Important Questions

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom

Very Short Answer Type Questions

Question 1.
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron
Answer:
Neutron is a neutral particle. Hence, it will not be deflected on passing through an electric field.

Question 2.
What is the nuclear radius of an atom whose mass number is 125?
Answer:
Nuclear radius, r = R₀A^1/3 where, R₀ = 1.4 x 10^-15 m,
∴ r = (1.4 x 10^-15 m) x (125)^1/3 = 7.0 x 10^-15 m.

Question 3.
The magnitude of charge on the electron is 4.8 x 10^-10 esu. What is the charge on the nucleus of a helium atom?
Answer:
Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2) x 4.8 x 10^-10 = + 9.6 x 10^-10 esu.

Question 4.
What is the difference in the origin of cathode rays and anode rays?
Answer:
Cathode rays originate from the cathode whereas anode rays do not originate from the anode. They are produced from the gaseous atoms by knock out of the electrons with high speed cathode rays.

Question 5.
What is the difference between atomic mass and mass number?
Answer:
Mass number is a whole number because it is the sum of number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass of its atom as compared with mass of an atom of C-12 isotope taken as 12.

Question 6.
What is the difference between a quantum and a photon?
Answer:
The smallest packet of energy of any radiation is called a quantum whereas that of light is called photon.

Question 7.
Arrange s, p and rf-subshells of a shell in the increasing order of , effective nuclear charge (Zeff) experienced by the electron present in them. [NCERT Exemplar]
Ans. s-orbital is spherical in shape, it shields the electrons from the nucleus more effectively than p-orbital which in turn shields more effectively than d-orbital. Therefore, the effective nuclear charge (Zeff) experienced by electrons present in them is d < p < s.

Question 8.
Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.
Answer:

Question 9.
Nickel atom can lose two electrons to form Ni ion. The atomic number of Ni is 28. From which orbital will nickel lose two electrons? [NCERT Exemplar]
Answer:
₂₈Ni = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²; Nickel will lose 2 electrons from 4s (outer most shell) to form Ni²⁺ ion.

Question 10.
Which of the following orbitals are degenerate?
3d_xy, 4d_xy, \(3 d_{z^{2}}\), 3d_yz, \(4 d_{z^{2}}\)
Answer:
The orbitals which belongs to same subshell and same shell are called degenerate orbitals. (3d_xy, \(3 d_{z} 2\), 3d_yz) and (4d_xy, 4dyz, 4d 2) are the two sets of degenerate orbitals.

Short Answer Type Questions

Question 1.
The Balmer series in the hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂ = 3, 4 … . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (R_H = 109677 cm1)
Answer:
From Rydberg formula,

Question 2.
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
Answer:
From de Broglie equation, wavelength, \(\lambda=\frac{h}{m v}\)
For same wavelength for two different particles, i.e., electron and proton, m₁v₁ = m₂v₂ (h is constant). Lesser the mass of the particle, greater will be the velocity. Hence, electron will have higher velocity.

Question 3.
Wavelengths of different radiations are given helow.
λ(A) = 300 nm, λ(B) = 300 pm, λ(C) = 3 nm, λ(D) = 30Å Arrange these radiations in the increasing order of their energies.
Answer:
(A) λ=3OOnm=3OO x 10^-9m
(B) λ =300µm=300 x 10^-6m
(C) λ =3nm = 3 x 10^-9 m
(D) λ = 30 = 30 x 10^-10m= 3 x 10^-9m
Energy, E = \(\frac{h c}{\lambda}\)
Therefore, E ∝ \(\frac{1}{\lambda}\)
Increasing order of energy is B

Question 4.
The electronic configuration of valence shell of Cu is 3d¹⁰4s¹ and not 3d⁹4s². How is this configuration explained?
Answer:
Configurations with completely filled and half-filled orbitals have extra stability. In 3d¹⁰4s¹, d-orbitals are completely filled and s-orbital is half-filled. Hence, it is a more stable configuration for Cu as compare to 3d⁹4s².

Question 5.
In each of the following pairs of salts, which one is more stable? (i) Ferrous and ferric salts (ii) Cuprous and cupric salts
Answer:
(i) Ferrous and ferric salts : In ferrous salts Fe²⁺, the configuration is 1s²,2s²,2p⁶,3s²,3p⁶,3d⁶. In ferric salts Fe³⁺, the configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵. As half-filled 3d⁵ configuration is more stable therefore ferric salts are more stable than ferrous salts.
(ii) Cuprous and cupric salts : In cuprous salts, the configuration of Cu⁺ is 1s²,2s², 2p⁶, 3s², 3p⁶, 3d¹⁰. In cupric salts the configuration of Cu²⁺, is, 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹. Although Cu+ has completely filled d-orbital, yet cuprous salts are less stable. This is because the nuclear charge is not sufficient enough to hold 18 electrons of Cu⁺ ion present in the outermost shell.

Long Answer Type Questions

Question 1.
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula \(\bar{v}=109677\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)
What points of Bohr’s model of an atom can he used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term. [NCERT Exemplar]
Answer:
The two important points of Bohr’s model that can be used to derive the given formula are as follows :
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state respectively.
Derivation : The energy of the electron in the nth stationary state is given by the expression,
\(E_{n}=-R_{\mathrm{H}}\left(\frac{1}{n^{2}}\right)\)
n = 1,2,3 …. ……(i)
where R_H is called Rydberg constant and its value is 2.18 x 10^-18 J.
The Energy of the lowest state, also called the ground state, is
E₁ = -2.18 x 10^-18\(\left(\frac{1}{1^{2}}\right)\) = 2.18 x 10^-18J ……. (ii)
The energy gap between the two orbits is given by the equation,
ΔE = E_f – E_i … (iii)
On combining equations (i) and (iii)

Question 2.
Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
Answer:

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Very short answer type questions

Question 1.
Write the product in the following reaction:

Answer:
CH₃– CH = CH-CH₂– CHO Pent-3-en-1-al

Question 2.
Complete the following reaction sequence: (NCERT Exemplar]

Answer:

Question 3.
Illustrate the following name reaction: Clemmensen reduction.
Answer:
Clemmensen reduction: The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.

Question 4.
Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃.
Name the reaction also. [NCERT Exemplar]
Answer:

Question 5.
Which acid of each pair shown here would you expect to be stronger?
(i) F-CH₂ -COOH or Cl-CH₂ -COOH

Answer:
(i) F-CH₂-COOH > Cl-CH₂-COOH, because of stronger -ve-I effect of F than Cl.

Question 6.
Identify the compounds A, B, and C in the following reaction. [NCERT Exemplar]

Answer:

Question 7.
Write the product(s) in the following reaction:

Answer:

Question 8.
Name the aldehyde which does not give Fehling’s solution test.
Answer:
Benzaldehyde.

Question 9.
Give the name of the reagent that brings the following transformation: But-2-ene to ethanal.
Answer:
O₃/H₂O-Zn dust

Question 10.
Write TUPAC names of the following structures: (NCERT Exemplar)

Answer:

Question 11.
(CH₃) ₃C-CHO does not undergo aldol condensation comment.
Answer:
Because no α-H is present.

Short answer type questions

Question 1.
Write the reactions involved in the following:
(i) Etard reaction
(ii) Stephen’s reduction
Answer:
(i) Etard reaction: Chromyl chloride oxidizes toluene to chromium complex which on hydrolysis gives benzaldehyde.

(ii) Stephen reductIon: Nitriles are reduced to corresponding imines with SnCl₂ in the presence of HCl, which on hydrolysis give the corresponding aldehyde.
SnCl₂ + 2HCl → SnCl₄ + 2[H]

Question 2.
Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2,5-dimethyl hexane-3-one. [NCERT Exemplar]

Answer:

Question 3.
Carboxylic acids contain carbonyl group but do not show the nucleophilic addition reaction like aldehydes or ketones. Why? [NCERT Exemplar]
Answer:
Carboxylic acids contain, carbonyl group but do not show nucleophilic addition reaction like aldehyde and ketone. Due to resonance as shown below the partial positive charge on the carbonyl carbon atom is reduced.

