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GNDU Question Paper-2024

Bachelor of Computer Application (BCA) (Hons.)

Semester (Batch 2024-28) (CBGS)

CHEMISTRY

(Inorganic Chemistry-IV)

Time Allowed: Three Hours Max. Marks:35

Note: Attempt Five questions in all, selecting at least One question from each section. The

Fifth question may be attempted from any section. All questions carry equal marks.

SECTION-A

1. (a) Explain the relationship between crystal field splitting and the pairing energy to

determine whether a given complex is high spin or low spin.

(b) Define C.F.S.E. and no. of unpaired electrons and C.F.S.E. for (i) Cr

(ii) Fe

(iii) Co

complexes?

2. (a) Discuss the crystal field splitting in square planar complexes.

(b) What factors affect the Crystal field splitting.

]. Explain on the basis of C.F.T.

SECTION-B

3. (a) What is Diamagnetic correction? Write short note on ferromagnetism and

antiferromagnetism.

(b) Calculate the spin magnetic value of Ni

and Fe

4. (a) What factors affect the stability of complexes ?

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(b) Derive rate law expression for nucleophilic substitution in square planar complexes.

two ?

SECTION-C

5. (a) What is Vibronic Coupling? Give one example which shows this phenomenon.

(b) What are Laporte selection rules for d-d transition? Under what condition these are

relaxed?

6. (a) Discuss the electronic transition in [Cr(H

]

(b) Discuss Orgel diagram for d¹. d in octahedral and tetrahedral complexes and explain

electronic transition between them.

SECTION-D

7. (a) What are Organometallic Compounds? Which are the different ways to classify

them? Discuss in detail.

(b) What are the applications of organo lithium compounds ?

8. (a) Discuss the preparation, properties of organo aluminium compounds.

(b) What do you mean by Homogeneous Hydrogenation ? Name the three homogeneous

hydrogenation catalyst used for hydrogenation of alke.

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GNDU Question Paper-2024

Bachelor of Computer Application (BCA) (Hons.)

Semester (Batch 2024-28) (CBGS)

CHEMISTRY

(Inorganic Chemistry-IV)

Time Allowed: Three Hours Max. Marks:35

Note: Attempt Five questions in all, selecting at least One question from each section. The

Fifth question may be attempted from any section. All questions carry equal marks.

SECTION-A

1. (a) Explain the relationship between crystal field splitting and the pairing energy to

determine whether a given complex is high spin or low spin.

(b) Define C.F.S.E. and no. of unpaired electrons and C.F.S.E. for (i) Cr

(ii) Fe

(iii) Co

complexes?

Ans: 1. (a) Relationship Between Crystal Field Splitting and Pairing Energy (High Spin and

Low Spin)

Crystal Field Theory (CFT) explains what happens when ligands (such as H₂O, NH₃, CN⁻, Cl⁻)

surround a transition metal ion. The ligands produce an electric field that causes the five d-

orbitals of the metal ion to split into two groups having different energies.

In an octahedral complex, the five d-orbitals split as:

Higher Energy

dz² dx²-y²

---------------------- Δo (Crystal Field Splitting

Energy)

t2g

dxy dxz dyz

Lower Energy

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Here:

t₂g orbitals have lower energy.

e_g orbitals have higher energy.

Δo (Delta o) is called Crystal Field Splitting Energy.

Another important energy is the Pairing Energy (P).

Pairing Energy is the energy needed to place two electrons in the same orbital.

Now compare Δo and P.

Case 1: Δo < Pairing Energy (P)

When crystal field splitting is smaller than pairing energy, electrons prefer to occupy

separate orbitals rather than pair up.

Result:

Maximum number of unpaired electrons.

Complex is called High Spin Complex.

Example:

d5 High Spin

eg ↑ ↑

t2g ↑ ↑ ↑

There are 5 unpaired electrons.

Case 2: Δo > Pairing Energy (P)

When crystal field splitting is larger than pairing energy, electrons prefer pairing in lower-

energy t₂g orbitals instead of moving to higher-energy e_g orbitals.

Result:

Minimum number of unpaired electrons.

Complex is called Low Spin Complex.

Example:

d5 Low Spin

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t2g ↑↓ ↑↓ ↑

Only 1 unpaired electron remains.

Easy Trick

Condition

Type of Complex

Δo < Pairing Energy

High Spin

Δo > Pairing Energy

Low Spin

Strong ligands like CN⁻, CO, NH₃ usually produce low-spin complexes, whereas weak ligands

like F⁻, Cl⁻, Br⁻, I⁻ usually produce high-spin complexes.

(b) Crystal Field Stabilization Energy (CFSE)

Definition

Crystal Field Stabilization Energy (CFSE) is the extra stability gained by a metal ion when

electrons occupy the lower-energy t₂g orbitals after crystal field splitting.

In an octahedral field:

Each electron in t₂g contributes −0.4 Δo.

Each electron in e_g contributes +0.6 Δo.

Formula:

CFSE = (Number of electrons in t₂g × −0.4Δo) + (Number of electrons in e_g × +0.6Δo)

(i) Cr³⁺

Atomic number of Cr = 24

Electronic configuration:

Cr = [Ar] 3d⁵ 4s¹

Cr³⁺ = 3d³

Electron arrangement:

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t2g ↑ ↑ ↑

t₂g electrons = 3

e_g electrons = 0

CFSE = 3(−0.4Δo) = −1.2Δo

Unpaired electrons = 3

(ii) Fe³⁺

Atomic number = 26

Fe = [Ar] 3d⁶ 4s²

Fe³⁺ = 3d⁵

For weak-field ligands (high spin):

eg ↑ ↑

t2g ↑ ↑ ↑

t₂g = 3

e_g = 2

CFSE

= 3(−0.4Δo) + 2(+0.6Δo)

= −1.2Δo + 1.2Δo

= 0

Unpaired electrons = 5

(iii) Co²⁺

Atomic number = 27

Co = [Ar] 3d⁷ 4s²

Co²⁺ = 3d⁷

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High-spin arrangement:

eg ↑ ↑

t2g ↑↓ ↑↓ ↑

t₂g electrons = 5

e_g electrons = 2

CFSE

= 5(−0.4Δo) + 2(+0.6Δo)

= −2.0Δo + 1.2Δo

= −0.8Δo

Unpaired electrons = 3

Summary Table

Metal Ion

d-Electrons

Electron Arrangement

Unpaired Electrons

CFSE

Cr³⁺

d³

t₂g³

−1.2Δo

Fe³⁺ (High Spin)

d⁵

t₂g³e_g²

Co²⁺ (High Spin)

d⁷

t₂g⁵e_g²

−0.8Δo

Complexes?

One of the biggest successes of Crystal Field Theory is that it explains why some transition

metal complexes are strongly attracted to a magnet while others are weakly attracted or

even not attracted at all.

The magnetic behavior depends entirely on the number of unpaired electrons.

