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Concept of pH
pH scale
The concept of pH and pOH plays a key role in acid-base functions. In an aqueous medium, it is usually represented as H₃O⁺.

A naked hydrogen ion has a vanishingly small size ( radius ~10⁻¹³ cm =10⁻¹⁵ Å) and therefore has a very high (charge/radius) ratio(~105 and is expected to be the most effective in polarizing other ions or molecules according to Fagan's rules.

In HO₃⁺ there are assumed to coordinate bonds from water oxygen to the proton, thus giving the proton a helium electronic configuration.

Evidence of pH concept

Perchloric acid (HClO₄) reacts vigorously with water and It gives a series of hydrates which are :
HClO₄ (112°C)

HClO₄, H₂O (+ 50°C)

HClO₄, 2H₂O (- 17.8°C)

HClO₄, 3H₂O (- 37°C )

HClO₄, 3.5H₂O (- 41.4°C)

Of these hydrates the most remarkable is the monohydrate, melting at the much higher temperature than the covalent anhydrous acid. It is very stable and can be heated to around 100⁰C without decomposition.

The monohydrate is about ten times vicious as the anhydrous acid. It has the same crystal lattice as the ionic ammonium perchlorate, showing that it is an ionic compound, [H₃O⁺][ClO₄].

Water as an acid and as a base

We know that water dissociates weakly to H⁺ and OH⁻ ions. Regardless of what other ions are present in water, there will always be an equilibrium between H⁺ and OH⁻ ions.

H₂O ⇄ H⁺ + OH⁻

The proton, however, will be solvated and is usually written as [H₃O⁺]. For simplicity, we will write H⁺ only. The above equilibrium will have its own equilibrium constant:

K = ([H⁺] × [OH⁻])/[H₂O]
or, K × [H₂O] = [H⁺] × [OH⁻]

The square brackets indicate concentrations. Recognizing the fact that in any dilute aqueous solution, the concentration of water molecules (55.5 moles/liter) greatly exceeds that of any other ion, [H₂O] can be taken as a constant. Hence,

K×[H₂O] = Kw = [H⁺]×[OH⁻]

Where Kw is the dissociation constant of water (ionic product of water). The value of H⁺ in pure water has been determined as 10⁻⁷ M so that Kw becomes,

Kw = [H + ] × [OH - ]
= 10⁻⁷ × 10⁻⁷
= 1.0 × 10⁻¹⁴ M
Concept of ionic product of water
Ionic product of water



The above relation tells us that in aqueous solution the concentration of H⁺ and OH⁻ are inversely proportional to each other.

If H⁺ concentration increases 100 fold, that of OH⁻ has to decrease 100 fold to maintain Kw constant.

Dissociation of water into H⁺ and OH⁻

Dissociation of water into H⁺ and OH⁻ ions are an endothermic reaction.

Endothermic reaction

The Reactions in which heat is absorbed by the system from the surroundings are known as endothermic reactions.
H₂O + 13.7 kcal → H⁺ + OH⁻

Le-Chatelier's Principle

If a system is in equilibrium, a change in any factors that determine the condition of equilibrium will cause the equilibrium to shift in such a way as to minimize the effect of this change.

Thus according to Le-Chatelier’s principle, increasing temperature will facilitate dissociation, thus giving higher values of Kw. The value of Kw at 20⁰ C, 25⁰ C, and 60⁰ C are 0.68×10⁻¹⁴, 1.00×10⁻¹⁴, and 9.55×10⁻¹⁴ respectively.

Concept of pH

The dissociation of water, Kw, has such low value that expressing the concentrations of H⁺ and OH⁻ ions of a solution in terms of such low figures is not much convenient and meaningful.

Such expressions necessarily have to involve the negative power of the base 10. Sorensen proposed the use of a term known as pH, defined as:

pH = - log[H⁺] = log(1/[H⁺])

Thus for a solution having H⁺ concentration, 10⁻¹ M has a pH = 1And for a solution having H⁺ concentration, 10⁻¹⁴ M has a pH = 14.

For such solutions having H⁺ concentration in the range of 10⁻¹ M to 10⁻¹⁴ M is more convenient and meaningful to express the acidity in terms of pH rather than H⁺ concentrations.

The use of small fractions or negative exponents can thus be avoided. For monobasic acid molarity and normality are the same while they are different for poly-basic acid.

Thus 0.1 M H₂SO₄ is really 0.2 N H₂SO₄ and the pH of the solution is,

pH = -log[H⁺] = -log(0.2)
= 0.699



It follows from these relations that the lower the pH, the more acidic the solution is. If the acidity of a solution goes down 100 fold its pH goes up by the two units. For example, a solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3.

Calculation of pH

0.002 M HCl solution

Since HCl is a strong electrolyte and completely dissociated.
Thus, [H⁺] = [HCl] = 0.002= 2×10⁻³ M
∴ pH = - log[H⁺]
= - log(2×10⁻³)
= (3 - log2)
= 2.7

0.002 M H₂SO₄ solution

For H₂SO₄, [H⁺] = 2[H₂SO₄]
= 2 × 0.002 = 4 × 10⁻³ M
pH = -log[H⁺] = - log(4×10⁻³)
= (3 - log4)
= 2.4

0.002 M acetic acid solution

For Acetic Acid [H⁺] = √(Ka× [CH₃COOH]
= √(2 × 10-5× 2 × 10⁻³)
= 2 × 10⁻⁴
Thus, pH = - log[H⁺]
= -log(2 × 10⁻⁴)
= (4 - log2)
= 3.7

Concept of pOH

The corresponding expression for the hydroxide ion is,
pOH = - log[OH-]

If the acidity of a solution goes down 100 fold its pH goes up by two units. A solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3. Taking the case of OH⁻ ions, the pOH will go down by two units (from 13 to 11).
The product of [H⁺] and [OH⁻] is 10⁻¹⁴ and that this has to remain constant.
Thus, [H⁺] [OH⁻] = 10⁻¹⁴
or, log[H⁺] [OH⁻] = log10⁻¹⁴
or, log[H⁺] + log[OH⁻] = -14
or, -log[H⁺] - log[OH⁻] = 14

pH + pOH = 14

We have from the definition,
pH = -log[H⁺] and pOH = -log [OH⁻]

Acidic, basic and neutral solution

We can now proceed to differentiate between neutral, acidic or basic on the basis of relative concentrations of H⁺ and OH⁻ ions on the one hand, and on the basis of pH on the other.

pH and pOH of a neutral solution

A neutral solution is one in which the concentrations of H+ and OH- ions are equal.
Thus, [H⁺] = [OH⁻] = 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] = 10⁻⁷ M
or, pH = 7

pH and pOH of acidic solution

[H⁺] 〉[OH⁻]
or, [H⁺] 〉 10⁻⁷ M
and [OH⁻] 〈 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] 〉 10⁻⁷ M
or, pH 〈 7

pH and pOH of basic solution

[H⁺]〈 [OH⁻]
or, [OH⁻] 〉10⁻⁷M
and [H⁺]〈 10⁻⁷M
In terms of pH, we have the following relations,
[H⁺] 〈 10⁻⁷ M
or, pH 〉 7

A mathematical definition of pH provides a negative value when [H⁺] exceeds 1 M. However pH measurements of such concentrated solutions are avoided as these solutions are not likely to be dissociated fully.

The concentration of such strongly acid solutions is best expressed in terms of molarity than in terms of pH.

Problems solutions

Problem

Calculate the [H⁺], [OH⁻] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.

Solution
[OH⁻] = (28×1000)/(56×200)
= 2.5M
(Molecular Weight of KOH = 56 gm)
[H⁺] = (1 × 10⁻¹⁴)/[OH⁻]
= (1 × 10⁻¹⁴)/(2.5)
= 4 × 10⁻¹⁵
pH = - log[H⁺]
= - log4 × 10⁻¹⁵
= (15 - log4)
Problem

Calculate the [H⁺], [OH⁻] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit.

Solution
[H⁺] = (20 × 0.1)/1000 = 0.002 = 2×10⁻³
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
= (1×10⁻¹⁴)/(2×10⁻³)
= 0.5 × 10⁻¹¹
pH = - log[H⁺]
= - log(2 × 10⁻³)
= (3 - log2)
= 2.7
Problem

The pH of a solution is 4.5. Calculate the concentration of H⁺ ion.

Solution
We have from the definition,
pH = - log[H⁺] = 4.5
or, log[H⁺] = - 4.5
∴ [H⁺] = 3.16 × 10⁻⁵
Concept of pH
pH scale
The concept of pH and pOH plays a key role in acid-base functions. In an aqueous medium, it is usually represented as H₃O⁺.

A naked hydrogen ion has a vanishingly small size ( radius ~10⁻¹³ cm =10⁻¹⁵ Å) and therefore has a very high (charge/radius) ratio(~105 and is expected to be the most effective in polarizing other ions or molecules according to Fagan's rules.