Question 4.
Give reasons:
(i) The α-hydrogen atoms of aldehydes and ketones are acidic in nature.
(ii) Propanone is less reactive than ethanal toward addition of HCN
(iii) Benzoic acid does not give Friedel Crafts reaction.
Answer:
(i) The acidity of a-hydrogen atom of carbonyl carbon is due to the strong withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) This is due to steric and electronic reasons. Sterically, the presence of two methyl groups in propanone hinders more the approach of nucleophile to carbonyl carbon than in ethanal having one methyl group. Electronically two methyl groups reduce the positivity of the carbonyl carbon more effectively in propanone than in ethanal.

(iii) Benzoic acid does not give Friedel Craft reaction because:

• the carboxyl group is strongly deactivating.
• the catalyst AlCl₃ which is a lewis acid gets bonded to the carboxyl group strongly.

Question 5.
An organic compound T having molecular formula C₄H₈O gives orange-red ppt. with a 2,4-DNP reagent.
It does not reduce Tollen’s reagent but gives yellow ppt. of iodoform on heating with NaOI.
Compound X on reduction with LiAlH₄ gives compound T’ which undergoes dehydration reaction on heating with -cone. H₂SO₄ to form but-2-ene. Identify the compounds X and Y.
Answer:

Reaction involved:

Long answer type questions

Question 1.
(a) Account for the following:
(i) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Out of CH₃CH₂ – CO – CH₂ – CH₃ and CH₃CH₂ – CH₂ –
CO – CH₃, which gives iodoform test?
Answer:
(a) (i) Because of -I effect of Cl atom in ClCH₂COOH and +I effect of CH₃ group in CH₃COOH the electron density in the O-H bond in ClCH₂COOH is much lower than CH₃COOH.
As a result O-H bond in ClCH₂COOH is much weaker than in CH₃COOH therefore loses a proton more easily than CH₃COOH.
Hence ClCH₂COOH acid is stronger acid than CH₃COOH.

(ii) Carboxylic acids are resonance hybrid of the following structures:

Similarly, a carbonyl group of aldehydes and ketones may regarded as resonance hybrid of following structures :

Because of contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. On the other hand, electrophilic character of carboxyl carbon is reduced due to contribution of structure (II). As carbonyl carbon of carboxyl group is less electropositive than carbonyl carbon in aldehydes and ketones, therefore, carboxylic acids do not give nucleophilic addition reactions of aldehydes and ketones.
(b) CH₃ – CH₂ – CH₂ – COCH₃.

Question 2.
Write the products formed when ethanal reacts with the following reagents:
(i) CH₃MgBr and then H₃O⁺
(ii) Zn-Hgyconc.HCl
(iii) C₆H₅CHO in the presence of dilute NaOH
Answer:

Question 3.
(a) Write the products of the following reactions :

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
Answer:

(iii) Cl-CH₂-COOH

(b)
(i) NaHCO₃ test
(ii) Iodoform test or Fehling test or Tollen’s test.

Question 4.
(a) Account for the following:
(i) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.
(ii) 2-Fluorobutanoic acid is a stronger acid than 3-Fluorobutanoic acid.

(b) Write the chemical equations to illustrate the following name reactions:
(i) Etard reaction.
(ii) Rosenmund’s reaction.
(c) Give the mechanism of cyanohydrin formation when carbonyl compounds react with HCN in the presence of alkali.
Answer:
(a)

• CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN due to steric and electronic factors.
• Because the inductive effect decreases with distance and hence the conjugate base of 2-Fluorobutanoic acid is more stable.

(b)

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry

Very short answer type questions

Question 1.
Why on dilution the ∧_m of CH₃COOH increases drastically while that of CH₃COONa increases gradually? [NCERT Exemplar]
Answer:
In the case of CH₃COOH, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.
CH₃COOH + H₂o → CH₃COO^– + H₃O⁺

Question 2.
Why is alternating current used for measuring resistance of an electrolytic solution? [NCERT Exemplar]
Answer:
Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.

Question 3.
Define electrochemical series.
Answer:
The arrangement of elements in the increasing or decreasing order of their standard reduction potentials is called electrochemical series.

Question 4.
What are secondary cells?
Answer:
Secondary cells are those cells which are rechargeable, i.e., the products can be changed back to reactants.

Question 5.
What is the necessity to use a salt bridge in a Galvanic cell?
Answer:
To complete the inner circuit and to maintain the electrical neutrality of the electrolytic solutions of the half-cells we use a salt bridge in a Galvanic cell.

Question 6.
Why does a dry cell become dead after a long time even if it has not been used?
Answer:
Even though not in use, a dry cell becomes dead after some time because the acidic NH4C1 corrodes the zinc container.

Question 7.
What does the negative sign in the expression \(\boldsymbol{E}_{\mathbf{Z n}^{2+} / \mathbf{Z n}}[latex] = -0.76 V mean?[NCERT Exemplar]
Answer:
It means that Zn is more reactive than hydrogen. When zinc electrode will be connected to SHE, Zn will get oxidised and H⁺ will get reduced.

Question 8.
Write the Nemst equation for the cell reaction in the Daniel cell. How will the £cell be affected when concentration of Zn2+ ions is increased? [NCERT Exemplar]
Answer:
Zn + Cu²⁺ > Zn²⁺ + Cu
E_cell = [latex]E_{\text {cell }}^{\ominus}\) – \(\frac{0.059}{2}\) log \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
E_cellcell decreases when concentration of Zn²⁺ ions, [Zn²⁺ ] increases.

Question 9.
What does the negative value of \(E_{\text {cell }}^{\ominus}\) indicate?
Answer:
Negative \(E_{\text {cell }}^{\ominus}\) value means Δ_rG^Θ will be +ve, and the cell will not work.

Question 10.
Can \(E_{\text {cell }}^{\ominus}\) or Δ_rG^Θ for cell reaction ever be equal to zero? [NCERT Exemplar]
Answer:
No.

Short answer type questions

Question 1.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer:
Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of cation and anion of the electrolyte.

In general, if an electrolyte on dissociation gives v₊ cations and v_– anions then its limiting molar conductivity is given by
\(\Lambda_{m}^{\ominus}=v_{+} \lambda_{+}^{0}+v_{-} \lambda_{-}^{0}\)

Where, \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the limiting molar conductivities of cations and anions respectively.

Conductivity of a solution decreases with dilution. This is due to the fact that the number of ions per unit volume that carry the current in a solution decreases with dilution.

Question 2.
Define the following terms:
(i) Fuel cell
(ii) Limiting molar conductivity (\(\))
Answer:
(i) A fuel cell is a device which converts the energy produced during the combustion of fuels like hydrogen, methanol, methane etc. directly into electrical energy. One of the most successful fuel cell is H₂ – O₂ fuel cell.

(ii) When concentration approaches zero, the molar conductivity is known
as limiting molar conductivity. It is represented by \(\Lambda_{m}^{\ominus}\).
\(\Lambda_{m}^{\ominus}\) (∧b)when → c

Question 3.
(i) Write two advantages of H₂ – O₂ fuel cell over ordinary cell.
(ii) Equilibrium constant (K_c) for the given cell reaction is 10. Calculate \(\boldsymbol{E}_{\text {cell }}^{\ominus}\).

Solution:
(i) The two main advantages of H₂ – O₂ fuel cell over ordinary cell are as follows:

• It has high efficiency of 60%-70%.
• It does not cause any pollution.

Question 4.
Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
Ag⁺(aq) + e^– → Ag(s) E^Θ = + 0.80 V
H⁺ (aq)+ e^– → \(\frac{1}{2}\)H₂(g) E^Θ = 0.00 V
On the basis of their standard reduction electrode potential (E^Θ) values, which reaction is feasible at the cathode and why?
Answer:
The reaction, Ag⁺ (aq) + e^– → Ag(s) is feasible at cathode as
cathodic reaction is one which has higher standard reduction electrode potential (\(E_{\text {red }}^{\ominus}\)).

