CBSE Sample Papers Class 11 Chemistry 2025

CBSE Sample Papers Class 11 Chemistry (2024-25)

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Class 11 – Chemistry
Sample Paper – 01 (2024-25)

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Maximum Marks: 70
Time Allowed: : 3 hours

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General Instructions:

1. There are 33 questions in this question paper with internal choice.
2. SECTION A consists of 16 multiple-choice questions carrying 1 mark each.
3. SECTION B consists of 5 very short answer questions carrying 2 marks each.
4. SECTION C consists of 7 short answer questions carrying 3 marks each.
5. SECTION D consists of 2 case-based questions carrying 4 marks each.
6. SECTION E consists of 3 long answer questions carrying 5 marks each.
7. All questions are compulsory.
8. The use of log tables and calculators is not allowed

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1. Section A
2. Consider the chemical reaction gven as,
   –CO (g) + — {tex}H_2{/tex}(g) → — {tex}C_8{/tex}H₁₈(l) + — {tex}H_2O{/tex}.
   This equation can be balanced by inserting the following in blank spaces

   a)8,2,1,17

   b)8,17,1,8

   c)8,8,1,17

   d)8,2,8,17

3. In an alpha scattering experiment, few alpha particles rebounded because

   a)The mass of the atom is concentrated in the centre

   b)Positive charge of the atoms very little space

   c)All the positive charge and mass of the atom is concentrated in small volume

   d)Most of the space in the atom is occupied

4. Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct?
   1. C (g) + 4 H (g) {tex} \to {\text{ C}}{{\text{H}}_{\text{4}}}\left( {\text{g}} \right);{\text{ }}{\Delta _{\text{r}}}{\text{H }} = {\text{ x kJ mo}}{{\text{l}}^{-{\text{1}}}}{/tex}
   2. C (graphite,s) + {tex}{\text{2}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{ }} \to {\text{ C}}{{\text{H}}_{\text{4}}}\left( {\text{g}} \right);{\text{ }}{\Delta _{\text{r}}}{\text{H }} = {\text{ y kJ mo}}{{\text{l}}^{-{\text{1}}}}{/tex}

   a)x = y

   b)x = 2y

   c)x < y

   d)x > y

5. Give the number of electrons in the species, {tex}O_2{/tex} and {tex}O_2^+{/tex}.

   a)16 and 8

   b)16 and 14

   c)16 and 15

   d)32 and 16

6. Which of the following always has a negative value?

   a)Heat of reaction

   b)Heat of solution

   c)Heat of formation

   d)Heat of combustion

7. What will be the Electronic configuration of the element having atomic number 24?

   a)1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s¹ 4p¹

   b)1s² 2s² 2p³ 3s² 3p⁶ 3d⁵ 4s¹ 4p³

   c)1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

   d)1s² 2s² 2p⁶ 3s² 3p³ 3d⁵ 4s¹ 4p³

8. On the reaction
   2Ag + 2H₂SO₄ {tex}\longrightarrow{/tex} Ag₂SO₄ + 2H₂O + SO₂ sulphuric acid acts as an?

   a)a reducing agent

   b)an oxidizing agent

   c)a catalyst

   d)an acid as well as an oxidant

9. The reaction
   {tex}C{H_3}CH = C{H_2} + HBr\xrightarrow{Peroxide} C{H_3}C{H_2}C{H_2}Br{/tex}  can be explained by

   a)carbanion formation

   b)free-radical mechanism

   c)carbocation formation

   d)electrophilic substitution

10. Which of the following organic materials damage DNA of our body?

    a)Tabacco

    b)Coal

    c)All of these

    d)Petroleum

11. In terms of period and group, where would you locate the element with Z = 114 in the modern periodic table?

    a)7th period and 14th group

    b)6th period 14th group

    c)6th period and 13th group

    d)7th period and 13th group

12. For the reaction at 298 K, 2A + B {tex}\to{/tex} C, {tex}\Delta{/tex}H = 400 kJ mol^-1 and {tex}\Delta {\text{S}}{/tex} = 0.2 kJ K^-1mol^-1. At what temperature will the reaction become spontaneous considering {tex}\Delta {\text{H}}{/tex} and {tex}\Delta {\text{S}}{/tex} to be constant over the temperature range.