1. Paramagnetic Complexes

If a complex contains one or more unpaired electrons, it is paramagnetic.

More unpaired electrons mean stronger attraction toward a magnetic field.

Examples:

Cr³⁺ → 3 unpaired electrons → Paramagnetic

Fe³⁺ (High Spin) → 5 unpaired electrons → Strongly Paramagnetic

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Co²⁺ (High Spin) → 3 unpaired electrons → Paramagnetic

2. Diamagnetic Complexes

If all electrons are paired, the complex is diamagnetic.

Diamagnetic substances are not attracted by a magnetic field.

Example:

A low-spin d⁶ complex like [Fe(CN)₆]⁴⁻ has all electrons paired, so it is diamagnetic.

Why Does This Happen?

The answer again depends on the comparison between Δo and Pairing Energy.

If Δo is small, electrons avoid pairing and occupy higher-energy orbitals. This creates

more unpaired electrons, making the complex high spin and strongly paramagnetic.

If Δo is large, electrons pair in the lower-energy t₂g orbitals. This reduces the number

of unpaired electrons, making the complex low spin and weakly paramagnetic or

diamagnetic.

Thus, Crystal Field Theory links the strength of the ligand, the crystal field splitting (Δo),

electron pairing, CFSE, and the number of unpaired electrons to explain the magnetic

properties of transition metal complexes.

Exam Points to Remember

Crystal Field Splitting (Δo): Energy difference between t₂g and e_g orbitals.

Pairing Energy (P): Energy required to pair two electrons in the same orbital.

Δo < P → High Spin Complex.

Δo > P → Low Spin Complex.

CFSE: Stability gained due to electrons occupying lower-energy t₂g orbitals.

Paramagnetic: One or more unpaired electrons.

Diamagnetic: All electrons paired.

Cr³⁺: 3 unpaired electrons, CFSE = −1.2Δo.

Fe³⁺ (High Spin): 5 unpaired electrons, CFSE = 0.

Co²⁺ (High Spin): 3 unpaired electrons, CFSE = −0.8Δo.

These concepts are closely connected: ligand strength determines crystal field splitting,

crystal field splitting determines electron pairing, electron pairing determines CFSE and

the number of unpaired electrons, and the number of unpaired electrons determines the

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magnetic behavior of the complex. This chain of ideas is the key to answering questions on

Crystal Field Theory in university examinations.

2. (a) Discuss the crystal field splitting in square planar complexes.

(b) What factors affect the Crystal field splitting.

]. Explain on the basis of C.F.T.

Ans: Introduction

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split into

different energy levels when ligands come close to the metal ion. This splitting happens

because the negatively charged ligands repel the electrons present in the metal's d-orbitals.

The amount of splitting depends on the arrangement of ligands around the metal ion.

In a square planar complex, four ligands are arranged in the same plane around the central

metal ion, forming the shape of a square.

Structure of a Square Planar Complex

Ligand

Ligand Metal Ligand

Ligand

All four ligands lie in the xy-plane.

Why Do d-Orbitals Split?

Before ligands approach, all five d-orbitals have the same energy (they are degenerate).

The five d-orbitals are:

dxy

dxz

dyz

dx²−y²

dz²

When ligands come close, orbitals pointing directly towards the ligands experience

maximum repulsion and their energy increases. Orbitals pointing away from ligands

experience less repulsion and remain at lower energy.

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Crystal Field Splitting in Square Planar Complex

The order of energy from highest to lowest is:

Highest Energy

dx²−y²

│

dxy

│

dz²

│

dxz = dyz

Lowest Energy

Why this order?

1. dx²−y² (Highest Energy)

Points directly toward all four ligands.

Faces maximum repulsion.

Therefore, it has the highest energy.

2. dxy

Lies in the xy-plane.

Faces ligands indirectly.

Receives moderate repulsion.

3. dz²

Points mainly along the z-axis.

Since there are no ligands above or below, it experiences less repulsion.

4. dxz and dyz (Lowest Energy)

Point between ligand directions.

Experience the least repulsion.

Hence they have the lowest energy.

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Importance of Square Planar Splitting

Most square planar complexes are formed by metals like:

Ni²⁺

Pd²⁺

Pt²⁺

These complexes generally have:

Large crystal field splitting

Pairing of electrons

Diamagnetic nature (no unpaired electrons)

2(b) Factors Affecting Crystal Field Splitting

The energy difference between split d-orbitals is called Crystal Field Splitting Energy (Δ).

Several factors determine how large or small this splitting will be.

1. Nature of the Metal Ion

A metal ion with higher positive charge attracts ligands more strongly.

Example:

Fe²⁺ < Fe³⁺

Fe³⁺ has greater crystal field splitting because it attracts ligands more strongly.

2. Oxidation State of the Metal

As oxidation state increases:

Attraction between metal and ligands increases.

Repulsion becomes stronger.

Splitting energy increases.

Example:

Co²⁺ < Co³⁺

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3. Nature of Ligands

Different ligands produce different amounts of splitting.

This is explained by the Spectrochemical Series.

Weak-field ligands:

I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O

Strong-field ligands:

NH₃ < en < NO₂⁻ < CN⁻ < CO

Weak ligands produce small splitting.

Strong ligands produce large splitting.

Example:

CN⁻ produces much greater splitting than H₂O.

4. Distance Between Metal and Ligands

Smaller distance → greater repulsion → larger splitting.

Larger distance → smaller splitting.

5. Geometry of the Complex

Different shapes produce different splitting.

General order:

Square Planar > Octahedral > Tetrahedral

Square planar complexes usually have the largest splitting.

6. Type of Metal (3d, 4d, 5d)

Transition metals of higher series show larger splitting.

3d < 4d < 5d

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Example:

Pt²⁺ shows greater splitting than Ni²⁺.

2(c) Why is [Ti(H₂O)₆]³⁺ Purple? (Crystal Field Theory)

Step 1: Identify the Metal Ion

The complex is:

[Ti(H₂O)₆]³⁺

Water is a neutral ligand.

Therefore,

Oxidation state of Ti = +3

Titanium atomic number = 22

Electronic configuration:

Ti = [Ar] 3d² 4s²

Ti³⁺ loses three electrons:

Ti³⁺ = [Ar] 3d¹

Thus, only one electron is present in the d-orbitals.

Step 2: Octahedral Splitting

Since six water molecules surround the metal ion, the complex has an octahedral geometry.

The d-orbitals split into two groups:

dx²−y² dz²

-----------------------

Δo

-----------------------

tg

dxy dxz dyz

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t₂g = lower energy

eg = higher energy

The single d-electron occupies the lower-energy t₂g orbital.

Step 3: Absorption of Light

When white light falls on the complex:

The d-electron absorbs energy equal to Δ₀ (the octahedral crystal field splitting

energy).

It jumps from the t₂g level to the eg level.

This process is called a dd electronic transition.

↑

│ Absorbs light (Δo)

│

------------------

tg

Step 4: Why Purple?