Question 5.
Calculate ∆G and log K_c for the following reaction at 298 K:
2Cr(8) + 3Fe²⁺(aq) > 2Cr³⁺ (aq) + 3Fe(s)
Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = 0.30 V
Solution:

Question 6.
Calculate the emf of the following cell at 298 K:
Cr(s)/Cr³⁺ (0.1M)//Fe<>2+ (0.01M)/(Fe(s) [Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = + 0.30 V]
Answer:
The cell reaction is as follows :
2Cr(s) + 3Fe²⁺(aq) > 3Fe(s) + 2Cr³⁺(aq)
For this reaction, n = 6
Now,
E_cell = \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) – \(\frac{2.303 R T}{n F}\) log \(\frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Fe}^{2+}\right]^{3}}\)
E_cell = 0.30 – \(\frac{0.059}{6}\) log \(\frac{\left[10^{-1}\right]^{2}}{\left[10^{-2}\right]^{3}}\)
E_cell = 0.26V

Question 7.
The conductivity of 10^-3 mol/L acetic acid at 25°C is 4.1 x 10^-5 S cm^-1. Calculate its degree of dissociation, if \(\Lambda_{m}^{0}\) for acetic acid at 25°C is 390.5 S cm² mol^-1.
Answer:
We know that ∧_m = \(\frac{1000 \mathrm{~K}}{\mathrm{C}}\)
∧_m = \(\frac{1000 \times 4.1 \times 10^{-5}}{10^{-3}}\)
= 41 S cm² mol^-1
Now, α = \(\frac{\Lambda_{m}^{c}}{\Lambda_{m}^{0}}\)
= \(\frac{41}{390.5}\) = 0.105 390.5

Question 8.
(i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
(ii) The products of electrolysis of aqueous NaCl at the respective electrodes are:
Cathode: H₂
Anode: Cl₂ and not 0₂. Explain.
Answer:
(i) ‘B’ is a strong electrolyte.
Because a strong electrolyte is already dissociated into ions, but on dilution inter ionic forces are overcome, ions are free to move. So, there is slight increase in molar conductivity on dilution.

(ii) On anode water should get oxidised in preference to Cl^– but due to overvoltage/over potential Cl” is oxidised in preference to water.

Long answer type questions

Question 1.
(a) The conductivity of 0.20 mol L^-1 solution of KC1 is 2.48 x 10^-2 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). Given λ⁰ (K⁺) = 73.5 S cm² mol^-1 and λ⁰ (Cl^–) = 76.5 S cm² mol^-1.

(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?
Answer:
(a) Substituting K =2.48 x 10^-2 S cm^-1, M = 0.20 molL^-1 in the expression ∧_m = \(\frac{K \times 1000}{M}[latex] , we get

Substituting ∧_m = 124 S cm² mol^-1, [latex]\Lambda_{m}^{\ominus}\) = 150 S cm² mol^-1 in the expression α = \(\frac{\Lambda_{m}}{\Lambda_{m}^{\ominus}}\), we get
Degree of dissociation, α = \(\frac{124 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{150 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}\) = 82666
α = 82.67%

(b) Primary cell. Mercury cell is more advantageous than dry cell because its cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life period.

Question 2.
(i) The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.905 x 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).
Given λ⁰ (H⁺) = 349.6 S cm² mol^-1 and λ⁰ (CH₃COO ) = 40.9 S cm² mol^-1

(ii) Define electrochemical cell. What happens if external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of electrochemical cell?
Solution:

(b) A device which is used to convert chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell.
If external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of
electrochemical cell, the reaction gets reversed and the electrochemical cell function as an electrolytic cell.

Question 3.
Calculate e.m.f and ΔG for the following cell at 298 K:
Mg(s) | Mg²⁺ (0.01 M) || Ag⁺ (0.0001 M) | Ag(s)
Given: \(\underset{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}{\boldsymbol{E}^{\circ}}\) = -2.37V, \(\boldsymbol{E}^{\ominus}{\left(\mathbf{A g}^{+} / \mathbf{A g}\right)}\) = + 0.80V
Solution:

E_cell = 3.17 – 0.0295 log 10⁶
E_cell = 3.17 – 0.177 V = 2.993 V
E_cell = 2.993 V
Substituting n = 2, F = 96500 C mol^-1, E_cell = 2.993 V in the
expression, ΔG = – nFE_cell, we get
ΔG = nFE_cell = -2 x 96500 C mol^-1 x 2.993V
ΔG = – 577649 J mol^-1 = – 577.649 kJ mol^-1

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 13 Amines

Very short answer type questions

Question 1.
CH₃NH₂ is more basic than C₆H₅NH₂ why?
Answer:
Aliphatic amines (CH₃NH₂) are stronger bases than aromatic amines (C₆H₅NH₂) because due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less easily available for protonation.

Question 2.
What is the role of pyridine in the acylation reaction of amines? [NCERT Exemplar]
Answer:
Pyridine and other bases are used to remove the side product, L e., HCl from the reaction mixture.

Question 3.
A primary amine, RNH₂ can be reacted with CH₃ – X to get secondary amine, R-NHCH₃ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH₂ forms only 2° mine? [NCERT Exemplar]
Answer:

Primary amines show carbylamine reaction in which two H-atoms attached to N-atoms of NH₂ are replaced by one C-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.

Question 4.
What is Hinsberg reagent? (NCERT Exemplar)
Answer:
Benzene sulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg reagent. It is used to distinguish between primary, secondary and tertiary amine.

Question 5.
The conversion of primary aromatic amines into diazonium salts is known as……
Answer:
Diazotisation.

Question 6.
Rearrange the following in an increasing order of their basic strengths:
C₆H₅NH₂,C₆H₅N(CH₃)₂,(C₆H₅)₂NH and CH₃NH₂
Answer:
(C₆H₅)₂NH < C₆H₅NH₂ <C₆H₅N(CH₃)₂ <CH₃NH₂

Question 7.
What is the best reagent to convert nitrile to primary amine? [NCERT Exemplar]
Answer:
Reduction of nitriles with sodium/alcohol or LiAlH₄ gives primary amine.

Question 8.
Suggest a route by which the following conversion can be accomplished: [NCERT Exemplar]

Answer:

Question 9.
What is the role of HNO₃ in the nitrating mixture used for the nitration of benzene? [NCERT Exemplar]
Answer:
HNO₃ acts as a base in the nitrating mixture and provides the electrophile NO₂.

Question 10.
Why is benzene diazonium chloride not stored and is used immediately after its preparation? [NCERT Exemplar]
Answer:
Benzene diazonium chloride is very unstable.

Short answer type questions

Question 1.
Write the structures of A, B and C in the following:

Answer:

Question 2.
Give a chemical test to distinguish between C₆H₅CH₂NH₂ and C₆H₅NH₂.
Answer:
C₆H₅CH₂NH₂ reacts with HNO₂ at 273-278 K to give diazonium salt, which being unstable decomposes with brisk evolution of N₂ gas.

whereas, C₆H₅NH₂ reacts with HNO₂ at 273-278 K to form stable benzene diazonium chloride, which upon treatment with an alkaline solution of f3-naphthol, gives an orange dye.

Question 3.
Complete the following reaction: [NCERT Exemplar]
Answer:
The reaction exhibits an azo-coupling reaction of phenols. Benzene diazonium chloride reacts with phenol in such a manner that the para position of phenol is coupled with diazonium salt to form p-hydroxy azobenzene.

Question 4.
A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this lg mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer. [NCERT Exemplar]
Answer:
The above-stated reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol generates phenoxide ion which is more electron-rich than phenol and more reactive for electrophilic attack. The electrophile in this reaction is aryldiazonium cation. As we know, stronger the electrophile faster is the reaction. p Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. So, nitrophenyl diazonium chloride couples preferentially with phenol

Question 5.
Under what reaction conditions (acidic, basic) the coupling reaction of aryl diazonium chloride with aniline is carried out? [NCERT Exemplar]
Answer:
In strongly basic conditions, benzene diazonium chloride is converted, into diazohydroxide and diazoate as both of which are not electrophilic and do not couple with aniline.

