Triangular faces of cis and mer octahedral complexes — JEE Advanced 2026 Paper 2 Q.13

01The question — verbatim

Section 3, numerical value — +4 for the correct value, 0 otherwise. Read it, then type what it gives you into the bench below.

Consider that the coordinating atoms of the ligands in cis-[Co(NH₃)₄Cl₂]Cl and mer-[Co(NH₃)₃Cl₃] octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangular faces in both the complexes having one N atom and two Cl atoms at their corners is ______.

Read the brackets carefully. Only what is inside the square brackets is bonded to the cobalt and reaches a vertex. In [Co(NH₃)₄Cl₂]Cl the third chloride sits outside them as the counter-ion, so the coordination sphere holds four N and two Cl, not three Cl. The bench below will refuse the run if you give it seven ligands.

What each part of the question is for

What you are given	Value	What it is for
Donor atoms of complex A, inside the brackets	4 N, 2 Cl	which label goes on each of the six numbered positions
Isomer of complex A	cis	the two chlorides are 90° apart, i.e. on an edge — the bench seats them at 1 and 5
Donor atoms of complex B	3 N, 3 Cl	a neutral complex — no counter-ion to discount
Isomer of complex B	mer	one chloride pair is trans, the third is cis to both — positions 1, 5 and 6
The pattern to count	1 N + 2 Cl	what the inspector looks for at the corners of each face
The solid	octahedron	6 positions, 12 edges, 8 triangular faces — 1–4 round the square plane, 5 and 6 axial, opposite pairs 1–3, 2–4, 5–6

02The bench

Type in each coordination sphere, press Run experiment, and watch the model get built, loaded and turned face by face. Sites are numbered 1–4 round the square plane and 5, 6 axial; the tally at the end is the answer.

1Enter the data 2Run the experiment 3Read the result waiting for your data

The modelno cage built

Matching faces found, live —

nothing built yet

The working, written as it is earned

Face ledger—

03Does it matter which vertices you pick?

The bench had to choose particular vertices for the chlorides. This panel checks, by exhaustion, that the choice cannot change the count — and keeps a log of your runs.

Every placement of the same isomer

Run the bench and this fills in.

Runs you have made

No runs yet — enter the data above and press Run experiment.

Lab notebook

The notebook writes itself as the bench settles things.

04Where it goes wrong

Four answers this question produces every year. Each is a single wrong move, and the bench above will reproduce all of them if you enter the data that way.

refused — the counter-ion on a vertex

[Co(NH₃)₄Cl₂]Cl has three chlorides written in it, but only two are ligands. Enter 4 N and 3 Cl and the bench refuses: seven donor atoms cannot sit on six vertices. That refusal is the check the question is really testing.

5 — reading mer as fac

Switch complex B to fac and the run gives 3 instead of 4, so the total comes out 5. In the facial isomer all three chlorides are mutually cis and one whole face is all chloride, which is not the pattern asked for.

4 — reading cis as trans

Switch complex A to trans and it contributes 0, not 2. Two trans chlorides sit on opposite sites — 5 and 6, say — they share no edge, and every face takes only one site from each opposite pair, so no face can hold both.

8 — counting the wrong pattern

Ask the inspector for 2 N and 1 Cl instead and it returns 4 + 4 = 8. The question is specific about which way round the corners go, and the two counts are genuinely different numbers.

05The record of the run

Filled in from your own run, quantity by quantity, as each one is established.

06Step-by-step solution

Complete a run on the bench and the derivation opens by itself.

The solution is still locked

Enter the composition of both complexes and let the inspection finish — the derivation and the answer box unlock the moment all sixteen faces have been checked.

Get the solid right first

Six donor atoms sit at the vertices of an octahedron. It has 6 vertices, 12 edges and 8 triangular faces, and every face takes exactly one vertex from each of the three opposite pairs — which is why there are 2 × 2 × 2 = 8 of them.

V − E + F = 6 − 12 + 8 = 2 (Euler)

Number the sites the usual way: 1, 2, 3, 4 going round the square plane and 5, 6 for the two axial sites. Then the three opposite (trans) pairs are 1–3, 2–4 and 5–6, and every other pair is cis.

cis = 90° apart, so the two share an edge
trans = 180° apart, opposite ends of one axis, so no face holds both
fac (facial) = three donors all mutually cis, capping one triangular face, e.g. 1, 2, 5
mer (meridional) = one pair trans and the third cis to both, so all three lie along a meridian, e.g. 1, 5, 6
Fill the vertices of the first complex — and count the ligands correctly

In cis-[Co(NH₃)₄Cl₂]Cl only the contents of the brackets are bonded to cobalt. The third chloride is the counter-ion.

