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Students can Download Chapter 6 Thermodynamics Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 6 Thermodynamics

Introduction
The study of energy transformations forms the subject matter or thermodynamics.

Thermodynamic Terms
The system and the surroundings
A system in thermodynamics refers to that part of universe in which observations are made and remain-ing universe constitutes the surroundings. The surroundings include everything other than the system. System and the surroundings together constitute the universe.

Types Of The System
1. Open System:
In an open system, there is exchange of energy and matter between system and surroundings

2. Closed System:
In a closed system, there is no exchange of matter, but exchange of energy is possible between system and the surroundings

3. Isolated System:
In an isolated system, there is no exchange of energy or matter between the system and the surroundings. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of an isolated system.

The State Of The System
The state of a system means the condition of the system when is macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. A process is said to occur when the state of the system changes.
The measurable properties required to describe the state of a system are called state variables or state functions. Temperature, pressure, volume, composition etc. are state variables.

The Internal Energy as a State Function
1. Work:
The system which can’t exchange heat between the system and surroundings through its boundary is called adiabatic system. The manner in which the state of such a system may be changed will be called adiabatic process. Adiabatic process is a process in which there is no transfer of heat between the system and surroundings.
Internal energy, U, of the system is a state function. The positive sign expresses that wad is positive when work is done on the system. Similarly, if the work is done by the system, wad will be negative.

2. Heat:
A system change its internal energy by ex-change of heat. The q is positive, when heat is transferred from the surroundings to the system and q is negative when heat is transferred from system to the surroundings. first law of thermodynamics, which states that The energy of an isolated system is constant.
i. e., ∆U = q + w
It is commonly stated as the law of conservation of energy i.e., energy can neither be created nor be destroyed.

Applications
Work
The work done due to expansion or compression of a gas against an opposing external pressure is called the pressure – volume type of work. It is a kind of mechanical work.

If Y is the initial volume and V_f is the final volume of a certain amount of gas and P_ex is the external pressure, then the work involved in the process is given by
w = – P_ex (V_f – V_i) or w = -P_ex ∆V

The negative sign of this expression is required to obtain conventional sign for w.

It must be noted that the above expression gives the work done by the gas in irreversible expansion or compression

Work done in isothermal reversible expansion (or compression) of a gas is given by the relation
w_rev =-2.303 nRT log \(\frac{V_{f}}{V_{i}}\)
Where n = the number of moles of the gas

Free expansion
Expansion of a gas in vaccum is called free expansion. Since P = 0 in vacuum, work done in free expansion = 0
Isothermal and free expansion of an ideal gas.
1. For isothermal irreversible change q= -w = p_ex (V_f – V_i)
2. For isothermal reversible change
q = -w = nRT In \(\frac{V_{f}}{V_{i}}\)
= 2.303 nRT log \(\frac{V_{f}}{V_{i}}\)
3. For adiabatic change q = 0, ∆U = w_ad

Enthalpy, H
1. A useful new state function:
We know that the heat absorbed at constant volume is equal to change in the internal dnergy i.e., ∆U= q_p

We may write equation as ∆U=q_p – p∆V at constant pressure, where q_p is heat absorbed by the system and -pAV represent expansion work done by the system.

We can rewrite the above equation as U₂ -U₁ = q_p – p(V₂ – V₁)

On rearranging, we get q_p = (U₂ +pV₂) = (U₁ +pV₁) Now we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as :
H=U+PV

so, equation becomes q_p = H₂ – H₁ = ∆H

Although q is a path dependent function, H is a state function because it depends on U, p and V, all of which are state functions.

Therefore, ∆H is independent of path. Hence,q_p is also independent of path.

For finite changes at constant pressure, we can write ∆H = ∆U + ∆pV

It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy. Remember ∆H = q_p heat absorbed by the system at constant pressure.

∆H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions which absorb heat from the surroundings.

2. Extensive and Intensive Properties:
An extensive property is a property whose value depends on the quantity or size of matter present, in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure, etc.

3. Heat Capacity:
The heat required to rise the temperature of the system in case of heat absorbed by the system.

The increase of temperature is proportional to the heat transferred q = coeff × ∆T

The magnitude of the coefficient depends on the size, composition, and nature of the system. We can also write it as q = C ∆T

The coefficient, C is called the heat capacity. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature. C is directly proportional to amount of substance. The molar heat capacity of a substance, C_m = \(\frac{C}{n}\), is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree Celsius (or one kelvin).

Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one kelvin). q = c × m × ∆T

4. The relationship between C_p and C_v for an ideal gas:
At constant volume, the heat capacity, C is denoted by C_v and at constant pressure, this is denoted by C_p. Let us find the relationship between the two. We can write equation for heat, q at constant volume as q_v=C_v ∆T = ∆U at constant pressure as q_p = C_p∆T = ∆H

The difference between C_p and C_v can be derived for an ideal gas as:
For a mole of an ideal gas, ∆H = ∆U + ∆(pV)
= ∆U + ∆(RT)
= ∆U + R∆T
On putting the values ∆H of ∆H and we have
C_p∆T = C_v∆T
C_p = C_v +R
C_p – C_v = R

Measurement Of ∆U And ∆H: Calorimetry
We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry.

1. ∆U measurements:
Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ∆V = 0.

2. ∆H measurements:
Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter at constant p

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings.

Therefore, q_p will be negative and ∆_rH will also be negative. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆_rH will be positive.

Enthalpy Change, ∆_rH Of A Reaction – Reaction Enthalpy
The enthalpy change accompanying a reaction is called the reaction enthalpy.
The reaction enthalpy change is denoted by ∆_rH
∆_rH = (sum of enthalpies of products) – (sum of enthalpies of reactants)

1. Standard enthalpy of reactions:
The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.
The standard state of a substance at a specified temperature is its pure form at 1 bar.

2. Enthalpy changes during phase transformations:
The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion ∆_fusH°.

Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization ∆_vapH°. And that of sublimation is called Standard enthalpy of sublimation, ∆_subH°.

3. Standard enthalpy of formation
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation( ∆_fH°).

Hess’s Law of Constant Heat Summation Hess’s Law:
If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Enthalpy Calculator is a free online tool that displays the Enthalpy for the given equation.

Enthalpies For Different Types Of Reactions
1. Standard enthalpy of combustion
Enthalpy of combustion of a substance is defined as the enthalpy change accompanying the complete combustion of one mole of the substance in excess of air or oxygen.

Standard enthalpy of combustion is defined as the enthalpy change accompanying the complete combustion of one mole of the substance in excess of air or oxygen when all the reactants and products are tin their standard states at the specified temperature. It is denoted as ∆_cH°.
For example, the complete combustion of one mole of methane evolves 890.3 kJ of heat. Thus, the enthalpy of combustion of methane is- 890.3 kJ mol^-1.

Combustion reactions are always accompanied by the evolution of heat. Hence enthalpy of combustion is always negative.

2. Enthalpy of atomization (symbol: ∆_aH°)
It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

3. Bond Enthalpy (symbol: ∆_bondH°)
The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.

Lattice Enthalpy
The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

Spontaneity
A process which has an urge or a natural tendency to occur under a given set of conditions is known as a spontaneous process.
Some of the spontaneous process need no initiation, i.e., they take place by themselves. Dissolution of common salt in water, evaporation of water in an open vessel, combination of NO and oxygen to form NO₂, neutralisation reaction between NaOH and HCl, etc. are examples of such processes. But some other spontaneous processes need initiation. For example, hydrogen reacts with oxygen to form water only when initiated by passing an electric spark. Once initiated, it occurs by itself.

1. Is decrease in enthalpy a criterion for spontaneity?
By analogy, we may be tempted to state that a chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions.lt becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.

2. Entropy and spontaneity
Entropy(S) is the measure of the degree of randomness or disorder in the system. The greater the disorder in an isolated system, the higher is the entropy. The crystalline solid state is the state of lowest entropy (most ordered), The gaseous state is state of highest entropy. ∆S is independent of path.

