol L−1 min−1. When the initial concentration of the reactant is 1.0 M, the half life of the reaction is

  •  6.932 min
  • KEY100 min
  •  0.6932 × 10−3 min
  •  0.6932 × 10−2 min
Given
  • Order = 1.
  • Initial rate = 0.6932 × 10−2 mol L−1 min−1.
  • Initial concentration [A]0 = 1.0 mol L−1.
Asked
Half life t½.
Concept to use
A rate and a rate constant are different things — the rate depends on concentration, the rate constant does not. Get k from the rate law first, then use the half-life formula. For first order, t½ is independent of the starting concentration, so once k is known the answer needs nothing else.
Formula to use
Rate = k[A]  →  k = Rate / [A]   |   t½ = 0.693 / k
Baby steps
  1. k = Rate ÷ [A] = (0.6932 × 10−2 mol L−1 min−1) ÷ (1.0 mol L−1).
  2. k = 6.932 × 10−3 min−1. Note the concentration units cancel and leave min−1, which is the correct unit for a first-order k.
  3. t½ = 0.6932 / k = 0.6932 ÷ (6.932 × 10−3).
  4. The 6932 digits cancel exactly: t½ = 0.1 × 103 = 100 min.
  5. Units: k is already per minute, so the answer is in minutes — no conversion to seconds is needed.
Answer
100 min
Shortcut
When [A]0 = 1 M, the rate constant is numerically equal to the initial rate. And the appearance of 0.6932 in the data is a deliberate signal that it will cancel against ln 2 = 0.693. Spotting both means the answer is 1/(10−2) = 100 with no long division at all.
This is a 30-second question. The three distractors are all what you get by forgetting to divide by [A] or by using the rate directly as k.
Q79 First order — fractional conversion Not attempted

The reaction A → B follows first-order kinetics. If 0.8 mol of A produces 0.6 mol of B in 10 min, the time required for 0.6 mol of A to produce 0.45 mol of B is:

  •  20 min
  • KEY10 min
  •  30 min
  •  40 min
Given
  • First-order reaction A → B (1 : 1 stoichiometry).
  • Case 1: start 0.8 mol, 0.6 mol converted, time 10 min.
  • Case 2: start 0.6 mol, 0.45 mol converted, time = ?
Asked
Time for the second case.
Concept to use
The single most useful property of first-order kinetics: the time taken to reach a given fraction of completion does not depend on how much you started with. Half-life is just the special case of that at 50 %. So instead of computing k, compare the fractions converted.
Formula to use
t = (1/k) ln[ 1 / (1 − f) ]   where f = fraction reacted — no [A]0 appears
Baby steps
  1. Case 1 fraction converted: f1 = 0.6 / 0.8 = 0.75, i.e. 75 % complete.
  2. Case 2 fraction converted: f2 = 0.45 / 0.6 = 0.75, i.e. also 75 % complete.
  3. Same fraction, same order → same time. Answer = 10 min.
  4. Cross-check the long way: t = (1/k) ln(1/0.25) = (1/k) ln 4 in both cases — identical, and [A]0 never enters.
Answer
10 min
Shortcut
Divide product by starting reactant for each case. If the two fractions match, the two times match — write the answer and move on. No k, no logarithms, roughly fifteen seconds.
75 % completion is also exactly two half-lives, so 10 min = 2 t½ → t½ = 5 min. Useful if a follow-up part asks for k.

Mechanism, catalysis and temperature

4 questions · 1 wrong · 3 blank
Q82 Mechanism — slow step with fast pre-equilibrium Not attempted

The reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) has been assigned to follow the following mechanism:
I. NO + NO ⇌ N2O2 (fast)
II. N2O2 + H2 → N2O + H2O (slow)
III. N2O + H2 → N2 + H2O (fast)
The rate constant of step II is 1.2 × 10−4 mol−1 L min−1, while the equilibrium constant of step I is 1.4 × 10−2. What is the rate of reaction when the concentration of NO and H2 each is 0.5 mol L−1?

