Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical): Exercises

Before we start, two words you must own (from Werner's Theory and Coordination Sphere):

Level 1 — Recognition

Problem 1.1

Name the type of isomerism shown by each pair:

(a) $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ and $[C o (N H_{3})_{5} (O N O)]^{2 +}$
(b) $[C o (N H_{3})_{5} B r] S O_{4}$ and $[C o (N H_{3})_{5} S O_{4}] B r$
(c) $[C o (N H_{3})_{6}] [C r (C N)_{6}]$ and $[C r (N H_{3})_{6}] [C o (C N)_{6}]$
(d) $[C r (H_{2} O)_{6}] C l_{3}$ and $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$

Recall Solution 1.1

(a) Linkage isomerism. WHAT changed: the $N O_{2}^{-}$ group stays inside the sphere on both, but binds through N (nitro) in one and through O (nitrito) in the other. WHY it's linkage: same ligand, same fence, only the donor atom flipped — this needs an ambidentate ligand (see Ambidentate vs Polydentate Ligands).

(b) Ionization isomerism. WHAT changed: $B r^{-}$ and $S O_{4}^{2 -}$ swapped fence-sides. In isomer 1, $B r^{-}$ is inside and $S O_{4}^{2 -}$ is the free counter-ion; in isomer 2 it's reversed. Different free ion → different solution chemistry.

(c) Coordination isomerism. WHAT changed: both cation and anion are complex ions, and the $N H_{3}$ / $C N^{-}$ ligands are distributed differently between the two metals ( $C o$ , $C r$ ).

(d) Hydrate (solvate) isomerism. WHAT changed: a water molecule moved from inside the sphere (coordinated) to outside (lattice water), and a $C l^{-}$ moved the other way. It is a special ionization isomerism where the swapping species is $H_{2} O$ .

Level 2 — Application

Problem 2.1

$[C o (N H_{3})_{5} (S C N)]^{2 +}$ has an ambidentate $S C N^{-}$ ligand. Write both linkage isomers and name the donor atom in each. Which is called thiocyanato and which isothiocyanato?

Recall Solution 2.1

$S C N^{-}$ has lone pairs on both S and N, so either can donate to the metal.

Bonds through S: $[C o (N H_{3})_{5} (S C N)]^{2 +}$ → called thiocyanato-S (write the metal-bound atom first: $M - S C N$ ).
Bonds through N: $[C o (N H_{3})_{5} (N C S)]^{2 +}$ → called isothiocyanato-N ( $M - N C S$ ).

Rule to read the name: the atom written next to the metal is the donor. $M - S C N$ = S-bound; $M - N C S$ = N-bound.

Problem 2.2

A solution of $C r C l_{3} \cdot 6 H_{2} O$ is treated with excess $A g N O_{3}$ and gives 2 mol of AgCl precipitate per mol of complex. Which of the three hydrate isomers is it? How many water molecules are coordinated?

Recall Solution 2.2

Only free (outside-sphere) $C l^{-}$ precipitates instantly with $A g N O_{3}$ . We got 2 mol AgCl ⇒ 2 free chlorides ⇒ 1 chloride must be inside the sphere.

Total Cl = 3, inside = 1, outside = 2.
Total $H_{2} O$ = 6, one lattice slot is taken by... let's count the sphere: it holds 5 waters + 1 Cl (6 donor sites, octahedral).

Structure: $[Cr (H_{2} O)_{5} Cl] C l_{2} \cdot H_{2} O$ — 5 coordinated waters, 1 lattice water, colour blue-green.

Level 3 — Analysis

Problem 3.1

For square-planar $[P t (N H_{3})_{2} C l_{2}]$ : how many geometrical isomers exist, name them, and state which is the anticancer drug and why. Draw the two arrangements. (See Cisplatin and Bioinorganic Chemistry.)

Figure — Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical)

Recall Solution 3.1

Square planar = 4 corners of a square around $P t$ (see figure, left = cis, right = trans).

cis: the two $C l$ sit at adjacent corners (90° apart, red arrows).
trans: the two $C l$ sit at opposite corners (180° apart).

So 2 geometrical isomers. cis = cisplatin, the anticancer drug: because the two $C l$ leaving groups are 90° apart, after they leave $P t$ can bind two adjacent DNA bases and kink the DNA. trans = transplatin, inactive: its $C l$ are 180° apart, too far to reach two neighbouring bases.

Problem 3.2

Explain, using symmetry, why tetrahedral $[Z n (N H_{3})_{2} C l_{2}]$ shows no cis–trans isomerism, while square-planar $[P t (N H_{3})_{2} C l_{2}]$ does.