Dilution factor formula. After the first tube, each tube is the dilution of the previous dilution tube.

∆S is related with q and T for a reversible reaction as: ∆S = \(\frac{q_{\text {rev }}}{T}\)

When a system is in equilibrium, the entropy is maximum, and the change in entropy, ∆S = 0.

3. Gibbs energy and spontaneity
we define a new thermodynamic function the Gibbs energy or Gibbs function, G, as G = H – TS

Gibbs energy change is a better parameter to determine the spontaneity or feasibility of a process. It can be summarised as follows.
i) If ∆G is negative (i.e., <0) the precess will be spontaneous. ii) If ∆G is zero, the precess is in equilibrium state. iii) If ∆G is positive (i.e., >0), the process is non- spontaneous in the forward direction. The reverse process may be spontaneous.

Ncert Supplementary Syllabus

Enthalpy of Dilution
It is known that enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure. This argument can be applied to any solvent with slight modification. Enthalpy change for dissolving one mole of gaseous hydrogen chloride in 10 mol of water can be represented by the following equation.
HCl(g) + 10 aq. → HCl. 10 aq. ∆H = -69.01 kJ/mol

Let us consider the following set of enthalpy changes:
(S- 5) HCl(g) + 25 aq. → HCl.25 aq. ∆H = -72.03 kJ/mol
(S-2) HCtlgi + 40 aq. → HCl.40 aq. ∆H = -72.79 kJ/mol
(S-3) HCl(g) + ∞ aq. → HCl. ∞aq. ∆H = -74.85 kJ/mol

The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of AH is given above in equation (S-3).
If we subtract the first equation (equation S-1) from the second equation (equation S-2) in the above set of equations, we obtain

This value (-0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the
HCl.25 aq. + 15 aq. → HCl.40 aq.
∆H = [ -72.79 – (-72.03)] kJ/mol
= -0.76 kJ/mol

This value (-0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the surroundings when additional solvent is added to the solution. The enthalpy of dilution of a solution is dependent on the original concentration of the solution and the amount of solvent added.

Theoretical Yield Calculator is a free online tool that displays the amount of product predicted with the complete utilisation of the limiting reactant.

Entropy and Second Law of Thermodynamics
We know that for an isolated system the change in energy remains constant. Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact, is the second law of thermodynamics. Like first law of thermodynamics, second law can also be stated in several ways. The second law of thermodynamics explains why spontaneous exothermic reactions are so common. In exothermic reactions, heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous.

Absolute Entropy and Third Law of Thermodynamics
Molecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively. When temperature of the system rises, these motions become more vigorous and entropy increases. On the other hand, when temperature is lowered, the entropy decreases. The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics. This is so because there is perfect order in a crystal at absolute zero.

The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0 K. The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone. For a pure substance, this can be done by summing \(\frac{q_{\text {rev }}}{T}\) increments from 0 K to 298 K. Standard entropies can be used to calculate standard entropy changes by a Hess’s law type of calculation.

Students can Download Chapter 4 Chemical Bonding and Molecular Structure Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 4 Chemical Bonding and Molecular Structure

Introduction
Matter is made up of different type of elements. The attractive force which holds the constituents together is called a chemical bond.

Kossel-Lewis Approach To Chemical Bonding
The bond between constituents are formed by the sharing of a pair of electrons or their transfer. G.N. Lewis introduced simple notations to represent these outer shell electrons in an atom. These notations are called Lewis symbols.

Significance of Lewis Symbols :
The number of dots around the symbol represents the number of valence electrons. This number of valence electrons helps to calculate the common or group valence of the element. The group valence of the elements is generally either equal to the number of dots in Lewis symbols or8 minus the number of dots or valence electrons.

How to Calculate Bond Order? Introduction to Bond Order.

Kossel, in relation to chemical bonding, drew attention to the following facts:
The bond formed, as a result of the electrostatic attraction between the positiveand negative ions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge (s) on the ion.
In terms of Lewis structures

1. According to electronic theory of chemical bond¬ing, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.

2. Covalent Bond, Langmuir in 1919 refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet, and by introducing the term covalent bond.

By Lewis – Langmuir theory the formation of chlorine molecule is as follows :

In water molecule covalent bond is as follows:

when two atoms share one electron pair they are said to be joined by a single covalent bond. If two atoms share two pairs of electrons, the covalent bond between them is called a double bond. And when combining atoms share three electron pairs as in the case of N₂ molecule a triple bond a triple bond is formed.

• The total number of electrons required for writing the structures are obtained by adding the valence electrons of the combining atoms.
• For anions, each negative charge would mean addition of one electron. For cations, each positive charge would result in subtraction of one electron from the total number of valence electrons.
• The least electronegative atom occupies the central position in the molecule/ion.

Formal charge
Formal charge (F.C.) on an atom in a Lewis structure = total number of valence electrons in the free atom— total number of non bonding (lone pairjelectrons—(1/2) total number of bonding(shared)electrons.

Let us consider the ozone molecule (O₃).
The Lewis structure of O₃ may be drawn as:

The atoms have been marked as 1,2 and 3. The formal charge on:

• The central O atom marked 1 = 6 – 2- \(\frac{1}{2}\)(6) = +1
• The end O atom marked 2 = 6 – 4 – \(\frac{1}{2}\)(4) = o
• The end O atom marked 3 = 6 – 6 – \(\frac{1}{2}\)(2) = -1

Hence, we represent O₃ along with the formal changes as follows:

Limitations of the Octet Rule
There are three types of exceptions to the octet rule. The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons.
Some compounds are BCl₃, AlCl₃ and BF₃.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO₂, the octet rule is not satisfied for all the atoms.

The expanded octet
In a number of compounds of these elements there are more than eight valence electrons around the central atom.Some examples are given below.

Ionic Or Electrovalent Bond
The formation of a positive ion involves ionization, i.e., removal of electrons from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.

A qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.

Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituentions.

Bond Parameters

Bond Length
It may be defined as the equilibrium distance between the centres of the nuclei of the two bonded i atoms in a molecule. Bond length are measured by spectroscopic, X-ray diffraction and electron diffraction techniques. It is usually expressed in Angstrom units (A°) or picometres (pm)
1 A° = 10^-10m and 1 pm = 10^-12m

Bond Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule.

Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

Bond Order:
In the Lewis description of covalent bond, the bond order is given by the number of bonds between the two atoms in a molecule. For example, the bond order in H₂ is one, in O₂ is two and in N₂ is three. Isoelectronic molecules and ions have identical bond orders. For example N₂, CO and NO⁺ have bond order 3. It is found that as the bond order increases, bond enthalpy increases and bond length decreases.

Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. As in the case of O₃.

O₃ is represented by the above 3 structures. These are called canonical structures.

Experimentally determined oxygen-oxygen bond lengths in the O₃ molecule are same (128 pm). Thus the oxygen-oxygen bonds in the O₃ molecule are intermediate between a double and a single bond. According to the concept of resonance, the canonical structures of the hybrid describes the molecule accurately.

Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.

Polarity of Bonds
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character. As a result of polarisation, the molecule possesses the dipole moment. Which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter Mathematically, it is expressed as follows: Dipole moment (µ) = change (Q) X distance of separation (r)

In case of polyatomic molecules, the dipole moment not only depend upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule.lt is due to the shifting of electrons to the side of more eletro negative element. For example,

The shifting of electrons is represented by an arrow. In case of H₂O the resultant dipole moment is given by the following figure:

Fajans Rules:
Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:

• The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
• The greater the charge on the cation, the greater the covalent character of the ionic bond.
• For cations of the same size and charge, the one, with electronic configuration (n-1)d”ns°, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.

The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond.

The Valence Shell Electron Pair Repul-Sion (Vspert) Theory
The main postulates of VSEPR theory are as follows:

• The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.
• Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
• These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them.
• The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.
• A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
• Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.

