CHAPTER 5 Chemistry Class 12: COORDINATION COMPOUNDS (QUICK NOTES & QUESTION BANK)

PART 1: CHAPTER OVERVIEW

1. Important Terms & Werner's Theory
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Coordination Compound: A compound where a central metal atom or ion is bonded to a fixed number of ions or molecules via coordinate (dative) bonds.

• Werner's Theory: Central metals possess two distinct types of valencies:

1. Primary Valency: Ionizable, corresponds to the oxidation state of the metal, and is satisfied only by negative ions.

2. Secondary Valency: Non-ionizable, corresponds to the Coordination Number (CN), and is satisfied by neutral molecules or negative ions. It has a fixed spatial orientation (geometry).

• Ligands: Ions or molecules that donate a pair of electrons to the central metal.

• Unidentate: Donates one electron pair (e.g., Cl-, H2O, NH3).

• Didentate: Donates two electron pairs simultaneously (e.g., oxalate 'ox' [C2O4]^2-, ethane-1,2-diamine 'en').

• Polydentate: EDTA is a classic hexadentate ligand (donates 6 electron pairs).

• Chelating Ligands: Di- or polydentate ligands that bind to a single metal ion forming a stable ring structure. Chelated complexes are exceptionally stable.

• Ambidentate Ligands: Unidentate ligands that can bind to the metal through two different donor atoms (e.g., -ONO vs -NO2, -SCN vs -NCS).

2. Nomenclature & Isomerism

• IUPAC Rules:

Name the ligands first in alphabetical order (with prefixes like di, tri, or bis, tris for complex ligands), followed by the metal.

• If the complex ion is an Anion (negative), the metal name must end in "-ate" (e.g., Cobaltate, Ferrate, Platinate).

• The oxidation state of the metal is always written at the end in Roman numerals inside parentheses.

• Isomerism Types:

• Structural Isomerism: Includes Linkage (due to ambidentate ligands), Coordination (exchange of ligands between cationic and anionic entities), Ionization (gives different ions in solution), and Solvate/Hydrate isomerism.

• Stereoisomerism:

1. *Geometrical Isomerism:

• Found in square planar and octahedral complexes. Exists as 'cis' (same ligands adjacent at 90 degrees) and 'trans' (same ligands opposite at 180 degrees). Also exists as facial (fac) and meridional (mer) in Ma3b3 octahedral types. (Note: Tetrahedral complexes never show geometrical isomerism).

1. Optical Isomerism: Chiral complexes that form non-superimposable mirror images (dextro and laevo forms). Highly common in cis-octahedral complexes with didentate chelating ligands (e.g., [Co(en)3]^3+). Trans isomers generally have a plane of symmetry and are optically inactive.

3. Valence Bond Theory (VBT)

• Explains bonding using hybridization of metal atomic orbitals (s, p, d) to accommodate ligand electron pairs.

• Coordination Number4:

• sp3 Hybridization -> Tetrahedral geometry (usually weak field ligands, e.g., [NiCl4]^2-).

• dsp2 Hybridization -> Square Planar geometry (strong field ligands, e.g., [Ni(CN)4]^2-).

• Coordination Number 6 (Octahedral):

• d2sp3 Hybridization -> Inner Orbital Complex / Low Spin Complex (uses inner 3d orbitals; usually strong field ligands).

• sp3d2 Hybridization -> Outer Orbital Complex / High Spin Complex (uses outer 4d orbitals; usually weak field ligands).

• Magnetic Nature: If unpaired electrons remain after filling, the complex is Paramagnetic. If all are paired up, it is Diamagnetic.

4. Crystal Field Theory (CFT)

• Treats the metal-ligand bond as purely electrostatic. The approach of negative ligands causes the five degenerate d-orbitals of the metal to split into levels with different energies.

• Octahedral Splitting: Split into a lower energy triplet t2g (d_xy, d_yz, d_zx) and a higher energy doublet eg (d_x2-y2, d_z2). The energy gap is denoted as do (Crystal Field Splitting Energy).

• Spectrochemical Series: Arrangement of ligands in increasing order of their crystal field splitting power:

I- < Br- < Cl- < F- < OH- < H2O < nh3 < en < CN- < CO

• Strong Field Ligands (like CN-, CO, en) cause a large splitting gap (do).

• Weak Field Ligands (like Cl-, F-) cause a small splitting gap (do).

Electron Pairing Rules for d4 configuration:

• If do > P (Pairing Energy) -> Strong Field Case: The 4th electron pairs up in the lower t2g level. Configuration becomes (t2g)4 (eg)0.

