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Q1. When both positive and negative ions of a salt are complex ions and the two isomers differs in the distribution of ligands in the cation and the anion; it is a type of:

• 1) Optical isomerism
• 2) Coordination isomerism
• 3) Linkage isomerism
• 4) Geometrical isomerism

Solution

When both positive and negative ions of a salt are complex ions and the two isomers differs in the distribution of ligands in the cation and the anion; it is a type of Coordination isomerism.    

Q2. What is meant by unidentate and bidentate ligands? Give example.

Solution

Ligand may contain one or more than one donor atoms. If only one donor atom is present in its molecule which can coordinate then it is unidentate. For example, H₂O, NH₃. If the ligand contains two donor atoms placed in a way that a five or six membered ring is formed with the metal ion then it is called as bidentate. Example is oxalate ion.

Q3. Write step wise stability constant expressions K₁ and K₂ for the reaction Cu ²⁺  + 4NH₃  ⇌  [Cu(NH₃)₄]²⁺

Solution

 

Q4. How will you distinguish between the two enantiomers of [PtCl₂(en)₂]⁺

Solution

Among the two enantiomers of [PtCl₂(en)₂]⁺, one isomer is dextrorotatory while the other is laevorotatory.    

Q5. From the cis and trans isomer of [CrCl₂(ox)₂]^3-which is chiral and optically active?

Solution

The two entities are represented as, i.e. from the cis and trans isomer of [CrCl₂(ox)₂]^3- optically active.

Q6. What type of isomerism is exhibited by [Cr(H₂O)₅SCN]³⁺.

Solution

[Cr(H₂O)₅SCN]³⁺is a type of linkage isomer, as here more than one atom of a monodentate ligand may function as a donor. So the possible forms are [Cr(H₂O)₅SCN]³⁺ and [Cr(H₂O)₅NCS]². Linkage isomerism arises in a coordination compound containing ambidentate ligand.

Q7. How would you account for the following:[Ni(CO)₄] has tetrahedral geometry while [Ni(CO)₄]^2- has a square planar geometry?

Solution

Nickel in [Ni(CO)₄] is in the zero oxidation state and has electronic configuration of [Ar]3d⁸4s² or 3d¹⁰.sp³ hybrid orbitals accommodate four pair of electrons from four CO groups and the resulting tetrahedral complex is diamagnetic due to the absence of unpaired electrons.Nickel in [Ni(CO)₄]^2- is in the +2 oxidation state and has electronic configuration of [Ar]3d¹⁰4s². dsp² hybrid orbitals accommodate four pair of electrons from four CO groups and the resulting square planer complex is diamagnetic due to the absence of unpaired electrons.

Q8. What is the IUPAC name of (NH₄)₃[Cr(SCN)₆] complex?

Solution

IUPAC name of (NH₄)₃[Cr(SCN)₆] is Ammonium hexathiocyanato-S-chromate(III).

Q9. What is ionisation isomerism? Give example.

Solution

Compounds which gives different ions in solution although they have same composition are called ionisation isomerism. This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄. Another example is, [Pt(NH₃)₄(OH)₂]SO₄ and [Pt(NH₃)₄SO₄](OH)₂.

Q10. Draw the structure of [Cr(CO)₆].

Solution

[Cr(CO)₆] is octahedral in shape.

Q11. Write the formula of the co-ordination compound dichlorobis(ethylenediammine)cobalt(III) sulphate.

Solution

The formula of the co-ordination compound is [Co(en)₂Cl₂]SO₄.

Q12. Draw the geometrical isomers of [Co(NH₃)₄Cl₂]⁺.

Solution

Octahedral complex [Co(NH₃)₄Cl₂]⁺exists as cis and trans isomer

Q13. Explain the structure of [Co(NH₃)₆]³⁺.

Solution

The cobalt ion is in +3 oxidation state and has an electronic configuration of 3d⁶. Six pairs of electron one from each ammonia molecule occupy the six hybrid orbitals, hence the molecule is diamagnetic and octahedral in shape.

Q14. Which ligand has six donor atoms?

Solution

EDTA is a hexadentate ligand with six donor atoms, i.e, two nitrogen atoms and four oxygen atoms of the carboxylic acid groups capable of bonding to the metal atom.

Q15. Why [Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is dimagnetic?

Solution

Chromium in [Cr(NH₃)₆]³⁺ is in +3 oxidation state and has electronic configuration of [Ar]3d⁵4s¹. Cr³⁺ ion provides six empty orbitals to accommodate six pairs of electrons from six molecules of ammonia. The resulting complex involves d²sp³ hybridisation and is octahedral. Due to the presence of these unpaired electrons in the complex it is paramagnetic in nature. Nickel in [Ni(CN)₄]^2- is in +2 oxidation state and has electronic configuration of [Ar]3d¹⁰4s². dsp² hybrid orbitals accommodate four pairs of electrons from four CN groups and the resulting square planner complex is diamagnetic due to the absence of unpaired electrons.

Q16. Write the structure of Zeises salt, Tetracarbonynickel and Ferrocene.

Solution

Zeisses salt, Tetracarbonynickel and Ferrocene