Similarly, in highly acidic conditions, aniline gets converted into an anilinium ion. From this, result aniline is no longer nucleophilic acid and hence will not couple with diazonium chloride. Hence, the reaction is carried out under mild conditions, i.e., pH 4-5

Question 6.
An organic aromatic compound ‘A’ with the molecular formula C₆H₇N is sparingly soluble in water. ‘A’ on treatment with dil. HCl gives a water-soluble compound ‘B’ ‘A’ also reacts with chloroform in presence of alcoholic KOH to form an obnoxious smelling compound ‘C’. ‘A’ reacts with benzene sulphonyl chloride to form an alkali-soluble compound ‘D’ ‘A’ reacts with NaNO₂ and HCl to form a compound ‘E’ which on reaction with phenol forms an orange-red dye ‘F’ Elucidate the structures of the organic compounds from ‘A’ to ‘F’
Answer:

Long answer type questions

Question 1.
Predict the reagent or the product in the following reaction sequence : [NCERT Exemplar]

Answer:
A correct sequence can be represented as follows including all reagents:

Hence,

(v) 5 = H₃PO₂/H₂O

Question 2.
A hydrocarbon ‘A’ (C₄H₈) on reaction with HC1 gives a compound ‘B’, (C₄H₉Cl), which on reaction with 1 mol of NH₃ gives compound ‘C’, (C₄H₁₁N). On reacting with NaNO₂ and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’ Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ to ‘D\ Explain the reactions involved. [NCERT Exemplar]
Answer:
(i)Addition of HCl to compound A shows that compound A is alkene.
Compound ‘B’ is C₄H₉Cl.
(ii) Compound‘B’reacts with NH₂. It forms amine‘C’.

(iii) ‘C’ gives diazonium salt with NaNO₂/HCl, which yields an optically active alcohol.
So, ‘C’ is aliphatic amine.
(iv) ‘A on ozonolysis produces 2 moles of CH₃CHO. So, A is CH₃– CH =CH-CH₃ (But-2-ene).
Reactions

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Very short answer type questions

Question 1.
Define mole fraction.
Answer:
Mole fraction of a component in a solution may be defined as the ratio of moles of that compdnent to the total number of moles of all the components present in the solution.

Question 2.
What is the similarity between Raoult’s law and Henry’s law?
Answer:
The similarity between Raoult’s law and Henry’s law is that both state that the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Question 3.
Why is the vapour pressure of a solution of glucose in water lower than that of water? (NCERT Exemplar)
Answer:
This is due to decrease in the escaping tendency of the water molecules from the surface of solution as some of the surface area is occupied by non-volatile solute, glucose particles.

Question 4.
What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone?
Answer:
Positive deviation.
Minimum boiling azeotropes.

Question 5.
State how does osmotic pressure vary with temperature.
Answer:
Osmotic pressure increases with increase in temperature.

Question 6.
What are isotonic solutions?
Answer:
The solutions of the same osmotic pressure at a given temperature are called isotonic solutions.

Question 7.
Define van’t Hoff factor.
Answer:
van’t Hoff factor may be defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property.

Question 8.
Why are aquatic species more comfortable in cold water in comparison to warm water? (NCERT Exemplar)
Answer:
Solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

Question 9.
What is semipermeable membrane? (NCERT Exemplar)
Answer:
Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules (water etc.) can pass, but solute molecules of bigger size cannot pass are called semipermeable membrane.

Question 10.
Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis. (NCERT Exemplar)
Answer:
Cellulose acetate.

Short answer type questions

Question 1.
State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid?
Answer:
It states that the partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p ∝ χ or p = K_Hχ where K_H is the
Henry’s constant.
Application of Henry’s law:
To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

Effect of temperature on solubility:
As dissolution is an exothermic process, therefore, according to Le Chatelier’s principle solubility should decrease with rise in temperature.

Question 2.
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Answer:
The addition of a non-volatile solute to a volatile solvent lowers its vapour pressure. In order to boil the solution, i.e., to make its vapour pressure equal to atmospheric pressure, the solution has to be heated at a higher temperature. In other words, the boiling point of solution becomes higher than solvent.

As elevation in boiling point depends on the number of moles of solute particles and independent of their nature, therefore, it is a colligative property.

Question 3.
Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water? Explain.
Answer:
No, the elevation in boiling point is not the same. NaCl, being an electrolyte, dissociates almost completely to give Na⁺ and Cl^– ions whereas glucose, being non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than 0.1 M glucose solution. Elevation in boiling point being a colligative property, is therefore, nearly twice for 0.1 M NaCl solution than for 0.1 M glucose solution.

Question 4.
Explain the solubility rule ‘like dissolves like’ in terms of intermolecular forces that exist in solutions. (NCERT Exemplar)
Answer:
If the intermolecular interactions are similar in both constituents, i.e., solute and solvent then solute dissolves in the solvent. e.g., polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, the statement ‘like dissolved like’ proves to be true.

Question 5.
Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature. However, molarity is a function of temperature. Explain.
(NCERT Exemplar)
Answer:
Molarity is defined as the number of moles of solute dissolved per litre of a solution. Since, volume depends on temperature and changes with change in temperature, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, so other concentration terms given in the question also do not do so. According to the definition of all these terms, mass of solvent used for making the solution is related to the mass of solute.

Question 6.
Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃ [Fe(CN)₆] Justify.
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.

Justification:
For 0.05 MAl₂(SO₄) ₃,i = 5
Number of particles = i × concentration
= 5 × 0.05
= 0.25 moles of ions
For 0.1MK₃ [Fe(CN)₆], i = 4
Number of particles = i × concentration
= 4 × 0.1
= 0.4 moles of ions
We know that, ΔT_f. Number of particles
Hence, 0.05 M Al₂(SO₄)₃ has higher freezing point because it has lower number of particles than 0.1 M K₃ [Fe (CN)₆].

Question 7.
The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (K_f for benzene = 5.12 K kg mol^-1)
Answer:
Given, ΔT_f = 2.12 K, K_f = 5.12 K kg mol^-1
We know that, ΔT_f = i K_f m
2.12 = \(\frac{i \times 5.12 \times 2.5 \times 1000}{122 \times 25}\)
or i = 0.505
For association,
i = 1 – \(\frac{\alpha}{2}\)
0.505 = 1 – \(\frac{\alpha}{2}\)
or α =0.99
Hence, percentage association of benzoic acid is 99%.

Long answer type questions

Question 1.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL^-1)
Answer
(a) A mixture of ethanol and acetone shows positive deviation from Raoult’s law.
In pure ethanol hydrogen bond exist between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of . ethanol thus breaking some of the hydrogen bonds and weakening molecular interactions considerably. Weakening of molecular interactions leads to increase in vapour pressure resulting in positive deviation from Raoult’s law.

(b) Let the mass of solution = 100 g
∴ Mass of glucose = 10 g
Number of moles of glucose \(=\frac{\text { Mass of glucose }}{\text { Molar mass }}\)
= \(\frac{10 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.056 mol

Question 2.
Discuss biological arid industrial importance of osmosis.
(NCERT Exemplar)
Answer:
The process of osmosis is of great biological and industrial importance as is evident from the following examples:

• Movement of water from soil into plant roots and subsequently into upper portion of the plant occurs partly due to osmosis.
• Preservation of meat against bacterial action by adding salt.
• Preservation of fruits against bacterial action by adding sugar.
  Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
• Reverse osmosis is used for desalination of water.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 14 Biomolecules

Very short answer type questions

Question 1.
What are oligosaccharides?
Answer:
Carbohydrates which on hydrolysis give two to ten molecules of monosaccharides are called oligosaccharides e. g., sucrose.

Question 2.
How do you explain the presence of all the six carbon atoms in glucose in a straight chain? (NCERT Exemplar)
Answer:
On prolonged heating with HI, glucose gives n-hexane.

Question 3.
Write the product obtained when D-glucose reacts with H₂N—OH.
Answer:

Question 4.
What are anomers?
Answer:
A pair of stereoisomers such as α-D-(+) glucose and β-D-(+) glucose which differ in configuration only around C₁ are called anomers.

Question 5.
What type of linkage is present in proteins?
Answer:

Question 6.
What are Vitamins?
Answer:
Vitamins are generally regarded as organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.

Question 7.
Why must vitamin C be supplied regularly in diet? [NCERT Exemplar]
Answer:
Vitamin C is water-soluble hence, they are regularly excreted in urine and can not be stored in our body, so, they are supplied regularly in diet.

Question 8.
Of the two bases named below, which one is present in RNA and which one is present in DNA? Thymine, Uracil.
Answer:

1. Thymine is present in DNA.
2. Uracil is present in RNA.

Question 9.
The activation energy for the acid-catalyzed hydrolysis of sucrose is 6.22 kJ mol^-1, while the activation energy is only 2.15 kJ mol-1 when hydrolysis is catalyzed by the enzyme sucrase. Explain. [NCERT Exemplar]
Answer:
Enzymes, the biocatalysts reduce the magnitude of activation energy by providing alternative paths. In the hydrolysis of sucrose, the enzyme sucrase reduces the activation energy from 6.22 kJ mol^-1 to 2.15 kJ mol^-1.