  • KEY2.1 × 10−7 mol L−1 min−1
  •  3.2 × 10−6 mol L−1 min−1
  •  3.5 × 10−4 mol L−1 min−1
  •  3.2 × 10−5 mol L−1 min−1
Given
  • Three-step mechanism; step II is the slow step.
  • k2 = 1.2 × 10−4 mol−1 L min−1.
  • K1 = 1.4 × 10−2 (equilibrium constant of the fast step).
  • [NO] = [H2] = 0.5 mol L−1.
Asked
Rate of the overall reaction under these conditions.
Concept to use
The slow step is the rate-determining step, so the rate law comes from step II alone. But step II contains N2O2, which is an intermediate — it is neither a reactant nor a product of the overall reaction, and its concentration is not something you can measure or are given. It must be eliminated using the fast pre-equilibrium of step I.
Formula to use
Rate = k2[N2O2][H2]   and   K1 = [N2O2] / [NO]2  →  [N2O2] = K1[NO]2
Baby steps
  1. From the slow step: Rate = k2 [N2O2] [H2].
  2. Eliminate the intermediate using step I: K1 = [N2O2]/[NO]2, so [N2O2] = K1[NO]2. The square comes from the two NO molecules — do not drop it.
  3. Substitute: Rate = k2 K1 [NO]2 [H2].
  4. Numbers: [NO]2 = (0.5)2 = 0.25 and [H2] = 0.5, so the concentration block = 0.25 × 0.5 = 0.125.
  5. Constants: k2K1 = (1.2 × 10−4)(1.4 × 10−2) = 1.68 × 10−6.
  6. Rate = 1.68 × 10−6 × 0.125 = 2.1 × 10−7 mol L−1 min−1.
Answer
2.1 × 10−7 mol L−1 min−1
Shortcut
Collapse the whole mechanism to one line before touching numbers: Rate = k2K1[NO]2[H2]. Then multiply the constants first (1.2 × 1.4 = 1.68, exponents −4 + −2 = −6) and the concentration block second (0.125). Keeping constants and concentrations separate stops exponent slips.
Fast steps after the slow step (step III here) never appear in the rate law. Only the slow step, plus whatever is needed to remove intermediates from it.
Q83 Catalyst — what it changes and what it doesn’t Marked wrong

Incorrect statements among the following are
A) A catalyst lowers activation energy of a reaction.
B) A catalyst alters gibbs energy.
C) A catalyst alters heat of reaction.
D) A catalyst alters rate constant.

  • KEYB, C Only
  • MARKEDB, D Only
  •  C, D Only
  •  A, B, C Only
Given
  • Four claims about what a catalyst does.
Asked
Which claims are incorrect. (Read the question stem carefully — it asks for the false ones.)
Concept to use
Split every statement into one of two boxes. Kinetics box — activation energy, rate constant, rate, half-life: a catalyst changes all of these, because it supplies an alternative pathway with a lower energy barrier. Thermodynamics box — ΔH, ΔG, ΔS, equilibrium constant K: a catalyst changes none of these, because they depend only on the initial and final states, and the catalyst alters neither. It speeds up forward and backward reactions equally, so equilibrium arrives sooner but at the same position.
Formula to use
k = A e−Ea/RT  —  lower Ea ⇒ larger k
Baby steps
  1. A) Lowers activation energy. This is the definition of catalysis. Correct statement.
  2. B) Alters Gibbs energy. ΔG is a state function fixed by reactants and products. A catalyst is regenerated and appears on neither side. Incorrect statement.
  3. C) Alters heat of reaction. ΔH is likewise a state function — the energy levels of reactants and products are untouched; only the hump between them is lowered. Incorrect statement.
  4. D) Alters rate constant. From k = A e−Ea/RT, if Ea drops then k rises. Correct statement.
  5. The incorrect ones are B and C.
Answer
B, C Only
Shortcut
A and D are two halves of the same fact — lowering Ea is raising k, via the Arrhenius equation. So they must be marked the same way. Any option that includes exactly one of them (like B,D or C,D) is self-contradictory and can be eliminated without reading the chemistry.
Where it went wrong
D was marked as incorrect. But statement A, which grants that the catalyst lowers Ea, forces D to be true — you cannot lower the exponent’s barrier and leave k unchanged. Watch also for the reversed stem: this question asks for the incorrect statements, and options are deliberately arranged so the ‘correct statements’ answer is also on the list.
Q85 Temperature coefficient Not attempted

The Temperature coefficient of reaction is 2.5. If its rate constant at T1 K is 2.5 × 10−3 sec−1, the rate constant at T2 K in sec−1 is (T2 = 10 + T1)

