This particular Inorganic Chemistry Practice Worksheet is designed to serve as a study guide for both students and interested parties as an educational challenge with questions about coordination compounds, isomerism, and molecular geometry. This will assist in preparing for exams or in learning the rudiments of inorganic chemistry. The problem book contains various types of problems which are supposed to check the knowledge of inorganic chemistry for your exams.
Category 1: Coordination Compound Formula Identification
(Often referred to by their common names such as “blue vitriol” etc.)
- Question: What is the formula of blue vitriol?
- (a) CuSO₄·4H₂O
- (b) CuSO₄·5H₂O
- (c) CuSO₄·6H₂O
- (d) CuSO₄·7H₂O
Answer: (b) CuSO₄·5H₂O
Explanation: Blue vitriol is the common name for copper(II) sulfate pentahydrate.
- Question: What is the formula of white vitriol (zinc sulfate)?
- (a) ZnSO₄·5H₂O
- (b) ZnSO₄·6H₂O
- (c) ZnSO₄·7H₂O
- (d) ZnSO₄·8H₂O
Answer: (c) ZnSO₄·7H₂O
Explanation: White vitriol is the hydrated form of zinc sulfate, usually as the heptahydrate.
- Question: What is the formula of green vitriol (iron(II) sulfate)?
- (a) FeSO₄·5H₂O
- (b) FeSO₄·6H₂O
- (c) FeSO₄·7H₂O
- (d) FeSO₄·8H₂O
Answer: (c) FeSO₄·7H₂O
Explanation: Green vitriol is iron(II) sulfate heptahydrate.
- Question: Mellow vitriol is another name for manganese(II) sulfate. Which of the following represents its common hydrated form?
- (a) MnSO₄·H₂O
- (b) MnSO₄·3H₂O
- (c) MnSO₄·5H₂O
- (d) MnSO₄·7H₂O
Answer: (b) MnSO₄·3H₂O
Explanation: Manganese(II) sulfate is commonly encountered as the trihydrate.
- Question: What is the formula of potassium ferricyanide?
- (a) K₃[Fe(CN)₆]
- (b) K₄[Fe(CN)₆]
- (c) K₂[Fe(CN)₆]
- (d) K₆[Fe(CN)₆]
Answer: (a) K₃[Fe(CN)₆]
Explanation: Potassium ferricyanide contains three potassium ions per ferricyanide ion.
- Question: Identify the formula of nickel(II) sulfate commonly encountered in hydrate form.
- (a) NiSO₄·4H₂O
- (b) NiSO₄·5H₂O
- (c) NiSO₄·6H₂O
- (d) NiSO₄·7H₂O
Answer: (c) NiSO₄·6H₂O
Explanation: Nickel(II) sulfate is most often isolated as the hexahydrate.
- Question: What is the formula for cobalt(II) sulfate when it is in its heptahydrate form?
- (a) CoSO₄·6H₂O
- (b) CoSO₄·7H₂O
- (c) CoSO₄·8H₂O
- (d) CoSO₄·5H₂O
Answer: (b) CoSO₄·7H₂O
Explanation: Cobalt(II) sulfate is usually found as the heptahydrate.
- Question: What is the formula for ferrous ammonium sulfate, also known as Mohr’s salt?
- (a) Fe(NH₄)₂(SO₄)₂·6H₂O
- (b) Fe(NH₄)₂(SO₄)₂·4H₂O
- (c) Fe(NH₄)₂(SO₄)₂·7H₂O
- (d) Fe(NH₄)₂(SO₄)₂·8H₂O
Answer: (a) Fe(NH₄)₂(SO₄)₂·6H₂O
Explanation: Mohr’s salt is the hexahydrate of ferrous ammonium sulfate.
- Question: What is the formula for copper(II) chloride dihydrate?
- (a) CuCl₂·H₂O
- (b) CuCl₂·2H₂O
- (c) CuCl₂·3H₂O
- (d) CuCl₂·4H₂O
Answer: (b) CuCl₂·2H₂O
Explanation: Copper(II) chloride typically crystallizes as the dihydrate.
- Question: Which formula represents chromium(III) sulfate?
- (a) Cr₂(SO₄)₃
- (b) Cr₂(SO₄)₃·12H₂O
- (c) Cr₂(SO₄)₃·7H₂O
- (d) Cr₂(SO₄)₃·3H₂O
Answer: (b) Cr₂(SO₄)₃·12H₂O
Explanation: Chromium(III) sulfate is commonly isolated as a dodecahydrate.