    a)3500 K

    b)2000 K

    c)1500 K

    d)2500 K

13. Soda lime is a mixture of

    a)NaOH + Ca(OH)₂

    b)NaOH + MgO

    c)NaOH + Mg(OH)₂

    d)NAOH + CaO

14. Assertion (A): Allyl and benzyl carbonium ions are stable than propyl corbonium ions.
    Reason (R): Electron releasing groups stabilize cabonium ions.

    a)Both A and R are true and R is the correct explanation of A.

    b)Both A and R are true but R is not the correct explanation of A.

    c)A is true but R is false.

    d)A is false but R is true.

15. Assertion (A): Trans-2-butene on reaction with Br₂ gives a meso-2,3-dibromobutane.
    Reason (R): The reaction involves the syn-addition of bromine.

    a)Both A and R are true and R is the correct explanation of A.

    b)Both A and R are true but R is not the correct explanation of A.

    c)A is true but R is false.

    d)A is false but R is true.

16. Assertion (A): Threshold frequency is a characteristic for a metal.
    Reason (R): Threshold frequency is the maximum frequency required for the ejection of electrons from the metal surface.

    a)Both A and R are true and R is the correct explanation of A.

    b)Both A and R are true but R is not the correct explanation of A.

    c)A is true but R is false.

    d)A is false but R is true.

17. Assertion (A): Under similar conditions of temperature and pressure, gases combine in a simple ratio of their volume but, not always.
    Reason (R): Gases deviate from ideal behaviour.

    a)Both A and R are true and R is the correct explanation of A.

    b)Both A and R are true but R is not the correct explanation of A.

    c)A is true but R is false.

    d)A is false but R is true.

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    Section B

18. Calculate the solubility of A₂X₃ in pure water, assuming that neither kind of ion reacts with water. The solubility product of A₂X₃, K_sp = 1.1 {tex}\times{/tex} 10^–23.
19. Helium has an electronic configuration of 1s² but it is placed in p-block in group 18. Explain.
20. How many moles of methane are required to produce 22g CO₂ (g) after combustion?
21. Explain why the system are not aromatic.
    

    OR

    What happens when benzene is oxidized at 770K in presence of V₂O₅? Give chemical equation.

