CBSE Class 12 Chemistry Coordination Compounds Case Study Questions Set 02

Case Study 1

Coordination compounds are the compounds in which the central metal atom is linked to a number of ions or neutral molecules by coordinate bonds and the donor atoms, molecules or anions which donate a pair of electrons to the metal atom or ion and form a coordinate bond with it are called ligands. Coordination number is the number of atoms or ions immediately surrounding a central atom in a complex or a crystal.
Coordination number of Ni²⁺ is 4 and it forms two complexes A and B as given below:
NiCl₂ + KCN → A (Cyano complex)
NiCl₂ + conc. HCl → B (Chloro complex)

Question 1. Write the IUPAC name of complex A.
Answer:
Potassium tetracyanonickelate(II).

Teacher's Note:
1. A is K₂[Ni(CN)₄]; an anionic complex takes the ending “-ate” in its name.

Question 2. Write the formula of the complex B.
Answer:
K₂[NiCl₄]

Teacher's Note:
1. With coordination number 4 and four Cl⁻ ligands, the complex ion carries a 2− charge: [NiCl₄]²⁻.

Question 3. Predict the magnetic nature of A.
Answer:
Diamagnetic.

Teacher's Note:
1. Strong CN⁻ pairs both unpaired 3d electrons of Ni²⁺, leaving none unpaired.

Question 4. Write the state of hybridization of the complex B.
Answer:
sp³.

Teacher's Note:
1. Weak Cl⁻ ligands do not cause pairing, so no 3d orbital is free and 4s4p³ orbitals are used.

Question 5. What will be the shape of the complex A?
Answer:
Square planar.

Teacher's Note:
1. dsp² hybridisation gives square planar shape, sp³ gives tetrahedral – do not mix the two.

Case Study 2

The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine. Formation of coordination compounds is largely used in analytical chemistry for the qualitative detection and quantitative estimation of metal ions. Coordination compounds also find several important applications in the field of medicine. Several coordination compounds are also used as antidote to poisoning caused by the ingestion of poisonous metals by human beings.

Question 6. Which complexing material is added to vegetable oils to remove the ill effects of undesired metal ions?
Answer:
EDTA

Teacher's Note:
1. EDTA is a hexadentate ligand that binds metal ions strongly, which makes it useful wherever metal ions must be removed.

Question 7. How would you detect the presence of nickel in a food sample?
Answer:
With the help of dimethylglyoxime which forms a red complex with Ni²⁺ ions.

Teacher's Note:
1. The red (rosy) precipitate with dimethylglyoxime is the standard test for Ni²⁺.

Question 8. What is chelate therapy?
Answer:
Chelate therapy is used for the removal of excess of metal ions present in toxic amounts in the body.

Teacher's Note:
1. Give an example in the exam: EDTA for lead poisoning, D-penicillamine for excess copper.

Question 9. Which complex is used in the treatment of cancer?
Answer:
Cisplatin

Teacher's Note:
1. Cisplatin is cis-[Pt(NH₃)₂Cl₂]; only the cis isomer is active.

Question 10. Name a vitamin which is a complex of cobalt.
Answer:
Vitamin B₁₂

Teacher's Note:
1. Vitamin B₁₂ (cyanocobalamin) is a cobalt complex, chlorophyll a magnesium complex, haemoglobin an iron complex.

Case Study 3

One of the most important studies about coordination compounds is the type of bonding present in them. Alfred Werner was the first to study the bonding in coordination compounds. He introduced the concept of primary valency of metal atom which is same as oxidation state and secondary valency of metal atom which is same as coordination number. Primary valencies are satisfied by negative ions whereas secondary valencies are satisfied by neutral and negative ligands. Further, primary valencies are ionizable and non-directional whereas secondary valencies are non-ionizable and directional and decide the geometry of the complex. However, some of the characteristics of coordination compounds could not be explained by Werner's theory. Hence, many new approaches were put forward. One of these is the Valence Bond Theory (VBT). According to this theory, the metal atom or ion under the influence of ligands can use its (n − 1) d, ns and np or ns, np and nd orbitals for hybridisation to yield an equivalent number of hybrid orbitals. Ligands donate electron pairs to these empty hybrid orbitals forming bonds. The type of hybridisation involved decides the geometry of the complex. This theory satisfactorily explains the magnetic properties of the coordination compounds.

