PRAYOGX
ADV-2026-P1-CHE-Q14

VSEPR Shape Lab — count the lone pairs, let the domains repel, name the shape

JEE Advanced 2026 Paper 1 · Q.14 Chemistry VSEPR · lone pairs · molecular shape Matching list · single option +4 / 0 / −1 Press start · watch it happen
Answer
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The answer appears when the investigation finishes.

QMission brief — the question, verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 4 — matching list set, one correct option: +4 if it is chosen, 0 if unanswered, −1 otherwise.

Q.14  Consider the following species:

SOCl2, XeOF4, ClF3, ClF5, XeF5+, SO32–, XeF3+, SF4

List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List-IList-II
(P) See-saw(1) one
(Q) T-Shaped(2) two
(R) Trigonal Planar(3) three
(S) Square Pyramidal(4) four
(5) zero
  1. (A)  P → 1; Q → 2; R → 5; S → 3
  2. (B)  P → 5; Q → 4; R → 2; S → 3
  3. (C)  P → 3; Q → 2; R → 1; S → 4
  4. (D)  P → 1; Q → 3; R → 5; S → 4
P →
?
Q →
?
R →
?
S →
?
OPTION
?
The mystery. Eight species, three central atoms, some neutral, some charged. The shape of each depends on how many lone pairs sit on the central atom and where they go. The lab counts the electrons, lets the electron domains push each other apart on a sphere, reads each shape from the angles between the bonded atoms, and tallies them.

01Interactive experiment

Eight species on eight cards, every one in 3-D. Press START EXPERIMENT: load → count electrons → domains repel → read the shapes → tally → List-II → test options A–D → answer.

Ready
01 Load02 Count e⁻03 Repel04 Shapes 05 Tally06 List-II07 Options08 Final answer
Set your values or use the question defaults.
Answer——

VSEPR shape lab

step 1 of 7

Eight species, four shapes to count. The question's species are loaded — press START EXPERIMENT.

✓ EXPERIMENT COMPLETE
✓ QUESTION VALUES LOADED

Customize experiment — optional

Instrument panel

step
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species
—
progress
0 %
valence e⁻
—
lone pairs
—
steric no.
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shape
—
P · Q · R · S
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option
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Pink lobes are lone pairs; coloured balls are bonded atoms (F light green, Cl green, O red). The shape is named from the bonded atoms only, from the angles between them.

Nothing needs to be chosen. Press START EXPERIMENT.

Every bond angle in every species

measured on the relaxed models — clusters at 90°, 107°, 120° and 175°

How many species have each shape

the four List-I shapes and the rest

Option analysis — claim → evidence → verdict

(A)P → 1; Q → 2; R → 5; S → 3

not yet examined

UNEVALUATED
(B)P → 5; Q → 4; R → 2; S → 3

not yet examined

UNEVALUATED
(C)P → 3; Q → 2; R → 1; S → 4

not yet examined

UNEVALUATED
(D)P → 1; Q → 3; R → 5; S → 4

not yet examined

UNEVALUATED
🔢Electron Counter
🌐Domain Master
💗Lone-Pair Whisperer
📐Shape Spotter
🧮Tally Keeper
🏆Chemistry Master
Reasoning log — input → action → change → observation → conclusion

Nothing investigated yet.

    Evidence table (0)

    02VSEPR builder

    Choose a central atom, how many F and O atoms it carries and the charge. The builder counts the lone pairs, relaxes the domains (trying every lone-pair placement) and names the shape.

    03Detailed solution

    The complete worked solution — every structure, step and calculation.

    Understand the concept

    Count the lone pairs

    The central atom's valence electrons, minus one for each F or Cl bond and two for each O (a double bond), minus the charge, leave the lone-pair electrons.

    LP = (V − e⁻ in bonds − charge) / 2

    Steric number

    Bonded atoms plus lone pairs: the number of electron domains, which fixes the electron geometry.

    SN = atoms + LP

    Lone pairs push harder

    LP–LP > LP–BP > BP–BP. In a trigonal bipyramid lone pairs go equatorial (only two neighbours at 90°); two lone pairs in an octahedron go trans.

    SN 5: LP equatorial

    Shape = atoms only

    The molecular shape is described by the positions of the atoms; lone pairs are invisible in it.