22. In an atom, an electron is moving with a speed of 600 ms^-1 with an accuracy of 0.005 %. Find the certainty with which the position of the electron can be located. (h = 6.6 {tex}\times{/tex} 10^-34 kg m²s^-1, mass of electron m_e = 9.1 {tex}\times{/tex} 10^-31 kg).
23. Section C
24. Draw the resonating structure of
    1. ozone molecule
    2. nitrate ion
25. Answer:
    1. Consider the same expansion, but this time against a constant external pressure of 1 atm.
    2. At what temperature entropy of a substance is zero?
    3. Predict the change in internal energy for an isolated system at constant volume.
    1. Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
    2. Under what conditions will the reaction occur, if
       1. both {tex}\Delta{/tex}H and {tex}\Delta{/tex}S are positive
       2. both {tex}\Delta{/tex}H and {tex}\Delta{/tex}S are negative
26. Write the anode reaction, the cathode reaction and the net cell reaction in the following cells. Which electrode would be positive terminal in each cell?
    1. Zn(s) | Zn²⁺ || Br₂ | Br^– | Pt(s)
    2. Cr(s) | Cr³⁺ || I₂ | I^– | Pt(s)
    3. Pt(s) | H₂(g) | H⁺ (aq) || Cu²⁺ | Cu(s)
27. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29 {tex}\times{/tex} 10¹⁵ (Hz) [1/3² – 1/n²] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
28. Among the elements B, AI, C and Si
    1. Which has the highest first ionization enthalpy?
    2. Which has the most negative electron gain enthalpy?
    3. Which has the largest atomic radius?
    4. Which has the most metallic character?
29. Calculate the molecular mass of the following.
    1. H₂O
    2. CO₂
    3. CH₄
30. Section D
31. Read the following text carefully and answer the questions that follow:
    Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. Finally, the purity of a compound is ascertained by determining its melting or boiling point. This is one of the most commonly used techniques for the purification of solid organic compounds. In crystallisation Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. In distillation Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Steam Distillation is applied to separate substances which are steam volatile and are immiscible with water. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points.
    1. Which method can be used to separate two compounds with different solubilities in a solvent?
    2. Distillation method is used to separate which type of substance?
    3. Which technique is used to separate aniline from aniline water mixture?
       OR
       Why chloroform and aniline are easily separated by the technique of distillation?
32. Read the following text carefully and answer the questions that follow:
    The molecular orbital theory is based on the principle of a linear combination of atomic orbitals. According to this approach when atomic orbitals of the atoms come closer, they undergo constructive interference as well as destructive interference giving molecular orbitals, i.e., two atomic orbitals overlap to form two molecular orbitals, one of which lies at a lower energy level (bonding molecular orbital). Each molecular orbital can hold one or two electrons in accordance with Pauli’s exclusion principle and Hund’s rule of maximum multiplicity. For molecules up to N₂, the order of filling of orbitals is:
    {tex}\sigma(1 s)_{\sigma}^{*}(1 s), \sigma(2 s)_{ \sigma}^{*}(2 s), \pi\left(2 p_{x}\right){/tex} {tex}=\pi\left(2 p_{y}\right), \sigma\left(2 p_{z}\right),_{\pi}^{*}\left(2 p_{x}\right){/tex}{tex}=\stackrel{*}{\pi}\left(2 p_{y}\right), \stackrel{*}{\sigma}\left(2 p_{z}\right){/tex}
    and for molecules after N₂, the order of filling is:
    {tex}\sigma(1 s)_{\sigma}^{*}(1 s), \sigma(2 s)_{ \sigma}^{*}(2 s), \sigma\left(2 p_{z}\right), \pi\left(2 p_{x}\right){/tex}{tex}=\pi\left(2 p_{y}\right),_{\pi}^{*}\left(2 p_{x}\right){/tex}{tex}=\stackrel{*}{\pi}\left(2 p_{y}\right), \stackrel{*}{\sigma}\left(2 p_{z}\right){/tex}
    Bond order = {tex}\frac{1}{2}{/tex}[bonding electrons – antibonding electrons] Bond order gives the following information:
    1. If bond order is greater than zero, the molecule/ion exists otherwise not.
    2. Higher the bond order, higher is the bond dissociation energy.
    3. Higher the bond order, greater is the bond stability.
    4. Higher the bond order, shorter is the bond length.
    1. Arrange the following negative stabilities of CN, CN⁺ and CN^– in increasing order of bond. (1)
    2. The molecular orbital theory is preferred over valence bond theory. Why? (1)
    3. Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so? (2)
       OR
       Bonding molecular orbital is lowered by a greater amount of energy than the amount by which antibonding molecular orbital is raised. Is this statement correct? (2)
33. Section E
34. Attempt any five of the following:
    1. Classify the hydrocarbons according to the carbon-carbon bond.
    2. Explain why alkynes are less reactive than alkenes towards addition of Br₂.
    3. What is electrophile in sulphonation?
    4. Why are Alkenes called olefins?
    5. n-propylmagnesium bromide on hydrolysis gives propane. Is there any other Grignard reagent which also gives propane? If so, give its name, structure and equation for the reaction.
    6. Write IUPAC name: {tex}\mathrm{CH}_{3} \mathrm{CH}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}{/tex}
    7. Which of the two trans-but-2-ene or trans-pent-2-ene is non-polar?
35. The value of K_p for the reaction,
    CO₂(g) + C(s) {tex} \rightleftharpoons {/tex} 2CO (g)
    is 3.0 at 1000 K. If initially, {tex}P_{CO_{2}}{/tex} = 0.48 bar and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.