Question 10.
Name the bases present in RNA. Which one of these is not present in DNA?
Answer:
The bases present in RNA are Adenine (A), Guanine (G), Cytosine (C) and Uracil (U). Uracil is not present in DNA.

Short answer type questions

Question 1.
What is essentially the difference between α-glucose and β-glucose? What is meany by pyranose structure of glucose?
Answer:
α -Glucose and β-Glucose differ only in the configuration of hydroxy group at C₁ and are called anomers and the C₁ carbon is called anomeric carbon. The six-membered cyclic structure of glucose is called pyranose (α-or β), in analogy with pyran. The cyclic structure of glucose is more correctly represented by Haworth structure as given below:

Question 2.
Describe the term D- and L-configuration used for amino acids with examples. [NCERT Exemplar]
Answer:
All naturally occurring a-amino acids (except glycine) are optically active due to the presence of chiral carbon atom. These have either D- or  L-configuration. D-form means that the amino (-NH₂) group is present towards the right-hand side. L-form shows the presence of -NH₂ group on the left-hand side.

Question 3.
Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration? [NCERT Exemplar]
Answer:
In case of cyclic structure of saccharide, if -OH group present at second last carbon is present at bottom side, then it is considered as D configuration (as shown above).

Question 4.
How do enzymes help a substrate to be attacked by the reagent effectively? [NCERT Exemplar]
Answer:
At the surface of enzyme, active sites are present. These active sites of enzymes hold the substrate molecule in a suitable position so that it can be attacked by the reagent effectively. This reduces the magnitude of activation energy.

Enzymes contains cavities of characteristics shape and possessing active groups known as active centre on the surface. The molecules of the reactant (substrate) having complementary shape, fit into these cavities. On account of these active groups, an activated complex is formed which then decomposes to yield the products.

Question 5.
Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypeptide chain in proteins? [NCERT Exemplar]
Answer:
α-amino acid forms a polypeptide chain by elimination of water molecules.

Question 6.
(i) Winch vitamin deficiency causes rickets?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with acetic acid gives glucose Penta acetate. What does it suggest about the structure of glucose?
Answer:
(i) Vitamin D
(ii) Uracil
(iii) 5-OH groups are present in glucose.

Long answer type questions

Question 1.
Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
Answer:
(i) 5′ and 3′ carbon atoms of pentose sugar.
(ii) Most probably the resemblance of with 2 esters (-COO)² groups joined together.

(iii) Phosphoric acid (H₃PO₄).
Nucleosides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of pentose sugar and a dinucleotide with phosphoric acid (CH₃PO₄) is formed.

Question 2.
Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecules? Draw a diagram to show pairing of nucleotide bases in double helix of DNA. [NCERT Exemplar]
Answer:
On complete hydrolysis of DNA, following fragments are formed-pentose sugar (β-D-2-deoxyribose), phosphoric acid (H₃PO₄) and bases (nitrogen-containing heterocyclic compounds).
Structures
(i) Sugar

(ii) Phosphoric acid

(iii) Nitrogen bases: DNA contains four bases.
Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).

A unit formed by the attachment of a base to 1′-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5′-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar. In DNA two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases.

The two strands are complementary to each other because hydrogen bonds are formed between specific pairs of base adenine form hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 15 Polymers

Very short answer type questions

Question 1.
Which of the following is a natural polymer? Buna-S, Proteins, PVC
Answer:
Proteins.

Question 2.
Can enzyme be called a polymer? [NCERT Exemplar]
Answer:
Enzymes are biocatalysts which are proteins and are thus polymers.

Question 3.

Answer:
Homopolymer.

Question 4.
Identify the type of polymer -A-B-B-A-A-A-B-A- [NCERT Exemplar]
Answer:
Copolymer.

Question 5.
Out of chain growth polymerisation and step-growth polymerisation, in which type will you place the following: [NCERT Exemplar]

Answer:
Chain growth polymerisation, as there is no loss of small molecules like water; methanol, etc.

Question 6.
What is the role off-butyl peroxide in the polymerisation of ethene?
Answer:
It acts as a free radical generating initiator in the chain initiation step of polymerisation of ethene.

Question 7.
Can nucleic acids, proteins and starch be considered as step growth polymers? [NCERT Exemplar]
Answer:
Yes, step-growth polymers are condensation polymers and they are formed by the loss of simple molecules like water leading to the formation of high molecular mass polymers.

Question 8.
Write the structure of the monomer used for getting the melamine-formaldehyde polymer.
Answer:
Melamine and formaldehyde

Question 9.
How is the following resin intermediate prepared and which polymer is formed by this monomer unit?

Answer:
Melamine and formaldehyde are starting materials for this intermediate. Its polymerisation gives melamine polymer.

Question 10.
Why does cis-polyisoprene possess elastic property? [NCERT Exemplar]
Answer:
cis-polyisoprene is also known’ as natural rubber. Its elastic property is due to the existence of weak van der Waals’ interactions between their various polymer chains.

Short answer type questions

Question 1.
Which of the following polymers soften on heating and harden on cooling? What are the polymers with this property collectively called? What are the structural similarities between such polymers? Bakelite, urea-formaldehyde resin, polythene, polyvinyls, polystyrene. [NCERT Exemplar]
Answer:
Polythene, polyvinyls and polystyrene soften on heating and harden on cooling. Such polymers are called thermoplastic polymers. These polymers are linear or slightly branched long-chain molecules. These possess intermolecular forces whose strength lies between strength of intermolecular forces of elastomers and fibres.

Question 2.
What is the role of benzoyl peroxide in addition polymerisation of alkenes? Explain its mode of action with the help of an example. [NCERT Exemplar]
Answer:
Role of benzoyl peroxide is to initiate the free radical polymerisation reaction which can be easily understood by taking an example of polymerisation of ethene of polythene.
(i) Chain initiation

(ii) Chain propagation

(iii) Chain terminator step

Question 3.
Low-density polythene and high-density polythene, both are polymers of ethene but there is marked difference in their properties. Explain. [NCERT Exemplar]
Answer:
Low density and high-density polythenes are obtained under different conditions. These differ in their structural features. Low-density polythenes are highly branched structures while high-density polythene consists of closely packed linear molecules. Close packing increases the density.

Question 4.
Differentiate between rubbers and plastics on the basis of intermolecular forces. [NCERT Exemplar]
Answer:
Rubber is a natural polymer which possess elastic properties. Natural polymer is a linear polymer of isoprene (2-methyl-1, 3-butadiene).
In natural rubber cis-polyisoprene molecules consists of various chains held together by weak van der Waals’ interaction and has coiling structure. So, it can be stretched like a spring.

Plastics are generally polymers of ethene known as polythene. Polythene is thermoplastic polymer which may be linear (HDP) or branched (LDP) these type of polymers. Possesses intermediate intermolecular forces of attraction. It has linear, structure that can be moulded but can’t be regained on its original shape after stretching.

Question 5.
A natural linear polymer of 2-methyl-l, 3-butadiene becomes hard on treatment with sulphur between 373 to 415 K and -S-S- bonds are formed between chains. Write the structure of the product of this treatment? [NCERT Exemplar]
Answer:
The product is called vulcanised rubber. Its structure is as follows :

Question 6.
Name the type of reaction involved in the formation of the following polymers from their respective monomers
(i) PVC.
(ii) Nylon6.
(iii) PHBV.
Answer:
(i) Addition
(ii) Condensation/Hydrolysis
(iii) Condensation.

Long answer type questions

Question 1.
Explain the following terms giving a suitable example for each:
(i) Elastomers
(ii) Condensation polymers
(iii) Addition polymers
Answer:
(i) Elastomers: These are the polymers having the weakest intermolecular forces of attraction between the polymer chains. The weak forces permit the polymer to be stretched. A few ‘cross links’ are introduced between the chains, which help the polymer to retract to its original position after the force is released as in vulcanised rubber. Elastomers thus possess an elastic character. For example, buna-S, buna-N, neoprene, etc.

(ii) Condensation polymers: The condensation polymers are formed by the repeated condensation reaction between different bi-functional or tri-functional monomer units usually with elimination of small molecules such as water, alcohol, hydrogen chloride, etc. For example, Nylon-6,6, nylon 6, terylene, etc.