In HO₃⁺ there are assumed to coordinate bonds from water oxygen to the proton, thus giving the proton a helium electronic configuration.

Evidence of pH concept

Perchloric acid (HClO₄) reacts vigorously with water and It gives a series of hydrates which are :
HClO₄ (112°C)

HClO₄, H₂O (+ 50°C)

HClO₄, 2H₂O (- 17.8°C)

HClO₄, 3H₂O (- 37°C )

HClO₄, 3.5H₂O (- 41.4°C)

Of these hydrates the most remarkable is the monohydrate, melting at the much higher temperature than the covalent anhydrous acid. It is very stable and can be heated to around 100⁰C without decomposition.

The monohydrate is about ten times vicious as the anhydrous acid. It has the same crystal lattice as the ionic ammonium perchlorate, showing that it is an ionic compound, [H₃O⁺][ClO₄].

Water as an acid and as a base

We know that water dissociates weakly to H⁺ and OH⁻ ions. Regardless of what other ions are present in water, there will always be an equilibrium between H⁺ and OH⁻ ions.

H₂O ⇄ H⁺ + OH⁻

The proton, however, will be solvated and is usually written as [H₃O⁺]. For simplicity, we will write H⁺ only. The above equilibrium will have its own equilibrium constant:

K = ([H⁺] × [OH⁻])/[H₂O]
or, K × [H₂O] = [H⁺] × [OH⁻]

The square brackets indicate concentrations. Recognizing the fact that in any dilute aqueous solution, the concentration of water molecules (55.5 moles/liter) greatly exceeds that of any other ion, [H₂O] can be taken as a constant. Hence,

K×[H₂O] = Kw = [H⁺]×[OH⁻]

Where Kw is the dissociation constant of water (ionic product of water). The value of H⁺ in pure water has been determined as 10⁻⁷ M so that Kw becomes,

Kw = [H + ] × [OH - ]
= 10⁻⁷ × 10⁻⁷
= 1.0 × 10⁻¹⁴ M
Concept of ionic product of water
Ionic product of water



The above relation tells us that in aqueous solution the concentration of H⁺ and OH⁻ are inversely proportional to each other.

If H⁺ concentration increases 100 fold, that of OH⁻ has to decrease 100 fold to maintain Kw constant.

Dissociation of water into H⁺ and OH⁻

Dissociation of water into H⁺ and OH⁻ ions are an endothermic reaction.

Endothermic reaction

The Reactions in which heat is absorbed by the system from the surroundings are known as endothermic reactions.
H₂O + 13.7 kcal → H⁺ + OH⁻

Le-Chatelier's Principle

If a system is in equilibrium, a change in any factors that determine the condition of equilibrium will cause the equilibrium to shift in such a way as to minimize the effect of this change.

Thus according to Le-Chatelier’s principle, increasing temperature will facilitate dissociation, thus giving higher values of Kw. The value of Kw at 20⁰ C, 25⁰ C, and 60⁰ C are 0.68×10⁻¹⁴, 1.00×10⁻¹⁴, and 9.55×10⁻¹⁴ respectively.

Concept of pH

The dissociation of water, Kw, has such low value that expressing the concentrations of H⁺ and OH⁻ ions of a solution in terms of such low figures is not much convenient and meaningful.

Such expressions necessarily have to involve the negative power of the base 10. Sorensen proposed the use of a term known as pH, defined as:

pH = - log[H⁺] = log(1/[H⁺])

Thus for a solution having H⁺ concentration, 10⁻¹ M has a pH = 1And for a solution having H⁺ concentration, 10⁻¹⁴ M has a pH = 14.

For such solutions having H⁺ concentration in the range of 10⁻¹ M to 10⁻¹⁴ M is more convenient and meaningful to express the acidity in terms of pH rather than H⁺ concentrations.

The use of small fractions or negative exponents can thus be avoided. For monobasic acid molarity and normality are the same while they are different for poly-basic acid.

Thus 0.1 M H₂SO₄ is really 0.2 N H₂SO₄ and the pH of the solution is,

pH = -log[H⁺] = -log(0.2)
= 0.699



It follows from these relations that the lower the pH, the more acidic the solution is. If the acidity of a solution goes down 100 fold its pH goes up by the two units. For example, a solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3.

Calculation of pH

0.002 M HCl solution

Since HCl is a strong electrolyte and completely dissociated.
Thus, [H⁺] = [HCl] = 0.002= 2×10⁻³ M
∴ pH = - log[H⁺]
= - log(2×10⁻³)
= (3 - log2)
= 2.7

0.002 M H₂SO₄ solution

For H₂SO₄, [H⁺] = 2[H₂SO₄]
= 2 × 0.002 = 4 × 10⁻³ M
pH = -log[H⁺] = - log(4×10⁻³)
= (3 - log4)
= 2.4

0.002 M acetic acid solution

For Acetic Acid [H⁺] = √(Ka× [CH₃COOH]
= √(2 × 10-5× 2 × 10⁻³)
= 2 × 10⁻⁴
Thus, pH = - log[H⁺]
= -log(2 × 10⁻⁴)
= (4 - log2)
= 3.7

Concept of pOH

The corresponding expression for the hydroxide ion is,
pOH = - log[OH-]

If the acidity of a solution goes down 100 fold its pH goes up by two units. A solution of pH 1 has [H⁺] which is 100 times greater than that of pH 3. Taking the case of OH⁻ ions, the pOH will go down by two units (from 13 to 11).
The product of [H⁺] and [OH⁻] is 10⁻¹⁴ and that this has to remain constant.
Thus, [H⁺] [OH⁻] = 10⁻¹⁴
or, log[H⁺] [OH⁻] = log10⁻¹⁴
or, log[H⁺] + log[OH⁻] = -14
or, -log[H⁺] - log[OH⁻] = 14

pH + pOH = 14

We have from the definition,
pH = -log[H⁺] and pOH = -log [OH⁻]

Acidic, basic and neutral solution

We can now proceed to differentiate between neutral, acidic or basic on the basis of relative concentrations of H⁺ and OH⁻ ions on the one hand, and on the basis of pH on the other.

pH and pOH of a neutral solution

A neutral solution is one in which the concentrations of H+ and OH- ions are equal.
Thus, [H⁺] = [OH⁻] = 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] = 10⁻⁷ M
or, pH = 7

pH and pOH of acidic solution

[H⁺] 〉[OH⁻]
or, [H⁺] 〉 10⁻⁷ M
and [OH⁻] 〈 10⁻⁷ M
In terms of pH, we have the following relations,
[H⁺] 〉 10⁻⁷ M
or, pH 〈 7

pH and pOH of basic solution

[H⁺]〈 [OH⁻]
or, [OH⁻] 〉10⁻⁷M
and [H⁺]〈 10⁻⁷M
In terms of pH, we have the following relations,
[H⁺] 〈 10⁻⁷ M
or, pH 〉 7

A mathematical definition of pH provides a negative value when [H⁺] exceeds 1 M. However pH measurements of such concentrated solutions are avoided as these solutions are not likely to be dissociated fully.

The concentration of such strongly acid solutions is best expressed in terms of molarity than in terms of pH.

Problems solutions

Problem

Calculate the [H⁺], [OH⁻] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.

Solution
[OH⁻] = (28×1000)/(56×200)
= 2.5M
(Molecular Weight of KOH = 56 gm)
[H⁺] = (1 × 10⁻¹⁴)/[OH⁻]
= (1 × 10⁻¹⁴)/(2.5)
= 4 × 10⁻¹⁵
pH = - log[H⁺]
= - log4 × 10⁻¹⁵
= (15 - log4)
Problem

Calculate the [H⁺], [OH⁻] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit.

Solution
[H⁺] = (20 × 0.1)/1000 = 0.002 = 2×10⁻³
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
= (1×10⁻¹⁴)/(2×10⁻³)
= 0.5 × 10⁻¹¹
pH = - log[H⁺]
= - log(2 × 10⁻³)
= (3 - log2)
= 2.7
Problem

The pH of a solution is 4.5. Calculate the concentration of H⁺ ion.

Solution
We have from the definition,
pH = - log[H⁺] = 4.5
or, log[H⁺] = - 4.5
∴ [H⁺] = 3.16 × 10⁻⁵

The f block elements appear in two series characterized by the filling of 4f and 5f orbitals in the respective third inner principal quantum level from outermost.

The 4f series contains fourteen elements cerium to lutetium with the atomic number from 58 to 71 and are called Lanthanides as they appear after lanthanum.

The 5f series contains fourteen elements thorium to lawrencium with the atomic number from 90 to 103 and are called actinides as they appear after actinium.

4f-block elements

The 4f block elements have been variously called rare earth, lanthanoids, and lanthanum. The lanthanide atoms and their trivalent ions have the following general electronic configuration.
Lanthanide atoms
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
where n has values 1 to 14
Lanthanide(M⁺³) ions
[Pd] 4fn 5S² 5P⁶
where n has values 1 to 14
f block elements electronic configuration
4f-block elements
In 4f-block inner transition elements with increasing atomic number electrons are added to the deep-seated 4f orbitals. The outer electronic configuration of 4f-elements is 6S² and inner orbitals contain f -electrons.