    OR

    One of the reactions that take place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.
    FeO(s) + CO(g) {tex}\rightleftharpoons{/tex} Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
    What is the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressures are: P_co = 1.4 atm and {tex}P_{CO_2}{/tex} = 0.80 atm?

36. Answer:
    1. 1. What is the basic principle of chromatography?
       2. What is the general molecular formula of saturated monohydric alcohols?
    2. OR
       1. What is the basic principle involved in the estimation of nitrogen by Dumas method.
       2. Explain hyperconjugation effect. How does hyperconjugation effect explain the stability of alkenes?
          

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Class 11 – Chemistry
Sample Paper – 01 (2024-25)

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Solution

                                                                                                                                    Section A

1. (b)

   8,17,1,8

   Explanation: The balanced equation for the given equation is,
   8CO(g) + 17H₂(g) {tex}\longrightarrow{/tex} C₈H₁₈(l) + 8H₂O

2. (b)

   Positive charge of the atoms very little space

   Explanation: Positive charge of the atoms very little space

3. (d)

   x > y

   Explanation: Here, x > y. Because in equation (B) all the reactants are in their most stable states of aggregation hence have zero standard molar enthalpies of formation.

4. (c)

   16 and 15

   Explanation: atomic number O has atomic number = 8 so number of electrons in {tex}O_2{/tex} = 16
   while in {tex}O_2^+{/tex} there is one unit positive charge so no. of electron =15.

5. (d)

   Heat of combustion

   Explanation: Combustion is an exothermic process. Hence heat of combustion has a negative value.

6. (c)

   1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

   Explanation: The atomic number 24 is of Cr. Due to half-filled orbital stability, it does not follow the Aufbau rule so its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹.

7. (d)

   an acid as well as an oxidant

   Explanation: sulfuric acid (H₂SO₄) is a strong acid, signifying that it fully dissociates into H₃O⁺ and HSO₄^– in an aqueous environment (the bisulfate ion is amphiprotic,  but usually behaved as a weak bronsted acid given it’s feeble alkalinity).
   Now, consider some oxidizing agents: F₂, O₂, Cl₂, etc. All of these species are driven by high electronegativities/electron affinities, a result of quantum mechanical effects that contribute to the properties of these agents. However, the H⁺ (H₃O⁺) ion is essentially a naked proton, resulting in an extremely dense positive charge; this will force it’s reduction by more electropositive species by either physical gaining of electrons or the sharing of a lone pair.
   Furthermore, the bisulfate/sulfate ion is capable of forming entropically preferable compounds due to the presence of oxygen, another powerful oxidizing agent.

8. (b)

   free-radical mechanism

   Explanation: In the presence of peroxide and light, the addition of HBr to unsymmetrical alkenes occurs contrary to Markovnikov’s rule. The chemistry follows a free-radical mechanism. Organic reactions, which proceed by homolytic fission are called free radical or homopolar or nonpolar reactions.

9. (c)

   All of these

   Explanation:

   Tabacco, coal and petroleum damages DNA of our body and causes cancer.

10. (a)

    7th period and 14th group

    Explanation: Elements with atomic numbers Z = (87 – 114) are placed in the 7th period.
    Thus,  the element with Z= 114 (Flerovium) is placed in the 7th period and 14th group of the modern periodic table.

11. (b)

    2000 K

    Explanation: Gibbs free energy, {tex}\Delta{/tex}G = {tex}\Delta{/tex}H – T{tex}\Delta{/tex}S
    At equilibrium {tex}\Delta{/tex}G = 0;  then T= {tex}\frac{\Delta H}{\Delta S}{/tex} = 2000K
    Therefore, above 2000K, the reaction will be spontaneous.