The repulsive interaction of electron pairs de-crease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)

Valence Bond Theory
Valence bond theory was introduced by Heitlerand London (1927) and developed further by Pauling and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition. First, we consider the formation of H2. When the attractive forces become greater than the repulsive forces, the molecule is formed and the system gets minimum energy. Because energy is released when a bond is formed. The energy so released is called bond enthalpy.

Orbital Overlap Concept
When two atoms approach each other, their atomic orbitals undergo partial interpenetration. This partial interpenetration of atomic orbitals is called overlapping of atomic orbitals. The electrons belonging to these orbitals are said to be shared and this results in the formation of a covalent bond. The main ideas of orbital of overlap concept of formation of covalent bonds are

• Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the atoms taking part in bonding.
• The orbitals undergoing overlapping must have electrons with opposite spins.
  Overlapping of atomic orbitals results in decrease of energy and formation of covalent bond.
• The strength of a covalent bond depends upon the extent of overlapping. The greater the overlapping, the stronger is the bond formed.

The above treatment of formation of covalent bond involving the overlap of half-filled atomic orbitals is called valence bond theory.

Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
(i) Sigma(σ) bond, and
(ii) pi(π) bond

(i) Sigma( σ) bond :
This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.
s-s overlapping:
In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below:

s-p overlapping:
This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.

p-p overlapping :
This type of overlap takes place between half filled p-orbitals of the two approaching atoms.

(ii) pi(π) bond :
In the formation of n bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to side wise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms.

Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double ortriple bonds).

Hybridisation
Hybridisationis defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.

The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.

These hybrid orbitals are stable due to their arrangement which provides minimum repulsion between electron pairs. Therefore, the type of hybridisation indicates the geometry of the molecules.

It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Types of Hybridisation
There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under:

(I) sp hybridisation:
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp- hybridised and linked directly to two other central atoms possesses linear geometry.The two sp hybrids point in the opposite direction which provides more effective overlapping resulting in the formation of stronger bonds.

Example of molecule having sp hybridisation BeCl₂:
The ground state electronic.configuration of Be is 1s²2s². In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its divalency. One 2s and one 2p-orbitalsget hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite direction forming an angle of 180°. Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be-Cl sigma bonds.

II) sp² hybridisation :
In this hybridisation there is involvement of one s and two p-orbitals in orderto form three equivalent sp² hybridised orbitals. For example, in BCl₂ molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s
electrons is promoted to vacant 2p orbital as a result boron has three unpaired electrons.

These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals. The three hybrid orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl₃ the geometry is trigonal planar with ClBCl bond angle of 120°

III) sp³ hybridisation:
This type of hybridisation can be explained by taking the example of CH₄ molecule in which there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp³ hybrid orbital of equivalent energies and shape.

There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron.

The angle between sp³ hybrid orbital is 109.5°

The structure of NH₃ and H₂O molecules can also be explained with the help of sp3hybridisation. In NH₃, the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s²\(p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) having three unpaired electrons in the sp³ hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N-H sigma bonds. Due to the force of repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramidal.

Other Examples of sp³, sp² and sp Hybridisation:
sp³ Hybridisation in C₂H₆ molecule:
In ethane molecule both the carbon atoms assume sp3 hybrid state. One of the four sp³ hybrid orbitals of carbon atom overlaps axially with similar orbitals of other atom to form sp³-sp³ sigma bond while the other three hybrid orbitals of each carbon atom are used in forming sp³-s sigma bonds with hydrogen atoms Therefore in ethane C-C bond length is 154 pm and each C-H bond length is 109 pm.

sp² Hybridisation in C₂H₄:
In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C-C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp²-s sigma bond with two hydrogen atoms. The unhybridised orbital (2p_x or 2p_y) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond, which consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms.

Thus, in ethene molecule, the carbon-carbon bond consists of one sp²-sp² sigma bond and one pi (π) bond between p orbitals which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length is 134 pm. The C-H bond is sp^s sigma with bond length 108 pm. The H-C-H bond angle is 117.6° while the H-C-C angle is 121°.

sp Hybridisation in C₂H₂:
In the formation of ethyne molecule, both the carbon atoms undergo sp- hybridisation having two unhybridised orbital i.e., 2p_y and 2p_x. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming σ bonds. Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds

Hybridisation of Elements involving d-Orbitals
The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the hybridisation involving either 3s, 3p, and 3d or 3d, 4s and 4p is possible. However, since the difference in energies of 3p and 4s orbitals is significant, no hybridisation involving 3p, 3d and 4s orbitals is possible.

1. Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below.

Now the five orbitals (i.eone s, three p, and one d orbitals) are available for hybridisation to yield a set of five sp3d hybrid orbitals which are directed towards the five comers of a trigonal bipyramidal.

Three sigma bond known as equatorial bonds lie in one plane and make an angle of 120° with each other.
The remaining two P-Cl bonds(called axial bonds)-one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane.

As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl₅ molecule more reactive.

2. Formation of SF₆ (sp³d² hybridisation):
In SF₆ the central sulphur atom has the ground state outer electronic configuration 3s²3p⁴. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp³d² hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF₆. These six sp³d² hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus SF₆ molecule has a regular octahedral geometry.

The structure of SF₆ is given below.

Molecular Orbital Theory
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of this theory are :

• The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.
• The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric.
• The number of molecular orbital formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
• The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
• Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
• The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’srule.

Formation of Molecular Orbitals
Linear Combination of Atomic Orbitals (LCAO)
The atomic orbitals of these atoms may be represented by the wave functions ψ_A and ψ_B. The formation of molecular orbitals is the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals as shown below.

Energy Level Diagram for Molecular orbitals

The increasing order of energies of various molecular orbitals for O₂ and F₂ is given below:
σ1s < σ*1s < σ2s < σ*2s < σ2p_x <(π2p_x = π2p_y) <(π*2p_x = π*2p_z) < σ*2p_x

This sequence of energy levels of molecular orbitals is not correct for the remaining molecules Li₂, Be₂, B₂, C₂, N₂. For molecules such as B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is
σ1s < σ*1s < σ2s < σ*2s <(π2p_x = π2p_y) < σ2p_x <(π*2p_x = π*2p_z) < σ*2p_z

The important characteristic feature of this order is that the energy of σ 2p_z molecular orbital is higher than that of π2p_x and π2p_y molecular orbitals. If the bonding influence is stronger a stable molecule results and if the antibonding influence is stronger,the molecule is unstable.

Bonding In Some Homonuclear Diatomic Molecules
Bond Order:
Bond order is defined as half of the difference between the number of electrons in the bonding molecular orbitals and that in the antibonding molecular orbitals.
Nh – N
i.e. Bond Order= \(\frac{N_{b}-N_{a}}{2}\). Where N_b is the number of electrons in the bonding molecular orbitals and Na is the number of electrons in the antibonding mo-lecular orbitals.

Significance of bond order:
Bond order conveys the following important informations about a molecule.
i) If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is negative or zero, the molecule is unstable and is not formed.
ii) Bond dissociation energy of a diatomic molecule is directly proportional to the bond order of the molecule. The greater the bond order, the higher is the bond dissociation energy.
iii) Bond order is inversely proportional to the bond length. The higherthe bond ondervalue, smaller is the bond length. For example, the bond length in N₂ molecule (having bond order 3) is less than that in O₂ molecule (having bond order 2).

Magnetic character:
If all the electrons in the mol-ecules of a substance are paired, the substance will be diamagnetic. On the other hand, if there are un-paired electrons in the molecule, the substance will be paramagnetic.

Hydrogen Bonding
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O orN) of another molecule. When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes, highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ⁺) while ‘X’ attain fractional negative charge (δ^–). This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as: H^δ+ – X^δ-

The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.