Only Cerium, Gadolinium, and Lutetium contain one electron in 5d orbital and the electronic configuration of the following elements are
Cerium
[Pd] 4f¹ 5S² 5P⁶ 5d¹ 6S²

Gadolinium
[Pd] 4f⁷ 5S² 5P⁶ 5d¹ 6S²

Lutetium
[Pd] 4f¹⁴ 5S² 5P⁶ 5d¹ 6S²
Electrons of similar spin developed an exchange interaction which leads to the stabilization of the system. For the electrons of similar spin, repulsion is less by an amount called exchange energy.

The greater the number of electrons with parallel spins the greater is exchanged interaction and the greater is the stability. This the basis of Hund's rules of maximum spin municipality.

For the f subshell, maximum stability will result if there are seven electrons with parallel spins in the seven f orbitals, each orbitals having one when the f subshell is half-filled.

Thus Gadolinium atom contains one electron in 5d orbital. 4f and 5d are very close in terms of energy levels. In such a case, the half-filled orbital is slightly more stable than orbital with one additional electron by increasing exchange energy. Thus when we add the next electron in half-filled 4f-orbital it may land on 5d-orbital.

In the Lutetium atom, the maximum capacity of electrons in 4f orbitals is 14. Thus when we come from Ytterbium to Lutetium the next electron land on 5d orbital.

5f-block elements

The 5f-block elements from thorium to lawrencium from the second series of inner transition elements are called man-made elements or Actinides. The Actinides atoms and their trivalent ions have the following general electronic configuration.
Actinides atoms
[Rn] 4fn 5d1-2 6S²
where n has values 1 to 14
Actinides(M⁺³) ions
[Rn] 4fn
where n has values 1 to 14
f block elements electronic configuration
5f-block elements


Questions answers

Question

Why is the +3 oxidation state so common and stable in lanthanides?

The nature of lanthanoids elements is such that three electrons are removed comparatively easy to give the normal trivalent state.

The ground state electronic configuration of the natural lanthanoids atoms is
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
The electronic configuration of trivalent ions with the reducing three electrons is
[Pd] 4fn 5S² 5P⁶
where n is 0 to 14 from Lanthanum to Lutetium.

As a consequence, the f electrons can not participate in the chemical reactions thus +3 oxidation state is common and stable in lanthanides.

Question

Why does praseodymium possess electronic configuration 4f³ 6s² instead of the expected one 4f² 5d¹ 6s²?

Answer

This can be explained by (n+l) rules, the orbital which has a higher value of (n+l) is the higher energy orbitals.

4f-orbital, (n+l) = 4+3 = 7
5d-orbital, (n+l) = 5+7 =7

Thus for the above case which has the highest number of principal quantum number n is higher energy orbital. Thus 5d orbital is the higher energy orbitals.

Again electrons are fed into orbitals in order of increasing energy until all the electrons have been accommodated.

Thus praseodymium posses electronic configuration 4f³ 6s².
The f block elements appear in two series characterized by the filling of 4f and 5f orbitals in the respective third inner principal quantum level from outermost.

The 4f series contains fourteen elements cerium to lutetium with the atomic number from 58 to 71 and are called Lanthanides as they appear after lanthanum.

The 5f series contains fourteen elements thorium to lawrencium with the atomic number from 90 to 103 and are called actinides as they appear after actinium.

4f-block elements

The 4f block elements have been variously called rare earth, lanthanoids, and lanthanum. The lanthanide atoms and their trivalent ions have the following general electronic configuration.
Lanthanide atoms
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
where n has values 1 to 14
Lanthanide(M⁺³) ions
[Pd] 4fn 5S² 5P⁶
where n has values 1 to 14
f block elements electronic configuration
4f-block elements
In 4f-block inner transition elements with increasing atomic number electrons are added to the deep-seated 4f orbitals. The outer electronic configuration of 4f-elements is 6S² and inner orbitals contain f -electrons.

Only Cerium, Gadolinium, and Lutetium contain one electron in 5d orbital and the electronic configuration of the following elements are
Cerium
[Pd] 4f¹ 5S² 5P⁶ 5d¹ 6S²

Gadolinium
[Pd] 4f⁷ 5S² 5P⁶ 5d¹ 6S²

Lutetium
[Pd] 4f¹⁴ 5S² 5P⁶ 5d¹ 6S²
Electrons of similar spin developed an exchange interaction which leads to the stabilization of the system. For the electrons of similar spin, repulsion is less by an amount called exchange energy.

The greater the number of electrons with parallel spins the greater is exchanged interaction and the greater is the stability. This the basis of Hund's rules of maximum spin municipality.

For the f subshell, maximum stability will result if there are seven electrons with parallel spins in the seven f orbitals, each orbitals having one when the f subshell is half-filled.

Thus Gadolinium atom contains one electron in 5d orbital. 4f and 5d are very close in terms of energy levels. In such a case, the half-filled orbital is slightly more stable than orbital with one additional electron by increasing exchange energy. Thus when we add the next electron in half-filled 4f-orbital it may land on 5d-orbital.

In the Lutetium atom, the maximum capacity of electrons in 4f orbitals is 14. Thus when we come from Ytterbium to Lutetium the next electron land on 5d orbital.

5f-block elements

The 5f-block elements from thorium to lawrencium from the second series of inner transition elements are called man-made elements or Actinides. The Actinides atoms and their trivalent ions have the following general electronic configuration.
Actinides atoms
[Rn] 4fn 5d1-2 6S²
where n has values 1 to 14
Actinides(M⁺³) ions
[Rn] 4fn
where n has values 1 to 14
f block elements electronic configuration
5f-block elements


Questions answers

Question

Why is the +3 oxidation state so common and stable in lanthanides?

The nature of lanthanoids elements is such that three electrons are removed comparatively easy to give the normal trivalent state.

The ground state electronic configuration of the natural lanthanoids atoms is
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
The electronic configuration of trivalent ions with the reducing three electrons is
[Pd] 4fn 5S² 5P⁶
where n is 0 to 14 from Lanthanum to Lutetium.

As a consequence, the f electrons can not participate in the chemical reactions thus +3 oxidation state is common and stable in lanthanides.

Question

Why does praseodymium possess electronic configuration 4f³ 6s² instead of the expected one 4f² 5d¹ 6s²?

Answer

This can be explained by (n+l) rules, the orbital which has a higher value of (n+l) is the higher energy orbitals.

4f-orbital, (n+l) = 4+3 = 7
5d-orbital, (n+l) = 5+7 =7

Thus for the above case which has the highest number of principal quantum number n is higher energy orbital. Thus 5d orbital is the higher energy orbitals.

Again electrons are fed into orbitals in order of increasing energy until all the electrons have been accommodated.

Thus praseodymium posses electronic configuration 4f³ 6s².

different kinds of crystalline solids
Crystalline solids
    Solids are characterized by their definite shape and also their considerable mechanical strength and rigidity. The rigidity due to the absence of translatory motion of the structural units (atoms, ions, etc) of the solids.
    These properties are due to the existence of very strong forces of attraction amongst the molecules or ions. It is because of these strong forces that the structural units (atoms, ions, etc) of the solids do not possess any translatory motion but can only have the vibrational motion about their mean position.
    Liquids can be obtained by heating solids up to or beyond their melting points. In solids, molecules do not possess any translatory energy but posses only vibrational energy. The forces of attraction amongst them are very strong.
    The effect of heating solids is to impart sufficient energy to molecules so that they can overcome these strong forces of attraction. Thus solids are less compressible than liquids and denser than the liquid.
    Solids are generally classified into two broad categories: crystals and amorphous substances.

What are Crystalline solids?

    The solids which posses a definite structure, sharp melting point, and the constituents may be atoms ions or molecules has definite or order arrangement of the constituents extends over a large distance on the solids (long-range order) is called crystalline solids.
    NaCl, KCl, Sugar, and Ice, quartz, etc are the examples of crystalline solids possess a sharp melting point. The pattern is such that having observed it in some small region of the crystal, it is possible to predict accurately the position of the particle in any region under observation.

Characteristics of crystalline solids

In the crystalline solid the constituents may be atoms ions or molecule.

  1. Crystalline solids are a sharp melting point, flat faces and sharp edges which is a well-developed form, are usually arranged symmetrically.
  2. Crystalline solids are definite and the ordered arrangement of the constituents extends over a large distance in the crystal and is called the long-range order.
  3. Crystalline solids (other than those belonging to the cubic class), on the other hand, are enantiotropic in nature. In this case, the magnitude of the property depends on the direction along which it is measured.