12. (d)

    NAOH + CaO

    Explanation: Soda lime = NaOH + CaO

13. (b)

    Both A and R are true but R is not the correct explanation of A.

    Explanation: The stability of carbonium ions is influenced by both resonance and inductive effects, allyl and benzyl carbonium ions can be stabilized by resonance but proply carbonium ion (CH₃CH₂CH₂⁺) has no resonating forms.

14. (c)

    A is true but R is false.

    Explanation: With trans-2-butene, the product of Br₂ addition is optically inactive due to the formation of symmetric meso compounds.
    

15. (c)

    A is true but R is false.

    Explanation: The threshold frequency is the minimum frequency required for the emission of electrons from the metal surface.

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    • Boost Exam Performance: With the right tools and resources, both students and teachers can achieve better results and improved learning outcomes.(d)

      A is false but R is true.

      Explanation: Gases do deviate from ideal behaviour but under the same conditions of pressure, temperature, etc., they combine in a simple ratio of volume fairly accurately.

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16. Section B
17. {tex}\mathrm{A}_{2} \mathrm{X}_{3} \rightarrow 2 \mathrm{A}^{3 +}+3 \mathrm{X}^{2-}{/tex}
    {tex}K_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^{2}\left[\mathrm{X}^{2-}\right]^{3}=1.1 \times 10^{-23}{/tex}
    If S = solubility of A₂X₃, then
    {tex}\left[\mathrm{A}^{3+}\right]=2 \mathrm{S} ;\left[\mathrm{X}^{2}\right]=3 \mathrm{S}{/tex}
    therefore, K_sp = (2S)²(3S)³ = 108S⁵
    = 1.1 {tex}\times{/tex} 10^–23
    thus, S⁵ = 1 {tex}\times{/tex} 10^–25
    S = 1.0 {tex}\times{/tex} 10^–5 mol/L
18. The electronic configuration of helium (He) is 1s^₂. It should be kept in s-block as the last electron enters in ‘s’ orbital. Due to the presence of completely filled valence shell (1s^₂ ) and similarities in properties with noble gases, it is placed in p-block i.e. group 18.
19. According to the chemical equation,
    CH₄ (g) + 2O₂(g) {tex}\to{/tex} CO₂(g) + 2H₂O (g)
    44g CO₂ (g) is obtained from 16 g CH₄ (g).
    [{tex}\therefore{/tex} 1 mol CO₂(g) is obtained from 1 mol of CH₄(g)] Number of moles of CO₂ (g)
    = 22 g CO₂ (g) {tex}\times{/tex}{tex}\frac{1 \mathrm{mol} \mathrm{CO}_{2}(\mathrm{g})}{44 \mathrm{gCO}_{2}(\mathrm{g})}{/tex}
    = 0.5 mol CO₂(g)
    Hence, 0.5 mol CO₂ (g) would be obtained from 0.5 mol CH₄ (g) or 0.5 mol of CH₄ (g) would be required to produce 22 g CO₂ (g).
20. For the given compound, the number of {tex}\pi{/tex}-electrons is 8.
    By Huckel’s rule,
    {tex}\Rightarrow{/tex} 4n + 2 = 8
    {tex}\Rightarrow{/tex} 4n = 6
    {tex}\Rightarrow{/tex} n = 3/2
    For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…).
    This is not true for the given compound as it is a fraction. Hence, it is not aromatic in nature.

    OR

    Benzene, when undergoes a process of a complete oxidation reaction by V₂O₅, gives Maleic Anhydride.
    