coordination sphere = 4 N + 2 Cl = 6 donor atoms ✓

Taking 4 N + 3 Cl would need seven vertices, which no octahedron has — a useful self-check that the brackets were read properly.
The cis complex: one Cl–Cl edge, two faces

cis puts the two chlorides 90° apart — say at 1 (square plane) and 5 (axial), which are 90° apart because every plane site is 90° from every axial one. They therefore share exactly one edge. Every edge of an octahedron is shared by exactly two faces, and the third corner of each of those two faces must be one of the four nitrogens.

faces with 1 N and 2 Cl in complex A = 2

Checked face by face on the bench: the other six faces carry either one chloride or none.
The mer complex: a trans pair does the work

In mer-[Co(NH₃)₃Cl₃] one pair of chlorides is trans and the third is cis to both — say 5 and 6 (the axial pair, opposite each other) plus 1 in the plane, which is 90° from both. Because the two opposite sites are both chloride, every one of the eight faces already carries one chloride from that pair.

the 4 faces through the third Cl pick up a second one → 4 faces with 1 N and 2 Cl

The remaining four faces run through position 3, the nitrogen opposite it, so they carry 2 N and 1 Cl.
Add the two

2 + 4 = 6
Why the particular vertices did not matter

All 12 cis placements of two chlorides give 2, and all 12 mer placements of three give 4 — the octahedron is symmetric enough that only the isomer label matters, never which numbered sites were used. Section 03 shows the exhaustive check.

Independent verification

Derived before anything was looked up. The eight faces, both isomer geometries and both counts were enumerated from first principles, and only then compared with a published solution.
The faces enumerated two ways. Taking one vertex from each opposite pair gives 8 faces; brute force over all C(6,3) = 20 triples and discarding those containing an opposite pair gives the same 8. Euler’s formula, 6 − 12 + 8 = 2, agrees.
Exhaustive over placements, not one drawing. All 15 ways of putting two chlorides on six vertices were classified — 12 cis, 3 trans — and every cis placement gives exactly 2 matching faces, every trans placement 0. All 20 ways of putting three were classified — 8 fac, 12 mer — and every mer placement gives exactly 4, every fac placement 3. So the answer cannot depend on which vertices were drawn.
The census closes. For the cis complex the eight faces split 2 / 4 / 2 / 0 across 0, 1, 2 and 3 chlorides; for the mer complex 0 / 4 / 4 / 0. Both sum to 8, so no face is double-counted or missed.
Structural cross-check. A face with two chlorides is exactly a Cl–Cl edge, and every edge lies in two faces: cis has one such edge, giving 2; mer has two, giving 4. This reproduces the enumeration by a different argument.
The distractors reproduced deliberately. fac instead of mer gives 2 + 3 = 5; trans instead of cis gives 0 + 4 = 4; counting 2 N + 1 Cl instead gives 4 + 4 = 8; putting the ionic chloride on a vertex needs seven vertices and is impossible. None of them is 6.
Answer key cross-check. A published worked solution for this paper argues from the same Cl–Cl edges and reports the same 6. IIT Roorkee’s official key was not retrievable at the time of writing.
Headless-browser run of this page, driving the form with the paper’s data and with deliberately wrong data, asserting that seven ligands are refused, that the answer is nowhere on the page until the inspection finishes, that all sixteen faces are checked and tallied live, and that the page never scrolls sideways at 390 px.

07Concepts and classification

What the question is made of, and where it sits in the syllabus.

What you need to know

octahedron: 6 positions · 12 edges · 8 triangular faces

numbering: 1 2 3 4 square plane · 5 6 axial
opposite (trans) pairs: 1–3, 2–4, 5–6

every face = one site from each opposite pair → 2³ = 8

cis = 90° apart, share an edge (e.g. 1 and 5)
trans = 180° apart, share no face (e.g. 5 and 6)

fac = three mutually cis, capping one face (e.g. 1, 2, 5)
mer = one pair trans, third cis to both (e.g. 1, 5, 6)

every edge is shared by exactly two faces

Only what is inside the square brackets is in the coordination sphere.

Classification

Exam	JEE (Advanced) 2026, Paper 2
Question	Q.13, Section 3 — numerical value
Marking	+4 for the correct value, 0 otherwise, no negative marking
Subject	Inorganic Chemistry
Chapter	Coordination Compounds
Topic	Geometrical isomerism in octahedral complexes
Also needs	The geometry of the octahedron; reading a coordination sphere off a formula
Difficulty	Moderate — no arithmetic, but three chances to misread
Answer	— run the experiment

01The question — verbatim

02The bench

03Does it matter which vertices you pick?

Every placement of the same isomer

Runs you have made

Lab notebook

04Where it goes wrong

05The record of the run

06Step-by-step solution

The solution is still locked

Get the solid right first

Fill the vertices of the first complex — and count the ligands correctly

The cis complex: one Cl–Cl edge, two faces

The mer complex: a trans pair does the work

Add the two

Why the particular vertices did not matter

Independent verification

07Concepts and classification

What you need to know

Classification