White light contains all colours.

The complex absorbs yellow-green light (whose energy matches Δ₀). The remaining

transmitted or reflected light is the complementary colour, which is purple (violet).

Thus, [Ti(H₂O)₆]³⁺ appears purple because of the dd transition predicted by Crystal Field

Theory.

Conclusion

Crystal Field Theory helps us understand how ligands influence the energies of the metal's

d-orbitals. In square planar complexes, the dx²−y² orbital has the highest energy because it

points directly toward the four ligands, while dxz and dyz have the lowest energy due to

minimal repulsion. The magnitude of crystal field splitting depends on factors such as the

metal ion, its oxidation state, the nature of the ligands, metalligand distance, geometry

of the complex, and whether the metal belongs to the 3d, 4d, or 5d series.

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The purple colour of [Ti(H₂O)₆]³⁺ is a direct application of Crystal Field Theory. Titanium(III)

has one d-electron, and when white light shines on the complex, this electron absorbs the

required energy and moves from the lower t₂g level to the higher eg level. Since yellow-

green light is absorbed, the complementary colour observed is purple. This simple

explanation shows how Crystal Field Theory not only predicts magnetic properties and

stability but also explains the beautiful colours of many transition metal complexes.

SECTION-B

3. (a) What is Diamagnetic correction? Write short note on ferromagnetism and

antiferromagnetism.

(b) Calculate the spin magnetic value of Ni

and Fe

Ans: Magnetism is one of the most interesting properties of matter. We see magnets

attracting iron nails, compass needles pointing towards the north, and magnetic strips in

electronic devices. But did you know that every substance has some magnetic property?

Some materials are strongly attracted by a magnet, some are weakly attracted, while others

are slightly repelled. Chemists study these magnetic properties because they help us

understand the arrangement of electrons in atoms and compounds.

(a) What is Diamagnetic Correction?

Every atom contains electrons. These electrons move around the nucleus and produce a tiny

magnetic field.

Some substances have all their electrons paired. Since every electron has an opposite-spin

partner, their magnetic effects cancel each other.

Such substances are called diamagnetic substances.

Examples:

Water (H₂O)

Sodium chloride (NaCl)

Benzene

Zinc ion (Zn²⁺)

Sometimes we measure the magnetic property of a compound that contains both paired

and unpaired electrons.

The paired electrons also produce a very small diamagnetic effect. This small effect reduces

the actual magnetic value.

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Therefore, before calculating the true magnetic property of a compound, we subtract this

small diamagnetic contribution.

This process is called Diamagnetic Correction.

Definition

Diamagnetic correction is the process of removing the small magnetic effect produced by

paired electrons so that the actual magnetic property of a substance can be determined

accurately.

Simple Diagram

Paired Electrons

↑↓

Magnetic effect = Cancelled

Overall = Diamagnetic

Ferromagnetism

Ferromagnetism is the strongest type of magnetism.

In ferromagnetic materials, all the tiny atomic magnets (called magnetic moments) point in

the same direction.

Because all of them work together, the material becomes strongly magnetic.

Even after removing the external magnet, these materials often remain magnetized.

Examples

Iron (Fe)

Cobalt (Co)

Nickel (Ni)

Diagram

Ferromagnetic Material

↑ ↑ ↑ ↑ ↑ ↑ ↑

All magnetic moments

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point in the same direction

Strong Magnet

Characteristics

Very strongly attracted by magnets.

Can become permanent magnets.

High magnetic susceptibility.

Used in transformers, motors, generators, speakers, and magnets.

Antiferromagnetism

Antiferromagnetism is different from ferromagnetism.

Here, neighbouring atoms have magnetic moments in opposite directions.

One points upward while the next points downward.

Because they cancel each other, the overall magnetism becomes almost zero.

Examples

Manganese oxide (MnO)

Nickel oxide (NiO)

Chromium oxide (Cr₂O₃)

Diagram

Antiferromagnetic Material

↑ ↓ ↑ ↓ ↑ ↓

Opposite directions

Net Magnetism = Nearly Zero

Characteristics

Weak magnetic behaviour.

Opposite alignment of spins.

Magnetic moments cancel each other.

Found in many transition metal oxides.

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Difference Between Ferromagnetism and Antiferromagnetism

Ferromagnetism

Antiferromagnetism

Magnetic moments are parallel

Magnetic moments are opposite

Strong magnetism

Very weak or zero magnetism

Attracted strongly by magnet

Weak attraction

Example: Iron, Nickel, Cobalt

Example: MnO, NiO

(b) Calculate the Spin Magnetic Value of Ni²⁺ and Fe²⁺

The magnetic moment due to electron spin is calculated using the Spin-only Formula:

Where:

μ = Magnetic moment (Bohr Magneton, BM)

n = Number of unpaired electrons

Step 1: Ni²⁺

Atomic number of Nickel = 28

Electronic configuration:

Ni =

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²

Ni²⁺ loses two electrons from the 4s orbital.

Configuration becomes:

3d⁸

In 3d⁸, there are 2 unpaired electrons.

So,

n = 2

Using the formula:

Answer

Spin magnetic moment of Ni²⁺ = 2.83 BM

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Step 2: Fe²⁺

Atomic number of Iron = 26

Electronic configuration:

Fe =

[Ar] 3d⁶ 4s²

Fe²⁺ loses two 4s electrons.

Configuration becomes:

3d⁶

A high-spin 3d⁶ ion has 4 unpaired electrons.

Thus,

n = 4

Using the formula:

Answer

Spin magnetic moment of Fe²⁺ = 4.90 BM

Summary Table

Ion

Unpaired Electrons

Magnetic Moment

Ni²⁺

2.83 BM

Fe²⁺

4.90 BM

Suppose you bring a material close to a magnet.

Some materials are attracted strongly, some weakly, and some are repelled.

The ability of a material to become magnetized in the presence of an external magnetic field

is called magnetic susceptibility.

Definition

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Magnetic susceptibility (χ) is the measure of how easily a substance becomes magnetized

when an external magnetic field is applied.

Large positive χ → strongly attracted (ferromagnetic or paramagnetic).

Small positive χ → weakly attracted (paramagnetic).

Negative χ → slightly repelled (diamagnetic).

Formula

Magnetic susceptibility is given by:

 





Where:

M = Magnetization of the material

H = Applied magnetic field

Variation of Magnetic Susceptibility with Temperature

Temperature affects the movement of electrons and the alignment of magnetic moments.

1. Diamagnetic substances

Almost no change with temperature.

Susceptibility remains nearly constant because all electrons are paired.

Temperature ↑

χ = Constant

2. Paramagnetic substances

As temperature increases, thermal motion disturbs the alignment of unpaired electrons.

Therefore:

Temperature ↑

Magnetic susceptibility ↓

This follows Curie's Law:

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 





where C is Curie's constant and T is the absolute temperature (Kelvin).

3. Ferromagnetic substances

At low temperatures, magnetic moments remain aligned, giving very high susceptibility.