(iii) Addition polymers: Addition polymers are formed by repeated addition of same or different monomer molecules. The monomers used are unsaturated compounds. For example, alkenes alkadienes and their derivatives. Polythene is an example of addition polymer.

Question 2.
Phenol and formaldehyde undergo condensation to give a polymer (A) which on heating with formaldehyde gives a thermosetting polymer (B). Name the polymers. Write the reactions involved in the formation of (A). What is the structural difference between two polymers? [NCERT Exemplar]
Answer:
Phenol and formaldehyde undergo condensation to give a polymer novolac (A) which on heating with formaldehyde gives bakelite (B) as a thermosetting polymer.
A sequence of the reaction can be written as follows :

Structural difference in between these two is that novolac is a linear polymer while bakelite is a cross-linked polymer.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Very short answer type questions

Question 1.
Give an example each of a molecular solid and an ionic solid.
Answer:
Molecular solids: CO₂, I₂, HCl
Ionic solids: NaCl, ZnS, CaF₂

Question 2.
Why does the window glass of the old buildings look milky?
Answer:
It is due to heating during the day and cooling at night, i.e., due to annealing over a number of years, glass acquires crystalline character.

Question 3.
What would be the nature of solid if there is no energy gap between valence band and conduction band?
Answer:
Conductor.

Question 4.
Express the relationship between atomic radius (r) and the edge length (a) in the bcc unit cell,
Answer:
Atomic radius (r) = \(\frac{\sqrt{3}}{4}\) a (edge length of unit cell).

Question 5.
Which point defect in its crystal units increases the density of a solid?
Answer:
Interstitial defect.

Question 6.
What is meant by the term ‘forbidden zone’ in reference to band theory of solids?
Answer:
The energy gap between valence band and conduction band is known as forbidden zone.

Question 7.
‘Crystalline solids are anisotropic in nature.’ What does this statement mean?
Answer:
It means that some of their physical properties like electrical conductivity, refractive index, etc., are diferent in different directions.

Question 8.
Why does the electrical conductivity of semiconductors increase with rise in temperature? [NCERT Exemplar)
Answer:
The gap between conduction band and valence band is small in semiconductors. Therefore, electrons from the valence band can jump to the conduction band on increasing temperature. Thus, they become more conducting as the temperature increases.

Question 9.
Why does table salt NaCl sometimes appear yellow in colour?
(NCERT Exemplar)
Answer:
Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic sites, known as F-centres. These electrons absorb energy from the visible region for the excitation which makes crystal appear yellow.

Question 10.
Why are liquids and gases categorised as fluids? (NCERT Exemplar)
Answer:
Liquids and gases have the tendency to flow, i.e., their molecules can move freely from one place to another. Therefore, they are known as fluids.

Short answer type questions

Question 1.
(i) What type of stoichiometric defect is shown by KC1 and why?
(ii) What type of semiconductor is formed when silicon is doped with As?
(iii) Which one of the following is an example of molecular solid : CO₂ or SiO₂?
Answer:
(i) KCl shows Schottky defect as the cation, K⁺ and anion, Cl^– are of almost similar sizes.
(ii) n-type semiconductor.
(iii) CO₂

Question 2.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm^-3. Use this information to calculate Avogadro’s number. (At. mass of Fe = 55.845 u)
Solution:
Given, a =286.65 pm = 286.65 × 10^-10 cm; M = 55.845 g mol^-1;
d = 7.874 g cm^-3
For bee unit cell, z = 2
Substituting the values in the expression, N_A = \(\frac{z \times M}{a^{3} \times d}\), we get
N_A = \(\frac{2 \times 55.845 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(286.65 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 7.874 \mathrm{~g} \mathrm{~cm}^{-3}}\)
N_A = 6.022 × 10²³ mol^-1

Question 3.
An element crystallises in a fee lattice with cell edge of 400 pm. The density of the element is 7 g cm^-3. How many atoms are present in 280 g of the element?
Solution:
Given, a = 400 pm = 400 × 10^-10 cm = 4 x× 10^-8 cm
Volume of the unit cell = a³
= (4 × 10^-8 cm)³ = 6.4 × 10^-23cm³

Since each fee unit cell contains 4 atoms, therefore, the total number of atoms in 280 g = 4 × 6.25 × 10²³ = 2.5 × 10²⁴ atoms

Question 4.
An element crystallises in a fee lattice with cell edge of 400 pm. Calculate the density if 200 g of this element contain 2.5 × 10²⁴ atoms.

Molar mass M = 48.18 g mol^-1
Here, z = 4, M = 48.18 g mol^-1, N_A = 6.022 × 10²³ mol^-1
a = 400 pm = 400 × 10^-10 cm = 4 × 10^-8 cm
Substituting these values in the expression,
d = \(\frac{z \times M}{a^{3} \times N_{A}}\) , we get
d = \(\frac{4 \times 48.18 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\) = 5 g cm^-3

Question 5.
Explain why does conductivity of germanium crystals increase on doping with galium?
Answer:
On doping germanium with galium some of the positions of lattice of germanium are occupied be galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearly germanium atom is not satisfied. The place remains vacant. This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position. Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Long answer type questions

Question 1.
A sample of ferrous oxide has actual formula Fe_0.93O_1.00In this
sample what fraction of metal ions are Fe²⁺ ions? What type of non-stoichiometric defect is present in this sample? (NCERT Exemplar)
Solution:
Let the formula of the sample be (Fe²⁺ )x (Fe³⁺ )_y O
On looking at the given formula of the compound
x + y = 0.93 ………….. (i)
Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen.
Therefore, 2x + 3y = 2 ……….. (ii)
x + \(\frac{3}{2}\)y = 1 ……………..(iii)
On subtracting equation (i) from equation (iii) we have
\(\frac{3}{2}\) y – y = 1 – 0.93 ⇒ \(\frac{1}{2}\)y = 0.07 ⇒ y = 0.14
On putting the value of y in equation (i), we get
x + 0.14 = 0.93
⇒ x = 0.93 – 0.14 ⇒ x = 0.79
Fraction of Fe2+ ions present in the sample = \(\frac{0.79}{0.93}\) = 0.849
Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.

Question 2.
(i) Following is the schematic alignment of magnetic moments:

Identify the type of magnetism. What happens when these substances are heated?
(ii) If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’ What is the relation between ‘r’ and ‘R’?
(iii) Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm. What is the radius of tungsten atom?
Answer:
(i) The given schematic alignment of magnetic moments shows ferrimagnetism. When these substances are heated they lose ferrimagnetism and become paramagnetic.

(ii) The radius of the octahedral void = r
The radius of the atoms in close packing = R
Relation between r and R is given as :
r = 0.414 R

(iii) Given, a = 316.5 pm
Vs
We know that for body centred cubic unit cell r = \(\frac{\sqrt{3}}{4}\) a
∴ r = \(\frac{\sqrt{3}}{4}\) × 316.5 pm = 136.88 pm

Question 3.
(i) Identify the type of defect shown in the following figure:

What type of substances show this defect?
(ii) A metal crystallises in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’?
(iii) An element with molar mass 63 g/ mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/cm 3. What is the nature of the cubic unit cell?
Answer:
(i) The given figure shows Schottky defect. This defect arises when equal number of cations and anions are missing from the lattice. It is a common defect in ionic compounds of high coordination number.
Where born cations and anion are of the same size, e.g., KCl, NaCl, KBr etc.

(ii) Edge length of the unit cell = a
Radius of the sphere = r
For body centred cubic structure ,
r = \(\frac{\sqrt{3}}{4}\)

(iii) We know that,
Density d = \(\frac{z \times M}{a^{3} \times N_{A}}\) or z = \(\frac{d \times a^{3} \times N_{A}}{M}\)
Given,
M = 63 g-mol^-1 6.3 × 10^-2 kg mol^-1
a = 360.8 pm = 360.8 × 10^-12 m = 3.608 × 10^-10m
d = 8.92 g/cm² = 0.892 kgm^-3
N_A = 6.022 × 10²³ mol^-1
On putting the given values in formula,

= 3.97 ≅ 4.
Since, 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centered.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 16 Chemistry in Everyday Life

Very Short Answer Type Questions

Question 1.
Where are receptors located? [NCERT Exemplar]
Answer:
Receptors are embedded in cell membrane.

