CHEM 361 / Test Bank / Isomerism
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Test Bank — Unit 3

Isomerism in Coordination Compounds

Practice problems drawn from Housecroft, Atkins, JD Lee, and others. Select an answer or click "Show Answer" to reveal the explanation.

Structural Isomerism

Questions 1–4
Housecroft & Sharpe, Chapter 20 Housecroft
Q1

Which of the following correctly lists the major types of structural isomerism found in coordination compounds?

  • a Geometric, optical, and conformational isomerism
  • b Ionization, hydrate, and optical isomerism
  • c Linkage, coordination, and geometric isomerism
  • d Ionization, hydrate, linkage, and coordination isomerism
Answer: (d)

Structural (or constitutional) isomers differ in how atoms are connected. In coordination chemistry, the four principal types are: ionization isomerism (different ions inside vs. outside the coordination sphere), hydrate (solvate) isomerism (water bound as ligand vs. lattice), linkage isomerism (different donor atom from an ambidentate ligand), and coordination isomerism (different distribution of ligands between two metal centres). Geometric and optical isomerism are forms of stereoisomerism, not structural isomerism.

Housecroft & Sharpe, Chapter 20 Housecroft
Q2

The compounds [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are isomers. What type of isomerism is this, and how would you distinguish them experimentally?

  • a Linkage isomerism; distinguished by IR spectroscopy
  • b Ionization isomerism; distinguished by adding BaCl2 or AgNO3 to aqueous solutions
  • c Hydrate isomerism; distinguished by thermogravimetric analysis
  • d Coordination isomerism; distinguished by magnetic measurements
Answer: (b)

These are ionization isomers — they produce different ions when dissolved in water. [Co(NH3)5Br]SO4 releases free SO42− ions (precipitated by BaCl2 as BaSO4), while [Co(NH3)5SO4]Br releases free Br ions (precipitated by AgNO3 as AgBr). The key diagnostic is that the counter-ion differs between the two isomers, so simple qualitative tests for the free anion identify each form.

JD Lee, Chapter 7 JD Lee
Q3

[Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2·H2O are hydrate isomers. When each is dissolved in water and treated with excess AgNO3, how many moles of AgCl precipitate per mole of complex?

  • a 3 mol AgCl from [Cr(H2O)6]Cl3 and 3 mol AgCl from [Cr(H2O)5Cl]Cl2·H2O
  • b 2 mol AgCl from [Cr(H2O)6]Cl3 and 3 mol AgCl from [Cr(H2O)5Cl]Cl2·H2O
  • c 3 mol AgCl from [Cr(H2O)6]Cl3 and 2 mol AgCl from [Cr(H2O)5Cl]Cl2·H2O
  • d 1 mol AgCl from [Cr(H2O)6]Cl3 and 2 mol AgCl from [Cr(H2O)5Cl]Cl2·H2O
Answer: (c)

In [Cr(H2O)6]Cl3, all three Cl ions are outside the coordination sphere as counter-ions, so all three are free to precipitate with AgNO3. In [Cr(H2O)5Cl]Cl2·H2O, one Cl is bound directly to Cr(III) and is not easily displaced, leaving only two free Cl ions available for precipitation. This difference in AgCl yield is the classic experimental test for hydrate isomers.

Atkins & Shriver, Chapter 7 Atkins
Q4

The complex [Co(NH3)5(NO2)]2+ exhibits linkage isomerism. One form has the nitro ligand (N-bound, κN) and the other has the nitrito ligand (O-bound, κO). Using HSAB theory, which form is more thermodynamically stable with Co(III)?