Types of Hydrogen Bonds
There are two types of hydrogen bonds

1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond

1. Intermolecular hydrogen bond:
It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

2. Intramolecular hydrogen bond:
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-Nitrophenol the hydrogen is in between the two oxygen atoms as shown below:

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Kerala Plus One Chemistry Notes Chapter 2 Structure of Atom

Plus One Chemistry Chapter 2 Notes Pdf Introduction
The atomic theory of matter was first proposed by John Dalton. His theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter.

Sub-Atomic Particles
Discovery Of Electron
The experiments of Michael Faraday in discharge tubes showed that when a high potential is applied to a gas taken in the discharge tube at very low pres-sures, certain rays are emitted from the cathode. These rays were called cathode rays.

The results of these experiments are summarised below:
1. The cathode rays start from cathode and move towards the anode.

2. In the absence of electrical or magnetic field, these rays travel in straight lines.ln the presence of electrical or magnetic field, they behave as negatively charged particles, i.e.,they consist of negatively charged particles, called electrons.

3. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.Thus, we can conclude that electrons are the basic constituent of all the atoms.

Charge To Mass Ratio Of Electron
In 1897, the British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m_e) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons.

From the amount of deviation of the particles from their path in the presence of electrical or magnetic field, the value of e/m was found to be 1.75882 × 10¹¹ coulomb per kg or approximately 1.75288 × 10⁸ cou-lomb per gram. The ratio e/m was found to be same irrespective of the nature of the gas taken in the dis-charge tube and the material used as the cathode.

Structure Of Atom Class 11 Notes Charge Of The Electron
Millikan (1868-1953) devised a method known as Oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10-¹⁹C.

Mass of the electron (m)

Discovery Of Protons And Neutrons
Electrical discharge earned out in the modified cathode ray tube led to the discovery of canal rays. The characteristics of these positively charged particles are listed below:

• unlike cathode rays, the e/m ratio of the particles depend upon the nature of gas present in the cathode ray tube.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for cathode rays.

The smallest and lightest positive ion was obtained from hydrogen and was called proton. Later, electrically neutral particles were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α – particles when electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named these particles as neutrons.

Atomic Models

Structure Of Atom Class 11 Notes Hsslive Thomson Model Of Atom
J.J. Thomson was the first to propose a model of the atom. According to him, the atom is a sphere in which positive charge is spread uniformly and the electrons are embedded in it so as to make the atom electrically neutral. This model is also known as “plumpudding model’. But this model was soon discarded as it could not explain many of the experimental observations.

Hsslive Structure Of Atom Notes Rutherford’s Nuclear Model of Atom
Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α – particles. The experiment is known as α -particle scattering experiment. On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom :

1. Most of the space in the atom is empty as most of the α -particles passed through the foil undeflected.

2. A few α – particles were deflected. Since the α – particles are positively charged, the deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α – particles.

3. Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom (after the discovery of protons). According to this model:
1.The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.

2. The electrons move around the nucleus with a very high speed in circular paths called orbits. Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.

3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Chemistry Notes For Class 11 Chapter 2 Atomic Numberand Mass Number
’ Knowing the atomic number Z and mass number A of an element, we can calculate the number of protons, electrons and neutrons present in the atom of the element.
Atomic Number (Z) = Number of protons = Number of electrons
Mass Number (A) – Atomic number (Z) = Number of neutrons

Isotopes, Isobars And Isotones
Isotopes are atoms of the same element having the same atomic number but different mass numbers. They contain different number of neutrons. For ex-ample, there are three isotopes of hydrogen having mass numbers 1,2 and 3 respectively. All the three isotopes have atomic number 1. They are represented as \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\) and \(_{ 1 }^{ 3 }{ H }\) and named as hydrogen or protium, deuterium (D) and tritium (T) respectively. Isobars are atoms of different elements which have the same mass number. For example, \(_{ 6 }^{ 14 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isobars.
Isotones may be defined as atoms of different elements containing same number of neutrons. For example \(_{ 6 }^{ 13 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isotones.

Developments Leading To The Bohr’S Model Of Atom
Neils Bohr improved the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were:

1. electromagnetic radiation possess both wave like and particle like properties(Dual character)
2. Experimental results regarding atomic spectra which can be explained only by assuming quantized electronic energy levels in atoms.

Wave Nature Of Electromagnetic Ra-Diation
Light is the form of radiation and it was supposed to be made of particles known as corpuscules.
As we know, waves are characterised by wavelength (λ), frequency (υ) and velocity of propagation (c) and these are related by the equation
c = vλ or v = \(\frac { c }{ \lambda } \)

The wavelengths of various electro magnetic radia-tions increase in the order.
γ rays < X-rays< uv rays < visible < IR < Microwaves < Radio waves

Particle Nature Of Electro Magnetic Radiation: Planck’S Quantum Theory
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner, a belief popular at that time. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (υ) and is expressed by the equation E = hυ

Class 11 Chemistry Chapter 2 Notes Photoelectric Effect
When a metal was exposed to a beam of light, electrons were emitted. This phenomenon is called photoelectric effect. Obseravations of the photoelectric effect experiment are the following:

• There is no time lag belween the striking of light beam and the ejection of electrons from the metal surface.
• The number of electrons ejected is proportional to the intensity or brightness of light.
• For each metal, there is a characteristic minimum frequency, u₀ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency u>u₀, the ejected electrons come out with certain kinetic energy.

The kinetic energies of these electrons increase with the increase of frequency of the light used.

Using Plank’s quantum theory Einstein explained photoelectric effect. When a light particle, photon with sufficient energy strikes an electron instantaneously to the electron during the collision and the electron is ejected without any time lag. Greater the energy of photon greater will be the kinetic energy of ejected electron and greater will be the frequency of radiation.

If minimum energy to eject an electron is hv0 and the photon has an energy equal to hv. Then kinetic en-ergy of photoelectron is given by, hv=hv₀ + 1/2 m_ev² where m_e is the mass of electron and hv₀ is called the work function.

Duel Behaviour Of Electromagnetic Ra-Diation
Light has dual behaviour that is it behaves either as a wave or as a particle. Due to this wave nature, it shows the phenomena of interference and diffraction.

Evidence For The Quantized Electronic Energy Levels : Atomic Spectra
It is observed that when a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. Since ordinary white light consists of waves with ail the wave-lengths in the visible range, a ray of white light is spread out into a series of coloured bands called spectrum. In a continuous spectrum light of different colours merges together. For example violet merges into blue, blue into green and soon.

Chapter 2 Class 11 Chemistry Notes Emission and absorption spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”.

A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. The study of emission or absorption spectra is referred to as spectroscopy Line spectra or atomic spectra is the spectra where emitted radiation is identified by the appearance of bright lines in the spectra.

Line spectrum of Hydrogen
The hydrogen spectrum consists of several series of lines named after their discoverers. Balmershowed in 1885 on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber (\(\overline { v } \)), then the visible lines of the hydrogen spectrum obey the following formula :
\(\overline { v } \) = 109,677 \(\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right] \mathrm{cm}^{-1}\)
where n = 3, 4, 5, ………….
The series of lines described by this formula are called the Balmer series.

The value 109,677cm^-1 is called the Rydberg constant for hydrogen. The first 5 series of lines correspond to n₁ = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively. Line specrum becomes more complex for heavier atoms.

Chapter 2 Chemistry Class 11 Notes Bhor’S Model For Hydrogen Atom
Bhors model for hydrogen atom says that
1. the energy of an electron does not change with time.
The diagram shows the Lyman, Balmer and Paschen series of transitions for hydrogen atom.

2. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by :
\(v=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\)
E₁ and E₂ are the energies of the lower and higher allowed energy states respectively.
The angular momentum of an electron in a given stationary state can be expressed as in equation,

Chemistry Chapter 2 Class 11 Notes Bohr’s theory for hydrogen atom:
1. The stationary states for electron are numbered n = 1,2,3. These integral numbers are known as Principal quantum numbers.
2. The radii of the stationary states are expressed as:
r_n = n² a₀
where a₀ = 52.9 pm

3. The most important property associated with the electron, is the energy of its stationary state. It is
given by the expression, \(E_{n}=-R_{H}\left(\frac{1}{n^{2}}\right)\)
where R_H is called Rydberg constant and its value is 2.18 × 10^-18 J. The energy of the lowest state, also called as the ground state, is
E₁ = -2.18 × 10^-18 \(\left(\frac{1}{1^{2}}\right)\) = -2.18 × 10^-18 J. The energy of the stationary state for n = ∝, will be :
E₂ = -2.18 × 10^-18 J\(\left(\frac{1}{2^{2}}\right)\) = -0.545 × 10^-18 J.

When the electron is free from the influence of nucleus(n = ∞), the energy is taken as zero. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign and depicts its stability relative to the reference state of zero energy and n = ∞

4. Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He⁺ Li²⁺, Be³⁺ and so on. The energies of the stationary states associated with these hydrogen-like species are given by the expression,

Structure Of Atom Class 11 Notes Pdf Explanation of Line Spectrum of Hydrogen
The frequency (v) associated with the absorption and emission of the photon can be evaluated by using equation,

Class 11 Chapter 2 Chemistry Notes Limitations of Bohr’s Model
Bohr’s model was too simple to account for the following points:
1. It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum. This model is also unable to explain the spectrum of atoms other than hydrogen Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Towards Quantum Mechanical Model Of The Atom
Two important developments which contributed significantly in the formulation of a more suitable and general model for atoms were:

1. Dual behaviour of matter
2. Heisenberg uncertainty principle

Structure Of Atom Class 10 Notes Pdf Dual Behaviour of Matter
The French physicist, de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i. e., both particle and wavelike properties. This means that just as the photon, electrons should. also have momentum as well as wavelength. de Broglie, from this analogy, gave the following relation between wavelength (λ) and momentum (p) of a material particle.
\(\lambda=\frac{h}{m v}=\frac{h}{p}\)

Heisenberg’s Uncertainty Principle
Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation,

∆x is the uncertainty in position and ∆p_x (or ∆v_x) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain ∆v_x is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v_x is small), then the position of the electron will be uncertain (∆x will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.

Significance of Uncertainty Principle
Heisenberg Uncertainty Principle rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.

Reasons for the Failure of the Bohr Model
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr.model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. There was no point in extending Bohr model to other atoms. In fact, an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.

Quantum Mechanical Model Of Atom
Quantum mechanics is a theoretical science that deals with the study of motions of microscopic objects such as electrons.

In quantum mechanical model of atom, the behaviour of an electron in an atom is described by an equation known as Schrodinger wave equation. Fora system, such as an atom or molecule whose energy does not change with time, the Schrodinger equation written as Hψ = Eψ where H is a mathematical operator, called Hamiltonian operator, E is the total energy and ψ is the amplitude of the electron wave called wave function.

Hydrogen Atom And The Schrodinger Equation
The wave function ψ as such has no physical significance. It only represents the amplitude of the electron wave. However ψ² may be considered as the probability density of the electron cloud. Thus, by determining ψ² at different distances from the nucleus, it is possible to trace out or identify a region of space around the nucleus where there is high probability of locating an electron with a specific energy.

According to the uncertainty principle, it is not possible to determine simultaneously the position and momentum of an electron in an atom precisely. So Bohr’s concept of well defined orbits for electron in an atom cannot hold good. Thus, in quantum mechanical mode, we speak of probability of finding an electron with a particular energy around the nucleus. There are certain regions around the nucleus where probability of finding the electron is high. Such regions are called orbitals. Thus an orbital may be defined as the region in space around the nucleus where there is maximum probability of finding an electron having a specific energy.

Orbitals and Quantum Numbers
Orbitals in an atom can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m_l

The principal quantum number n’ is a positive integer with value of n= 1, 2, 3 ……………

The principal quantum number determines the size and to large extent the energy of the orbital.

The principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by ‘n²’ Ait the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters
n= 1 2 3 4 ………………
Shell = K LM N ………………

Size of an orbital increases with increase of principal quantum number ‘n’. Since energy of the orbital will increase with increase of n.

Azimuthal quantum number, ‘F is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n – 1), that is, for a given value of n, the possible value of l are: l = 0, 1, 2, ……….. (n – 1)

Each shell consists of one or more subshells or sub-levels. The number of subshells in a principal shell is equal to the value of n. For example h the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l= 0, 1) in the second shell (n = 2), three l= 0, 1, 2) and so on. Each sub-shell is assigned an azimuths! quantum number (l). Sub-shells corresponding to different values of l are represented by the following symbols.
l : 0 1 2 3 4 5 …………….
Notation for sub-shell : s p d f g h …………….

Magnetic orbital quantum number. ‘m_l’ gives information about the spatial orientation of the or bital with respect to standard set of co-ordinate axis. For any sub-shell (defined by T value) 21+ 1 values of m,are possible and these values are given by:
m, = -l, -(l-1), (l-2)… 0, 1… (l-2), (l-1), l Thus for l = 0, the only permitted value of m,= 0, [2(0) + 1 = 1, one s orbital].

Electron spin ‘s’:
George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m_s). Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or -½. These are called the two spin states of the electron and are. normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different m_s values (one +½ and the other -½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.

Shapes of Atomic Orbitals
The orbital wave function or V for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

According to the German physicist, Max Bom, the square of the wave function (i.e., ψ²) at a point gives the probability density of the electron at that point.

For 1 s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it. The region where this probability I density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n.

These probability density variation can be visualised . in terms of charge cloud diagrams.

Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|² is constant. Boundary ‘ surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. The s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given distance is equal in all the directions.

unlike s-orbitals, the boundary surface diagrams of p orbitals are not spherical. Instead, each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z-axis, they are given the designations 2p_x, 2p_y, and 2p_z. It should be understood, however, that there is no simple relation between the values of m, (-1, 0 and+1) and the x, y and z directions. For our purpose, it is sufficient to remember that, because there are three possible values of m, there are, therefore, three p orbitals whose axes are mutually perpendicular. Like s orbitals, p orbitals increase in size and energy with increase in the principal quantum number

The number of nodes are given by (n -2), that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.

For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3 as the value of l cannot be greater than n-1. There are five m; values (-2, -1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The five d-orbitals are designated as d_xy, d_yz, d_xz, d_x²-y² and d_z². The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, d_z², is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d…) also have shapes similar to 3d orbital, but differ in energy and size.

Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of p_z orbital, xy-plane is a nodal plane, in case of d_xy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by T, i.e., one angular node for p orbitals, two angular nodes for cf orbitals and so on. The total number of nodes are given by (n-1), i.e., sum of I angular nodes and (n-l-1) radial nodes.

Energies Of Orbitals
The order of energy of orbitals in single electron sys-tem are given below:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f The orbitals having same energy are called degenerate.

Filling Of Orbitals In Atom
Aufbau principle: According to this principle in the ground state of an atom, an electron will occupy the orbital of lowest energy and orbitals are occupied by electrons in the order of increasing energy.

Pauli’s exclusiohn principle : Pauli’s exclusion principle states that ‘no two electrons in an atom can have the same values for all the four quantum numbers’

Since the electrons in an orbital must have the same n, I and m quantum numbers, if follows that an orbital can contain a maximum of two electrons provided their spin quantum numbers are different. This is an important consequence of Pauli’s exclusion principle which says that an orbital can have maximum two electrons and these must have opposite spins.

Hund’s rule of maximum multiplicity :
This rule states that electron pairing in orbitals of same energy will not take place until each available orbital of a given subshell is singly occupied (with parallel spin).
The rule can be illustrated by taking the example of carbon atom. The atomic number of carbon is 6 and its electronic configuration is 1s²2s²2p². The two electrons of the 2p subshell can be distributed in the following three ways.