Question 2.
Which site of an enzyme is called allosteric site? [NCERT Exemplar]
Answer:
Sites different from active site of enzyme where a molecule can bind and affect the active site is called allosteric site.

Question 3.
What is the harmful effect of hyperacidity? [NCERT Exemplar]
Answer:
Ulcer development in stomach.

Question 4.
Write the name of an antacid which is often used as a medicine.
Answer:
Ranitidine (Zantac).

Question 5.
What is the medicinal use of narcotic drugs? [NCERT Exemplar]
Answer:
Since narcotic drugs relieve pain and produce sleep, these are chiefly used for the relief of post-operative pain, cardiac pain and pain of terminal cancer and in childbirth.

Question 6.
Which type of drugs come under antimicrobial drugs? [NCERT Exemplar]
Answer:
Antiseptics, antibiotics and disinfectants.

Question 7.
What is the mode of action of antimicrobial drugs? [NCERT Exemplar]
Answer:
Antimicrobial drugs can kill the microorganism such as bacteria, virus, fungi or other parasites. They can, alternatively, inhibit the pathogenic action of microbes.

Question 8.
Which one of the following drugs is an antibiotic? Morphine, Equanil, Chloramphenicol, Aspirin
Answer:
Chloramphenicol.

Question 9.
What is meant by ‘narrow-spectrum antibiotics’?
Answer:
Antibiotics which are mainly effective against Gram-positive or Gram-negative bacteria are known as narrow-spectrum antibiotics. For , example, penicillin G.

Question 10.
Define the limited spectrum antibiotics.
Answer:
Antibiotics which are mainly effective against a single organism or disease, are called as limited spectrum antibiotics.

Short answer type questions

Question 1.
Why are certain drugs called enzyme inhibitors?[NCERT Exemplar]
Answer:
Enzymes have active sites that bind the substrate for effective and quick chemical reactions. The functional groups present at the active site of enzyme interact with functional groups of substrate via ionic bonding, hydrogen bonding, van der Waal’s interaction, etc. Some drugs interfere with this interaction by blocking the binding site of enzyme and prevent the binding of actual substrate with enzyme. This inhibits the catalytic activity of the enzyme, therefore, these are called inhibitors.

Question 2.
Sodium salts of some acids are very useful as food preservatives. Suggest a few such acids. [NCERT Exemplar]
Answer:
A preservative is naturally occurring or synthetically produced substance that is added to foods to prevent decomposition by microbial growth or by undesirable chemical changes. Sodium salts of some acids are very useful as food preservatives.

Some examples of such acids are as follows :

• Benzoic acid in the form of its sodium salts constitutes one of the most common food preservatives. Sodium benzoate is a common preservative in acid or acidified foods such as fruit, juices, pickles etc. Yeasts are inhibited by benzoate to a greater extent than are moulds and bacteria.
• Sorbic acid and its salts (sodium, potassium, and calcium) also have preservative activities but the applications of -sodium sorbate (C₆H₇NaO₂) are limited compared to that for potassium salt.
• Sodium erythorbate (C₆H₇NaO₆) is a food additive used predominate in meats, poultry and soft drinks.
• Sodium propanoate[Na(C₂H₅COO)] is used in bakery products as mould inhibitor.

Question 3.
What is the side product of soap industry? Give reactions showing soap formation. [NCERT Exemplar]
Answer:
Soaps are sodium or potassium salts of long-chain fatty acids such as stearic acid, oleic acid and palmitic acid. Soaps containing sodium salts are formed by heating fat (i. e., glyceryl ester of fatty acid) with aqueous sodium hydroxide solution.

This reaction is known as saponification. In this reaction, esters of fatty acids are hydrolyzed and the soap obtained remains in colloidal form. It is precipitated from the solution by adding NaCl. The solution left after removing the soap contains glycerol as side product.

Question 4.
Hair shampoos belong to which class of synthetic detergent? [NCERT Exemplar]
Answer:
Hair shampoos are made up of cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions, e. g., cetyltrimethylammonium bromide.

Question 5.
Explain why sometimes foaming is seen in river water near the place where sewage water is poured after treatment?
[NCERT Exemplar]
Answer:
Detergents have long hydrocarbon chains. If their hydrocarbon chain is highly branched, then bacteria cannot degrade this easily. Such detergents are non-biodegradable. Slow degradation of detergents leads to their accumulation.

These non-biodegradable detergents persist in water even after sewage treatment and cause foaming in rivers, ponds and their water get polluted. In order to overcome this issue branching of the hydrocarbon chain is controlled and kept to a minimum.

Long answer type questions

Question 1.
What are enzyme inhibitors? Classify them on the basis of their mode of attachment on the active site of enzymes. With the help of diagrams explain how do inhibitors inhibit the enzymatic activity? [NCERT Exemplar]
Answer:
Enzymes are responsible to hold the substrate molecule for a chemical reaction and they provide functional groups which will attack the substrate to carry out the chemical reaction. Drugs which inhibit any of the two activities of enzymes are called enzyme inhibitors.

Enzyme inhibitors can block the binding site thereby preventing the binding of the substrate to the active site and hence inhibiting the catalytic activity of the enzyme.
Drugs inhibit the attachment of natural substrate on the active site of enzymes in two different ways as explained below :
(i) Drugs which compete with natural substrate for their attachment on the active sites of enzymes are called competitive inhibitors.

(ii) Some drugs, however, do not bind to the active site but bind to a different site of the enzyme which is called allosteric site. This binding of the drug at allosteric site changes the shape of the active site of the enzyme in such a way that the natural substrate cannot recognize it. Such enzymes are called non-competitive inhibitors.

Question 2.
In what respect to prontosil and salvarsan resemble? Is there any resemblance between azo dye and prontosil? Explain. [NCERT Exemplar]
Answer:
Prontosil, also called sulfamide chrysoidine, trade name of the first synthetic drug used in the treatment of general bacterial infections in humans. Prontosil resulted from research, directed by German chemist and pathologist Gerhard Domagk, on the antibacterial action of azo dyes. A red azo dye of low toxicity, prontosil was shown by Domagk to prevent mortality in mice infected with Streptococcus bacteria.

The dye was also effective in controlling staphylococcus infections in rabbits. Within a relatively short period, it was demonstrated that prontosil was effective not only in combating experimental infections in animals but also against Streptococcal disease in humans, including meningitis and puerperal sepsis. Structural formula of prontosil is as follows :

From the structure of prontosil, it is very clear that it has -N = N- linkage. It was discovered that the part of the structure of prontosil molecule shown inbox, i.e., p-amino benzene sulphonamide has antibacterial activity.
Salvarsan is also known as arsphenamine. It was introduced at the beginning of 1910s as the first effective treatment for syphilis. It is an organoarsenic molecule and has -As = As- double bond.

Salvarsan and prontosil show similarity in their structure. Both of these drugs are antimicrobials. Salvarsan contains -As = As- linkage whereas prontosil has —N = N— linkage.

Prontosil (a red azo dye) and azo dye both have -N = N- linkage.

Question 3.
Ashwin observed that his friend Shubhain was staying aloof, not playing with friends, and becoming easily irritable for some weeks. Ashwin told his teacher about this, who, in turn, called Shubham’s parents and advised them to consult a doctor. The doctor after examining Shubham prescribed antidepressant drugs for him.
After reading the above passage, answer the following questions:
(i) Name two antidepressant drugs.
(ii) Mention the values shown by Ashwin.
(iii) How should Shubham’s family help him other than providing medicine?
(iv) What is the scientific explanation for the feeling of depression?
Answer:
(i) Equanil, Iproniazid, phenelzine (any two)
(ii) Empathetic, caring, sensitive.
(iii) They should talk to him, be a patient listener, can discuss the matter with the psychologist.
(iv) If the level of noradrenaline is low, then the signal sending activity becomes low and the person suffers from depression.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 8 Electromagnetic Waves Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 8 Electromagnetic Waves

Very short answer type questions

Question 1.
How are radiowaves produced?
Answer:
They are produced by rapid accelerations and deaccelerations of electrons in aerials.

Question 2.
How are microwaves produced?
Answer:
By using a magnetron.

Question 3.
Write two uses of microwaves.
Answer:
Uses of Microwaves

• In RADAR communication.
• In analysis of molecular and atomic structure.