  • a The nitro (κN) form, because Co(III) is a hard acid and N is a harder donor than O in NO2
  • b The nitrito (κO) form, because oxygen is more electronegative and bonds more strongly
  • c Both forms are equally stable because NO2 is a symmetric ligand
  • d The nitrito (κO) form, because Co(III) is a soft acid that prefers oxygen donors
Answer: (a)

Co(III) is a hard Lewis acid (small, highly charged d6 ion). In the NO2 ligand, the nitrogen lone pair is less electronegative and more polarisable than the oxygen lone pairs, but crucially the N-donor provides stronger σ-donation to the hard Co(III) centre. Experimentally, the nitro (κN) isomer is the thermodynamic product while the nitrito (κO) isomer is the kinetic product. The classic Jørgensen experiment demonstrated the irreversible conversion of the yellow nitrito form to the red-brown nitro form upon standing.

Geometric (cis-trans) Isomerism

Questions 5–8
Housecroft & Sharpe, Chapter 20 Housecroft
Q5

For a square planar complex of the type [Ma2b2] (where a and b are monodentate ligands), how many geometric isomers are possible?

  • a 1 — only one arrangement is possible
  • b 2 — cis (like ligands adjacent) and trans (like ligands opposite)
  • c 3 — cis, trans, and a meridional form
  • d 4 — all possible permutations of positions
Answer: (b)

In a square planar [Ma2b2] complex, two distinct geometric isomers exist. In the cis isomer, the two identical ligands (a) occupy adjacent positions (90° apart). In the trans isomer, they occupy opposite positions (180° apart). These isomers are non-superimposable and have different physical and chemical properties — a fact with profound consequences in medicinal chemistry (e.g., cisplatin vs. transplatin).

Housecroft & Sharpe, Chapter 20 Housecroft
Q6

For an octahedral complex of the type [MA2B2C2] (where A, B, and C are different monodentate ligands, each present twice), how many geometric isomers are possible?

  • a 3
  • b 4
  • c 6
  • d 9
Answer: (c)

An octahedral [MA2B2C2] complex has 6 geometric isomers. One systematic way to count: first choose which pair is trans — there are three ways to place one pair trans (A trans, B trans, or C trans). For each such choice the remaining four ligands (two pairs) occupy the equatorial plane, where they can be arranged in cis-cis or cis-trans patterns, giving 2 arrangements per trans-pair choice. This yields 3 × 2 = 6 distinct isomers. Some of these isomers may also be chiral.

Atkins & Shriver, Chapter 7 Atkins
Q7

Cisplatin (cis-[PtCl2(NH3)2]) is a potent anticancer drug, but the trans isomer (transplatin) is clinically inactive. What does this dramatic difference in biological activity tell us about the importance of geometry?

  • a Geometry does not matter; the difference is due to solubility
  • b The trans isomer is more reactive and is destroyed before reaching DNA
  • c Only the trans isomer can bind to DNA
  • d The cis geometry allows both Cl ligands to be replaced and cross-link adjacent guanine bases on DNA; the trans geometry cannot form this intrastrand cross-link
Answer: (d)

Cisplatin’s anticancer activity depends on its ability to form 1,2-intrastrand cross-links between adjacent guanine bases on the same DNA strand. The cis arrangement of the two labile Cl ligands (90° apart) provides the correct geometry to bridge two neighbouring guanines. In transplatin, the Cl ligands are 180° apart, making intrastrand cross-linking geometrically impossible. This case powerfully illustrates that geometric isomers, despite having identical composition, can have completely different biological and chemical behaviour.

JD Lee, Chapter 7 JD Lee
Q8

Can tetrahedral complexes exhibit cis-trans (geometric) isomerism?

  • a No — in a tetrahedron all positions are equivalent relative to each other, so no cis/trans distinction exists
  • b Yes — [MA2B2] has cis and trans forms in tetrahedral geometry
  • c Only if the metal is a d8 ion
  • d Only if chelating ligands are present
Answer: (a)

In a regular tetrahedron, every pair of vertices subtends the same angle at the centre (109.5°). There is no distinction between “adjacent” and “opposite” positions as there is in square planar or octahedral geometry. Therefore, a tetrahedral [MA2B2] complex has only one possible arrangement — no cis-trans isomerism is possible. Tetrahedral complexes can, however, exhibit optical isomerism if they have four different ligands (i.e., [Mabcd]).

