Every chemistry question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the rule, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
4Questions lost
3Attempted, wrong
1Left blank
16Marks at stake
The shape of this paper
Only four chemistry questions were flagged, and the attempt rate was high — three of
four answered. That is a marked change from the previous paper, where six of nine chemistry
questions were left blank.
All three wrong answers share a shape: a trend was known but applied in the wrong
direction, or an exception was allowed to spread too far. None was a gap in knowledge.
Four questions — two from group 15 and two from group 16 — grouped here by sub-topic. Three were attempted and missed, and all three turn on the direction of a trend or on how far an exception reaches.
Q 151Attempted · wrongCatenation · X–X bond strength in group 15
The catenation property of an element depends on X – X Bond strength, where 'X' represents an element. Identify correct order of X – X Bond strength of VA group elements.
N–N > P–P > As–As
Her answerAs–As > P–P > N–N
P–P > As–As > N–N
CorrectP–P > N–N > As–As
Given
Group 15 (VA) elements: N, P, As.
Single X–X bond enthalpies are being compared.
Asked
The correct order of single-bond strength.
Concept to use
The general trend down a group is that single bonds get weaker as atoms get larger, because the shared pair sits further from both nuclei. That predicts N–N > P–P > As–As. But nitrogen breaks the rule. It is so small that the two lone pairs on the bonded nitrogens are forced very close together, and their repulsion weakens the bond — dropping N–N below P–P. This anomaly is exactly why phosphorus catenates readily and nitrogen does not.
DiagramAnimatedThe nitrogen anomaly: small atom, lone-pair repulsion, weak N–N.
Formula to useN–N 163 · P–P 201 · As–As 146 kJ mol⁻¹
Baby steps
Start from the general trend: bond strength falls down the group as size increases.
Apply the nitrogen anomaly: N is so small that lone-pair repulsion across the short N–N bond weakens it substantially.
So N–N drops below P–P, but it still stays above the much larger As–As.
Order: P–P > N–N > As–As.
Consequence worth remembering: phosphorus forms P₄ chains and rings, while nitrogen's catenation stops at N₂H₄ and N₃⁻.
Answer
P–P > N–N > As–As
Why this option and not the others
Option
Verdict
Reason
N–N > P–P > As–As
rule out
The naive trend, ignoring the nitrogen anomaly. It is the answer you get by assuming smaller always means stronger.
As–As > P–P > N–N
rule out
The trend completely reversed. It would mean bonds strengthen as atoms grow, which is not true anywhere in the group.
P–P > As–As > N–N
rule out
Correctly puts P first, but over-corrects by pushing N below As. The anomaly drops N one place, not two.
P–P > N–N > As–As
keep
P–P is the strongest because of nitrogen's lone-pair repulsion; As–As is weakest because of size.
Shortcut
Remember it through the consequence rather than the numbers: phosphorus catenates and nitrogen does not, so P–P must be the stronger bond. That fixes first place. Arsenic is much the biggest of the three, so it takes last place on the ordinary size argument. Nitrogen is squeezed into the middle. The chemistry you already know about catenation gives you the order without any bond enthalpies at all.
Where it went wrong
The chosen order runs the trend entirely backwards — As–As strongest, N–N weakest. It looks as though the anomaly was remembered (N–N is unusually weak) and then applied as a full reversal of the whole series. The correction is that the anomaly affects nitrogen only: everything below nitrogen still follows the ordinary size trend, so P–P must beat As–As. One exception does not invert a group.
Q 155Attempted · wrongFirst ionisation enthalpy down group 15
The first ionization energy is the lowest for
CorrectBi
Sb
As
Her answerN
Given
Group 15 elements: N, As, Sb, Bi.
First ionisation enthalpy is being compared.
Asked
Which has the lowest first ionisation enthalpy.
Concept to use
Ionisation enthalpy measures how tightly the outermost electron is held. Going down a group the atom gets bigger and the inner shells screen the nucleus more effectively, so the outer electron is held more loosely and ionisation enthalpy falls steadily. The lowest value therefore belongs to the element furthest down the group.
DiagramAnimatedIonisation enthalpy falls steadily down the group to bismuth.