According to Hund’s rule, the configuration in which the two unpaired electron occupying 2px, and 2py orbitals with parallel spin is the correct configuration of carbon.

Exceptional configurations of chromium and copper
The electronic configuration of Cr (atomic number 24) is expected to be [Ar] 4s² 3d⁴, but the actual configuration is [Ar] 4s¹ 3d⁵. Similarly, the actual configuration of Cu (At. No. 29) is [Ar] 4s¹ 3d¹⁰ instead of the expected configuration [Ar] 4s² 3d⁹.

This is because of the fact that exactly half filled or completely filled orbitals (i.e., d⁵, d¹⁰, f⁷, f¹⁴) have lower energy and hence have extra stability.

Students can Download Chapter 2 Structure of Atom Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 2 Structure of Atom

Plus One Chemistry Structure of Atom One Mark Questions and Answers

Plus One Chemistry Structure Of Atom Questions Question 1.
Which of the following is not true for cathode rays?
a) They possess kinetic energy
b) They are electromagnetic waves
c) They produce heat
d) They produce mechanical pressure
Answer:
b) They are electromagnetic waves

Previous Hse Questions From The Chapter Structure Of Atom Question 2.
The mass of the electron = ________ kg
Answer:
9.11 × 10^-31 kg

Plus One Chemistry Chapter 2 Questions And Answers Question 3.
Bohr’s orbits are called stationary states because
a) Electrons in them are stationary
b) Their orbits have fixed radii
c) The electrons in them have fixed energy
d) The protons remain in the nuclei and are stationary
Answer:
c) The electrons in them have fixed energy

Structure Of Atom Class 11 Questions And Answers Pdf Question 4.
The metal which gives photoelectrons most easily is
a) Lithium
b) Sodium
c) Calcium
d) Cesium
Answer:
d) Cesium

Questions On Structure Of Atom Class 11 Question 5.
The orbitals having same energy are called ________ orbitals.
Answer:
degenerate

Questions On Quantum Numbers Class 11 Question 6.
Match the following:

1. Spherically symmetrical – d
2. Dumb-bell – s
3. Doubly dumb-bell – p

Answer:

1. Spherically symmetrical – s
2. Dumb-bell – p
3. Doubly dumb-bell – d

Class 11 Chemistry Chapter 2 Important Questions With Answers Question 7.
Which of the following set of quantum numbers is correct for an electron in 4/ orbital
a) n = 4 l = 4 ml = -4 ms = +14
b) m = 4 l = 3 ml = +4 ms = +14
c) n = 4 l = 3 ml = -3 ms =-14
d) n = 4 l = 3 ml = +4 ms = -14
Answer:
c) n = 4 l = 3 ml = -3 ms = -14

Structure Of Atom Class 11 Questions And Answers Question 8.
The limiting line of Balmer Series has the frequency of ________ .
Answer:
8.23 × 10¹⁴

Class 11 Chemistry Chapter 2 Important Questions Question 9.
The number of orbitals and the maximum number of electrons that can be accommodated in a principal quantum level are
Answer:
n² & 2n²

Structure Of Atom Important Questions Question 10.
If the uncertainty in position and momentum of a particle like electrons are equal the uncertainty in velocity is ________
Answer:
\(\triangle V=\frac { 1 }{ 2m } \sqrt { \frac { h }{ \pi } } \)

Important Questions Of Atomic Structure Class 11 Question 11.
The total energy of an electron in a Bohr orbit is given by ________
Answer:
\(\frac{-Z e^{2}}{8 \pi \varepsilon_{0} r}\)

Plus One Chemistry Structure of Atom Two Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2
Of the following which is/are correct? Give justification.
a) n = 2 l = 1 m = 0 s = +½
b) n = 3 l = 3 m = 2 s =-½
c) n = 4 l = 3 m = 1 s = +½
d) n = 3 l = 2 m = 3 s = +½
Answer:
a) n = 2 f = 1 m = 0 s = +½
c) n = 4 f = 3 m=1 s= -½
Option b) is wrong because when n = 3, ^ =0, 1, 2
Option d) is wrong because when l = 2, m = -2, -1, 0, 1, 2

Question 3.
Match the following:
1. Anode rays – Nucleus
2. Cathode rays – Plum-pudding model
3. J.J Thomson – Proton
4. Thea-particle scattering experiment – Electron
Answer:
1. Anode rays – Proton
2. Cathode rays – Electron
3. J.J Thomson – Plum-pudding model
4. Thea-particle scattering experiment – Nucleus

Question 4.
Calculate the uncertainty in the determination of velocity of a ball of mass 200 g, if the uncertainty in the determination of position is 1 A.
[h=6.626 × 10^-34 J s]
Answer:

Question 5.
Which among the following sets of quantum numbers is/are not possible?
a) n = 3, l = 2, m = 0, s = +½
b) n = 2, l= 1, m = 0, s = +½
c) n = 1, l= 0, m = 0, s = -½
d) n = 4, l = 2, m = 2, s = -½
Answer:
All sets are possible.

Question 6.
1. How many sub-shells are associated with n = 4?
2. How many electrons will be present in the sub-shells having m_s value of –\(\frac{1}{2}\) for n = 4?
Answer:
1. For n = 4, l can have values 0, 1, 2, 3. Thus, there are four sub-shells in n = 4 energy level.
These four sub-shells are 4s, 4p, 4d and 4f.

2. For n = 4, the number of orbitals = (4)² = 16.
Each orbital can have one electron with m_s = –\(\frac{1}{2}\).
Thus, there are 16 electrons in sub-shells having n = 4 and m_s = –\(\frac{1}{2}\)

Question 7.
i) Name the principle which restricts the pairing of electrons in degenerate orbitals. .
ii) How many electrons can be accomodated in the sub-shell having n = 4 and l = 2?
Answer:
i) Hund’s rule of maximum multiplicity,
ii) 10 electrons (4d sub-shell).

Question 8.
If the electron is to be located within 5 x 10^-5A°, what will be the uncertainty in its velocity?
(Mass of the electron = 9.1 x 10^-31 kg).
Answer:
According to Heisenberg’s uncertainty principle,

Question 9.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴per second, calculate the power of this laser.
Answer:

Plus One Chemistry Structure of Atom Three Mark Questions and Answers

Question 1.
Fill in the blanks suitably by studying the relationship of the given pairs:

1. Lyman : Ultraviolet:: Balmer: ……………..
2. s-subshell:spherical:: p-subshell: …………….
3. Rydberg’s formula: \(\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\) :: de Broglie relation:

Answer:

1. Balmer: Visible
2. psubshell: dumb-bell
3. de Broglie relation: \(\lambda=\frac{h}{m v}\)

Question 2.
Fill in the blanks:

Answer:

Question 3.
Filling of electrons in the orbitals on a ground state atom is governed by three rules.
a) Which are the three rules?
b) State any one of them.
Answer:
a) 1) Aufbau principle
2) Pauli’s exclusion principle
3) Hund’s rule of maximum multiplicity

b) Hund’s rule of maximum multiplicity – electron pairing in orbitals of same energy will not take place until each degenerate orbital of a given subshell is singly occupied.

Question 4.
During a class room discussion, one of your friends argued that, “we can’t determine both position and velocity of an electron”.

1. Is it true?
2. Which principle is behind your answer?
3. State it.

Answer:

1. Yes. It is true.
2. Heisenberg’s uncertainty principle.
3. Heisenberg’s uncertainty principle states that it is not possible to determine simultaneously both position and momentum of a microscopic moving particle such as electron with absolute accuracy.

