Dipole moments of BF₃, NH₄⁺, NF₃ and NH₃ — JEE Advanced 2026 Paper 1 Q.3

QThe question — verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 1 — single correct option, +3 for the correct option, 0 if unanswered, −1 otherwise.

Q.3 The correct order of dipole moments for the given species is

(A)BF₃ = NH₄⁺ < NF₃ < NH₃
(B)BF₃ < NH₄⁺ < NF₃ < NH₃
(C)NH₄⁺ < BF₃ < NH₃ < NF₃
(D)BF₃ < NH₄⁺ < NH₃ < NF₃

What is being asked. A dipole moment μ measures how unevenly charge is spread in a molecule: charge × separation, in debye (1 D = 3.336 × 10⁻³⁰ C m). Every polar bond has a small dipole of its own, and a lone pair adds one more. The molecule's dipole moment is the vector sum of all of them — so the shape decides whether they add up or cancel. The experiment below builds each species, adds its dipoles, and then measures the result the way a physicist would: by the twist (torque) an electric field puts on it.

01Interactive experiment

A sample holder sits between two charged plates. Press START — the experiment builds each of the four species, draws every bond dipole and lone-pair moment, adds them tip to tail, switches on the field and measures the torque. Then it ranks the four and finds the matching option.

Ready

PrepareBuildBond dipoles Add vectorsField onRankCalculateAnswer

Set your values or use the question defaults.

Dipole bench — sample between charged plates

no species yet

Four species wait in the sample tray. The standard data are already loaded — press START EXPERIMENT.

✓ USING QUESTION VALUES

Optional — change anything, or nothing

The question prints no numbers; these are the standard data it relies on — Pauling electronegativities, the lone-pair moment of N and the VSEPR bond angles.

Bond moment = 1 D × (difference in electronegativity), pointing to the more electronegative atom.

Live data

species

—

shape

—

each bond moment

— D

lone-pair moment

— D

sum of bonds

— D

resultant μ

— D

field E

off

angle to field θ

—

torque τ

—

Measured dipole moment of each species

μ = τ_max / E, read from the torque meter. After the last one the bars line up in order.

Torque against angle to the field

τ = μE sin θ — one curve per species. A flat line means no dipole: nothing to twist.

Calculation — written as the experiment produces the numbers

Nothing measured yet.

Measurement log (0)

Nothing needs to be typed. Press START EXPERIMENT.

02Detailed solution

Run the experiment — the solution opens with the answer.

The solution is still locked

The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.

Concept — dipole moment is a vector sum

A bond dipole appears when two bonded atoms differ in electronegativity: the more electronegative atom takes a slightly larger share of the bonding pair (δ−) and leaves the other δ+. Chemists draw it as an arrow ↦ from δ+ to δ−. A lone pair on the central atom also sits on one side of it, so it carries a moment of its own, pointing out into the lone pair.

μ_molecule = | Σ μ_bonds + μ_{lone pairs} | (vectors — direction matters)

In the experiment these were the blue bond arrows, the orange lone-pair arrow and the green resultant formed when they were laid tip to tail.
Shapes from VSEPR

BF₃
B: 3 bond pairs, 0 lone pairs → trigonal planar, 120°

NH₄⁺
N⁺: 4 bond pairs, 0 lone pairs → tetrahedral, 109.5°

NF₃
N: 3 bond pairs, 1 lone pair → trigonal pyramidal, ≈102°

NH₃
N: 3 bond pairs, 1 lone pair → trigonal pyramidal, ≈107°
BF₃ and NH₄⁺ — perfect cancellation

Each B–F bond is strongly polar (χ: F 3.98, B 2.04), but three equal vectors at 120° in a plane add to zero: the tip-to-tail chain closes into a triangle. Four equal vectors to the corners of a tetrahedron close in the same way. Neither has a lone pair on the central atom to spoil the symmetry.

μ(BF₃) = 0 μ(NH₄⁺) = 0 ⇒ BF₃ = NH₄⁺

NH₄⁺ is an ion, but its charge is spread symmetrically round N, so about its own centre it has no dipole. In the field it would drift as a whole (the holder kept it in place) but it felt no twist — the torque meter read zero, exactly as for BF₃.
NH₃ and NF₃ — same shape, opposite story

In a pyramid the sideways parts of the three bond moments cancel and only their components along the axis survive: 3μ_b cos β, where β is the angle between a bond and the axis.