Optical Isomerism & Chirality

Questions 9–12
Housecroft & Sharpe, Chapter 20 Housecroft
Q9

What is the rigorous symmetry criterion for a coordination compound to be optically active (chiral)?

  • a The complex must lack a C2 rotation axis
  • b The complex must possess a mirror plane (σ)
  • c The complex must lack any improper rotation axis (Sn), including σ (= S1) and i (= S2)
  • d The complex must have at least one asymmetric carbon atom
Answer: (c)

A molecule is chiral (optically active) if and only if it has no improper rotation axis Sn. Since a mirror plane is S1 and an inversion centre is S2, the absence of Sn automatically means no mirror planes and no inversion centre. Note that the concept of an “asymmetric carbon” is specific to organic chemistry; in coordination compounds, chirality arises from the spatial arrangement of ligands around the metal, not from carbon stereocentres.

Housecroft & Sharpe, Chapter 20 Housecroft
Q10

The complex [Co(en)3]3+ (en = ethylenediamine) exists as two enantiomers designated Δ and Λ. What do these labels describe?

  • a Δ rotates plane-polarised light to the right; Λ rotates it to the left
  • b Δ has a right-handed helical arrangement of the chelate rings when viewed along the C3 axis; Λ has a left-handed arrangement
  • c Δ is the cis isomer; Λ is the trans isomer
  • d Δ has the ligands in the equatorial plane; Λ has them axial
Answer: (b)

The Δ (delta) and Λ (lambda) labels describe the absolute configuration of tris-chelate octahedral complexes. When viewed along the principal C3 axis, the three chelate rings form a propeller-like arrangement. If this propeller follows a right-handed (clockwise) helix, the configuration is Δ; if left-handed (anticlockwise), it is Λ. These labels specify the spatial arrangement, not the direction of optical rotation — the relationship between absolute configuration and sign of rotation must be determined experimentally or computationally.

Atkins & Shriver, Chapter 7 Atkins
Q11

Consider the complex [Co(en)2Cl2]+. Can the cis isomer be optically active? What about the trans isomer?

  • a The cis isomer is optically active (no Sn); the trans isomer is not (it possesses a centre of inversion and mirror planes)
  • b Both the cis and trans isomers are optically active
  • c Neither isomer is optically active
  • d Only the trans isomer is optically active because it has lower symmetry
Answer: (a)

The cis-[Co(en)2Cl2]+ ion belongs to the point group C2, which contains only a C2 axis and no improper rotation elements. It is therefore chiral and exists as Δ and Λ enantiomers. The trans isomer, by contrast, possesses a centre of inversion (i = S2) and a mirror plane, placing it in the C2h point group. The presence of these Sn elements renders it achiral. This is a textbook example showing that geometric isomers can differ in their chirality.

Housecroft & Sharpe, Chapter 3 Housecroft
Q12

How does optical activity in metal complexes relate to the absence of a plane of symmetry (σ) and an inversion centre (i)?

  • a The absence of σ alone is sufficient for optical activity
  • b The absence of i alone is sufficient for optical activity
  • c Both σ and i must be present for optical activity
  • d Both σ and i must be absent (as special cases of Sn); the complete criterion is the absence of any Sn axis
Answer: (d)

A mirror plane is equivalent to S1 and an inversion centre to S2. The absence of both is necessary but, strictly, not always sufficient — the complete criterion is the absence of any improper rotation axis Sn. For most coordination compounds encountered in practice, checking for σ and i covers the common cases, but the rigorous test (relevant for groups like S4) requires checking all Sn elements. A molecule lacking all Sn is chiral: it cannot be superimposed on its mirror image and will rotate plane-polarised light.

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Isomerism in Coordination Compounds