Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Each of these is really asking the same question: get the geometry right first, then the prefix tells you where to put things. The prefix is meaningless until the shape is settled.
a & b. CoCl3Br3 — octahedral, six ligands
An octahedral MA3B3 complex has exactly two geometric isomers, and these are they.
fac (facial): the three Cl occupy one triangular face of the octahedron — all three mutually cis (90° to each other). The three Br do the same on the opposite face. Practical way to draw it: put Cl on two adjacent equatorial positions and the top axial position; Br fills the remaining three.
Cl—top axial, Cl—front equatorial, Cl—right equatorial | Br—bottom axial, Br—back equatorial, Br—left equatorial
mer (meridional): the three Cl lie along a meridian — a great circle through the metal. Their angles are 90°, 90°, 180°: two Cl trans to each other, the third cis to both. Draw Cl at both axial positions plus one equatorial; Br takes the other three equatorial sites.
Cl—top axial, Cl—bottom axial, Cl—front equatorial | Br—the other three equatorial
The fast test: count trans pairs among the three identical ligands. fac has zero, mer has exactly one. If you drew something with two or three trans pairs, you have made a mistake — there is no third isomer.
c. "cis-SCl2F" — check this formula against your problem set
As written this cannot have cis/trans isomerism. SCl2F has only three bonds to sulfur, and a three-coordinate centre has no way to arrange substituents into distinct cis and trans forms. It is also an odd-electron species.
The intended compound is almost certainly SCl2F2. Sulfur then has four bonds and one lone pair (AX4E) → trigonal bipyramidal electron geometry, see-saw molecular shape, with the lone pair in an equatorial position (lone pairs always take equatorial, where there is more room).
That leaves two axial and two equatorial bonding sites. Fluorine, being more electronegative, prefers axial:
trans-SCl2F2: both F axial (180° apart), both Cl equatorial.
cis-SCl2F2: one F axial and one F equatorial, so the two F are ~90° apart; the two Cl are split the same way.
Worth confirming the formula with your instructor before you submit — if the problem really does say SCl2F, that part of the question has a typo.
d. trans-Pd(NH3)2Cl2 — square planar, and this is the whole point
Pd(II) is d8, and d8 four-coordinate complexes are square planar, not tetrahedral. This matters enormously: a tetrahedral MA2B2 complex has no cis/trans isomerism at all — every pair of positions is equivalent. The question can only be asked because the geometry is square planar.
trans: the two NH3 directly opposite each other at 180°, and consequently the two Cl also opposite each other. In a square, Cl and NH3 alternate around the ring.
NH3
|
Cl—Pd—Cl
|
NH3
(The cis isomer would put the two NH3 on adjacent corners, 90° apart. For the platinum analogue that distinction is the difference between cisplatin, a chemotherapy drug, and transplatin, which is not — a nice reminder that geometric isomers are genuinely different substances, not different drawings of one.)
Summary of what decides each answer: 6 ligands → octahedral → fac/mer. 4 bonds + 1 lone pair → see-saw → cis/trans by axial vs equatorial. 4 ligands on d8 → square planar → cis/trans by 90° vs 180°.