Formula to usedown a group: size ↑ , screening ↑ , IE ↓
Baby steps
List the group in order: N, P, As, Sb, Bi — nitrogen at the top, bismuth at the bottom.
Down the group, atomic radius increases and shielding by inner shells increases.
Both effects reduce the pull on the outermost electron.
So ionisation enthalpy falls all the way down: N 1402 > P 1012 > As 947 > Sb 834 > Bi 703 kJ mol⁻¹.
The lowest is bismuth.
Answer
Bi
Why this option and not the others
Option
Verdict
Reason
Bi
keep
Lowest in the group, largest atom, most shielding — therefore the loosest hold on the outer electron.
Sb
rule out
Lower than As but still above Bi. Second lowest of those offered.
As
rule out
Middle of the group; its value is well above bismuth's.
N
rule out
The highest in the group, not the lowest. Nitrogen is small and its half-filled 2p³ shell is extra stable, so it holds its electrons more tightly than any other member.
Shortcut
For ionisation enthalpy, the answer to “lowest in a group” is always the bottom element, and “highest” is always the top one. Just read the position in the periodic table — no values needed. The only place this rule bends is across a period, where half-filled and fully-filled shells create small local reversals.
Where it went wrong
Nitrogen was chosen, and nitrogen is the exact opposite — it has the highest first ionisation enthalpy in the group, at 1402 against bismuth's 703. The likely cause is that nitrogen is the element most discussed in this chapter, and its stable half-filled 2p³ configuration is a much-emphasised fact, so it comes to mind first. The direction of the trend is the thing to fix: check whether the question wants highest or lowest, then simply read down or up the group.
Q 166Attempted · wrongGroup 16 · statement set
The physical properties of group-16 elements are (A) Po is a radio-active non-metal. (B) All group-16 elements show allotropy. (C) Po exhibits common oxidation states of −2, +2, +4, +6 The option with the correct set of properties is
A and B
Her answerB and C
A, B and C
CorrectB only
Given
Group 16 (the chalcogens): O, S, Se, Te, Po.
Three statements about their physical properties.
Asked
Which statements are correct.
Concept to use
Two facts about polonium settle two of the three statements. Polonium is a radioactive metal, not a non-metal — metallic character increases down the group, with O and S non-metals, Se and Te metalloids, and Po a metal. And because it is metallic it does not readily take up electrons, so it shows only +2 and +4, never −2. Statement B, meanwhile, is a standard group property: all the chalcogens show allotropy.
DiagramAnimatedPolonium is a radioactive metal, and it does not show −2.
Formula to useO, S non-metals → Se, Te metalloids → Po METAL (radioactive)
Baby steps
Statement A. Po is radioactive, which is right, but it is a metal, not a non-metal. Incorrect.
Statement B. Oxygen (O₂, O₃), sulphur (rhombic, monoclinic), selenium, tellurium and polonium all show allotropy. Correct.
Statement C. The group as a whole shows −2, +2, +4 and +6, but polonium shows only +2 and +4. A metal does not gain electrons to reach −2. Incorrect.
Only B survives.
Answer
B only
Why this option and not the others
Option
Verdict
Reason
A and B
rule out
B is right but A calls polonium a non-metal, which contradicts the metallic trend down the group.
B and C
rule out
B is right; C attributes the whole group's oxidation states to polonium alone, which shows only +2 and +4.
A, B and C
rule out
Fails on both A and C.
B only
keep
Allotropy is genuinely shown right across the group; both statements about polonium are wrong.
Shortcut
When a statement set repeatedly names the bottom element of a group, that is the signal to apply the metallic trend. Down any p-block group, non-metal becomes metalloid becomes metal — and metals do not take negative oxidation states. Testing statements A and C against that one principle disposes of both without recalling anything specific about polonium.
Where it went wrong
Statement C was accepted, and it is the harder of the two traps: −2, +2, +4 and +6 is the correct list for group 16 as a whole, so the list itself looks familiar and right. What changes it is the word Po — the statement attaches a group-wide property to one specific element that does not share it. This is the same pattern as several questions on the previous paper: a fact that is true in general, applied to a case it does not cover. The check is to ask whether the statement is about the group or about one element.