    SN 5, LP 1 → see-saw

    1. Step 1 — Count electrons for all eight species

      SpeciesVe⁻ in bondschargeLP = (V − bonds − charge)/2SNelectron geometryshape
      SOCl₂6O 2 + 2 Cl 1 = 4014tetrahedraltrigonal pyramidal
      XeOF₄8O 2 + 4 F 1 = 6016octahedralsquare pyramidal
      ClF₃73 F 1 = 3025trigonal bipyramidalT-shaped
      ClF₅75 F 1 = 5016octahedralsquare pyramidal
      XeF₅⁺85 F 1 = 5+116octahedralsquare pyramidal
      SO₃²⁻63 O 2 = 6−214tetrahedraltrigonal pyramidal
      XeF₃⁺83 F 1 = 3+125trigonal bipyramidalT-shaped
      SF₄64 F 1 = 4015trigonal bipyramidalsee-saw
      e.g. XeF₃⁺: (8 − 3 − 1)/2 = 2 LP, SN = 3 + 2 = 5  ·  SO₃²⁻: (6 − 6 + 2)/2 = 1 LP, SN = 4  ·  XeOF₄: (8 − 6 − 0)/2 = 1 LP, SN = 6

    2. Step 2 — Place the lone pairs, name the shapes

      FFSFFSF₄ — see-sawFFClFClF₃ — T-shapedFFFClFFClF₅ — square pyramidalOClSClSOCl₂ — trigonal pyramidal
      Lone pairs (pink lobes) take space but are not part of the shape. In a trigonal bipyramid they sit equatorial; in an octahedron one lone pair leaves a square pyramid.

      SN 5 with 1 LP (SF₄): the lone pair is equatorial → see-saw. SN 5 with 2 LP (ClF₃, XeF₃⁺): both equatorial → T-shaped. SN 6 with 1 LP (XeOF₄, ClF₅, XeF₅⁺) → square pyramidal. SN 4 with 1 LP (SOCl₂, SO₃²⁻) → trigonal pyramidal, not planar.

    3. Step 3 — Tally and match

      see-saw: SF₄ → 1 → (1)  ·  T-shaped: ClF₃, XeF₃⁺ → 2 → (2)
      trigonal planar: none → 0 → (5)  ·  square pyramidal: XeOF₄, ClF₅, XeF₅⁺ → 3 → (3)
      P → 1, Q → 2, R → 5, S → 3  ⇒  correct option = —
      OptionPQRSVerdict
      (A)1 ✓2 ✓5 ✓3 ✓correct
      (B)5 ✗4 ✗2 ✗3 ✓rejected
      (C)3 ✗2 ✓1 ✗4 ✗rejected
      (D)1 ✓3 ✗5 ✓4 ✗rejected

      Traps: treating SO₃²⁻ or SOCl₂ as trigonal planar (forgetting the lone pair) gives R → 1 or 2; putting the SF₄ lone pair axial gives a trigonal pyramid, not a see-saw.

    Independent verification

    • Electron counts. Recounted in a separate script, and cross-checked with the formal-charge Lewis structures (SO₃²⁻: three S–O, one lone pair).
    • Repulsion model. Domains relaxed on a sphere with a 1/r⁶ repulsion, LP weights larger; every lone-pair placement tried — the lowest-energy one reproduces all eight textbook shapes, and the control XeF₄ comes out square planar.
    • Shape from angles. Each shape is named only from the angles between bonded atoms (e.g. see-saw: one ≈180°, one ≈120°, four ≈90°).
    • Solution websites. MathonGo's published solution (their Q46) gives (A).
    • This page. Counts, relaxation, shapes, tally and option are all computed at run time; nothing is written in.
    • Answer key. Checked afterwards against the official final key printed with the paper: Q14 → A.

    04How to use this simulation

    For a student opening it for the first time. The whole investigation runs by itself after one press.

    STEP 1

    Read the mission brief

    Eight species; how many are see-saw, T-shaped, trigonal planar, square pyramidal? Which option matches?

    STEP 2

    Check the values

    The question's species are loaded. You do not need to change anything. The answer is shown above the experiment.

    STEP 3

    Optional: customize

    Swap SO₃²⁻ for SO₃, XeF₃⁺ for XeF₄ or SF₄ for SF₆ and watch the counts change. ↺ RESTORE QUESTION CONDITIONS puts everything back.

    STEP 4

    Start

    Press ▶ START EXPERIMENT under the experiment screen. Nothing else to click (about a minute; SLOW MOTION or 2× changes the pace).

    STEP 5

    Watch the counting and the repulsion

    Each card counts its electrons; then the domains spread over a sphere and the lone pairs (pink) take their places.

    STEP 6

    Shapes, tally, options

    The shapes are named from the bond angles, the species drop into bins and each option is tested. Badges unlock on the way.

    STEP 7

    Explore

    In the VSEPR builder make any species from Xe, Cl or S with F, O and a charge. Then read the Detailed solution.

    STEP 8

    Experiment again

    Press ▶ REPLAY EXPERIMENT, or change a species and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.

    Classification

    ExamJEE Advanced 2026, Paper 1, Chemistry, Section 4, Q.14
    ChapterChemical Bonding and Molecular Structure
    TopicVSEPR theory - lone pairs, steric number and molecular shapes of main-group and xenon species
    DifficultyModerate
    Answer— run the experiment