Amorphous solids

    The solids which do not possess a definite structure, sharp melting point, and the constituents may be atoms ions or molecules do not have definite or order arrangement of the constituents extends over a short distance on the solids is called amorphous solids.
    Amorphous solids such as glass, pitch, rubber, plastics, etc although possessing many characteristics of crystalline solids such as definite shape rigidity and hardness, do not have this ordered arrangement and melt gradually over a range of temperature.
    For this reason, they are not considered as solids but rather highly supercooled liquids.

Difference between crystalline and amorphous solids

  1. Crystalline solids possess definite structure, sharp melting point but amorphous solids that do not possess a definite structure, sharp melting point.
  2. In crystalline solids constituents(atoms ions or molecules) have definite or order arrangement of the constituents extends over a large distance on the solids (long-range order) but amorphous solids the constituents may be atoms ions or molecules do not have definite or order arrangement.

Classification of crystalline solids

    On the basis of nature of force operating between constituent particles(atoms, ions or molecules) of matter, crystalline solids are classified into four categories,

Molecular crystalline solids

    Forces that hold the constituents of molecular crystals are of Van der Waals types. These are weaker forces because of which molecular crystals are soft and possess low melting points.
    CO₂, CCl₄, Ar, and most of the organic compounds are examples of these types of crystals.
    This class further classified into three category
  1. Non-polar molecular crystalline solids
    The constituent particles of these types of crystalline solids are non-directional atom(H, He, etc.) or non-polar molecules(H₂, O₂, Cl₂, CO₂, CH₄, etc.). And the force operating between constituent particles(atoms or molecules) is a weak London force of attraction.
  2. Polar molecular crystalline solids
    The constituent particle of this type of crystalline solids is polar molecules(SO₂, NH₃, etc.)  and the force operating between constituent particles is the dipole-dipole attraction force.
  3. H-bonded molecular crystalline solids
    The constituent molecule of these types of crystalline solids is polar molecule and these molecules are bounded each other by hydrogen bonding. Ice is an example of this type of crystal.

Ionic crystalline solids

    The forces involved here are of electrostatic forces of attraction. These are stronger than the non-directional type. Therefore ionic crystals strong and likely to be brittle.
    They have little electricity with high melting and boiling point and can not be bent. The melting point of the ionic crystal increases with the decreasing size of the constituent particles.
    In ionic crystals, some of the atoms may be held together by covalent bonds to form ions having a definite position and orientation in the crystal lattice. CaCO₃ is an example of these types of crystalline solids.

Covalent crystalline solids

    The forces involved here are chemical nature (covalent bonds) extended in three dimensions. They are strong and consequently, the crystals are strong and hard with high melting points. Diamond, graphite, silicon, etc. are examples of these types of crystalline solid.

Metallic crystalline solids

    Electrons are held loosely in these types of crystals. Therefore they are good conductors of electricity. Metallic crystalline solids can be bent and are also strong.
    Since the forces have non-directional characteristics the arrangement ao atoms frequently correspond to the closet packing of the sphere.

Isotropic crystalline solids

    Carbon has several crystalline isotropic forms only two of them are common diamond and graphite. There are four other rare and poorly understood allotropes, β-graphite, Lonsdaleite or hexagonal diamond, Chaoite (a very rare mineral) and carbon VI.
    The last two forms appear to contain -C≡C-C≡C- and are closer to the diamond in their properties. 

Structure of graphite

    The various amorphous forms of carbon like carbon black, soot, etc. are all microcrystalline forms of graphite.
    Graphite consists of a layer structure in each layer the C-atoms are arranged in hexagonal planner arrangement with SP² hybridized with three sigma bonds to three neighbors and one π-bonds to one neighbor.
    The resonance between structures having an alternative mode of π bonding makes all C-C bonds equal, 114.5 pm equal, consistent with a bond order of 1.33. 
    The π electrons are responsible for the electrical conductivity of graphite. Successive layers of C-atoms are held by weak van der Waals forces at the separation of 335pm and can easily slide over one another.

Structure of diamond

    In diamond, each SP³ hybridized carbon is tetrahedrally surrounded by four other carbon atoms with C-C bond distance 154 pm. These tetrahedral belong to the cubic unit cell.
    Natural diamond commonly contains traces of nitrogen or sometimes very rarely through traces of al in blue diamonds.
    An extremely rare Lonsdalete allotrope found in certain meteorites, the tetrahedral units are stacked to form a hexagonal wurtzite types lattice.
different kinds of crystalline solids
Crystalline solids
    Solids are characterized by their definite shape and also their considerable mechanical strength and rigidity. The rigidity due to the absence of translatory motion of the structural units (atoms, ions, etc) of the solids.
    These properties are due to the existence of very strong forces of attraction amongst the molecules or ions. It is because of these strong forces that the structural units (atoms, ions, etc) of the solids do not possess any translatory motion but can only have the vibrational motion about their mean position.
    Liquids can be obtained by heating solids up to or beyond their melting points. In solids, molecules do not possess any translatory energy but posses only vibrational energy. The forces of attraction amongst them are very strong.
    The effect of heating solids is to impart sufficient energy to molecules so that they can overcome these strong forces of attraction. Thus solids are less compressible than liquids and denser than the liquid.
    Solids are generally classified into two broad categories: crystals and amorphous substances.

What are Crystalline solids?

    The solids which posses a definite structure, sharp melting point, and the constituents may be atoms ions or molecules has definite or order arrangement of the constituents extends over a large distance on the solids (long-range order) is called crystalline solids.
    NaCl, KCl, Sugar, and Ice, quartz, etc are the examples of crystalline solids possess a sharp melting point. The pattern is such that having observed it in some small region of the crystal, it is possible to predict accurately the position of the particle in any region under observation.

Characteristics of crystalline solids

In the crystalline solid the constituents may be atoms ions or molecule.

  1. Crystalline solids are a sharp melting point, flat faces and sharp edges which is a well-developed form, are usually arranged symmetrically.
  2. Crystalline solids are definite and the ordered arrangement of the constituents extends over a large distance in the crystal and is called the long-range order.
  3. Crystalline solids (other than those belonging to the cubic class), on the other hand, are enantiotropic in nature. In this case, the magnitude of the property depends on the direction along which it is measured.

Amorphous solids

    The solids which do not possess a definite structure, sharp melting point, and the constituents may be atoms ions or molecules do not have definite or order arrangement of the constituents extends over a short distance on the solids is called amorphous solids.
    Amorphous solids such as glass, pitch, rubber, plastics, etc although possessing many characteristics of crystalline solids such as definite shape rigidity and hardness, do not have this ordered arrangement and melt gradually over a range of temperature.
    For this reason, they are not considered as solids but rather highly supercooled liquids.

Difference between crystalline and amorphous solids

  1. Crystalline solids possess definite structure, sharp melting point but amorphous solids that do not possess a definite structure, sharp melting point.
  2. In crystalline solids constituents(atoms ions or molecules) have definite or order arrangement of the constituents extends over a large distance on the solids (long-range order) but amorphous solids the constituents may be atoms ions or molecules do not have definite or order arrangement.

Classification of crystalline solids

    On the basis of nature of force operating between constituent particles(atoms, ions or molecules) of matter, crystalline solids are classified into four categories,

Molecular crystalline solids

    Forces that hold the constituents of molecular crystals are of Van der Waals types. These are weaker forces because of which molecular crystals are soft and possess low melting points.
    CO₂, CCl₄, Ar, and most of the organic compounds are examples of these types of crystals.
    This class further classified into three category
  1. Non-polar molecular crystalline solids
    The constituent particles of these types of crystalline solids are non-directional atom(H, He, etc.) or non-polar molecules(H₂, O₂, Cl₂, CO₂, CH₄, etc.). And the force operating between constituent particles(atoms or molecules) is a weak London force of attraction.
  2. Polar molecular crystalline solids
    The constituent particle of this type of crystalline solids is polar molecules(SO₂, NH₃, etc.)  and the force operating between constituent particles is the dipole-dipole attraction force.
  3. H-bonded molecular crystalline solids
    The constituent molecule of these types of crystalline solids is polar molecule and these molecules are bounded each other by hydrogen bonding. Ice is an example of this type of crystal.

Ionic crystalline solids

    The forces involved here are of electrostatic forces of attraction. These are stronger than the non-directional type. Therefore ionic crystals strong and likely to be brittle.
    They have little electricity with high melting and boiling point and can not be bent. The melting point of the ionic crystal increases with the decreasing size of the constituent particles.
    In ionic crystals, some of the atoms may be held together by covalent bonds to form ions having a definite position and orientation in the crystal lattice. CaCO₃ is an example of these types of crystalline solids.

Covalent crystalline solids

    The forces involved here are chemical nature (covalent bonds) extended in three dimensions. They are strong and consequently, the crystals are strong and hard with high melting points. Diamond, graphite, silicon, etc. are examples of these types of crystalline solid.

Metallic crystalline solids

    Electrons are held loosely in these types of crystals. Therefore they are good conductors of electricity. Metallic crystalline solids can be bent and are also strong.
    Since the forces have non-directional characteristics the arrangement ao atoms frequently correspond to the closet packing of the sphere.