When the temperature reaches the Curie temperature, this alignment breaks down.

Above the Curie temperature:

Ferromagnetic substances behave like paramagnetic substances.

Susceptibility decreases sharply.

Simple Temperature Trend

Diamagnetic

│──────────────

│

└────────────── Temperature

Paramagnetic

│\

│ \

└────────── Temperature

Ferromagnetic

│\

│ \

│ \____

│ \

└──────────── Temperature

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Curie Temperature

Final Revision Points

Diamagnetic correction removes the small magnetic effect of paired electrons to

obtain the true magnetic property.

Ferromagnetism: magnetic moments align in the same direction, producing strong

magnetism (examples: Fe, Co, Ni).

Antiferromagnetism: neighbouring magnetic moments align in opposite directions,

nearly canceling each other (examples: MnO, NiO).

Spin-only magnetic moment is calculated using  



󰇛  󰇜BM.

o Ni²⁺: 2 unpaired electrons → 2.83 BM

o Fe²⁺: 4 unpaired electrons → 4.90 BM

Magnetic susceptibility (χ) measures how easily a substance becomes magnetized in

an external magnetic field.

Temperature effect:

o Diamagnetic: almost no change.

o Paramagnetic: susceptibility decreases with increasing temperature (Curie's

Law).

o Ferromagnetic: high at low temperatures but falls sharply above the Curie

temperature due to loss of magnetic ordering.

4. (a) What factors affect the stability of complexes ?

(b) Derive rate law expression for nucleophilic substitution in square planar complexes.

two ?

Ans: 4(a) Factors Affecting the Stability of Complexes

A complex (coordination compound) is formed when a central metal ion is surrounded by

molecules or ions called ligands. These ligands donate a pair of electrons to the metal ion

and form coordinate bonds.

Stability of a complex means how strongly the metal ion and ligands remain attached to

each other. A more stable complex is difficult to break, while a less stable complex

dissociates easily.

Simple Diagram

NH3

NH3 Cu2+ NH3

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NH3

Metal Ion = Cu2+

Ligands = NH3

Complex = [Cu(NH3)4]2+

Factors Affecting Stability of Complexes

1. Nature of the Metal Ion

The type of metal ion plays an important role.

Smaller metal ions hold ligands more strongly.

Highly charged metal ions attract ligands more strongly.

Example: Fe³⁺ forms more stable complexes than Fe²⁺ because Fe³⁺ has a higher positive

charge.

2. Nature of the Ligand

Different ligands have different strengths.

Some ligands donate electrons more effectively and form stronger bonds.

Strong ligands include:

CN⁻

NH₃

Weak ligands include:

H₂O

F⁻

The stronger the ligand, the more stable the complex.

3. Chelate Effect

A chelating ligand attaches to the metal ion through two or more donor atoms.

Example: EDTA

Normal Ligand

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M NH3

Chelating Ligand

N M N

Chelating ligands form ring structures, making the complex much more stable.

This phenomenon is called the Chelate Effect.

4. Size of the Metal Ion

Small metal ions bring ligands closer together and form stronger coordinate bonds.

Hence,

Smaller ion → Greater stability

5. Oxidation State of Metal

Higher oxidation state means stronger attraction towards ligands.

Example:

Fe³⁺ > Fe²⁺ in complex stability.

6. Crystal Field Stabilization Energy (CFSE)

Some electronic arrangements are naturally more stable because electrons occupy lower-

energy orbitals.

Greater CFSE means a more stable complex.

7. Temperature and Solvent

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Very high temperature may break coordinate bonds.

The solvent also affects stability because some solvents compete with ligands for binding to

the metal ion.

4(b) Rate Law Expression for Nucleophilic Substitution in Square Planar Complexes

Square planar complexes are commonly formed by metals such as Pt²⁺, Pd²⁺, and Au³⁺.

A typical complex looks like this:

NH3

Cl Pt NH3

Suppose a nucleophile (Y) replaces one chloride ion.

Reaction:

[PtCl4]2− + Y

↓

[PtCl3Y]2− + Cl−

Mechanism

The substitution generally follows the Associative (A) Mechanism.

Step 1 (Slow Step)

The nucleophile first approaches the metal ion.

[PtCl4]2− + Y

↓

Activated Intermediate

This is the rate-determining (slow) step.

Step 2 (Fast Step)

One chloride ion leaves.

Intermediate

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↓

[PtCl3Y]2− + Cl−

Derivation of Rate Law

Let

Complex = C

Nucleophile = Y

Reaction:

C + Y → Product

Since the first step is slow,

Rate depends on both the complex and nucleophile.

Therefore,

Rate = k[C][Y]

where:

Rate = speed of reaction

k = rate constant

[C] = concentration of complex

[Y] = concentration of nucleophile

This is called a second-order rate law, because the reaction depends on the concentrations

of two reactants.

Flow Diagram

Complex + Nucleophile

│

▼

Slow Association Step

│

▼

Intermediate

│

Fast Loss of Leaving Group

▼

Product

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Conclusion: In square planar complexes undergoing nucleophilic substitution by the

associative mechanism, the reaction rate is proportional to both the concentration of the

complex and the nucleophile:

Rate = k[Complex][Nucleophile]

4(c) Stepwise and Overall Stability Constant

When a metal ion binds ligands one at a time, each binding step has its own stepwise

stability constant.

The stability after all ligands have attached is described by the overall stability constant.

Stepwise Stability Constant (K₁, K₂, K₃...)

Suppose ammonia ligands attach one by one.

First Step

Cu2+ + NH3  [Cu(NH3)]2+

K1 = [Cu(NH3)2+] / ([Cu2+][NH3])

Second Step

[Cu(NH3)]2+ + NH3  [Cu(NH3)2]2+

K2 = [Cu(NH3)2 2+] / ([Cu(NH3)2+][NH3])

Third Step

[Cu(NH3)2]2+ + NH3  [Cu(NH3)3]2+

This has constant K₃.

Similarly,

Fourth step has K₄.

Each equilibrium constant represents the stability of one individual ligand addition.

Overall Stability Constant (β)

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Instead of considering each step separately, we can consider the complete reaction.

Cu2+ + 4NH3  [Cu(NH3)4]2+

The equilibrium constant for this overall reaction is called the overall stability constant,

represented by β (beta).

It measures the stability of the fully formed complex.

Relationship Between Stepwise and Overall Stability Constants

The overall stability constant is equal to the product of all the stepwise stability constants:

  



 



 



 



This means that the stability of the complete complex depends on the stability of each

individual ligand-binding step.

Diagram

Cu2+

│

▼

[Cu(NH3)]2+

│

▼

[Cu(NH3)2]2+

│

▼

[Cu(NH3)3]2+

│

▼

[Cu(NH3)4]2+

Overall Stability Constant:

β = K1 × K2 × K3 × K4

Summary

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Complex stability depends on factors such as the nature, size, and charge of the

metal ion, the strength and type of ligand (especially chelating ligands), crystal field

stabilization, temperature, and the solvent.

