πŸ§ͺ CHEMISTRY β€’ CHAPTER 5 β€’ TN BOARD

12th Chemistry Coordination Chemistry

Prepare for the Tamil Nadu Class 12 Chemistry Public Examination with carefully selected Chapter 5 – Coordination Chemistry Important 3 Mark Questions. These chapter-wise questions are compiled from the last six years of 1st Mid-Term, Quarterly, 2nd Mid-Term, Half-Yearly, Revision Exam 1, Revision Exam 2, and Tamil Nadu State Board Public Examination question papers to help students revise important concepts, practise descriptive answers, improve accuracy, and score high marks in the board examination.

05

Chapter

3 Marks

Important Questions

TN Board

Latest Syllabus

πŸ”₯ MOST REPEATED

Most Repeated 3 Mark Questions

These questions have appeared repeatedly in previous Tamil Nadu Class 12 Chemistry examination papers. Students should give these questions priority during revision.

01
Give the difference between double salts and coordination compounds? πŸ”₯ Most Repeated
Double Salt Co-ordination Compound
Dissociates to give simple ions in solutions. Never dissociates to give simple ions.
These exist only in solid state and dissociate into constituent species in their solutions. They retain their identity in solid as well as in solution state.
Lose its identity in dissolved state. Does not lose its identity in dissolved state.
Positive and negative ions are present. Simple and complex ions are present.
Example:
Potash alum: K2SO4Β·Al2(SO4)3Β·H2O 		Example:
K4[Fe(CN)6]
02
In the complex identify the following? ⭐ Compulsory Question πŸ”₯ Most Repeated
A : [Pt(NH3)3(NO2)]Cl
A) Central metal atom/ion: Pt (Platinum)
B) Ligands: NH3, NO2
C) Coordination Number: 4
D) IUPAC Name: Triamminetrichloroplatinum(II) chloride
E) Coordination Entity: Pt(NH3)3(NO2)
B : [Pt(NO2)(H2O)(NH3)2]Br
A) Central metal atom/ion: Pt
B) Ligands: NO2 – Negative ligand, H2O – Neutral ligand, NH3 – Neutral ligand
C) Coordination Entity: Pt(NO2)(H2O)(NH3)2
D) Oxidation Number of central metal ion:
X + (βˆ’1) = +1
X = +2
E) Coordination Number: 4
1 NO2 + 1 H2O + 2 NH3 = 4 ligands attached
C : [Ag(NH3)2]+
i) Ligand: NH3 β€” Ammine
ii) Central metal atom: Ag β€” Silver (Ag)
iii) IUPAC Name: Diamminesilver(I) ion
iv) Coordination Number: 2
v) Magnetic Properties: Diamagnetic
03
In an octahedral crystal, draw and explain the figure to show splitting of d-orbitals? πŸ”₯ Most Repeated
Splitting of d-orbitals in an octahedral crystal field

Splitting of d-orbitals in an octahedral crystal field

During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals dxΒ²βˆ’yΒ² and dzΒ² (represented as eg orbitals) will increase by 3/5 Ξ”o while that of the other three orbitals dxy, dyz and dxz (represented as t2g orbitals) decrease by 2/5 Ξ”o.

Here, Ξ”o represents the crystal field splitting energy in the octahedral field.

04
Give one test to differentiate [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl? πŸ”₯ Most Repeated
(i) [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionisation isomers.
(ii) When these isomers are dissolved in water, they ionise to give different ions in solution which react differently with different reagents.
[Co(NH3)5Cl]SO4 β†’ [Co(NH3)5Cl]2+ + SO42βˆ’
[Co(NH3)5SO4]Cl β†’ [Co(NH3)5SO4]+ + Clβˆ’
For Eg. 1: Barium Chloride test
[Co(NH3)5Cl]SO4 + BaCl2 β†’ BaSO4↓
(White precipitate)
[Co(NH3)5SO4]Cl + BaCl2 β†’ No reaction
For Eg. 2: Silver nitrate test
[Co(NH3)5SO4]Cl + AgNO3 β†’ AgCl↓
(Curdy white precipitate)
[Co(NH3)5Cl]SO4 + AgNO3 β†’ No reaction
πŸ“Œ
Exam Tip: Questions 1–4 are identified as highly repeated questions. Focus especially on the compulsory complex-identification question, the crystal-field splitting diagram and the ionisation isomerism test.
πŸ“š APPEARED ONCE

Other Important 3 Mark Questions

These questions have appeared once in the analysed question papers. They are still important for complete chapter revision.

05
Write short note on – Ο€-Back bonding in metal carbonyl?
  • In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.
  • The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of the central metal atom.
  • This electron pair donation forms a Οƒ-bond.
    M Οƒ bond CO
  • This sigma bond formation increases the electron density in metal d orbitals and makes the metal electron rich.
  • In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty Ο€* orbital on the carbonyl ligand and transfers the added electron density back to the ligand.
  • This second component is called Ο€-back bonding.
πŸ“˜ COORDINATION CHEMISTRY β€’ CHAPTER 5

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