Question 4.
To which part of the electromagnetic spectrum does a wave of frequency 3 × 10¹³ Hz belong?
Answer:
The frequency of 3 × 10¹³ Hz belongs to the infrared waves.

Question 5.
Name the electromagnetic waves, which (i) maintain the Earth’s warmth and (ii) are used in aircraft navigation.
Answer:
(i) Infrared rays
(ii) Microwaves

Question 6.
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiation. Name the radiations and write the range of their frequency.
Answer:
Welders wear special goggles or face mask with glass windows to protect their eyes from ultraviolet rays. The range of UV rays is 4 × 10^-7 m (400 nm) to 6 x 10^-10 m (0.6 nm).

Question 7.
How are X-rays produced?
Answer:
X-rays are produced when high energetic electron beam is made incident on a metallic target of high melting point and high atomic weight.

Question 8.
Write two uses of X-rays.
Answer:
Uses of X-rays

• In medical diagnosis as they pass through the muscles not through the bones.
• In detecting faults, cracks, etc. in metal products.

Question 9.
A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency? (NCERT Exemplar)
Answer:
On decreasing the frequency, reactance X_C = \(\frac{1}{\omega C}\) will increase which will lead to decrease in conduction current. In this case I_d = I_c, hence displacement current will decrease.

Question 10.
Do electromagnetic waves carry energy and momentum?
Answer:
Yes. Electromagnetic waves carry energy and momentum.

Question 11.
Why is the orientation of the portable radio with respect to broadcasting station important? (NCERT Exemplar)
Answer:
As electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric/magnetic part of the wave.

Question 12.
The charge on a parallel plate capacitor varies as q = q₀ cos 2πvt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor? (NCERT Exemplar)
Answer:
Conduction current I_C = Displacement current I_D
I_C = I_D = \(\frac{d q}{d t}\) = \(\frac{d}{d t}\) (q₀ cos 2π vt) = -2πcq₀vsin2πvt

Question 13.
Professor C.V. Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of electromagnetic waves was he exhibiting? Give one more example of this property. (NCERT Exemplar)
Answer:
Electromagnetic waves exert radiation pressure. Tails of comets are due to solar radiation.

Short answer type questions

Question 1.
Write the generalised expression for the Ampere’s circuital law in terms of the conduction current and the displacement current. Mention the situation when there is
(i) only conduction current and no displacement current,
(ii) only displacement current and no conduction current.
Answer:
Generalised Ampere’s Circuital Law
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c + μ₀ε₀\(\frac{d \phi_{E}}{d t}\)
Line integral of magnetic field over closed loop is equal to p 0 times sum of conduction current and displacement current.

(i) In case of steady electric field in a conducting wire, electric field does not change with time, conduction current exists in the wire but displacement current may be zero.
So \(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c

(ii) In large region of space, where there is no conduction current, but there is only a displacement current due to time varying electric field (or flux).
So Φ \(\vec{B} \cdot \overrightarrow{d l}\) = μ₀ ε₀ \(\frac{d \phi_{E}}{d t}\)

Question 2.
How are infrared waves produced? Why are these referred to as ‘heat waves’? Write their one important use.
Answer:
Infrared waves are produced by hot bodies and molecules. Infrared waves are sometimes referred to as heatwaves. This is because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is they heat up and heat their surroundings.
Infrared lamps are used in physical therapy and in remote control of devices.

Question 3.
(i) Arrange the following electromagnetic waves in the descending order of their wavelength.
(a) Microwaves
(b) Infrared rays
(c) Ultraviolet radiation
(d) γ-rays
(ii) Write one use each of any two of them.
Answer:
(i) The decreasing order ofwavelength of electromagnetic waves are Microwaves > Infrared > Ultraviolet > y-rays

(ii) Microwaves: They are used in RADAR devices,
γ-rays: It is used in radio therapy.

Question 4.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is
i = ε₀\(\frac{d \phi_{\boldsymbol{E}}}{d t}\)
Where Φ_E is the electric flux produced during charging of the capacitor plates.
Answer:
Ampere’s circuital law is given by
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c
For a circuit containing capacitor, during its charging or discharging the current within the plates of the capacitor varies producing displacement current I_d Hence, Ampere’s circuital law is generalised by Maxwell, given as
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c + μ₀I_d
The electric flux (Φ_E) between the plates of capacitor changes with time, producing current within the plates which is proportional to (\(\frac{d \phi_{E}}{d t}\))
Thus, we get,
I_c = ε₀ \(\frac{d \phi_{E}}{d t}\)ε

Question 5.
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves.
Answer:
Electromagnetic wave produced by oscillating charged particle. Mathematical expression for electromagnetic wave travel along z-axis:
E_x = E₀ sin(kz – ωt) [For electric field]
B_y = B₀ sin(kz – ωt) [For magnetic field]
Properties
(i) Have oscillating electric perpendicular direction.
(ii) Transverse nature.

Question 6.
Electromagnetic waves with wavelength
(i) λ₁ is used in satellite communication.
(ii) λ₂ is used to kill germs in water purifier.
(iii) λ₃ is used to detect leakage of oil in underground pipelines.
(iv) λ₄ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each. (NCERTExemplar)
(a) λ₁ → Microwave, λ₂ → UV
λ₃ → X rays, λ₄ → Infrared

(b) λ₃ < λ₂ <λ₄ < λ₁

(c) Microwave-RADAR
UV-LASIK eye surgery
X-ray-Bone fracture identification (bone scanning)
Infrared-Optical communicatio

Long answer type questions

Question 1.
Draw a labelled diagram of Hertz’s experiment. Explain how electromagnetic radiations are produced using this set-up.
Answer:
Hertz Experiment: Hertz’s experiment was based on the fact that an oscillating electric charge radiates electromagnetic waves and these waves carry energy which is being supplied at the cost of K.E. of the oscillating charge.

Hertz Apparatus: The experimental arrangement used by Hertz for the production and detection of electromagnetic waves in the laboratory, is shown in fig. His experimental arrangement consists of two metal sheets P₁ and P₂. These sheets are connected to a source of very high voltage (i.e. an induction coil, which can supply a potential difference of several thousand volts). S₁ and S₂ are two metal spheres connected to the metal sheets P₁ and P₂ . The distance between the metal sheets is kept nearly 60 cm and that between the sphere is normally from 2 cm to 2.5 cm.

The two plates P₁ and P₂ form a capacitor of very low capacitance (C). The circuit containing P₁ and P₂ (being completed by conducting wire), has also some low value of inductance L. It thus forms an LC circuit. Detector (D) consisting of a coil to the ends of which two other small metal spheres S₁‘and S₂‘ are connected.

Working of Hertz Apparatus: Due to existence of very high voltage, air present in the gap across the plates of spheres S₁ and S₂ gets ionised. Due to presence of the ions or charged particles, the path between the spheres S₁ and S₂ become conducting. As a result of this, very high time- varying current flows across the gap between S₁ and S₂ (as plates P₁ and P₂ form an LC circuit). Due to this a spark is produced. Since, sheets P₁ ,
P₂ form an LC-circuit, hence, electromagnetic waves of frequency f = \(\frac{1}{2 \pi} \sqrt{\frac{1}{L C}}\)

Function of the Detector D: Hertz detected the electromagnetic waves by means of a detector D kept at suitable distance from the conducting spheres S₁, S₂ Detector D is made of two similar conducting spheres S₁‘ and S₂‘ joined to the ends of a coil to form another LC circuit.

The frequency of this LC circuit is made equal to the frequency of electromagnetic waves reaching it. The frequency can be adjusted by changing the diameter of the coil of the detector and by changing the distance between S₁‘ and S₂‘. Hertz placed the detector in such a way that the magnetic lines of force produced by the oscillating electric field across the gap between S₁‘ and S₂‘ are normal to the plane of coil (C). When magnetic lines of force cut the detector coil, an emf is induced in it. Hence, air in the gap between S₁‘ and S₂‘ gets ionised. A conducting path becomes available for the induced current to flow across the gap. Thus, the spark is produced between S₁‘ and S₂‘. Hertz also observed that the spark across S₁‘ and S₂‘ was greatest when the S₁‘ S₂‘ and S₁ S₂ were parallel to each other. This clearly established that electromagnetic waves produced were polarise i.e., \(\vec{E}\) and \(\vec{B}\) always lie in one plane.