Q 170Left blankGroup 16 · matching by electronic structure
Match column I with Column II for 16th group elements. A. O I. Melting point is highest among Chalcogens. B. Se II. The K shell contains 8 electrons. C. S III. M shell contains 5 empty orbitals D. Te IV. M shell contains 9 electrons with s = +½ V. Ionization enthalpy is the highest among Chalcogens.
CorrectA–V, B–IV, C–III, D–I
A–II, B–IV, C–V, D–III
A–V, B–III, C–II, D–I
A–V, B–IV, C–I, D–II
Given
A = O (Z = 8), B = Se (Z = 34), C = S (Z = 16), D = Te (Z = 52).
Five properties, of which four are to be used.
Asked
The correct matching.
Concept to use
Every one of these properties can be read straight off an electron configuration or off a group trend. Write the shell structures first and the matching falls out. Note that property II — a K shell with 8 electrons — is impossible, since the K shell holds at most 2. It is a decoy, and spotting that removes two options at once.
DiagramAnimatedShell structures and group trends — everything needed is here.
A = O. Oxygen is at the top of the group, so it has the highest ionisation enthalpy among the chalcogens. A–V.
D = Te. Melting point rises down the group, so tellurium has the highest melting point of those listed. D–I.
C = S is 2, 8, 6 — its M shell holds 3s²3p⁴. The M shell has 9 orbitals in all (one 3s, three 3p, five 3d); 3s and 3p are occupied, leaving the five 3d orbitals empty. C–III.
B = Se is 2, 8, 18, 6 — its M shell is complete with 18 electrons, which pair up as 9 with s = +½ and 9 with s = −½. B–IV.
Property II is left over, and rightly so: no K shell can hold 8 electrons.
Answer: A–V, B–IV, C–III, D–I.
Answer
A–V, B–IV, C–III, D–I
Why this option and not the others
Option
Verdict
Reason
A-V, B-IV, C-III, D-I
keep
Every pairing checks out against the shell structures and the group trends, and the impossible property II is correctly left unused.
A-II, B-IV, C-V, D-III
rule out
Gives oxygen the impossible K-shell-of-8 property, and hands sulphur the highest ionisation enthalpy, which belongs to oxygen.
A-V, B-III, C-II, D-I
rule out
A and D are right, but B and C are swapped and C is given the impossible property II.
A-V, B-IV, C-I, D-II
rule out
Gives sulphur the highest melting point — it belongs to tellurium — and again uses the impossible property II.
Shortcut
Spot the impossible option first. The K shell can never hold more than 2 electrons, so property II cannot be matched to anything — and three of the four answer options use it. Eliminating those leaves exactly one, without checking a single melting point or shell structure. In any matching question with a spare property, ask whether that spare is a genuine leftover or a planted impossibility.
What the four have in common
Reading the paper as a whole
Trends known, directions lost
Q
What was known
What went wrong
155
That ionisation enthalpy varies down group 15
The direction. Nitrogen has the highest IE in the group, not the lowest — it
is the most-discussed element in the chapter, so it comes to mind first.
151
That N–N is anomalously weak
How far the anomaly reaches. It drops nitrogen one place, below P–P. It does
not reverse the whole group, so P–P still beats As–As.
166
The oxidation states of group 16 as a whole
Which element they apply to. −2, +2, +4, +6 is right for the group and wrong for
polonium, which shows only +2 and +4.
The common thread is scope: a true statement applied more widely than it holds. That is
a different failure from not knowing the chemistry, and it responds to a different check —
ask whether the claim is about the group or about one element.
Two filters worth carrying
For “highest/lowest in a group”, just read the position. Ionisation
enthalpy is highest at the top and lowest at the bottom, every time. No values needed. That
answers Q155 in about five seconds.
Down any p-block group: non-metal → metalloid → metal. And metals do not
take negative oxidation states. Applying that single principle disposes of two of the three
statements in Q166 without recalling anything specific about polonium.
And for Q170, the blank one: look for the impossible option first. A K shell can never
hold 8 electrons, so property II cannot be matched to anything — and three of the four
answer options use it. That eliminates them without checking a single shell structure.