Recall Solution 3.2

In a tetrahedron (figure, right) every vertex is $109.5°$ from every other vertex — all four positions are equivalent. There is no "opposite" corner; any two ligands are always "adjacent." So placing the two $C l$ anywhere gives the same molecule → only 1 arrangement → no geometrical isomerism.

In a square plane (figure, left) two corners are adjacent (90°) and two are opposite (180°) — these are genuinely different positions, so cis and trans become distinct molecules.

Rule: cis–trans needs a geometry with distinguishable "adjacent" vs "opposite" — that means square planar or octahedral, never tetrahedral.

Level 4 — Synthesis

Problem 4.1

For octahedral $[C o (e n)_{2} C l_{2}]^{+}$ (where $e n$ = ethylenediamine, a bidentate chelating ligand): find the number of geometrical isomers, then for each determine whether it is optically active (chiral). Give the total number of stereoisomers. (Chirality background: Chirality in Organic Chemistry.)

Recall Solution 4.1

Step 1 — geometrical isomers. The two $C l$ can be cis (adjacent, 90°) or trans (opposite, 180°) ⇒ 2 geometrical isomers.

Step 2 — chirality test for each. A molecule is chiral if it is non-superimposable on its mirror image; the quick test is: does it have a plane or centre of symmetry? If yes → achiral; if no → chiral (exists as two enantiomers).

trans- $[C o (e n)_{2} C l_{2}]^{+}$ : the two axial $C l$ and the flat $e n$ rings give it a plane of symmetry ⇒ superimposable on its mirror image ⇒ optically inactive (achiral). → 1 form.
cis- $[C o (e n)_{2} C l_{2}]^{+}$ : it has no plane or centre of symmetry ⇒ non-superimposable mirror images ⇒ optically active, existing as a pair of enantiomers $Δ$ and $Λ$ (figure, the two propeller-twist hands). → 2 forms.

Step 3 — total. trans (1) + cis- $Δ$ (1) + cis- $Λ$ (1) = $3$ stereoisomers total.

Counting rule: total = (geometrical count) with each chiral geometrical isomer counted twice. Here: trans(×1) + cis(×2) = 3.

Level 5 — Mastery

Problem 5.1

Consider $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ . Classify every type of isomerism it can display and justify each. Then decide whether the cis form of the inner sphere $[C r (H_{2} O)_{4} C l_{2}]^{+}$ is optically active.

Recall Solution 5.1

Work through the categories systematically.

(1) Hydrate / ionization isomerism — YES. It is one member of the trio with $[C r (H_{2} O)_{6}] C l_{3}$ and $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ . Here inside-sphere = $4 H_{2} O + 2 C l$ ; free $C l^{-} = 1$ (gives 1 mol AgCl); lattice water = 2.

(2) Geometrical (cis–trans) — YES. Inner sphere is octahedral $M A_{4} B_{2}$ ( $A = H_{2} O$ , $B = C l$ ) ⇒ the two $C l$ can be cis (90°) or trans (180°).

(3) Optical — NO (for both cis and trans here). In $M A_{4} B_{2}$ with monodentate ligands, the cis form still possesses a plane of symmetry (the plane containing the two $C l$ and $C r$ ), so it is superimposable on its mirror image ⇒ achiral. (Contrast Problem 4.1, where the $e n$ chelate rings removed that plane.) The trans form is obviously achiral too.

Summary: hydrate/ionization ✔, geometrical ✔, optical ✘ (cis achiral because a symmetry plane survives with monodentate ligands).

Problem 5.2

For octahedral $[C o (e n)_{3}]^{3 +}$ (three bidentate $e n$ ligands, formula type $M (AA)_{3}$ ): how many geometrical isomers, and how many total stereoisomers? Justify.

Recall Solution 5.2

Geometrical isomers: with three identical symmetric bidentate ligands wrapping the octahedron, there is only 1 way to arrange them (all three chelate spans are equivalent) ⇒ 1 geometrical isomer.

Optical activity: the three $e n$ rings twist around the metal like a three-bladed propeller. This arrangement has no plane and no centre of symmetry ⇒ chiral ⇒ two enantiomers, the left-handed $Λ$ and right-handed $Δ$ propellers.

Total stereoisomers $= 1 \times 2 = 2$ (the $Δ/Λ$ enantiomeric pair). This is the textbook example of a purely optically-isomeric coordination compound.

Recall One-line self-test

Total stereoisomers of $[C o (e n)_{2} C l_{2}]^{+}$ ::: 3 (trans + cis- $Δ$ + cis- $Λ$ ) Free chlorides in $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ ::: 2 Geometrical isomers of tetrahedral $M A_{2} B_{2}$ ::: 0 Total stereoisomers of $[C o (e n)_{3}]^{3 +}$ ::: 2 ( $Δ$ and $Λ$ )