- Question: What is the formula for sodium hexanitrocobaltate(III) when hydrated?
- (a) Na₃[Co(NO₂)₆]·6H₂O
- (b) Na₂[Co(NO₂)₆]·4H₂O
- (c) Na₃[Co(NO₂)₆]·5H₂O
- (d) Na₄[Co(NO₂)₆]·6H₂O
Answer: (a) Na₃[Co(NO₂)₆]·6H₂O
Explanation: This is a standard representation for the hydrated form of sodium hexanitrocobaltate(III).
- Question: Identify the formula of ammonium dichromate.
- (a) (NH₄)₂Cr₂O₇
- (b) (NH₄)₂CrO₄
- (c) NH₄Cr₂O₇
- (d) (NH₄)₃Cr₂O₇
Answer: (a) (NH₄)₂Cr₂O₇
Explanation: Ammonium dichromate contains two ammonium ions per dichromate ion.
- Question: What is the formula for potassium hexacyanoferrate(II)?
- (a) K₃[Fe(CN)₆]
- (b) K₄[Fe(CN)₆]
- (c) K₂[Fe(CN)₆]
- (d) K₆[Fe(CN)₆]
Answer: (b) K₄[Fe(CN)₆]
Explanation: Four potassium ions balance the -4 charge on the hexacyanoferrate(II) ion.
- Question: What is the formula for silver dichromate?
- (a) Ag₂Cr₂O₇
- (b) Ag₂CrO₄
- (c) AgCr₂O₇
- (d) Ag₃Cr₂O₇
Answer: (a) Ag₂Cr₂O₇
Explanation: Two silver ions are required to balance the dichromate ion’s charge.
- Question: Identify the formula for the tetrahydrate of magnesium sulfate.
- (a) MgSO₄·H₂O
- (b) MgSO₄·3H₂O
- (c) MgSO₄·4H₂O
- (d) MgSO₄·5H₂O
Answer: (c) MgSO₄·4H₂O
Explanation: Magnesium sulfate is known in many hydrates; the tetrahydrate is one common form.
- Question: What is the formula for barium chromate?
- (a) BaCrO₄
- (b) Ba₂CrO₄
- (c) BaCr₂O₇
- (d) Ba₂Cr₂O₇
Answer: (a) BaCrO₄
Explanation: Barium chromate is a yellow precipitate with the formula BaCrO₄.
- Question: What is the formula for potassium permanganate?
- (a) KMnO₄
- (b) K₂MnO₄
- (c) KMnO₃
- (d) K₃MnO₄
Answer: (a) KMnO₄
Explanation: Potassium permanganate contains one potassium ion per permanganate ion.
- Question: Which formula represents sodium thiosulfate pentahydrate?
- (a) Na₂S₂O₃·3H₂O
- (b) Na₂S₂O₃·4H₂O
- (c) Na₂S₂O₃·5H₂O
- (d) Na₂S₂O₃·6H₂O
Answer: (c) Na₂S₂O₃·5H₂O
Explanation: Sodium thiosulfate is most commonly encountered as the pentahydrate.
- Question: What is the formula for ferrous sulfate typically found in its heptahydrate form?
- (a) FeSO₄·6H₂O
- (b) FeSO₄·7H₂O
- (c) FeSO₄·8H₂O
- (d) FeSO₄·5H₂O
Answer: (b) FeSO₄·7H₂O
Explanation: Ferrous sulfate usually crystallizes as the heptahydrate.
- Question: What is the formula for copper(I) chloride dihydrate?
- (a) CuCl·H₂O
- (b) CuCl·2H₂O
- (c) CuCl₂·H₂O
- (d) CuCl₂·2H₂O
Answer: (b) CuCl·2H₂O
Explanation: Copper(I) chloride may crystallize as a dihydrate, indicating one copper(I) ion per chloride with two water molecules.
Category 2: Isomerism in Coordination Compounds
- Question: Define geometrical isomerism in coordination compounds.
Answer: Geometrical isomerism occurs when coordination compounds with the same formula have different spatial arrangements of ligands around the central atom (e.g., cis- vs. trans- forms).
Explanation: The differences in ligand positioning can lead to distinctly different chemical and physical properties. - Question: What is optical isomerism?
Answer: Optical isomerism occurs when compounds are non-superimposable mirror images of each other.
Explanation: These enantiomers rotate plane-polarized light in opposite directions. - Question: Explain linkage isomerism with an example.
Answer: Linkage isomerism arises when a ligand capable of coordinating through different atoms (ambidentate ligand) binds in alternate modes. For example, nitrite can bind through nitrogen (nitro) or oxygen (nitrito).