Nucleophilic substitution in square planar complexes usually follows an associative

mechanism, where the nucleophile first attaches to the metal. The rate law is Rate =

k[Complex][Nucleophile], showing a second-order reaction.

Stepwise stability constants (K₁, K₂, K₃, ) describe the stability of each successive

ligand-binding step, while the overall stability constant (β) represents the stability of

the final complex. Their relationship is:

  



 



 



 

A larger value of β indicates a more stable coordination complex.

SECTION-C

5. (a) What is Vibronic Coupling? Give one example which shows this phenomenon.

(b) What are Laporte selection rules for d-d transition? Under what condition these are

relaxed?

Ans: Introduction

When light falls on a transition metal complex, electrons absorb energy and jump from one

energy level to another. However, not every electronic transition is allowed according to the

selection rules. Surprisingly, some "forbidden" transitions are still observed in practice. One

important reason for this is Vibronic Coupling.

What is Vibronic Coupling?

The word Vibronic comes from two words:

Vibrational (Vibra) = Movement or vibration of atoms in a molecule.

Electronic (Tronic) = Movement of electrons between energy levels.

Definition:

Vibronic coupling is the interaction between the electronic motion of electrons and the

vibrations of the molecule. This interaction allows some electronic transitions that are

normally forbidden to occur with weak intensity.

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In simple words, imagine a person trying to jump over a wall that is normally too high. If the

ground beneath the person suddenly moves upward (because of vibration), the jump

becomes possible. Similarly, molecular vibrations help electrons make transitions that were

otherwise forbidden.

Simple Diagram

Without vibration

Higher Energy Level

------------------- ✖ Transition Forbidden

Lower Energy Level

-------------------

With vibration (Vibronic Coupling)

Higher Energy Level

------------------- ↑ Allowed (Weak)

Molecular

Vibrations

Lower Energy Level

-------------------

The vibration temporarily changes the symmetry of the molecule, making the forbidden

transition partially allowed.

Why is Vibronic Coupling Important?

It helps explain why weak absorption bands appear in the spectra of many transition metal

complexes.

Without vibronic coupling:

Many d-d transitions would never appear.

With vibronic coupling:

Weak absorption bands become visible in UV-Visible spectroscopy.

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Example of Vibronic Coupling

A common example is the octahedral complex [Ti(H₂O)₆]³⁺.

Normally, its d-d transition is weak because it does not fully satisfy the selection rules.

However, the water molecules continuously vibrate. These vibrations temporarily distort

the octahedral shape, allowing the electronic transition to occur.

As a result, a weak absorption band appears in its spectrum.

Other examples include complexes of:

Mn²⁺

Cr³⁺

Co²⁺

where weak d-d bands are observed due to vibronic coupling.

Key Points

Vibronic coupling means interaction between molecular vibrations and electronic

transitions.

It relaxes selection rules.

Forbidden transitions become weakly allowed.

It produces weak absorption bands in UV-Visible spectra.

5(b) What are Laporte Selection Rules for d-d Transition? Under what condition are these

relaxed?

What is the Laporte Selection Rule?

The Laporte Selection Rule tells us whether an electronic transition is allowed based on the

symmetry of a molecule.

It mainly applies to molecules having a center of symmetry, such as most octahedral

complexes.

Statement of Laporte Rule

The rule says:

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Electronic transitions are allowed only when there is a change in parity.

Parity means symmetry.

There are two types:

g (gerade) = Symmetric with respect to the center.

u (ungerade) = Unsymmetric with respect to the center.

Allowed transitions:

g → u ✔ Allowed

u → g ✔ Allowed

Forbidden transitions:

g → g ✖ Forbidden

u → u ✖ Forbidden

d-d Transition

The d-orbitals in an octahedral complex are g orbitals.

A d-d transition means:

d(g) → d(g)

Since both orbitals have the same symmetry (g → g), the transition is forbidden by the

Laporte Rule.

Therefore, octahedral complexes usually show weak absorption bands.

Simple Diagram

Octahedral Complex

Higher d orbital (eg)

--------------------

↑

d → d

Forbidden

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(g → g)

--------------------

Lower d orbital (t2g)

When is the Laporte Rule Relaxed?

Although the transition is forbidden, it can still occur under certain conditions.

1. Vibronic Coupling

The molecule vibrates continuously.

These vibrations temporarily remove the center of symmetry, making the transition partly

allowed.

This is the most common reason.

2. Distortion of Structure

If the octahedral complex becomes distorted (for example due to the JahnTeller effect),

the center of symmetry is partially lost.

As a result, d-d transitions become more intense.

3. Tetrahedral Complexes

Tetrahedral complexes do not have a center of symmetry.

Therefore, the Laporte rule does not apply strictly.

Their d-d transitions are much more allowed and hence more intense.

Key Points

Laporte rule applies mainly to centrosymmetric complexes.

d(g) → d(g) transitions are forbidden.

Vibronic coupling relaxes the rule.

Structural distortion also relaxes it.

Tetrahedral complexes naturally escape this restriction.

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5(c) Discuss why tetrahedral complexes give intense spectra.

Introduction

When transition metal complexes absorb visible light, electrons move between d-orbitals.

The intensity of the absorption depends on whether the transition is allowed or forbidden.

One interesting observation is that tetrahedral complexes show much stronger and more

intense colours than octahedral complexes of the same metal ion.

Why are Tetrahedral Complexes More Intense?

1. No Center of Symmetry

The biggest reason is that tetrahedral complexes do not possess a center of symmetry.

Because of this, the Laporte Selection Rule is not strictly applicable.

As a result:

d-d transitions become partially allowed.

More light is absorbed.

Stronger colours are produced.

2. Greater Mixing of Orbitals

In tetrahedral complexes, the d-orbitals can mix more easily with p-orbitals.

This orbital mixing increases the probability of electronic transitions.

Therefore, the absorption bands become stronger.

3. Higher Probability of Electronic Transition

Since the transitions are less restricted, electrons can move more easily from lower-energy

d-orbitals to higher-energy d-orbitals.

Greater probability means greater absorption intensity.

Simple Comparison Diagram

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Octahedral Complex

Center of Symmetry

✔

d → d Transition

Weak

Light absorbed: Less

Colour: Pale

-------------------------------

Tetrahedral Complex

No Center of Symmetry

✖

d → d Transition

More Allowed

Light absorbed: More

Colour: Bright and Intense

Example

Consider cobalt(II):

[Co(H₂O)₆]²⁺ (Octahedral) → Pale pink colour with weak absorption.

[CoCl₄]²⁻ (Tetrahedral) → Deep blue colour with much stronger absorption.

The tetrahedral complex appears much more intensely coloured because the Laporte rule is

relaxed due to the absence of a center of symmetry.