Isotropic crystalline solids

    Carbon has several crystalline isotropic forms only two of them are common diamond and graphite. There are four other rare and poorly understood allotropes, β-graphite, Lonsdaleite or hexagonal diamond, Chaoite (a very rare mineral) and carbon VI.
    The last two forms appear to contain -C≡C-C≡C- and are closer to the diamond in their properties. 

Structure of graphite

    The various amorphous forms of carbon like carbon black, soot, etc. are all microcrystalline forms of graphite.
    Graphite consists of a layer structure in each layer the C-atoms are arranged in hexagonal planner arrangement with SP² hybridized with three sigma bonds to three neighbors and one π-bonds to one neighbor.
    The resonance between structures having an alternative mode of π bonding makes all C-C bonds equal, 114.5 pm equal, consistent with a bond order of 1.33. 
    The π electrons are responsible for the electrical conductivity of graphite. Successive layers of C-atoms are held by weak van der Waals forces at the separation of 335pm and can easily slide over one another.

Structure of diamond

    In diamond, each SP³ hybridized carbon is tetrahedrally surrounded by four other carbon atoms with C-C bond distance 154 pm. These tetrahedral belong to the cubic unit cell.
    Natural diamond commonly contains traces of nitrogen or sometimes very rarely through traces of al in blue diamonds.
    An extremely rare Lonsdalete allotrope found in certain meteorites, the tetrahedral units are stacked to form a hexagonal wurtzite types lattice.

    Le-Chatelier principle predicts quantitatively the effect on the system at equilibrium when some of the variables such as temperature, pressure, and concentration of the equilibrium of a chemical reaction.

Le-Chatelier principle definition

    If a system at equilibrium is subjected to change, the system will react in such a way so as to oppose or reduce the change if this is possible that is the system tends to balance or counteract the effects of any imposed stress.

Effect of pressure

    According to the Le-Chatelier principle with the increase of pressure, the reaction will shift in a direction where the no of moles is reduced thus the system will try to lower the pressure.
N₂ + 3 H₂ ⇆ 2 NH₃
    The increase of pressure is to shift the reaction in a direction where the sum of the stoichiometric number of gaseous molecules is lowered thus lowering the pressure.
    In other words, an increase in pressure shifts the equilibrium to the low volume side of the rection whereas a decrease of pressure shifts it to the high volume side.

Effect of temperature

    According to the Le-Chatelier principle with the increase of temperature, the equilibrium will shift in the endothermic direction that is shifted to the high enthalpy side.
    If the reaction proceeds from low enthalpy side to high enthalpy side heat is absorbed and it is for this reason this direction is known as endothermic direction.
N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal
where ΔH = ΣHproduct - ΣHreactant
    Thus the enthalpy of the reactants in the above reaction higher than that of the products. Thus with the increases in temperature backward reaction favors and thus the equilibrium shifted to the higher enthalpy side and the production of ammonia is decreased.
    With the decrease of temperature, the equilibrium will shift in the exothermic direction that is shifted to the low enthalpy side.
    If the reaction proceeds from high enthalpy side to low enthalpy side heat is released and it is for this reason this direction is known as exothermic direction.
    Thus with the decreases in temperature forward reaction favors and thus the equilibrium shifted to the low enthalpy side and the production of ammonia is increased.
N₂ + O₂ ⇆ 2 NO ΔH = +44 kcal
    Enthalpy of the reactants in the above reaction lower than that of the products. Thus with the increases in temperature forward reaction favors and thus the equilibrium shifted to the lower enthalpy side and the production of NO- is increasing.
    Decreases in temperature for the above reaction backward reaction favors and the equilibrium shifted to the high enthalpy side and the production of NO is decreased.

Addition of inert gas

    Addition of inert gas(He, Ne, Ar, etc) at constant temperature by two way
Constant volume
    The addition of inert gas at constant volume can not affect the equilibrium since the concentration of the total reacting components remain unchanged.
Constant pressure
    When inert gas is added to the system at equilibrium at constant pressure the volume of the reacting system is increased and thereby total concentration is decreased.
    According to the Le-Chatelier principle, the system will move in the direction in which no of moles is increases and thereby the concentration of the system is also increased.


Le-Chatelier principle and addition of inert gas on equilibrium
Le-Chatelier principle

Effect of catalyst on equilibrium

    Catalyst can speed up the reaction it does not affect the equilibrium of the reaction. A reversible reaction, the equilibrium state is one in which the forward and reverse reaction rates are equal.
    The presence of a catalyst, speeds up both the forward and backward reaction, thereby allowing the system to reach equilibrium faster.
    This is very important that the addition of a catalyst has no effect on the final equilibrium position of the reaction. Thus we can not increases the production of the product.
    Catalysts can be lowering the transition state and the reaction proceeds faster rate. It can be lowering the energy of the transition state(rate-limiting step), catalysts reduce the required energy of activation to allow the reaction proceeds faster rate and reach equilibrium more rapidly.

Properties of the reacting system

    The le-chatelier principle provides the reacting system with some special features.
  1. For example, if the volume of the nonreactive system is decreased by a specific amount the pressure rises correspondingly.
    In the reactive system, the equilibrium shifted to the low volume sides (if ΔV ≠ 0), so the pressure increases become less than in the non-reactive system.
    The response of the system is moderate in the shift in the equilibrium position makes the reactive system higher compressibility than the non-reactive one.
  2. Similarly, if the fixed quantity of the heat is supplied to the non-reacting system temperature is corresponding increases.
    In a reactive system, such amount of heat supplied does not increase the temperature so much as the non-reacting system. Since the equilibrium is a shift to the higher enthalpy side and the temperature is less increased.
    This shift of equilibrium makes the heat capacity much higher than the non-reactive system. This is useful since the reacting system is chosen as a heat storage medium.

Question answers

Problem
    Why does vapour pressure of a liquid decrease with the addition of a nonvolatile solid solute?
Answer
    The pure solve is the mole faction x1 = 1 but when the non-volatile solute is added to the solvent the mole fraction of the solvent is decreased from 1 that is x1 ㄑ1.
    To reduce the effect according to the Le-Chatelier principle the solvent is less vapourised and the mole fraction of the solvent in a solution is thus improved. Thus there occurring a lowering of vapour pressure.
Problem
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal what would most likely happen if increasing the pressure of the reaction?
Answer
    According to the Le-Chatelier principle yield of NH₃ is increased if pressure is increasing.
Question
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal equilibrium shifted to forward direction when
(A) the concentration of NH3 increases
(B) pressure is decreasing
(C) the concentration of N2 and H2 decrease
(D) pressure increases and temperature decreases
Answer
    (D) pressure increases and temperature decreases
    Le-Chatelier principle predicts quantitatively the effect on the system at equilibrium when some of the variables such as temperature, pressure, and concentration of the equilibrium of a chemical reaction.

Le-Chatelier principle definition

    If a system at equilibrium is subjected to change, the system will react in such a way so as to oppose or reduce the change if this is possible that is the system tends to balance or counteract the effects of any imposed stress.

Effect of pressure

    According to the Le-Chatelier principle with the increase of pressure, the reaction will shift in a direction where the no of moles is reduced thus the system will try to lower the pressure.
N₂ + 3 H₂ ⇆ 2 NH₃
    The increase of pressure is to shift the reaction in a direction where the sum of the stoichiometric number of gaseous molecules is lowered thus lowering the pressure.
    In other words, an increase in pressure shifts the equilibrium to the low volume side of the rection whereas a decrease of pressure shifts it to the high volume side.

Effect of temperature

    According to the Le-Chatelier principle with the increase of temperature, the equilibrium will shift in the endothermic direction that is shifted to the high enthalpy side.
    If the reaction proceeds from low enthalpy side to high enthalpy side heat is absorbed and it is for this reason this direction is known as endothermic direction.
N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal
where ΔH = ΣHproduct - ΣHreactant
    Thus the enthalpy of the reactants in the above reaction higher than that of the products. Thus with the increases in temperature backward reaction favors and thus the equilibrium shifted to the higher enthalpy side and the production of ammonia is decreased.
    With the decrease of temperature, the equilibrium will shift in the exothermic direction that is shifted to the low enthalpy side.
    If the reaction proceeds from high enthalpy side to low enthalpy side heat is released and it is for this reason this direction is known as exothermic direction.
    Thus with the decreases in temperature forward reaction favors and thus the equilibrium shifted to the low enthalpy side and the production of ammonia is increased.
N₂ + O₂ ⇆ 2 NO ΔH = +44 kcal
    Enthalpy of the reactants in the above reaction lower than that of the products. Thus with the increases in temperature forward reaction favors and thus the equilibrium shifted to the lower enthalpy side and the production of NO- is increasing.
    Decreases in temperature for the above reaction backward reaction favors and the equilibrium shifted to the high enthalpy side and the production of NO is decreased.