Explanation: The different binding sites result in distinct isomers with unique properties. - Question: What is ionization isomerism?
Answer: Ionization isomerism involves the exchange of a ligand in the coordination sphere with an ion outside the coordination sphere.
Explanation: This results in compounds that give different ions in solution even though their overall composition is similar. - Question: Describe coordination isomerism.
Answer: Coordination isomerism occurs in compounds containing complex cations and anions, where the ligands exchange between the two metal centers.
Explanation: This type of isomerism is common in compounds with two or more metal ions. - Question: How does the presence of a chiral center lead to isomerism in coordination compounds?
Answer: When a coordination compound contains a chiral metal center or asymmetric ligand arrangement, it can form enantiomers that are non-superimposable mirror images.
Explanation: Such compounds exhibit optical activity. - Question: Differentiate between cis- and trans- isomers with respect to square planar complexes.
Answer: In square planar complexes, cis isomers have identical ligands adjacent to each other, while trans isomers have identical ligands opposite each other.
Explanation: This spatial difference affects properties like reactivity and color. - Question: What type of isomerism is illustrated by [Co(NH₃)₅(NO₂)]Cl₂ versus [Co(NH₃)₅(ONO)]Cl₂?
Answer: Linkage isomerism
Explanation: The nitrite ligand can bind via nitrogen or oxygen, leading to two different linkage isomers. - Question: In an octahedral complex, what is the difference between facial (fac) and meridional (mer) isomers?
Answer: In fac isomers, three identical ligands occupy one face of the octahedron, while in mer isomers, they lie in a plane passing through the center.
Explanation: The different arrangements lead to distinct chemical behavior. - Question: How does the presence of different oxidation states affect isomerism in coordination compounds?
Answer: Different oxidation states can lead to varied coordination numbers and ligand arrangements, which may result in different isomeric forms.
Explanation: The change in electron count can alter geometry and bonding. - Question: What is stereoisomerism?
Answer: Stereoisomerism is the general term for isomers that have the same connectivity of atoms but differ in the three-dimensional arrangement of atoms.
Explanation: It includes both geometrical and optical isomerism. - Question: Explain why [Pt(NH₃)₂Cl₂] exhibits isomerism.
Answer: This square planar complex can exist as either cis or trans isomers depending on the positions of the chloride ligands relative to each other.
Explanation: The relative position changes the chemical and physical properties. - Question: What is the significance of isomerism in pharmaceuticals?
Answer: Isomers, particularly optical isomers, can have vastly different biological activities, with one isomer often being therapeutically active and the other inactive or even harmful.
Explanation: Drug safety and efficacy are closely tied to the specific isomer present. - Question: Identify the type of isomerism in the complex [Cr(ox)₃]³⁻ where ox is oxalate.
Answer: Optical isomerism
Explanation: The tris(oxalato)chromate(III) complex can form enantiomers that are mirror images of each other. - Question: How can isomerism affect the color of coordination compounds?
Answer: Different ligand arrangements can alter the d-orbital splitting in the metal center, leading to changes in the wavelengths of light absorbed and, therefore, the observed color.
Explanation: This phenomenon is often exploited in colorimetric sensors. - Question: What type of isomerism is observed in [Co(NH₃)₄Cl₂]⁺?
Answer: Geometrical isomerism
Explanation: Depending on whether the chlorides are adjacent (cis) or opposite (trans), the complex can exist in two geometrical forms. - Question: Why might a coordination compound show both linkage and geometrical isomerism?
Answer: If a compound contains an ambidentate ligand and additional ligands that can arrange in different spatial orientations, it may display both types of isomerism.
Explanation: Each type of isomerism arises from different aspects of the coordination environment. - Question: Describe ionization isomerism in the context of [Co(NH₃)₅Br]SO₄.
Answer: In this complex, if the bromide ion and the sulfate ion exchange positions (one inside and one outside the coordination sphere), two ionization isomers can be formed.
Explanation: The different ions released in solution lead to distinct chemical behavior. - Question: What is the effect of chelation on isomerism in coordination compounds?
Answer: Chelation, by binding multiple sites on a metal, can reduce the number of possible isomers because the chelating ligand fixes the geometry.
Explanation: This “locking” effect minimizes variations in ligand positions. - Question: How can temperature influence the distribution of isomers in a coordination compound?
Answer: Temperature can affect the equilibrium between different isomeric forms; some isomers may be more stable at lower temperatures while others predominate at higher temperatures.