21.  Given, Mass of electron m_e = 9.1 × 10^-31 kg and h = 6.6 × 10^-34 kg m²s^-1Uncertainty in speed, {tex}\Delta v=600 \times \frac{0.005}{100}=0.03 \mathrm{ms}^{-1}{/tex}Using Heisenberg’s uncertainty principle, {tex}\Delta x=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03}{/tex}{tex}=1.92 \times 10^{-3} \mathrm{m}{/tex}.
    .
22. Section C
    1. The resonating structures of O₃ are shown below:
       
    2. Nitrate ion, {tex}NO^-_3{/tex}
       
       
23. Answer:
    1. We have q = -w = p_ex (8) = 8 litre-atm
    2. The Third Law of Thermodynamics states, “The entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K).
    3. For an isolated system at constant volume, there is no transfer of energy in the form of heat or work.
       So, {tex}\Delta U = q + W{/tex}
       {tex}\Rightarrow \Delta U = 0+0{/tex}
       {tex}\Rightarrow \Delta U =0{/tex}
    1. A substance has perfectly ordered arrangement of its constituent particles only at absolute zero. When the element from itself. This means no heat change.
       Thus {tex}\Delta{/tex}_fH = 0
       1. If both {tex}\Delta{/tex}H and {tex}\Delta{/tex}S are positive {tex}\Delta{/tex}G can be -ve only in magnitude. Thus the temperature should be high.
       2. If both {tex}\Delta{/tex}H and {tex}\Delta{/tex}S are negative {tex}\Delta{/tex}G can be negative only T{tex}\Delta{/tex}S < {tex}\Delta{/tex}H is magnitude. Thus the value of T should be low.
    1. At anode: Zn {tex}\rightarrow{/tex} Zn²⁺ + 2e^– …(i)
       At cathode: Br₂ + 2e^– {tex}\rightarrow{/tex} 2Br^– ….(ii)
       on adding (i) and (ii)
       Zn + Br₂ {tex}\rightarrow{/tex} Zn²⁺ + 2 Br^–
    2. At anode: [Cr {tex}\rightarrow{/tex} Cr³ + 3e^– ] {tex}\times{/tex} 2 …(iii)
       At cathode: [I2 + 2e^– {tex}\rightarrow{/tex} 2I^–] {tex}\times{/tex} 3 …(iv)
       on adding (iii) and (iv)
       2Cr + 3I₂ {tex}\rightarrow{/tex} 6I^– + 2Cr³⁺
    3. At anode: H₂(g) {tex}\rightarrow{/tex} 2H⁺ (aq) + 2e^– …(v)
       At cathode: Cu²⁺ (aq) + 2e^– {tex}\rightarrow{/tex} Cu(s) …(vi)
       on adding (v) and (vi)
       H₂(g) + Cu²⁺ (aq) {tex}\rightarrow{/tex} Cu(s) + 2H⁺ (aq)
       Cathode will be positive terminal in each cell.
24. v = (3.29 {tex}\times{/tex}10¹⁵ Hz) {tex}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right){/tex}
    {tex}\lambda{/tex} = 1285 nm = 1285 {tex}\times{/tex} 10^-9 m = 1.285 {tex}\times{/tex}10^-6 m
    v = {tex}\frac{c}{\lambda}=\frac{\left(3 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(1.285 \times 10^{-6} \mathrm{m}\right)}{/tex} = 2.3346 {tex}\times{/tex} 10¹⁴s^-1
    2.3346 {tex}\times{/tex}10¹⁴ = 3.29 {tex}\times{/tex}10¹⁵ {tex}\left[\frac{1}{3^{2}}-\frac{1}{\mathrm{n}^{2}}\right]{/tex}
    {tex}\frac{2.3346}{32.9}=\frac{1}{3^{2}}-\frac{1}{\mathrm{n}^{2}}{/tex}or 0.71 = {tex}\frac{1}{9}-\frac{1}{n^{2}}{/tex}
    {tex}\frac{1}{n^{2}}=\frac{1}{9}{/tex}– 0.071 = 0.111 – 0.071 = 0.04
    n² = {tex}\frac{1}{0.04}{/tex}= 25 or n = 5
    Paschen series lies in infrared region of the spectrum.
    1. C has the highest first ionization enthalpy.
    2. C has the most negative electron gain enthalpy.
    3. AI  has the largest atomic radius.
    4. AI  has the most metallic character.
    1. Molecular mass of H₂O = 2 {tex}\times{/tex}H + 1 {tex}\times{/tex} O = 2 {tex}\times{/tex} 1.0079u + 1 {tex}\times{/tex} 16.00u = 18.0158 u
    2. Molecular mass of CO₂ =1{tex}\times{/tex} C + 2 {tex}\times{/tex} O = 1 {tex}\times{/tex}12.01 u + 2{tex}\times{/tex} 16.22 u = 44.01 u
    3. Molecular mass of CH₄ = 1 {tex}\times{/tex} C+ 4 {tex}\times{/tex} H = 1 {tex}\times{/tex}12.01 u + 4 {tex}\times{/tex} 1.0079 u = 16.0416 u
25. Section D
    1. Fractional crystallizationis used to separate two compounds with different solubilities in a solvent.
       • volatile liquids from nonvolatile impurities.
       • the liquids having sufficient difference in their boiling points.
    2. Aniline is separated from aniline water mixture by steam distillation as one of the substances in the mixture is water and the other, a water insoluble substance.
       OR
       Chloroform and aniline are easily separated by the technique of distillation because chloroform and aniline have sufficient difference in their boiling points.
    1. The increasing order of negative stabilities of CN, CN⁺ and CN^–  is CN⁺ > CN > CN^–.
    2. The molecular orbital theory is preferred over valence bond theory because molecular orbital theory explains the magnetic nature of the molecule.
    3. In ethyne, hydrogen atoms are connected to sp hybridized carbon atoms, but in ethene, they are attached to sp² hybridized carbon atoms and in ethane, they are attached to sp³ hybridized carbons.
       OR
       The given statement is not correct because the bonding molecular orbital is lowered by a lesser amount of energy than the amount by which antibonding molecular orbital is raised.
26. Section E
27. Attempt any five of the following:
    1. Hydrocarbons are categorized into three categories according to the carbon-carbon bond that exists between them:
       1. Saturated hydrocarbon (In which carbon-carbon single bond are present)
       2. Unsaturated hydrocarbon (In which carbon-carbon double and triple bonds are present)
       3. Aromatic hydrocarbon (In which alternate single and double bond and (4n+2){tex}\pi{/tex} electrons are present)
    2. The triple bonds of alkynes, because of its high electron density, are easily attacked by electrophiles, but less reactive than alkenes due to the compact C-C electron cloud. The three-membered ring bromonium ion formed from the alkyne (A) has a full double bond causing it to be more stained and less stable than the one from the alkene (B),
       Also, the carbon’s of (A) that are part of the bromonium ion has more s-character than (B), further making (A) less stable than (B).
    3. SO₃
    4. Alkenes are commonly known as olefins because the lower members form oily products on treatment with chlorine or bromine.
    5. Iso-propylmagnesium bromide, (CH₃)₂CHMgBr,
       (CH₃)₂CHMgBr + H₂O {tex}\longrightarrow{/tex} CH₃CH₂CH₃ + Mg(OH)Br
    6. 2-methylbutane
    7. In trans-but-2-ene, the dipole moments of the two C—CH₃ bonds are equal and opposite and therefore, they cancel out each other.
       Hence, trans-2-butene is non-polar.
       