Addition of inert gas

    Addition of inert gas(He, Ne, Ar, etc) at constant temperature by two way
Constant volume
    The addition of inert gas at constant volume can not affect the equilibrium since the concentration of the total reacting components remain unchanged.
Constant pressure
    When inert gas is added to the system at equilibrium at constant pressure the volume of the reacting system is increased and thereby total concentration is decreased.
    According to the Le-Chatelier principle, the system will move in the direction in which no of moles is increases and thereby the concentration of the system is also increased.


Le-Chatelier principle and addition of inert gas on equilibrium
Le-Chatelier principle

Effect of catalyst on equilibrium

    Catalyst can speed up the reaction it does not affect the equilibrium of the reaction. A reversible reaction, the equilibrium state is one in which the forward and reverse reaction rates are equal.
    The presence of a catalyst, speeds up both the forward and backward reaction, thereby allowing the system to reach equilibrium faster.
    This is very important that the addition of a catalyst has no effect on the final equilibrium position of the reaction. Thus we can not increases the production of the product.
    Catalysts can be lowering the transition state and the reaction proceeds faster rate. It can be lowering the energy of the transition state(rate-limiting step), catalysts reduce the required energy of activation to allow the reaction proceeds faster rate and reach equilibrium more rapidly.

Properties of the reacting system

    The le-chatelier principle provides the reacting system with some special features.
  1. For example, if the volume of the nonreactive system is decreased by a specific amount the pressure rises correspondingly.
    In the reactive system, the equilibrium shifted to the low volume sides (if ΔV ≠ 0), so the pressure increases become less than in the non-reactive system.
    The response of the system is moderate in the shift in the equilibrium position makes the reactive system higher compressibility than the non-reactive one.
  2. Similarly, if the fixed quantity of the heat is supplied to the non-reacting system temperature is corresponding increases.
    In a reactive system, such amount of heat supplied does not increase the temperature so much as the non-reacting system. Since the equilibrium is a shift to the higher enthalpy side and the temperature is less increased.
    This shift of equilibrium makes the heat capacity much higher than the non-reactive system. This is useful since the reacting system is chosen as a heat storage medium.

Question answers

Problem
    Why does vapour pressure of a liquid decrease with the addition of a nonvolatile solid solute?
Answer
    The pure solve is the mole faction x1 = 1 but when the non-volatile solute is added to the solvent the mole fraction of the solvent is decreased from 1 that is x1 ㄑ1.
    To reduce the effect according to the Le-Chatelier principle the solvent is less vapourised and the mole fraction of the solvent in a solution is thus improved. Thus there occurring a lowering of vapour pressure.
Problem
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal what would most likely happen if increasing the pressure of the reaction?
Answer
    According to the Le-Chatelier principle yield of NH₃ is increased if pressure is increasing.
Question
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal equilibrium shifted to forward direction when
(A) the concentration of NH3 increases
(B) pressure is decreasing
(C) the concentration of N2 and H2 decrease
(D) pressure increases and temperature decreases
Answer
    (D) pressure increases and temperature decreases

Bohr's atomic theory

    According to Rutherford's atomic theory, the entire mass of an atom was concentrated in a tiny positively charged nucleus around which the extranuclear electrons are moving in constant motion in the orbit.
    But a moving charged particle emits radiation, will then loss kinetic energy and eventually hit the nucleus.
    To resolve the anomalous position Bohr's atomic theory was proposed which is based on several novel postulates.
Atomic theory questions and answers
Atomic theory

    The following formulas will be widely used in solving the questions of Bohr's atomic theory.
Question
    The energy of an electron in the first Bohr orbit of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?
Answer
The energy of an electron in nth orbit,
En = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²
= - 30.6 eV
Question
    What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Å.
Answer

The radius of an electron in nth orbit,
rn = r₁ × n²

For second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,
r₂ = 0.529 × 2²
= 2.12 Å

Quantum number orbitals

    For the size shape and orientation of an orbital, we need to know about the four quantum numbers and these quantum numbers are designed as, principal quantum number(n), azimuthal quantum number(l), magnetic quantum number(m) and spin quantum number(s).
    The following rules will be widely used in solving the questions of quantum number orbitals.
Principal quantum number is denoted by n

It can have integral values from 1 to ∞

The azimuthal quantum number denoted by l
For a given value of n

l = 0, 1, 2, 3, ........ (n -1)

Total number of different values of l equal to n

The magnetic quantum number is denoted by ml
For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)
Question
    Write the correct set of four quantum numbers for the valence electron of Rubidium.
Answer
    The atomic number of Rubidium atom is 37. Thus the electronic configuration of rubidium is,
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹
or, [Kr]³⁶ 5S¹
    So the valence electron of rubidium means 5S¹ electron of a rubidium atom. Thus for 5S¹ electron, principal quantum number = 5, azimuthal quantum number = 0, magnetic quantum number = 0, and spin quantum number = +½.
    So the correct set of four quantum numbers for the valence electron of Rubidium atom are,
5, 0, 0, +½
Question
    What are the four quantum numbers of the 19th electron of Cr atom?
Answer
    The four quantum numbers of the 19th electron of Cr atom are 4, 0, 0, +½
Question
    How many electrons in an atom can have the following quantum numbers n=4 and l=1?
Answer
    n = 4 means principal quantum number 4 and l = 1 means the azimuthal quantum number is 1, that is P orbital.
    Thus when n =1 and l = 0 we can describe 4P orbital. For P orbital magnetic quantum numbers = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.
    So (2 × 3) = 6 electrons in an atom can have the following quantum numbers n =4 and l =1.
Question
    How many possible orbitals are there for n = 4?
Answer
    n = 4 means principal quantum number 4.
    Thus when n =4, l = 0, 1, 2, 3. That is 4S, 4P, 4d, and 4f orbitals. S orbital has one orientation of space, p orbitals have three orientation in space, d orbitals have five orientation in space, and f orbitals have seven orientation in space. 
    Thus the number of possible orbitals when n = 4 is (1 + 3 + 5 + 7) = 16
Question
    How many possible orbitals are there for n = 3, l =1, and ml =0 ?
Answer
    n = 3 means principal quantum number 3, l = 1 means azimuthal quantum number = 0, and ml = 0 means magnetic quantum number zero that is one orientation in space.
    Thus the number of orbitals is 1, 3S orbital.
Question
    An electron with the highest value of the magnetic quantum number as 3, what is its principal quantum number?
Answer
    If the highest value of the magnetic quantum number is 3, then the highest value of the azimuthal quantum number is 3.
    Thus the principal quantum number of this orbital is (l+1) = 3+1 = 4.
Question
    What is the wavelength of the Hɑ line of the Balmer series of hydrogen?
Answer
    The wavenumber of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Again for Hɑ line n2 = 3
Thus the wavelength for Hɑ - line, 1/λ = 109677 [(1/2²) - (1/3²)]
= 109677 × (5/36)
∴ λ = 36/(5 × 109677)
= 6.564 × 10⁻⁵ cm
= 656.5 nm
Question
    The lowest wavelength of the Lyman series of the hydrogen atom is x. What is the highest wavelength of the Paschen series of the He⁺² ion?
Answer
    The highest wavenumber of the Lyman lines, ⊽max = 1/λmin = R[(1/1²) - (1/∞²)]
or, 1/x = R
or, R = 1/x
    The lowest wavenumber of the Paschen series of He⁺² ion is,
⊽min = 1/λmax = R Z² [(1/3²) - (1/4²)]
or, λmax = (9 × 16)/(4 × 7R)
∴ λmax = 36x/7
Question
    How many photons of light having the wavenumber a is necessary to provide 3 J of energy?
Answer
    From the plank theory,
E = nhν
Where n is the number of photons and E is the energy of the photon source.
3 = n h c ⊽ where ⊽ = wave number
∴ n = 3/hca
Question
    What is the De-Broglie wavelength of an electron orbiting in the n = 6 state of a hydrogen atom? (r₀ = Bohr's radius)
Answer
We know that 2πr = nλ

or, λ = 2πr/n
where r = 6² × r₀ =36 r₀
∴ λ = (2π ×36 r₀)/6 = 12πr₀
Question
    Find out the number of unpaired electrons of an ion M+x(Z of M = 25) with the magnetic moment √15 B. M. and also find out the value of x?
Answer
We know that magnetic moment = √n(n+2)
Thus √15 = √n(n+2)
or, √3(3+2) = √n(n+2)
∴ n = 3
Again the electronic configuration of M is [Ar]¹⁸ 4S² 3d⁵
    We find out that the number of unpaired electrons of M+x ion is 3. For this electronic configuration, we need to remove four electrons from its valence shell.
Thus the electronic configuration of M is
[Ar]¹⁸ 3d³
Thus x = 4

Bohr's atomic theory

    According to Rutherford's atomic theory, the entire mass of an atom was concentrated in a tiny positively charged nucleus around which the extranuclear electrons are moving in constant motion in the orbit.
    But a moving charged particle emits radiation, will then loss kinetic energy and eventually hit the nucleus.
    To resolve the anomalous position Bohr's atomic theory was proposed which is based on several novel postulates.
Atomic theory questions and answers
Atomic theory