Explanation: Thermal energy can overcome the energy barriers between isomeric states, altering the distribution.
Category 3: Molecular Geometry Classification (Matching Questions)
For the following matching questions, match List I (compounds) with List II (their corresponding molecular geometries).
- Match List:
List I:
A. XeF₂
B. SF₄
C. PCl₅
D. IF₇
List II:
I. Linear
II. See-saw
III. Trigonal bipyramidal
IV. Pentagonal bipyramidal
Answer: A – I, B – II, C – III, D – IV
Explanation: XeF₂ is linear, SF₄ adopts a see-saw shape, PCl₅ is trigonal bipyramidal, and IF₇ is pentagonal bipyramidal. - Match List:
List I:
A. CH₄
B. NH₃
C. H₂O
D. BF₃
List II:
I. Tetrahedral
II. Trigonal pyramidal
III. Bent
IV. Trigonal planar
Answer: A – I, B – II, C – III, D – IV
Explanation: CH₄ is tetrahedral, NH₃ is trigonal pyramidal, H₂O is bent, and BF₃ is trigonal planar. - Match List:
List I:
A. SF₆
B. ClF₃
C. XeF₄
D. PF₅
List II:
I. Octahedral
II. T-shaped
III. Square planar
IV. Trigonal bipyramidal
Answer: A – I, B – II, C – III, D – IV
Explanation: SF₆ is octahedral, ClF₃ is T-shaped, XeF₄ is square planar, and PF₅ is trigonal bipyramidal. - Match List:
List I:
A. ClF₅
B. BrF₅
C. IF₇
D. TeF₆
List II:
I. Square pyramidal
II. Pentagonal bipyramidal
III. Distorted octahedral
IV. Octahedral
Answer: A – I, B – I, C – II, D – IV
Explanation: ClF₅ and BrF₅ typically adopt square pyramidal geometries, IF₇ is pentagonal bipyramidal, and TeF₆ is octahedral. - Match List:
List I:
A. CO₂
B. H₂S
C. O₃
D. NO₂
List II:
I. Linear
II. Bent
III. Angular
IV. V-shaped
Answer: A – I, B – II, C – II, D – II
Explanation: CO₂ is linear, while H₂S, O₃, and NO₂ (in its common form) exhibit bent geometries due to lone pairs. - Match List:
List I:
A. PF₃
B. AsF₃
C. SbF₃
D. BiF₃
List II:
I. Trigonal pyramidal
II. Tetrahedral
III. Seesaw
IV. Distorted trigonal pyramidal
Answer: A – I, B – I, C – I, D – IV
Explanation: PF₃, AsF₃, and SbF₃ typically show trigonal pyramidal shapes; BiF₃ may be distorted due to the inert pair effect. - Match List:
List I:
A. NH₄⁺
B. CH₄
C. SiH₄
D. GeH₄
List II:
I. Tetrahedral
II. Trigonal planar
III. Octahedral
IV. Linear
Answer: A – I, B – I, C – I, D – I
Explanation: All these cations and molecules have a tetrahedral geometry. - Match List:
List I:
A. ClO₄⁻
B. PF₆⁻
C. SF₆
D. MoF₆
List II:
I. Tetrahedral
II. Octahedral
III. Square planar
IV. Trigonal bipyramidal
Answer: A – I, B – II, C – II, D – II
Explanation: ClO₄⁻ is tetrahedral; PF₆⁻, SF₆, and MoF₆ adopt octahedral geometries. - Match List:
List I:
A. XeO₄
B. SO₄²⁻
C. PO₄³⁻
D. SiO₄⁴⁻
List II:
I. Tetrahedral
II. Square planar
III. Octahedral
IV. Trigonal bipyramidal
Answer: A – I, B – I, C – I, D – I
Explanation: All these oxyanions have a tetrahedral arrangement of oxygen atoms. - Match List:
List I:
A. BrF₇
B. IF₇
C. ClF₇
D. SbF₇
List II:
I. Pentagonal bipyramidal
II. Octahedral
III. Square pyramidal
IV. Distorted pentagonal bipyramidal
Answer: A – I, B – I, C – IV, D – IV
Explanation: Heptacoordinated complexes such as BrF₇ and IF₇ typically have a pentagonal bipyramidal shape, though distortions may occur in heavier elements. - Match List:
List I:
A. SF₄
B. SeF₄
C. TeF₄
D. PoF₄
List II:
I. Seesaw
II. T-shaped
III. Square planar
IV. Distorted seesaw
Answer: A – I, B – I, C – IV, D – IV
Explanation: Lighter chalcogen tetrafluorides (SF₄, SeF₄) adopt a seesaw geometry, with heavier analogues showing distortions. - Match List:
List I:
A. ClF₃
B. BrF₃
C. IF₃⁺
D. AtF₃
List II:
I. T-shaped
II. Trigonal planar
III. Bent
IV. Linear
Answer: A – I, B – I, C – II, D – I
Explanation: ClF₃ and BrF₃ are typically T-shaped; IF₃⁺ often forms a planar structure; AtF₃ tends to be T-shaped. - Match List:
List I:
A. [Co(NH₃)₆]³⁺
B. [Fe(CN)₆]³⁻
C. [Cr(H₂O)₆]³⁺
D. [Ni(H₂O)₆]²⁺
List II:
I. Octahedral
II. Tetrahedral
III. Square planar
IV. Trigonal bipyramidal
Answer: A – I, B – I, C – I, D – I
Explanation: All these complexes are octahedral. - Match List:
List I:
A. [PtCl₄]²⁻
B. [Ni(CN)₄]²⁻
C. [Ag(NH₃)₂]⁺
D. [AuCl₄]⁻
List II:
I. Square planar
II. Tetrahedral
III. Linear
IV. Octahedral
Answer: A – I, B – II, C – III, D – I
Explanation: [PtCl₄]²⁻ and [AuCl₄]⁻ are square planar, [Ni(CN)₄]²⁻ is tetrahedral, and [Ag(NH₃)₂]⁺ is linear. - Match List:
List I:
A. [Cr(ox)₃]³⁻
B. [Co(en)₃]³⁺
C. [Fe(ox)₃]³⁻
D. [Ru(bpy)₃]²⁺
List II:
I. Octahedral (optically active)
II. Square planar
III. Tetrahedral
IV. Trigonal planar
Answer: A – I, B – I, C – I, D – I
Explanation: These coordination complexes are all octahedral; many, such as [Co(en)₃]³⁺, exhibit optical isomerism. - Match List:
List I:
A. [MnO₄]⁻
B. [ReO₄]⁻
C. [CrO₄]²⁻
D. [MoO₄]²⁻
List II:
I. Tetrahedral
II. Octahedral
III. Square planar
IV. Trigonal pyramidal
Answer: A – I, B – I, C – I, D – I
Explanation: The oxyanions listed are tetrahedral. - Match List:
List I:
A. [CuCl₄]²⁻
B. [ZnCl₄]²⁻
C. [CdCl₄]²⁻
D. [NiCl₄]²⁻
List II:
I. Tetrahedral
II. Square planar
III. Octahedral
IV. Linear
Answer: A – II, B – I, C – I, D – II
Explanation: [CuCl₄]²⁻ and [NiCl₄]²⁻ may adopt square planar geometry (especially for d⁹ and d⁸ configurations), while [ZnCl₄]²⁻ and [CdCl₄]²⁻ are typically tetrahedral. - Match List:
List I:
A. [FeBr₆]³⁻
B. [CoF₆]³⁻
C. [CrCl₆]³⁻
D. [RuCl₆]³⁻
List II:
I. Octahedral
II. Tetrahedral
III. Square planar
IV. Pentagonal bipyramidal
Answer: All – I
Explanation: These hexahalometallate complexes are octahedral. - Match List:
List I:
A. [Pt(NH₃)₂Cl₂]
B. [Pt(NH₃)₄]²⁺
C. [PdCl₂]
D. [AuCl₃]
List II:
I. Square planar
II. Tetrahedral
III. Octahedral
IV. Linear
Answer: A – I, B – II, C – I, D – I
Explanation: Many platinum and palladium complexes are square planar; [Pt(NH₃)₄]²⁺ can be tetrahedral in rare cases, but is more often octahedral when additional ligands are present. (Acceptable answers may vary with context.) - Match List:
List I:
A. [Cr(CO)₆]
B. [Fe(CO)₅]
C. [Ni(CO)₄]
D. [Mo(CO)₆]
List II:
I. Octahedral
II. Trigonal bipyramidal
III. Tetrahedral
IV. Square planar
Answer: A – I, B – II, C – III, D – I
Explanation: [Cr(CO)₆] and [Mo(CO)₆] are octahedral, [Fe(CO)₅] is trigonal bipyramidal, and [Ni(CO)₄] is tetrahedral.
The questions assess a basic knowledge of either formula identification, isomerism, or molecular geometry classification in coordination chemistry. Feel free to modify options or explanations as needed for your teaching.
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