Conclusion

Vibronic coupling is the interaction between molecular vibrations and electronic transitions,

allowing weakly forbidden transitions to occur. The Laporte Selection Rule states that g → g

transitions, such as d-d transitions in octahedral complexes, are forbidden because they do

not involve a change in parity. However, molecular vibrations (vibronic coupling), structural

distortions, or the absence of a center of symmetry can relax this rule. Tetrahedral

complexes naturally lack a center of symmetry, so their d-d transitions are much more

allowed. Consequently, they absorb light more strongly and display brighter, more intense

colours than comparable octahedral complexes. Understanding these concepts helps

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explain the colours and UV-Visible spectra of transition metal complexes in coordination

chemistry.

6. (a) Discuss the electronic transition in [Cr(H

]

(b) Discuss Orgel diagram for d¹. d in octahedral and tetrahedral complexes and explain

electronic transition between them.

Ans: 6(a) Discuss the electronic transition in [Cr(H₂O)₆]³⁺

The complex [Cr(H₂O)₆]³⁺ is called hexaaquachromium(III) ion. In this complex, one

chromium ion (Cr³⁺) is surrounded by six water (H₂O) molecules. Since six ligands are

present, the complex has an octahedral geometry.

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Step 1: Electronic Configuration

Atomic number of Chromium = 24

Electronic configuration of Cr = [Ar] 3d⁵ 4s¹

Cr³⁺ loses three electrons, so its configuration becomes:

Cr³⁺ = [Ar] 3d³

Thus, there are 3 electrons in the d-orbitals.

Step 2: Splitting of d-Orbitals

When six water molecules approach the chromium ion, the five d-orbitals do not remain at

the same energy. Due to the crystal field created by the ligands, they split into two groups:

Lower energy orbitals (t₂g) → dxy, dxz, dyz

Higher energy orbitals (e_g) → dx²−y², dz²

The energy difference between these two sets is called Crystal Field Splitting Energy (Δ₀).

Higher Energy

↑ ↑

------------

Δo

------------

↑ ↑ ↑

t2g t2g t2g

Lower Energy

Since Cr³⁺ has three electrons, all three occupy the lower-energy t₂g orbitals according to

Hund's rule.

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Configuration:

t₂g³ e_g⁰

Step 3: Electronic Transition

When white light falls on the complex, some light is absorbed.

If the absorbed light has energy equal to Δ₀, one electron jumps from t₂g to e_g.

Before Absorption

------

t2g ↑ ↑ ↑

After Absorption

eg ↑

------

t2g ↑ ↑

This movement of an electron is called a dd electronic transition.

The transition may be written as:

t₂g³ → t₂g²e_g¹

Step 4: Why is the Complex Colored?

White light contains all colours.

The complex absorbs mainly yellow-green light. The remaining colours combine and our

eyes observe the solution as violet.

Therefore,

Absorbed colour → Yellow-Green

Observed colour → Violet

Hence [Cr(H₂O)₆]³⁺ appears violet because of dd electronic transitions.

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Step 5: Selection Rule

Although the transition is spin allowed (the number of unpaired electrons does not change),

it is Laporte forbidden because the complex has a centre of symmetry.

Due to vibrations of ligands (vibronic coupling), the transition becomes partially allowed, so

the absorption band is weak but visible.

Key Points

Geometry → Octahedral

Metal ion → Cr³⁺

Electronic configuration → d³

Ground state → t₂g³

Excited state → t₂g²e_g¹

Transition → dd transition

Colour → Violet

6(b) Discuss the Orgel Diagram for d¹ and d⁹ in Octahedral and Tetrahedral Complexes

The Orgel diagram is a simple diagram used to understand how the energy levels of d-

orbitals split in weak-field complexes and how electronic transitions occur.

It is mainly used for high-spin complexes.

(i) d¹ Configuration

Examples:

Ti³⁺

V⁴⁺

Only one electron is present in the d-orbitals.

In an Octahedral Complex

Higher Energy

_____

Δo

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_____

t2g ↑

The single electron remains in the lower-energy t₂g orbital.

When light is absorbed:

t2g → eg

Only one electronic transition is possible.

In a Tetrahedral Complex

In tetrahedral geometry the splitting is reversed.

Higher Energy

_____

Δt

_____

e ↑

The electron occupies the lower-energy e orbital.

On absorbing light:

e → t2

Again, only one transition occurs.

(ii) d⁹ Configuration

Example:

Cu²⁺

Electronic arrangement:

Octahedral:

eg ↑↓ ↑

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Δo

t2g ↑↓ ↑↓ ↑↓

There is one empty position (hole) in the e_g level.

Electronic transition mainly occurs between:

t2g → eg

Only one main absorption band is generally observed.

Difference Between Octahedral and Tetrahedral

Octahedral

Tetrahedral

Lower level = t₂g

Lower level = e

Higher level = e_g

Higher level = t₂

Splitting = Δ₀

Splitting = Δₜ

Δ₀ is larger

Δₜ is smaller (≈4/9 Δ₀)

Importance of Orgel Diagram

Predicts electronic transitions.

Explains the colour of coordination compounds.

Helps estimate crystal field splitting.

Useful only for weak-field (high-spin) complexes.

6(c) What is Microstate? Calculate Microstates for p¹ Configuration

A microstate is one possible arrangement of electrons in the available orbitals while

considering both the orbital and spin of each electron.

In simple words, imagine each electron can choose:

Which orbital it occupies.

Whether its spin is up (↑) or down (↓).

Each unique arrangement is one microstate.

p-Orbitals

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A p-subshell contains three orbitals:

px py pz

Each orbital can contain:

↑ (spin +½)

↓ (spin ½)

Therefore,

Each orbital has 2 possible spin states.

Total available positions:

3 orbitals × 2 spins = 6 positions

p¹ Configuration

There is only one electron.

It may occupy any one of the six available positions.

Possible arrangements:

1. px ↑

2. px ↓

3. py ↑

4. py ↓

5. pz ↑

6. pz ↓

Hence,

Total Microstates = 6

Formula for Number of Microstates

Microstates 



 󰇛  󰇜

where:

n = total available electron positions

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r = number of electrons

For p¹:

n = 6

r = 1

Therefore,

Total microstates = 6

Final Revision (Exam Points)

[Cr(H₂O)₆]³⁺ is an octahedral d³ complex. Electrons occupy the t₂g orbitals, and a dd

transition (t₂g → e_g) occurs on absorbing light, giving the complex its violet colour.

Orgel diagrams show how d-orbital energy levels split in weak-field complexes. For

d¹ and d⁹ configurations, there is one principal electronic transition in both

octahedral and tetrahedral geometries, though the order of energy levels differs.

A microstate is a unique arrangement of electrons considering both orbital and spin.

A p¹ configuration has 6 microstates, because one electron can occupy any of the six

available spin-orbital positions.

SECTION-D

7. (a) What are Organometallic Compounds? Which are the different ways to classify

them? Discuss in detail.

(b) What are the applications of organo lithium compounds ?