Question 11. The compound PtCl₄.2 HCl gives no precipitate with excess of AgNO₃ solution. What are the primary and secondary valencies of the metal in the compound ?
Answer:
The compound PtCl₄.2 HCl is written in the complex form as H₂[PtCl₆]. It does not give precipitate with AgNO₃ because it does not ionize to give Cl⁻ ions. If x is primary valency of Pt, then
2 × (+1) + x + 6 × (−1) = 0 or 2 + x − 6 = 0 or x = +4
Secondary valency of Pt is its coordination number which is 6.

Teacher's Note:
1. No AgCl precipitate means no Cl⁻ outside the bracket, so all six Cl must be ligands.

Teacher's Note:
1. Two complexes can have the same hybridisation (sp³) but different magnetism - Ni(CO)₄ is diamagnetic, [NiCl₄]²⁻ paramagnetic.

Question 13. What are outer orbital and inner orbital complexes ? Give one example of each.
Answer:
When the octahedral complex formed involves the inner (n−1) d orbitals for d²sp³ hybridisation, the complex formed is called inner orbital complex, e.g., [Fe(CN)₆]⁴⁻.
When the octahedral complex formed involves the use of outer orbital nd-orbitals for hybridisation (called sp³d²), the complex formed is called outer orbital complex, e.g. [CoF₆]³⁻.

Teacher's Note:
1. Inner orbital complexes are usually low spin; outer orbital complexes are high spin.

Teacher's Note:
1. Learn the spin-only values: n = 1, 2, 3, 4, 5 give 1.73, 2.83, 3.87, 4.90, 5.92 BM.

Case Study 4

In octahedral complexes, when the ligands approach the metal ion, the d-orbitals split into two sets one with lower energy consisting of d_xy, d_yz and d_xz called t_2g orbitals and the other with higher energy consisting of d_x²−y² and d_z² called e_g orbitals. The difference of energy between the two sets is called crystal field splitting energy (CFSE) and is represented by Δ₀. It can be calculated from the number of electrons occupying t_2g orbitals and the number occupying e_g orbitals. For complexes upto d³ configuration, the three t_2g orbitals of lower energy are first singly occupied. From d⁴ to d⁷, whether the electron will pair up in t_2g or enter into e_g depends upon whether Δ₀ > P or < P where P is called pairing energy. Accordingly, we may get high spin or low spin complexes. The crystal field theory also helps to explain the properties of complexes like magnetism, colour etc. In the square planar complexes and in the tetrahedral complexes, the splitting of d-orbitals takes place in a different way than in the octahedral complexes.

Question 15. What is Bari centre ? In terms of CFSE, how the splitting of d-orbitals of tetrahedral complexes takes place with respect to Bari centre ?
Answer:
The mean value of energy of the perturbed d-orbitals in the presence of a crystal field is called 'Bari centre'.
d-orbitals of tetrahedral complexes split up into two sets, one consisting of d_x²−y² and d_z² orbitals (called e-orbitals) with lower energy and the other consisting of d_xy, d_yz and d_xz (called t₂) with higher energy. t₂ is 0.4 Δ_t and e is − 0.6 Δ_t.

Teacher's Note:
1. Tetrahedral splitting is the reverse of octahedral, and the labels lose the 'g' because a tetrahedron has no centre of symmetry.

Question 16. In terms of CFSE (Δ₀) and pairing energy (P), explain when do we get low spin complexes and when do we get high spin complexes for d⁴ to d⁷ configuration ? What type of ligands are responsible for their formation ?
Answer:
If Δ₀ < P, we get high spin complexes. The ligands for which Δ₀ < P are called weak field ligands.
If Δ₀ > P, we get low spin complexes. Ligands for which Δ₀ > P are called strong field ligands.

Teacher's Note:
1. Only d⁴ to d⁷ ions have a high spin/low spin choice in octahedral complexes; d¹-d³ and d⁸-d¹⁰ have one arrangement only.

Question 17. Why [Ti(H₂O)₆] Cl₃ is coloured but on heating it becomes colourless ?
Answer:
[Ti(H₂O)₆]³⁺ is coloured (violet) because its single d-electron absorbs green-yellow light and undergoes d-d transition from t_2g to e_g.
On heating, the ligands H₂O are lost and we know that in the absence of ligands, crystal field splitting does not occur and, therefore, no d-d transition can occur. Hence, it becomes colourless.