• If do < P -> Weak Field Case: The 4th electron jumps up to the higher eg level. Configuration becomes (t2g)3 (eg)1.

• Color in Complexes: Attributed to electronic transitions between the split d-orbital levels (t2g to eg d-d transitions) by absorbing specific visible wavelengths. Anhydrous CoCl2 or CuSO4 are white/colorless because without ligands, d-orbital splitting does not occur.

PART 2: INTEGRATED QUESTION BANK

SECTION A: MULTIPLE CHOICE QUESTIONS

Q1. Which of the following coordination entities will be completely optically active (exist as non-superimposable mirror images)?

(a) trans-[Co(en)2Cl2]+
(b) cis-[Co(en)2Cl2]+
(c) [Cr(NH3)5Cl]2+
(d) [Ni(CN)4]2-

Answer: (b)

Explanation: The trans-[Co(en)2Cl2]+ isomer has a plane of symmetry, making it achiral and optically inactive. The cis-[Co(en)2Cl2]+ isomer lacks a plane of symmetry, making it chiral. Its mirror image cannot be superimposed, meaning it is highly optically active.

Q2 [NEET].

What is the correct IUPAC name for the coordination complex: [Ag(NH3)2][Ag(CN)2]?
(a) Diamminesilver(I) dicyanoargentate(I)
(b) Dicyanosilver(I) diammineargentate(I)
(c) Diamminesilver(II) dicyanoargentate(II)
(d) Diamminesilver(I) dicyanosilver(I)

Answer: (a)
Explanation: The complex contains a cationic part [Ag(NH3)2]+ and an anionic part [Ag(CN)2]-. Both silver ions have an oxidation state of +1. The metal in the positive part retains its common name "Silver", while the metal in the negative part takes the suffix "-ate", becoming "Argentate".

Q3 [JEE Main]. Using Crystal Field Theory, write out the correct electronic configuration and find the number of unpaired electrons for an octahedral [Fe(H2O)6]2+ complex. (Atomic number of Fe = 26)
(a) (t2g)4 (eg)2 with 4 unpaired electrons
(b) (t2g)6 (eg)0 with 0 unpaired electrons
(c) (t2g)3 (eg)3 with 6 unpaired electrons
(d) (t2g)5 (eg)1 with 4 unpaired electrons

*Answer:

(a)
Explanation: Fe2+ has a 3d6 valence configuration. H2O is a weak field ligand, which means the crystal field splitting energy is smaller than the pairing energy (do < P). Electrons fill up singly across both levels before pairing. The first 3 go into t2g, the next 2 go into eg, and the 6th electron pairs up in t2g. This gives a configuration of (t2g)4 (eg)2, leaving exactly 4 unpaired electrons.

SECTION B: SHORT ANSWER & CONCEPTUAL QUESTIONS

Q4 [CBSE].

Describe the underlying chemical cause of 'Linkage Isomerism' in coordination chemistry. Give one real-world example pair.

Answer:
Linkage isomerism arises in a coordination compound when it contains an ambidentate ligand. An ambidentate ligand possesses two distinct potential donor atoms but can only coordinate through one atom at a time to the metal.

• Example Pair:

1. [Co(NH3)5(NO2)]Cl2 -> Pentaamminenitrocobalt(III) chloride (Ligand binds via the Nitrogen atom: M-NO2).

2. [Co(NH3)5(ONO)]Cl2 -> Pentaamminenitritocobalt(III) chloride (Ligand binds via the Oxygen atom: M-ONO).

Q5 [NEET]. Explain why [Ni(CN)4]2- is square planar and diamagnetic, whereas [NiCl4]2- is tetrahedral and highly paramagnetic, even though both contain Ni2+. (Atomic number of Ni = 28).

Answer:
In both complexes, Nickel is in the +2 state with a 3d8 configuration.

• In [Ni(CN)4]2-, CN- is a strong field ligand. It forces the two unpaired 3d electrons to pair up, leaving one inner 3d orbital vacant. The metal undergoes dsp2 hybridization, forming a square planar geometry. Since all electrons are paired, it is diamagnetic.

• In [NiCl4]2-, Cl- is a weak field ligand and cannot force the 3d electrons to pair up. The 3d8 configuration keeps its 2 unpaired electrons. The metal utilizes outer orbitals to undergo sp3 hybridization, resulting in a tetrahedral geometry. The 2 unpaired electrons make it paramagnetic.

SECTION C: NUMERICAL CHALLENGES

Q6 [CBSE].