    The following formulas will be widely used in solving the questions of Bohr's atomic theory.
Question
    The energy of an electron in the first Bohr orbit of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?
Answer
The energy of an electron in nth orbit,
En = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²
= - 30.6 eV
Question
    What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Å.
Answer

The radius of an electron in nth orbit,
rn = r₁ × n²

For second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,
r₂ = 0.529 × 2²
= 2.12 Å

Quantum number orbitals

    For the size shape and orientation of an orbital, we need to know about the four quantum numbers and these quantum numbers are designed as, principal quantum number(n), azimuthal quantum number(l), magnetic quantum number(m) and spin quantum number(s).
    The following rules will be widely used in solving the questions of quantum number orbitals.
Principal quantum number is denoted by n

It can have integral values from 1 to ∞

The azimuthal quantum number denoted by l
For a given value of n

l = 0, 1, 2, 3, ........ (n -1)

Total number of different values of l equal to n

The magnetic quantum number is denoted by ml
For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)
Question
    Write the correct set of four quantum numbers for the valence electron of Rubidium.
Answer
    The atomic number of Rubidium atom is 37. Thus the electronic configuration of rubidium is,
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹
or, [Kr]³⁶ 5S¹
    So the valence electron of rubidium means 5S¹ electron of a rubidium atom. Thus for 5S¹ electron, principal quantum number = 5, azimuthal quantum number = 0, magnetic quantum number = 0, and spin quantum number = +½.
    So the correct set of four quantum numbers for the valence electron of Rubidium atom are,
5, 0, 0, +½
Question
    What are the four quantum numbers of the 19th electron of Cr atom?
Answer
    The four quantum numbers of the 19th electron of Cr atom are 4, 0, 0, +½
Question
    How many electrons in an atom can have the following quantum numbers n=4 and l=1?
Answer
    n = 4 means principal quantum number 4 and l = 1 means the azimuthal quantum number is 1, that is P orbital.
    Thus when n =1 and l = 0 we can describe 4P orbital. For P orbital magnetic quantum numbers = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.
    So (2 × 3) = 6 electrons in an atom can have the following quantum numbers n =4 and l =1.
Question
    How many possible orbitals are there for n = 4?
Answer
    n = 4 means principal quantum number 4.
    Thus when n =4, l = 0, 1, 2, 3. That is 4S, 4P, 4d, and 4f orbitals. S orbital has one orientation of space, p orbitals have three orientation in space, d orbitals have five orientation in space, and f orbitals have seven orientation in space. 
    Thus the number of possible orbitals when n = 4 is (1 + 3 + 5 + 7) = 16
Question
    How many possible orbitals are there for n = 3, l =1, and ml =0 ?
Answer
    n = 3 means principal quantum number 3, l = 1 means azimuthal quantum number = 0, and ml = 0 means magnetic quantum number zero that is one orientation in space.
    Thus the number of orbitals is 1, 3S orbital.
Question
    An electron with the highest value of the magnetic quantum number as 3, what is its principal quantum number?
Answer
    If the highest value of the magnetic quantum number is 3, then the highest value of the azimuthal quantum number is 3.
    Thus the principal quantum number of this orbital is (l+1) = 3+1 = 4.
Question
    What is the wavelength of the Hɑ line of the Balmer series of hydrogen?
Answer
    The wavenumber of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Again for Hɑ line n2 = 3
Thus the wavelength for Hɑ - line, 1/λ = 109677 [(1/2²) - (1/3²)]
= 109677 × (5/36)
∴ λ = 36/(5 × 109677)
= 6.564 × 10⁻⁵ cm
= 656.5 nm
Question
    The lowest wavelength of the Lyman series of the hydrogen atom is x. What is the highest wavelength of the Paschen series of the He⁺² ion?
Answer
    The highest wavenumber of the Lyman lines, ⊽max = 1/λmin = R[(1/1²) - (1/∞²)]
or, 1/x = R
or, R = 1/x
    The lowest wavenumber of the Paschen series of He⁺² ion is,
⊽min = 1/λmax = R Z² [(1/3²) - (1/4²)]
or, λmax = (9 × 16)/(4 × 7R)
∴ λmax = 36x/7
Question
    How many photons of light having the wavenumber a is necessary to provide 3 J of energy?
Answer
    From the plank theory,
E = nhν
Where n is the number of photons and E is the energy of the photon source.
3 = n h c ⊽ where ⊽ = wave number
∴ n = 3/hca
Question
    What is the De-Broglie wavelength of an electron orbiting in the n = 6 state of a hydrogen atom? (r₀ = Bohr's radius)
Answer
We know that 2πr = nλ

or, λ = 2πr/n
where r = 6² × r₀ =36 r₀
∴ λ = (2π ×36 r₀)/6 = 12πr₀
Question
    Find out the number of unpaired electrons of an ion M+x(Z of M = 25) with the magnetic moment √15 B. M. and also find out the value of x?
Answer
We know that magnetic moment = √n(n+2)
Thus √15 = √n(n+2)
or, √3(3+2) = √n(n+2)
∴ n = 3
Again the electronic configuration of M is [Ar]¹⁸ 4S² 3d⁵
    We find out that the number of unpaired electrons of M+x ion is 3. For this electronic configuration, we need to remove four electrons from its valence shell.
Thus the electronic configuration of M is
[Ar]¹⁸ 3d³
Thus x = 4

Alkenes nomenclature
Alkenes nomenclature

Unsaturated hydrocarbon alkenes

    The compounds contain at least one pair of adjacent carbon atoms linked by multiple bonds, then that compound is said to be unsaturated.

Ethylene

    Ethylene contains a double bond. There are only four univalent hydrogen atoms present in ethylene, therefore ethylene said to be an unsaturated compound.
H₂C = CH₂

Acetylene

    Acetylene contains a triple bond and there are only two univalent hydrogen atoms.
HC ☰ CH

Alkenes

    Alkenes are the unsaturated hydrocarbons that contain one double bond. They have the general formula CnH2n, as they contain two hydrogen atoms less than the alkanes, alkenes are called unsaturated hydrocarbons.
    The double bond is called 'olefinic bond or 'ethylenic bond'. The name olefin arose from the fact that ethylene was called 'olefiant gas'( oil-forming gas) since it forms oily liquids when treated with chlorine or bromine.
    The original name given to this homologous series was olefine, but it was later decided to reserve the suffix - ine for basic substances only.

Nomenclature of alkenes

Common naming of alkenes

    In the common naming system of an olefin is named according to the following rules,
    The total number of carbon atom in the olefin is counted and the name of the corresponding alkane is determined.
H2C=CH2
the corresponding alkane is ethane
    By changing the name of the corresponding alkane, the suffix -ane of the latter into - ylene.
For ethane changing the suffix -ane of the latter into - ylene.
    The position of the double bond is indicated by number 1, 2, 3, 4...., or Greek Letters α, β, ⋎, ઠ, ...., the end carbon atom nearest to the double bond is denoted by 1, next 2,and so on or α, next β, and so on.
    These letters are then known as locants. The locants of the double bond carbon atom are then placed before the name of the olefin as obtain from rules 1 and 2.
    A hyphen is written in between the locants and the name. The locants are used only to name alkenes containing more than three carbon atoms. alkenes of low molecular weights only have common names.


Problem
    Write down the common names of the following compounds: (i) CH3CH2CH2CH=CH2 (ii) CH3CH=CHCH2CH3.
Answer
(i) CH₃ - CH₂ - CH₂ - CH=CH2
Parent alkane is pentane

Thus the name of the compound is,
1 - pentylene or α - pentylene

(ii) CH3 - CH = CH - CH2 - CH3
Parent alkane is pentane

Thus the name of the compound is
2 -pentylene or β - pentylene

Substituted or derived naming of alkenes

    Another method of nomenclature is to consider ethylene as the parent substance and higher member is derivatives of ethylene.
    If the compound is mono-substituted then no difficulty arises in naming. But the compound is a disubstituted derivative of ethylene isomerism is possible. Since the alkyl groups are of attached the same or different carbon atoms.
    When the groups are attached to the same carbon atom of the olefins named as the asymmetrical compound.
    When the groups are attached to the different carbon atom of the olefins named as the symmetrical compound.
CH3 ㄧCH = CH2
Methylethylene
CH3ㄧCH2ㄧCH = CH2
Ethylethylene
CH3 ㄧ(H3C)C = CH2
as - dimethyl ethylene
CH3ㄧCH = CHㄧCH3
sym - dimethyl ethylene