Question 5.
The arguments of two students is as given:
Student 1 : “We need four quantum numbers to represent an electron in a multi-electron atom.”
Student 2 : “We need only first three quantum numbers to represent an electron in a multi-electron atom.”
a) Which are the four quantum numbers?
b) Who is correct? Why?
c) Write the possible four quantum numbers of the valence electron of Na atom.
Answer:
a) The four quantum numbers are:
i) Principal quantum number (n)
ii) Azimuthal quantum number (l)
iii) Magnetic quantum number (m_l)
iv) Spin quantum number (m_s)

b) The argument of student 1 is correct. If there are two electrons in the same subshell the first three quantum numbers become the same. Hence we need fourth quantum number to identify the electron.

c) n = 3, l = 0, m = 0, s = +½

Question 6.
a) Identify the experiment associated with the following figure:

b) Explain the experiment done by Rutherford and give its observations.
c) Write the concludions of the experiment.
Answer:
a) It is the figure of α -particle scattering experiment (gold foil experiment).

b) In this experiment, a stream of high energy α – particles from a radioactive source was directed at a thin foil of gold metal which had a circular fluorescent zinc sulphide screen around it. Whenever α-particles struck the screen, a tiny flash of light was produced at that point. The α-particles striking the gold foil were analysed. It was observed that:

1. Most of the α – particles passed through gold foil undeflected.
2. A small fraction of the α – particles are deflected by small angles.
3. A very few α – particles(1 in 20,000) bounced back i.e., deflected by nearly 180°.

c) Rutherford drew the following conclusions regard¬ing the structure of atom from this experiment:

1. Most of the space in the atom is empty as most of the α – particles passed through the foil undeflected.
2. The positive charge of the atom is concentrated in a very small volume (called nucleus) that repelled and deflected the positively charged α – particles.
3. Volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

Question 7.
The 4s subshell has more energy than 3p subshell.
a) Is it true? Justify your answer.
b) StateAufbau principle.
Answer:
a) Yes. For both 4s and 3p subshells the (n+l) value is 4. But 4s, with high value of ‘n’ has higher en¬ergy.
b) Aufbau principle – In the ground state of the atoms, the orbitals are filled in order of their increasing energies.

Question 8.
Two students were analysing the electronic configurations of the first 30 elements of the Periodic Table as part of an assignment. They found that two elements showed difference from other twenty eight elements.

1. Which are the two elements?
2. Write their electronic configurations.
3. Why they show this anomalous behaviour?

Answer:
1. ₂₄Cr and ₂₉Cu.
2. ₂₄Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ and ₂₉Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
3. This is due to the fact that exactly half filled and completely filled orbitals (i.e., d⁵, d¹⁰) have extra stability due to symmetrical distribution of electrons and maximum exchange energy.

Question 9.
Three box diagrams of 2p³ configuration are given below:
i)

1. Which one is correct?
2. Name the principle behind your answer.
3. State the principle.

Answer:

1. The correct one is (ii).
2. Hund’s rule of maximum multiplicity.
3. Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.

Question 10.
J.J. Thomson proposed his atom model in 1898.

1. Explain Thomson’s model of atom.
2. Why Thomson’s atom model is called plum pudding model or watermelon model?
3. What is the limitation of Thomson’s atom model?

Answer:
1. J.J. Thomson proposed that an atom possess a spherical shape in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. The mass of the atom is assumed to be uniformly distributed over the atom. This model explained the overall neutrality of the atom.

2. Thomson’s model of atom can be visualised as a pudding or watermelon of positive charge with electrons embedded into it like the plums or seeds.

3. Thomson’s model was not consistent with the results of later experiments. It failed to explain the observations of Rutherford’s α-particle scattering experiment.

Question 11.
A student argued that the 3d orbitals will be filled only after the 4s orbital is completely filled in accordance with aufbau principle. Then another student opposed by saying that it is not true for certain elements like Cr and Cu.
a) Whose argument is correct?
b) Write electronic configurations of ₂₄Crand ₂₉Cu. Justify your answer.
Answer:
a) Both arguments are correct.
b) Cr and Cu have anomalous electronic configurations. This is because half filled and completely filled sub-shells have extra stability due to the symmetrical distribution of electrons maximum exchange energy.
₂₄Cr = 1 s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ OR [Ar]3d⁵ 4s¹ This is due to the extra stability of half filled 3d⁵ sub-shell.

₂₉Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ OR [Ar]3d¹⁰ 4s¹ This is due to the extra stability of completely filled 3d¹⁰ sub-shell.

Question 12.
a) What are the atomic numbers of elements whose
outermost electronic configurations are given by
i) 3s¹
ii) 3p⁵?
b) Which of the following are isoelectronic species?
Na⁺, K⁺, Mg²⁺,Ca²⁺, S^2-,Ar
c) What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms^-1? m
Answer:
a) i) 3s¹-Atomic number is 11 (Na)
ii) 3p⁵ -Atomic number is 17 (Cl)

b) Na⁺, Mg²⁺ (both of them have same no. of electrons, i.e., 10 each)

Question 13.
Quantum numbers are a set of four numbers used to designate electron in an atom.

1. How many electrons in an atom can have the following quantum numbers, n = 1, l = 0 ?
2. Give the quantum numbers of the valence electron of an atom with atomic number 13.
3. Draw the shape of orbital having n = 1 and l = 0.

Answer:
1. 2 electrons (i.e., 1s orbital)
2. The element with atomic number 13 is aluminium. ₁₃Al ⇒ [Ne] 3s²3p¹⁺
The valence electron is in the 3p orbital. Hence, the quantum numbers for the valence electron are n = 3, l= 1, m = -1, 0 +1
3. It is the 1s orbital

Question 14.
a) i) What is meant by line spectra or atomic spectra?
ii) Name the series of lines in the hydrogen spectrum belonging to the visible region.
iii) What is the wave length of light emitted when the electron in a hydrogen atom undergoes transmission from n = 4 to n = 2? (RH= 109677 cm-1)
b) State the principles/rules for filling of orbitals in atoms.
Answer:
a) i) Line spectra or atomic spectra are the spectra obtained from excited atoms due to emission of radiation. The emitted radiation is identified by the appearance of bright lines.
ii) Balmer series
iii) n₁ = 2, n₂ = 4

b)

1. Aufbau principle: In the ground state of the atoms, the orbitals are filled in order of their increasing energies.
2. Pauli’s exclusion principle: No two electrons in an atom can have the same set of four quantum numbers.
3. Hund’s rule of maximum multiplicity: Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.

Question 15.
Line emission spectra are often called finger print of atoms.
a) Justify the above statement.
b) Yellow light emitted from a sodium lamp has a wave length (X) of 580 nm. Calculate the frequency and wave number of this yellow light.
Answer:
a) Each element has a unique line emission spectrum. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms in the same way as finger prints are used to identify people.

Question 16.
1. How many orbitals are possible in a p-subshell. Which are they?
2. What is the shape of p-orbital?
3. Sketch the boundary surface diagrams of 2p orbitals.
Answer:
1. Three.
These are p_x, p_y and p_z orbitals.
2. Dumb-bell shaped.
3.

Question 17.
Electronic configuration of an element written by a student is given below:

a) Which rule is violated here?
b) Give the correct configuration.
c) What is the uncertainty in position of an electron if the uncertainty in its velocity is 1.159×107 m/s?
Answer:
a) Hund’s rule of maximum multiplicity.

Plus One Chemistry Structure of Atom Four Mark Questions and Answers

Question 1.
Bohr’s model of hydrogen atom is a modification of
Rutherford’s model.
a) Write any two merits of Bohr’s model.
b) Write any two demerits of Bohr’s model.
Answer:
a)

1. Bohr’s model could explain the stability of an atom.
2. Bohr’s model could explain the atomic spectrum of hydrogen.

b)

1. Failed to explain the finer details of hydrogen atom spectrum observed by using sophisticated spectroscopic techniques.
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Question 2.
Consider the statement, “The two electrons of He atom have the same set of quantum numbers.”

1. Do you agree?
2. Name the principle applied here.
3. State the principle.
4. Write the all quantum numbers of outer electrons of the atom.