One mole of an aqueous coordination compound with the empirical formula CoCl3 . 5NH3 reacts with an excess of AgNO3 solution to precipitate exactly 2 moles of AgCl(s) cream precipitate. Deduce the structural coordination formula, coordination number of the metal, and its IUPAC name.

Answer:* Step 1: Find the number of ionizable chloride ions.

Since 1 mole of the complex yields 2 moles of AgCl precipitate, there must be exactly 2 chloride ions present outside the coordination sphere as counter ions.

• Step 2: Formulate the inner coordination sphere.

Out of the 3 total Cl atoms, 2 are outside. This leaves 1 Cl atom and all 5 NH3 molecules inside the bracket attached to the central Cobalt atom: [Co(NH3)5Cl]Cl2.

• Step 3: Determine the Coordination Number (CN).

The central Cobalt atom is directly bonded to 5 ammine ligands and 1 chloro ligand. Total bonds = 5 + 1 = 6. So, CN = 6.

• Step 4: Write out the IUPAC Name.

Oxidation state of Co: x + 5(0) + 1(-1) + 2(-1) = 0 -> x = +3.

Name: Pentaamminenitrocobalt(III) chloride or Pentaamminechloridocobalt(III) chloride.
Answer: Formula is [Co(NH3)5Cl]Cl2, CN is 6, and the IUPAC name is Pentaamminechloridocobalt(III) chloride.

Q7 [JEE Main]. Calculate the Crystal Field Stabilization Energy (CFSE) in kJ/mol for a high-spin octahedral d4 metal ion complex if the value of do is given as 15000 cm-1. (Conversion: 1 cm-1 = 0.01196 J/mol).

Answer:

1. Determine electron distribution for high-spin d4:

High-spin means weak field splitting (do < P), so the 4th electron goes into the higher eg orbital.
Configuration = (t2g)3 (eg)1

1. Apply the CFSE formula for an octahedral field:
   CFSE = [ -0.4 x (number of electrons in t2g) + 0.6 x (number of electrons in eg) ] x do
   CFSE = [ -0.4 x 3 + 0.6 x 1 ] x do
   CFSE = [ -1.2 + 0.6 ] x do = -0.6 x do

2. Substitute the do value:
   CFSE = -0.6 x 15000 cm-1 = -9000 cm-1

3. Convert to kJ/mol:
   Energy = -9000 cm-1 x 0.01196 J/mol = -107.64 J/mol = -0.108 kJ/mol.
   Answer: The CFSE value is -0.108 kJ/mol (stabilization energy magnitude is 0.108 kJ/mol).

SECTION D: INTEGRATED CASE STUDY BASED QUESTION
Q8 [4 Marks].

Medical oncology treatments rely heavily on specialized heavy metal coordination complexes like Cisplatin, cis-[Pt(NH3)2Cl2], to bind to cancer cell DNA and halt tumor growth. Conversely, inside the human respiratory system, the biological complex Hemoglobin uses an iron-centered porphyrin ring to reversibly bind oxygen. In industrial analytical extractions, adding the hexadentate chelating ligand EDTA helps bind and trap harmful heavy metal impurities from hard water systems.

(a) What structural feature makes EDTA an excellent antidote for heavy metal poisoning? (1 Mark)

(b) Draw or describe the structural difference between Cisplatin and its trans-isomer. Why can trans-platin not be used as an effective anti-cancer drug? (1 Mark)
(c) When 1 mole of cis-[Pt(NH3)2Cl2] dissolves completely in water, how many total ions are produced? Explain based on Werner's formulation. (2 Marks)

Answer:
(a) EDTA is a hexadentate chelating ligand. It wraps around a toxic heavy metal ion using 6 donor atoms (2 Nitrogen and 4 Oxygen atoms) to form a highly stable, ring-shaped chelate complex. This renders the metal non-toxic and allows it to be safely excreted from the body.

(b) In Cisplatin (the cis-isomer), the two identical chloro ligands sit right next to each other at a 90-degree angle. In trans-platin, they sit directly opposite each other at a 180-degree angle. Trans-platin cannot be used as a drug because its opposite geometry prevents it from binding effectively to two adjacent nitrogen atoms on a DNA strand.

(c) Let's analyze the complex based on Werner's theory:

• The formula is cis-[Pt(NH3)2Cl2].

• All ligands (2 NH3 and 2 Cl) are inside the coordination sphere, satisfied by the non-ionizable secondary valency.

• There are zero counter-ions outside the coordination brackets.

• When dissolved in water, the coordination sphere remains completely intact as a single neutral molecule and does not break apart.
  Answer:

Exactly 0 ions are produced (total ion count is zero because it is a non-electrolyte).