  •  10 × 10−3
  •  1 × 10−2
  •  6.25 × 10−2
  • KEY6.25 × 10−3
Given
  • Temperature coefficient μ = 2.5.
  • k at T1 = 2.5 × 10−3 s−1.
  • T2 = T1 + 10 K.
Asked
k at T2.
Concept to use
The temperature coefficient is defined as the factor by which the rate constant multiplies for a 10 K rise — conventionally k(308)/k(298). So the only thing to work out is how many 10-degree blocks separate the two temperatures, and raise μ to that power.
Formula to use
μ = k(T+10) / k(T)   →   k(T + 10n) = μn × k(T)
Baby steps
  1. Temperature rise = T2 − T1 = 10 K → exactly one block, so n = 1.
  2. k2 = μ1 × k1 = 2.5 × (2.5 × 10−3).
  3. 2.5 × 2.5 = 6.25, and the power of ten is untouched.
  4. k2 = 6.25 × 10−3 s−1.
Answer
6.25 × 10−3 s−1
Shortcut
Count the blocks: n = ΔT / 10, then multiply by μn. For a 20 K rise it would be 2.52 = 6.25×, for 30 K it would be 2.53. One multiplication here.
The distractor 6.25 × 10−2 has the right digits and the wrong exponent — it is there to catch a slipped power of ten. Multiply the mantissas and copy the exponent across unchanged.
Q88 Reaction-coordinate diagram — Ea and ΔH Not attempted

The following reaction-coordinate diagram represents the reaction A → P. The energy values are given in kJ mol−1. Based on the diagram, the activation energy (Ea) and enthalpy change (ΔH) for the forward reaction, respectively (in kJ mol−1) are:
From the graph: reactant A sits at 10, the peak sits at 25, product P sits at 5.

  •  5, −15
  • KEY15, −5
  •  25, +5
  •  10, −25
Given
  • Energy of reactant A = 10 kJ mol−1.
  • Energy of the transition state (peak) = 25 kJ mol−1.
  • Energy of product P = 5 kJ mol−1.
Asked
Forward activation energy Ea, and enthalpy change ΔH.
Concept to use
Activation energy is a height above the reactant line, not a height above the axis. The value at the peak is the threshold energy; subtract the reactant energy from it to get Ea. Enthalpy change is the difference between the two flat levels, products minus reactants, and its sign tells you exothermic or endothermic.
Formula to use
Ea(forward) = Epeak − Ereactant   |   ΔH = Eproduct − Ereactant
Baby steps
  1. Read the three levels off the graph: A = 10, peak = 25, P = 5.
  2. Ea = 25 − 10 = 15 kJ mol−1.
  3. ΔH = 5 − 10 = −5 kJ mol−1.
  4. The minus sign is expected: the product line lies below the reactant line, so the reaction is exothermic.
  5. Answer: 15, −5.
Answer
15, −5
Shortcut
Two checks kill three options before any arithmetic. (1) The product is drawn lower than the reactant, so ΔH must be negative — that removes ‘25, +5’. (2) Ea can never equal the raw peak value unless the reactant sits at zero — that removes it again. Then Ea = 25 − 10 picks the survivor.
For the reverse reaction, Ea(reverse) = 25 − 5 = 20 kJ mol−1, and indeed Ea(f) − Ea(r) = 15 − 20 = −5 = ΔH. That identity is a fast way to check any answer to this style of question.

What the twenty have in common

Read together rather than one at a time, the errors fall into four groups — and three of the four are habits, not gaps.

1 · Trend direction reversed — Q46, Q50, Q70
The fact was known; the direction was not. Basicity of group-15 hydrides falls down the group (Q46). Single-bond energy does not fall smoothly, because N–N is anomalously weak (Q50). The lighter dioxide reduces and the heavier one oxidises, not the reverse (Q70). Fix: for each trend, write the direction as an arrow once and recite the exception beside it.

2 · Default rule applied without checking for the exception — Q59, Q63
Oxygen was assigned −2 in OF2 out of reflex, when fluorine forces it to +2. S2Cl2 was picked in Q59 without assigning its oxidation number, which would have shown S at +1 in five seconds. Fix: in any oxidation-state question, write the number above every central atom before reading the options.

3 · Statement questions judged too fast — Q47, Q83, Q89
All three were lost on a qualifier. Q47 said “in its oxides”, which excludes −3. Q83 asked for the incorrect statements and contains two options that are logically inconsistent. Q89 puts the restriction on molecularity, not order. Fix: mark each statement T or F in the margin before looking at any option.

4 · Kinetics skipped rather than attempted — ten questions
This is the biggest single item on the page and it is not a chemistry problem. Q78, Q79, Q81, Q85 and Q86 are each one or two lines. Q88 is a subtraction. Q87 is two ratios. Together those seven are worth 28 marks and about eight minutes. Fix: a timed set of twenty one-line kinetics problems, answer only, no working written out — the aim is to make starting automatic.

The three shortcuts worth memorising from this paper

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. Reaction-coordinate values in Q88 are read from the graph in the original paper.