Ans: (a) Organometallic Compounds

Organometallic compounds are one of the most important topics in chemistry because they

connect organic chemistry (compounds containing carbon) with inorganic chemistry

(metals and their compounds).

Imagine that carbon and a metal become close partners by forming a direct chemical bond.

Such compounds behave differently from ordinary organic or inorganic compounds and are

widely used in industries, medicines, fuel production, and chemical manufacturing.

What are Organometallic Compounds?

An organometallic compound is a chemical compound that contains at least one direct

bond between a carbon atom of an organic group and a metal atom.

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In simple words:

Organic part = Carbon-containing group (like CH₃, C₂H₅, C₆H₅)

Metal part = Lithium (Li), Magnesium (Mg), Zinc (Zn), Mercury (Hg), Iron (Fe), etc.

When carbon is directly attached to the metal, the compound is called an

organometallic compound.

General Formula

R M

Where:

R = Organic group (alkyl or aryl group)

M = Metal

Examples:

CH₃Li (Methyllithium)

C₂H₅MgBr (Ethyl magnesium bromide)

Zn(CH₃)₂ (Dimethyl zinc)

Simple Diagram

Organic Group Metal

CH3 ───────── Li

(Carbon) Direct Bond (Metal)

Organometallic Compound

The direct CarbonMetal bond is the most important feature.

Characteristics of Organometallic Compounds

1. They contain a carbon-metal bond.

2. Most are highly reactive.

3. Many react quickly with water and air.

4. They are powerful reagents in organic synthesis.

5. They help prepare alcohols, acids, medicines, plastics, and many industrial chemicals.

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Classification of Organometallic Compounds

Organometallic compounds can be classified in different ways.

1. Classification Based on the Nature of Carbon-Metal Bond

(i) Ionic Organometallic Compounds

In these compounds, the bond is mainly ionic because the metal is highly electropositive.

Examples:

CH₃Li

C₂H₅Na

Features:

Very reactive

React violently with water

Strong bases

(ii) Covalent Organometallic Compounds

Here the carbon-metal bond is mainly covalent.

Examples:

Zn(CH₃)₂

Hg(CH₃)₂

Features:

Comparatively stable

Less reactive than ionic compounds

(iii) Electron-deficient Organometallic Compounds

These compounds do not have enough electrons for normal bonding.

Example:

Al₂(CH₃)₆

Features:

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Form bridge structures

Common in aluminium compounds

(iv) Transition Metal Organometallic Compounds

These contain transition metals like iron, cobalt, nickel, chromium, etc.

Examples:

Ferrocene

Nickel carbonyl

Features:

Very important in catalysis

Used in industries

2. Classification Based on Type of Metal

Alkali Metal Compounds

Metal = Lithium, Sodium, Potassium

Examples:

CH₃Li

C₂H₅Na

Alkaline Earth Metal Compounds

Metal = Magnesium

Example:

C₂H₅MgBr (Grignard reagent)

Transition Metal Compounds

Examples:

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Ferrocene

Nickel Carbonyl

Main Group Metal Compounds

Examples:

Aluminium compounds

Zinc compounds

Tin compounds

Importance of Organometallic Compounds

They are used in:

Organic synthesis

Pharmaceutical industry

Plastic manufacturing

Fuel production

Polymer industries

Agriculture

Catalysts

Petrochemical industries

Because of these uses, organometallic chemistry is considered one of the most valuable

branches of modern chemistry.

(b) Applications of Organolithium Compounds

Organolithium compounds are organometallic compounds in which carbon is directly

bonded to lithium (Li).

General Formula:

RLi

Examples:

CH₃Li (Methyllithium)

C₂H₅Li (Ethyllithium)

n-Butyllithium

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They are among the strongest reagents used in laboratories and industries.

Structure

CH3 ─── Li

Carbon directly attached

to Lithium

Applications of Organolithium Compounds

1. Preparation of Alcohols

Organolithium compounds react with aldehydes and ketones to produce alcohols.

This reaction is widely used in organic chemistry laboratories.

2. Preparation of Carboxylic Acids

They react with carbon dioxide (CO₂) and form carboxylic acids after hydrolysis.

This is an important industrial method for preparing organic acids.

3. Strong Base in Organic Reactions

Organolithium compounds remove hydrogen atoms from many organic compounds.

Hence, they are used as very strong bases in chemical reactions.

4. Preparation of Pharmaceuticals

Many medicines require organolithium compounds during manufacturing.

They help prepare complex medicinal molecules efficiently.

5. Polymer Industry

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They are used as initiators in polymerization reactions.

This helps manufacture:

Synthetic rubber

Plastics

Special polymers

6. Preparation of Other Organometallic Compounds

Organolithium compounds react with many metal salts to prepare new organometallic

compounds.

Example:

Organozinc compounds

Organocopper compounds

7. Industrial Organic Synthesis

They help manufacture:

Fine chemicals

Agrochemicals

Dyes

Perfumes

Specialty chemicals

8. Research Laboratories

Scientists use organolithium compounds to build new carbon-carbon bonds.

This makes them essential in modern research and advanced organic synthesis.

Advantages of Organolithium Compounds

Very strong nucleophiles.

Highly reactive reagents.

Useful in making complex organic molecules.

Essential for industrial chemical production.

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Important in pharmaceutical and polymer industries.

Precautions

Since organolithium compounds are extremely reactive:

They react rapidly with water and moisture.

They may catch fire on exposure to air.

They are stored under dry nitrogen or argon gas.

Special laboratory techniques are required for handling them safely.

Conclusion

Organometallic compounds are compounds that contain a direct carbon-metal bond,

combining the properties of organic and inorganic chemistry. They are classified based on

the nature of the carbon-metal bond (ionic, covalent, electron-deficient, and transition-

metal compounds) and also according to the type of metal involved. These compounds play

a vital role in chemical industries, catalysis, and organic synthesis.

Among them, organolithium compounds are particularly important because they are highly

reactive and are widely used for preparing alcohols, carboxylic acids, pharmaceuticals,

polymers, and many other valuable chemicals. Their unique reactivity makes them

indispensable tools in both industrial manufacturing and modern chemical research.

8. (a) Discuss the preparation, properties of organo aluminium compounds.

(b) What do you mean by Homogeneous Hydrogenation ? Name the three homogeneous

hydrogenation catalyst used for hydrogenation of alke.

Ans: 8(a) Discuss the Preparation and Properties of Organo Aluminium Compounds

Organo aluminium compounds are chemical compounds in which aluminium (Al) is directly

bonded to carbon (C) atoms of organic groups such as methyl (CH₃), ethyl (C₂H₅), or propyl

(C₃H₇).

The most common example is Triethyl Aluminium (Al(C₂H₅)₃).

These compounds are extremely important in organic chemistry, polymer chemistry, and

industrial manufacturing because they are used as catalysts for making plastics like

polyethylene and polypropylene.

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What are Organo Aluminium Compounds?