28. For the reaction,
    let ‘x’ be the decrease in pressure of CO₂, then
    CO₂(g) + C(s) {tex} \rightleftharpoons {/tex} 2CO(g)
    Initial
    pressure: 0.48 bar 0
    At equilibrium: (0.48 – x)bar 2x bar
    {tex}K_{p}=\frac{p_{C O}^{2}}{p_{C O_{2}}}{/tex}
    K_p = (2x)²/(0.48 – x) = 3
    4x² = 3(0.48 – x)
    4x² = 1.44 – x
    4x² + 3x – 1.44 = 0
    a = 4, b = 3, c = –1.44
    x = {tex}\frac{(-b \pm \sqrt{b^{2}-4 a c})}{2 a}{/tex}
    {tex}x = {-3 \pm \sqrt{3^2-4(4)(-1.44)} \over 2(4)}{/tex}
    = (–3 {tex}\pm{/tex} 5.66)/8
    = (–3 + 5.66)/8 (as value of x cannot be negative hence we neglect that value)
    x = 2.66/8 = 0.33
    The equilibrium partial pressures are,
    {tex}P_{CO_2}{/tex} = 2x = 2 {tex}\times{/tex} 0.33 = 0.66 bar
    {tex}P_{CO_2}{/tex} = 0.48 – x = 0.48 – 0.33 = 0.15 bar