I.U.P.A.C. naming of alkenes

    According to the I.U.P.A.C. system of nomenclature, the class suffix of the olefins is - ene, and so the series becomes the alkene series.
  1. The longest carbon chain containing the double bond is chosen as the parent alkene.
    CH₃C(CH₃)₂CH₂CH(CH₃)CH=C(CH₃)CH₂CH₃
    The parent part is here base chain, it consists of 8 carbons. The basename, therefore, is to be derived from octane.
  2. The position of the double bond and side chains are indicated by numbers, the lowest number possible being given to the double bond, and this is placed before the suffix.
  3. IUPAC naming alkenes or Olifins
    IUPAC naming alkenes
  4. The name of which is obtained by changing the suffix - ane of the corresponding alkanes into - alkenes.
    To give the lowest number of possible double-bonded carbon is numbered 3.
  5. There are four branches, One methyl branch on carbon atom number 3, three methyl branches on 5th and 7th carbon atoms.
    These are to be indicated as prefixes to the base name. Their names with their locants are 3, 5, 7, 7 - tetramethyl.
Hence the full name is,
3, 5, 7, 7 - tetramethyl - 3 - octene
Problem
    What are the names of the following compounds in the IUPAC system ? (i) CH3 - CH2 - CH = CH2 (ii) C(CH3)2 = CH2 (iii) CH3 - CH = C(CH3) - CH2 - CH3 (iv) CH2 = C(C2H5) - CH(CH3)2
Answer
(i) CH₃ - CH₂ - CH = CH₂
but-1-ene

(ii) (CH₃)₂C = CH₂
2-methylprop-1-ene

(iii) CH₃ - CH = C(CH₃) - CH₂ - CH₃
3-methylpent-2-ene

(iv) CH₂ = C(C₂H₅) - CH(CH₃)₂
2-ethyl-3-methylbut-1-ene
Problem
    Write out the (ignoring stereochemistry) of the isomeric pentanes, and name them by the IUPAC system. Give the structures of the products formed from each on ozonolysis.
Answer
    The molecular formula of the pentene is C₅H₁₂.
    Now take each one in turn and introduce one double bond, starting at the least substituted end and shifting the double bond inwards.
CH₃CH₂CH₂CH=CH₂
pent-1-ene

CH₃CH₂CH=CHCH₃
pent-2-ene

CH₃CH(CH₃)CH=CH₂
3-methylbut-1-ene

CH₃C(CH₃)=CHCH₃
2-methylbut-2-ene

CH₃CH₂C(CH₃)=CH₂
2-methylbut-1-ene
    The product obtained from the ozonide depends on the nature of the reagents used. Here we small use of Zn and acid to give aldehyde and/or ketones.
CH₃CH₂CH₂CH=CH₂

CH₃CH₂CH₂CHO + HCHO

CH3CH2CH=CHCH3

CH3CH2CHO + CH3CHO

CH3CH(CH3)CH=CH2

CH3(CH3)CHCHO + HCHO

CH3CH2C(CH3)=CH2

HCHO + CH3COCH2CH3

CH3C(CH3)=CHCH3

CH3(CH3)C=O + CH3CHO
Alkenes nomenclature
Alkenes nomenclature

Unsaturated hydrocarbon alkenes

    The compounds contain at least one pair of adjacent carbon atoms linked by multiple bonds, then that compound is said to be unsaturated.

Ethylene

    Ethylene contains a double bond. There are only four univalent hydrogen atoms present in ethylene, therefore ethylene said to be an unsaturated compound.
H₂C = CH₂

Acetylene

    Acetylene contains a triple bond and there are only two univalent hydrogen atoms.
HC ☰ CH

Alkenes

    Alkenes are the unsaturated hydrocarbons that contain one double bond. They have the general formula CnH2n, as they contain two hydrogen atoms less than the alkanes, alkenes are called unsaturated hydrocarbons.
    The double bond is called 'olefinic bond or 'ethylenic bond'. The name olefin arose from the fact that ethylene was called 'olefiant gas'( oil-forming gas) since it forms oily liquids when treated with chlorine or bromine.
    The original name given to this homologous series was olefine, but it was later decided to reserve the suffix - ine for basic substances only.

Nomenclature of alkenes

Common naming of alkenes

    In the common naming system of an olefin is named according to the following rules,
    The total number of carbon atom in the olefin is counted and the name of the corresponding alkane is determined.
H2C=CH2
the corresponding alkane is ethane
    By changing the name of the corresponding alkane, the suffix -ane of the latter into - ylene.
For ethane changing the suffix -ane of the latter into - ylene.
    The position of the double bond is indicated by number 1, 2, 3, 4...., or Greek Letters α, β, ⋎, ઠ, ...., the end carbon atom nearest to the double bond is denoted by 1, next 2,and so on or α, next β, and so on.
    These letters are then known as locants. The locants of the double bond carbon atom are then placed before the name of the olefin as obtain from rules 1 and 2.
    A hyphen is written in between the locants and the name. The locants are used only to name alkenes containing more than three carbon atoms. alkenes of low molecular weights only have common names.


Problem
    Write down the common names of the following compounds: (i) CH3CH2CH2CH=CH2 (ii) CH3CH=CHCH2CH3.
Answer
(i) CH₃ - CH₂ - CH₂ - CH=CH2
Parent alkane is pentane

Thus the name of the compound is,
1 - pentylene or α - pentylene

(ii) CH3 - CH = CH - CH2 - CH3
Parent alkane is pentane

Thus the name of the compound is
2 -pentylene or β - pentylene

Substituted or derived naming of alkenes

    Another method of nomenclature is to consider ethylene as the parent substance and higher member is derivatives of ethylene.
    If the compound is mono-substituted then no difficulty arises in naming. But the compound is a disubstituted derivative of ethylene isomerism is possible. Since the alkyl groups are of attached the same or different carbon atoms.
    When the groups are attached to the same carbon atom of the olefins named as the asymmetrical compound.
    When the groups are attached to the different carbon atom of the olefins named as the symmetrical compound.
CH3 ㄧCH = CH2
Methylethylene
CH3ㄧCH2ㄧCH = CH2
Ethylethylene
CH3 ㄧ(H3C)C = CH2
as - dimethyl ethylene
CH3ㄧCH = CHㄧCH3
sym - dimethyl ethylene

I.U.P.A.C. naming of alkenes

    According to the I.U.P.A.C. system of nomenclature, the class suffix of the olefins is - ene, and so the series becomes the alkene series.
  1. The longest carbon chain containing the double bond is chosen as the parent alkene.
    CH₃C(CH₃)₂CH₂CH(CH₃)CH=C(CH₃)CH₂CH₃
    The parent part is here base chain, it consists of 8 carbons. The basename, therefore, is to be derived from octane.
  2. The position of the double bond and side chains are indicated by numbers, the lowest number possible being given to the double bond, and this is placed before the suffix.
  3. IUPAC naming alkenes or Olifins
    IUPAC naming alkenes
  4. The name of which is obtained by changing the suffix - ane of the corresponding alkanes into - alkenes.
    To give the lowest number of possible double-bonded carbon is numbered 3.
  5. There are four branches, One methyl branch on carbon atom number 3, three methyl branches on 5th and 7th carbon atoms.
    These are to be indicated as prefixes to the base name. Their names with their locants are 3, 5, 7, 7 - tetramethyl.
Hence the full name is,
3, 5, 7, 7 - tetramethyl - 3 - octene
Problem
    What are the names of the following compounds in the IUPAC system ? (i) CH3 - CH2 - CH = CH2 (ii) C(CH3)2 = CH2 (iii) CH3 - CH = C(CH3) - CH2 - CH3 (iv) CH2 = C(C2H5) - CH(CH3)2
Answer
(i) CH₃ - CH₂ - CH = CH₂
but-1-ene

(ii) (CH₃)₂C = CH₂
2-methylprop-1-ene

(iii) CH₃ - CH = C(CH₃) - CH₂ - CH₃
3-methylpent-2-ene

(iv) CH₂ = C(C₂H₅) - CH(CH₃)₂
2-ethyl-3-methylbut-1-ene
Problem
    Write out the (ignoring stereochemistry) of the isomeric pentanes, and name them by the IUPAC system. Give the structures of the products formed from each on ozonolysis.
Answer
    The molecular formula of the pentene is C₅H₁₂.
    Now take each one in turn and introduce one double bond, starting at the least substituted end and shifting the double bond inwards.
CH₃CH₂CH₂CH=CH₂
pent-1-ene

CH₃CH₂CH=CHCH₃
pent-2-ene

CH₃CH(CH₃)CH=CH₂
3-methylbut-1-ene

CH₃C(CH₃)=CHCH₃
2-methylbut-2-ene

CH₃CH₂C(CH₃)=CH₂
2-methylbut-1-ene
    The product obtained from the ozonide depends on the nature of the reagents used. Here we small use of Zn and acid to give aldehyde and/or ketones.
CH₃CH₂CH₂CH=CH₂

CH₃CH₂CH₂CHO + HCHO

CH3CH2CH=CHCH3

CH3CH2CHO + CH3CHO

CH3CH(CH3)CH=CH2

CH3(CH3)CHCHO + HCHO

CH3CH2C(CH3)=CH2

HCHO + CH3COCH2CH3

CH3C(CH3)=CHCH3

CH3(CH3)C=O + CH3CHO

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