NH₃: N is more electronegative than H → bond arrows point toward N, the same way as the lone pair → they add

NF₃: F is more electronegative than N → bond arrows point toward F, away from the lone pair → they oppose

μ(NH₃) = 3μ_N–H cos β + μ_lp > μ(NF₃) = | μ_lp − 3μ_N–F cos β |

With the page's data: μ(NH₃) ≈ 1.94 D and μ(NF₃) ≈ 0.24 D.

This is why the torque meter jumped for NH₃ but barely moved for NF₃, even though the N–F bond is the more polar of the two.
Order and answer

measured: BF₃ 0, NH₄⁺ 0, NF₃ 0.23 D (0.80 × 10⁻³⁰ C m), NH₃ 1.47 D (4.90 × 10⁻³⁰ C m)

BF₃ = NH₄⁺ < NF₃ < NH₃

Answer: —
Why the other options are wrong

(B) and (D) put BF₃ below NH₄⁺, and (C) puts NH₄⁺ below BF₃. Both are exactly zero by symmetry, so any strict order between them is wrong. (C) and (D) also put NF₃ above NH₃ — the trap of comparing only bond polarities (N–F 0.94 > N–H 0.84) and forgetting the lone pair. Set the lone-pair moment to 0 in the experiment and you will see NF₃ overtake NH₃.

Independent verification

Symbolic. A computer-algebra system summed the bond vectors for each VSEPR shape with the bond moment left as a symbol: BF₃ and NH₄⁺ give exactly zero for any value, and the pyramid gives 3m cos β + L (NH₃-type) and L − 3m cos β (NF₃-type).
Model. With Pauling electronegativities, 1 D per unit difference and a 1.0 D lone pair: NH₃ 1.94 D, NF₃ 0.24 D.
Robustness. NF₃ < NH₃ holds for any lone-pair moment above 0.15 D and for every bond-angle pair tried (NH₃ 100–110°, NF₃ 98–106°). With no lone pair it flips — option (D)'s trap.
Measured. Gas-phase values BF₃ 0, NH₄⁺ 0, NF₃ 0.23 D, NH₃ 1.47 D give the same order.
Quantum chemistry. A B3LYP/aug-cc-pVDZ calculation at the experimental geometries gives BF₃ 0.000, NH₄⁺ 0.000, NF₃ 0.197, NH₃ 1.522 D — and puts the negative end of NF₃ on the fluorine side, as the vector model says.
This page. The experiment computes each resultant from the vectors, reads it back from the torque meter, ranks the four and matches the order against the printed options at run time — the answer is not written into the page.
Answer key. Checked afterwards against the official final key printed with the paper: Q3 → (A).

03How to use this simulation

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

STEP 1

Read the question

Four species — BF₃, NH₄⁺, NF₃, NH₃. Which has the biggest dipole moment, which the smallest, and are any equal?

STEP 2

Check the values

The standard data — electronegativities, the lone-pair moment of N and the bond angles — are already filled in. You do not need to enter anything.

STEP 3

Optional: change a value

Change any one value, or several. An empty box means "use the standard value". The badge switches to ⚙ USING CUSTOM VALUES; ↺ RESET TO QUESTION VALUES puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT. There is nothing else to click — the experiment runs from beginning to end on its own (under a minute; 2× makes it faster).

STEP 5

Watch

Each species is built in 3D. Its bond dipoles appear as blue arrows and its lone pair as an orange one. The arrows slide tip to tail and the green resultant appears — or the chain closes and there is none. Then the plates are charged and the torque meter shows how hard the field twists it.

STEP 6

Answer reveal

After the four measurements the bars line up in order, the experiment says ✓ EXPERIMENT COMPLETE and reveals the answer. The header answer box unlocks at the same time.

STEP 7

Understand

Scroll to Detailed solution. Each step points back to what you saw in the experiment.

STEP 8

Experiment again

Press ▶ REPLAY EXPERIMENT, or change a value — set the lone-pair moment to 0, or make χ(B) almost equal to χ(F) — and press ⚙ RUN WITH CUSTOM VALUES. Which species never move, whatever you do? CLASSROOM MODE enlarges everything for a projector.

Classification

Exam	JEE Advanced 2026, Paper 1, Chemistry, Section 1, Q.3
Chapter	Chemical Bonding and Molecular Structure
Topic	Dipole moment — vector sum of bond and lone-pair moments
Difficulty	Easy-Moderate
Answer	— run the experiment