    OR

    
    {tex}{Q_p} = \frac{{pc{o_2}}}{{pco}} = \frac{{(0.8\;atm)}}{{(1.4\;atm)}} = 0.571{/tex}
    Since Q_p > k_p (0.265), this means that the reaction will move in the backward direction to attain the equilibrium. Therefore, the partial pressure of CO₂ will decrease while that of CO will increase so that the equilibrium may be attained again. Let p atm be the decrease in the partial pressure of CO₂. Therefore, the partial pressure of CO will increase by the same magnitude i.e. p atm.
    {tex}P_{CO_2}{/tex} = (0.8 – p) atm; {tex}P_{CO}{/tex}(g) = (1.4 + p) atm
    At equilibrium {tex}{K_p} = \frac{{p_{co_2}}}{{p_{co}}} = \frac{{(0.8 – p)\;atm}}{{(1.4 + p)\;atm}} = \frac{{(0.8 – p)}}{{(1.4 + p)}}{/tex}
    or {tex}0.265 = \frac{{(0.8 – p)}}{{(1.4 + p)}}{/tex}
    0.371 + 0.265 p = 0.8 – p or 1.265 p = 0.8 – 0.371 = 0.429
    p = 0.429 / 1.265 = 0.339 atm
    The equilibrium partial pressure of CO is ({tex}P_{CO}{/tex})_eq = (1.4 + 0.339) = 1.739 atm
    The equilibrium partial pressure of CO₂ is ({tex}P_{CO_2}{/tex})eq = (0.8 – 0.339) = 0.461 atm

29. Answer:
    1. 1. Chromatography is based on the principle of different adsorption.
       2. Monohydric alcohols are the compounds derived from an alkane by replacing one H by – OH group.
          Example:
          {tex}\mathop {{\text{C}}{{\text{H}}_4}}\limits_{{\text{Methane}}} \xrightarrow{ replacing\ H \ with \ OH }\mathop {{\text{C}}{{\text{H}}_{\text{3}}} – {\text{OH}}}\limits_{{\text{Methanol}}} {/tex}
          Therefore, the general molecular formula of saturated monohydric alcohols is C_nH_2n+1OH.
    2. OR
       1. This method is based upon the fact that nitrogenous compound is heated with copper oxide in an atmosphere of carbondioxide yield free nitrogen.{tex}\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{N}_{z}+\mathrm{CuO} \longrightarrow x \mathrm{CO}_{2}+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}+\frac{z}{2} \mathrm{N}_{2}+(\mathrm{Cu}){/tex}
       2. Hyperconjugation: The relative stability of various classes of carbonium ions may be explained by the number of no-bond resonance structures that can be written for them. Such structures are arrived by shifting the bonding electrons from an adjacent C – H bond to the electron – deficient carbon. In this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no-bond character in the adjacent C – H bond is called Hyper conjugation or No-bond Resonance. The greater the hyperconjugation, the greater will be the stability of the compound. The increasing order of stability can be shown as:
          

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