The word "Organo" means an organic (carbon-containing) group, while "Aluminium" refers

to the metal aluminium.

So, an organo aluminium compound is simply a compound where aluminium is directly

attached to carbon atoms.

General Formula

AlR₃

Where:

Al = Aluminium

R = Alkyl group (CH₃, C₂H₅, C₃H₇, etc.)

Example:

Al(CH₃)₃ → Trimethyl aluminium

Al(C₂H₅)₃ → Triethyl aluminium

Preparation of Organo Aluminium Compounds

There are mainly two methods of preparing organo aluminium compounds.

1. Reaction of Aluminium with Alkyl Halides

Aluminium reacts with alkyl halides (RX) in the presence of suitable conditions.

Reaction

2Al + 3RX → Al2R6 + 3X

Example

2Al + 3C2H5Cl

↓

Al2(C2H5)6

This product exists as a dimer, meaning two aluminium molecules join together.

Simple Diagram

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Ethyl Chloride

│

▼

Aluminium Metal

│

▼

Triethyl Aluminium

2. Reaction of Grignard Reagent

Grignard reagents react with aluminium chloride.

Reaction

3RMgX + AlCl3

↓

AlR3 + 3MgClX

Example

3C2H5MgBr + AlCl3

↓

Al(C2H5)3

This method is commonly used in laboratories.

Properties of Organo Aluminium Compounds

These compounds possess several unique properties.

1. Colourless Liquid

Most organo aluminium compounds are colourless liquids.

Example:

Triethyl aluminium is a colourless liquid.

2. Highly Reactive

They react very quickly with many substances.

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They react with

Water

Oxygen

Alcohol

Acids

This is why they must be stored carefully.

3. Catch Fire in Air (Pyrophoric Nature)

One of their most important properties is that they ignite automatically when exposed to

air.

Air

│

▼

Triethyl Aluminium

│

▼

󹻦󹻧 Fire

Hence, they are always stored under nitrogen or argon gas.

4. React with Water

Reaction with water is violent.

Al(C2H5)3 + H2O

↓

Aluminium Hydroxide + Ethane Gas

Hydrogen or hydrocarbon gases may also be released depending on the compound.

5. Lewis Acid Nature

Aluminium has an incomplete octet.

Therefore, it accepts an electron pair.

Hence organo aluminium compounds act as Lewis acids.

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This property makes them excellent catalysts.

6. Dimer Formation

Many organo aluminium compounds exist as dimers.

Instead of remaining single molecules, two molecules combine together.

Diagram

C2H5 C2H5

\ /

Al------Al

/ \

C2H5 C2H5

The aluminium atoms are connected through bridging alkyl groups.

7. Catalyst in Polymer Industry

This is their most important industrial property.

They are used with transition metal compounds to prepare ZieglerNatta catalysts, which

help convert small alkene molecules into long polymer chains.

Examples:

Polyethylene

Polypropylene

Uses of Organo Aluminium Compounds

Some important applications are:

Manufacture of polyethylene and polypropylene.

Catalyst in organic synthesis.

Production of pharmaceuticals.

Preparation of alcohols and hydrocarbons.

Used in research laboratories.

Important reagents in petrochemical industries.

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Easy Memory Tip

Remember "FAR CAT"

F → Fire in air (Pyrophoric)

A → Acts as Lewis Acid

R → Reacts with water rapidly

C → Catalyst

A → Aluminium bonded with carbon

T → Used in making plastics

8(b) What is Homogeneous Hydrogenation?

Hydrogenation means adding hydrogen (H₂) to an unsaturated compound such as an

alkene or alkyne to convert it into a saturated compound.

When the catalyst and the reactants are present in the same phase (usually all dissolved in

a liquid solution), the process is called homogeneous hydrogenation.

Simple Definition

Homogeneous hydrogenation is the process of adding hydrogen to an alkene or alkyne

using a catalyst that is present in the same liquid phase as the reactants.

Why is it called "Homogeneous"?

The word homogeneous means "same phase."

Imagine sugar dissolved completely in water. Everything forms one uniform solution.

Similarly,

Alkene + Catalyst + Hydrogen

(All dissolved)

Same Solution

│

▼

Homogeneous Hydrogenation

No solid catalyst is present.

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Example

Ethene reacts with hydrogen.

CH2 = CH2 + H2

│

Homogeneous Catalyst

│

▼

CH3 CH3

(Ethane)

The double bond breaks, and hydrogen atoms are added.

How Does the Process Work?

The catalyst first binds with hydrogen gas.

Then it binds with the alkene.

Finally, hydrogen atoms are transferred to the alkene.

The catalyst is regenerated and can be used again.

Flow Diagram

Hydrogen (H2)

│

▼

Catalyst binds H2

│

▼

Catalyst + Alkene

│

▼

Hydrogen Added

│

▼

Alkane Formed

Advantages of Homogeneous Hydrogenation

High selectivity.

Faster reactions.

Mild temperature and pressure.

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Excellent product purity.

Useful in pharmaceutical and fine chemical industries.

Three Homogeneous Hydrogenation Catalysts

1. Wilkinson's Catalyst

Formula

RhCl(PPh3)3

Contains rhodium.

Most famous homogeneous hydrogenation catalyst.

Used for hydrogenation of alkenes under mild conditions.

2. Crabtree's Catalyst

An iridium-based catalyst.

More reactive than Wilkinson's catalyst.

Can hydrogenate sterically hindered (crowded) alkenes that are difficult to react.

3. RhodiumPhosphine Catalyst

Example:

RhH(CO)(PPh3)3

or other soluble rhodiumphosphine complexes.

These catalysts are widely used in industrial hydrogenation because they are efficient and

highly selective.

Comparison Between Homogeneous and Heterogeneous Hydrogenation

Homogeneous Hydrogenation

Heterogeneous Hydrogenation

Catalyst and reactants are in the same liquid

phase

Catalyst is in a different phase (usually a

solid)

Very selective

Less selective

Easy to control reaction

Difficult to control precisely

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Common catalysts: Wilkinson's, Crabtree's

Common catalysts: Nickel, Palladium,

Platinum

Key Points for Exam

Organo aluminium compounds contain a direct aluminiumcarbon bond and have

the general formula AlR₃.

They are prepared by reaction of aluminium with alkyl halides or Grignard reagents

with aluminium chloride.

Important properties include high reactivity, pyrophoric nature, Lewis acidity,

dimer formation, and use as polymerization catalysts.

Homogeneous hydrogenation is the addition of hydrogen to unsaturated

compounds using a catalyst in the same liquid phase as the reactants.

Three important homogeneous hydrogenation catalysts are Wilkinson's catalyst

(RhCl(PPh₃)₃), Crabtree's catalyst (Ir-based), and rhodiumphosphine catalysts.

These catalysts provide high selectivity and are widely used in organic synthesis and

industry.

This paper has been carefully prepared for educational purposes. If you notice any mistakes or

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