Teacher's Note:
1. Same idea explains why CuSO₄.5H₂O is blue but anhydrous CuSO₄ is white.

Question 18. Calculate the CFSE (Δ₀) for a complex with d⁵ configuration when the ligand is a weak field ligand.
Answer:
For d⁵ configuration for a complex with a weak field ligand, we will have high spin state, i.e., t_2g³ e_g²
CFSE = (− 0.4 x + 0.6 y) Δ₀ = (− 0.4 × 3 + 0.6 × 2) Δ₀ = 0.0

Teacher's Note:
1. High spin d⁵ (and d⁰, d¹⁰) always has zero CFSE because the gain in t_2g is exactly cancelled by the loss in e_g.

Case Study 5

As in case of organic compounds, coordination compounds also show two types of isomerism viz. structural isomerism and stereo isomerism. There are four types of structural isomerism shown by the coordination compounds viz. ionization isomerism, hydrate isomerism, linkage isomerism and coordination isomerism. Similarly, there are two types of stereoisomerism viz. geometrical isomerism and optical isomerism. Geometrical isomerism is shown by square planar as well as octahedral complexes whereas optical isomerism is shown only by octahedral complexes.

Question 19. Write the coordination isomer of [Cu(NH₃)₄] [PtCl₄].
Answer:
[Pt(NH₃)₄] [CuCl₄]

Teacher's Note:
1. Swap the complete ligand sets between the two metals; both metals keep their +2 oxidation states.

Question 20. Name the ionization isomer of [Cr(H₂O)₅Br] SO₄.
Answer:
[Cr(H₂O)₅SO₄] Br, i.e., pentaaquasulphatochromium (III) bromide.

Teacher's Note:
1. SO₄²⁻ inside the sphere still occupies one coordination position here, keeping CN = 6.

Question 21. Write IUPAC name of the linkage isomer of
(i) [Co(NH₃)₅NO₂]Cl₂
(ii) [Cr(H₂O)₅SCN]²⁺
Answer:
(i) [Co(NH₃)₅ONO]Cl₂, pentaamminenitrito-O-cobalt (III) chloride
(ii) [Cr(H₂O)₅NCS]²⁺, pentaaquaisothiocyanatochromium (III) ion.

Teacher's Note:
1. Bonded through S it is thiocyanato (thiocyanato-S); bonded through N it is isothiocyanato (thiocyanato-N).

Question 22. Draw the structures of isomers of [Pt(NH₃)₂Cl₂]

Answer:
Two geometrical isomers (cis and trans) :

Teacher's Note:
1. Square planar complexes have a plane of symmetry (the molecular plane), so they never show optical isomerism.

Case Study 6

It is well known that transition metals form complexes. In modern terminology, these complexes are called coordination compounds. These compounds are present in biological systems also like chlorophyll, haemoglobin and vitamin B₁₂. Some non-transition metals also form complexes like chlorophyll. Though complexes are formed from two or more simple salts, yet they differ from double salts in a number of ways especially in their dissociation. In coordination compounds, a number of neutral or anionic species are linked to the central metal atom by coordinate bonds. Such species are called ligands. Ligands may be monodentate or bidentate or polydentate. Bidentate and polydentate ligands generally form chelates. Some unidentate ligands may have more than one coordinating atom. These are called ambidentate ligands. The number of unidentate ligands attached to the central metal atom is called its coordination number. The net charge carried by the metal atom is called its oxidation number. IUPAC has recommended some definite rules to write the names of coordination compounds when formula is given or vice versa.

Case Study 7

The homoleptic carbonyls (compounds containing carbonyl ligands only) are formed by most of the transition metals. These carbonyls have simple, well-defined structures. Such compounds are called organometallic compounds. They contain at least one metal-carbon or metalloid-carbon bond. The metal-carbon bond possesses both σ and π-character. A weak M–C σ-bond is formed by donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. A strong M–C π-bond is formed by the back donation of a pair of electrons from a filled orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens bond between CO and the metal. Based on the nature of metal-carbon bond we have σ-bonded and π-bonded organometallic compounds. All metal carbonyls have well defined structures.