Answer:

1. No.
2. Pauli’s exclusion principle.
3. No two electrons in an atom can have same set of four quantum numbers.
4. Forthe1st electron: n = 2, l = 0, m = 0, s = +½.
   For the 2nd electron: n = 2, l = 0, m = 0, s = -½

Question 3.
Complete the following table with respect to the va¬lence electron of each element.

Answer:

Question 4
Analyse the following figure showing transitions of electrons in the hydrogen atom.

a) Name the series (a), (b), (c), (d) and (e).
b) Mention the region of the spectrum in which each series belongs to.
c) Explain how they are obtained.
d) Calculate the wave length of the first line of series (b).
[R_H = 109677 cm^-1]
Answer:
(a) = Lyman series,
(b) = Balmer series,
(c) = Paschen series,
(d) = Brackett series,
(e) = Pfund series

b)
Lyman series – UV region
Balmer series – Visible region
Paschen series – Infrared region
Brackett series – Infrared region
Pfund series – Infrared region

c) In hydrogen atom there is one electron which is present in first orbit in ground state. When energy is supplied this electron may be excited to some higher energy level. Since in a sample of hydrogen there are large number of atoms, the electrons in different atoms absorb different amounts of energies and are excited to different higher energy levels. Now, from excited states, the electron may return to ground state in one or more jumps. These different downward jumps are associated with different amounts of energies and hence result in the emission of radiations of different wavelengths which appear as different lines in the hydrogen spectrum.
Series – Obtained when electron jumps from any of the higher energy levels to
Lyman series – 1^st energy level
Balmer series – 2^nd energy level
Paschen series – 3^rd energy level
Brackett series – 4^th energy level
Pfund series – 5^th energy level

d) n₁ = 2, n₂ = 3, R_H = 109677 cm^-1

Question 5.
Quantum numbers are the address of an electron in an atom. Justify the statement by explaining different quantum numbers.
Answer:
Quantum numbers are certain numbers which are used to identify an electron in an atom.
The following four quantum numbers are used for this purpose.
1. Principal quantum number (n):
It determines the size and to large extent the energy of the orbital. It also identifies the shell. The value of ‘n’ ranges from 1 to a. With increase in the value of ‘n’, the number of allowed orbitals increases and are given by ‘n2’. All the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by letters K (n = 1), L (n = 2), M (n = 3), N (n = 4) etc. The size and energy of the orbital will increase with increase of ‘n’.

2. Azimuthal/Orbital angular momentum/Subsidiary quantum number (l): It defines the three dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n-1). It gives an idea regarding the subshell in which the electrons are present.

3. Magnetic quantum number (m_l):
It gives information about the spatial orientation of the orbital with respect to standard set of coordinate axis. For any sub-shell, (2l+1) values of m_l are possible ranging from -l to +l including zero. The permitted values of m_l gives the number of orbitals in that sub-shell.

4. Spin quantum number (m_s) :
It refers to the orientation of the spin of the electron. An orbital can have two electrons. If an electron is spinning in the clockwise direction, it is given a spin quantum number value of+1/2 and if an electron is spinning in the anti-clockwise direction it is given spin quantum number value of -1/2. These are called two spin states of the electron and are represented by two arrows ↑ (spin up) and ↓. (spin down). Thus, the two electrons in an orbital should have opposite spins.

Question 6.
Dual nature of matter was proposed by Louis de Broglie.
a) Calculate the de Broglie wavelength associated with an electron with velocity equal to that of light.
b) State Heisenberg’s uncertainty principle and give its mathematical expression.
Answer:

b) It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.
Mathematically, it can be given as in the equation

∆x ⇒ uncertainty in position of the particle
∆px ⇒ uncertainty in momentum of the particle
∆vx ⇒ uncertainty in velocity of the particle
m ⇒ mass of the particle and
h ⇒ the Planck’s constant

Question 7.
Rutherford’s atom model had strong similarity to a small scale solar system. (4)
a) What are the important features of Rutherford’s. nuclear model of atom?
b) What are the drawbacks of Rutherford’s model of atom?
Answer:
a) i) The positive charge and most of the mass of the atom is densely concentrated in extremely small region of the atom called nucleus.
ii) The nucleus is surrounded by electrons that ’ move around the nucleus with a very high speed in circular paths called orbits.
iii) Electrons and the nucleus are held together by electrostatic forces of attraction.

b) i) It failed to explain the stability of atom,
ii) It says nothing about the electronic structure of atoms.

Question 8.
a) Calculate the momentum of a particle which has de Broglie wave length of 250 pm. (4)
b) The distribution of electron into orbitals of an atom is called its electronic configuration.
i) Give the valence shell electronic configuration of chromium atom.
ii) Which among the following configurations is more.stable, d4 ord5? Justify your answer.
Answer:
a) According to the de Broglie matter wave equation

b) i) ₂₄Cr → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
ii) d⁵ is more stable.
The d⁵ configuration is half filled. It has symmetrical distribution of electrons and maximum exchange energy. Hence, it has extra stability compared to d⁴ configuration.

Question 9.
a) Name and state the principle, which restricts the maximum number of electrons in an orbital to be two.
b) Using s, p, d, f notation represent the sub-shell with the following quantum numbers.
i) n = 1, l = 0
ii) n = 4, l=3
c) The uncertainty in the position and velocity of a particle are 10 cm and 5.27 × 10³m/s respectively. Calculate the mass of the particle (h=6.626 × 10^-34 J s).
Answer:
a) Pauli’s exclusion principle
No two electrons in an atom can have the same set of four quantum numbers.
b) i) 1s ii) 4f
c) According to Heisenberg’s uncertainty principle,

Question 10.
The photoelectric effect was first observed by H.Hertz.
a) What is photoelectric effect?
b) What are the observations of photoelectric effect experiment?
Answer:
a) It is the phenomenon of ejection of electrons when certain soft metals like potassium, rubidium, cae-sium, etc. are exposed to a beam of light.
b) i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface of metal.
ii) The number of electrons ejected is proportional to the intensity or brightness of light.
iii) For each metal, there is a characteristic minimum frequency (υ₀), known as threshold frequency below which photoelectric effect is not observed.

Question 11.
A mathematical representation is given below:
\(\Delta x \times \Delta p_{x} \geq \frac{h}{4 \pi}\)
a) Which principle is illustrated by this equation?
b) If the position of the electron is measured within an accuracy of ±0.002 nm, calculate the uncertainty in the momentum of the electron.
c) Using s, p, d notation represent the sub-shell with the following quantum numbers:
i) n = 3 l = 2
ii) n = 5 l = 1
Answer:
a) Heisenberg’s uncertainty principle.

Plus One Chemistry Structure of Atom NCERT Questions and Answers

Question 1.

1. Calculate the number of electrons which will together weigh one gram.
2. Calculate the mass and charge of one mole of electrons.

Answer:
1. Mass of one electron = 9.11 × 10^-31 kg
∴ Number of electrons in one gram = \(\frac{10^{-3} \mathrm{kg}}{9.11 \times 10^{-31} \mathrm{kg}}=1.098 \times 10^{27}\)

2. Mass of one electron = 9.11 × 10^-31 kg
∴ Mass of 1 mole of electrons = 9.11 × 10^-31 kg × 6.022 × 10²³ = 5.486 x 10^-7 kg
Charge on one electron = 1.602 × 10^-19 C
∴ Charge on one mole of electrons
= (1.602 × 10^-19C) × (6.022 × 10²³)
= 9.65 × 10⁴ C.

Question 2.
Write the complete symbol forthe atom with the given atomic number (Z) and atomic mass (A):
i) Z = 17, A = 35
ii) Z = 92, A = 233
iii) Z = 4, A = 9
Answer:

Question 3.
Electro.magnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer:
Ionisation energy of sodium = Energy of one photon of radiation of wavelength.
Energy of photon = hυ

Question 4.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J energy?
Answer:
Suppose N photons of the light with wavelength 4000 pm can provide 1 J of energy.
Energy of N photons = Nhυ

Question 5.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷m S^-1. (2)
Answer:

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