HBSE Class 11 Chemistry Question Paper 2024 Answer Key

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HBSE Class 11 Chemistry Question Paper 2024 Answer Key

Section – A (1 Mark)

1. Number of significant figures in 2.005 is :
(a) Two
(b) Three
(c) Four
(d) Infinite
Answer – (c) Four

2. How many unpaired electrons are present in Nitrogen atom?
(a) 1
(b) 2
(c) 3
(d) 4
Answer – (c) 3

3. The elements with atomic Number 9, 17, 35, 53, 85 are all :
(a) Halogen’s
(b) Noble gas
(c) Alkali Earth metal
(d) Transition metal
Answer – (a) Halogen’s

4. Which compound does not show zero dipole moment?
(a) CO2
(b) BF3
(c) H2O
(d) CCl4
Answer – (c) H2O

5. What is the molar mass of H2O in gm/mol ?
(a) 44
(b) 18
(c) 17
(d) 60
Answer – (b) 18
H2O = 2×1 + 16 = 2 + 16 = 18

6. When water is added to quick lime the reaction is :
(a) Explosive
(b) Endothermic
(c) Exothermic
(d) Photochemical
Answer – (c) Exothermic

7. The oxidation number of Cl in KClO3 is :
(a) +5
(b) +3
(c) +1
(d) –1
Answer – (a) +5
+ 1 + x + 3(–2) = 0
1 + x – 6 = 0
x = 5

8. Molecular mass of volatile substance is determined by :
(a) Kjeldahl’s method
(b) Duma’s method
(c) Victor Mayer’s method
(d) Leibig’s method
Answer – (c) Victor Mayer’s method

9. Mathematical statement of first law of thermodynamics is :
(a) ∆U = q + w
(b) ∆U = q × w
(c) ∆U = – q – w
(d) ∆U = w – q
Answer – (a) ∆U = q + w

10. How many Sigma and Pi bonds are present in the Benzene?
(a) 12 σ and 3 π
(b) 11 σ and 2 π
(c) 12 σ and 1 π
(d) 13 σ and 3 π
Answer – (a) 12 σ and 3 π

11. Which orbital is dumb-bell shaped?
(a) s
(b) p
(c) d
(d) f
Answer – (b) p

12. The IUPAC name of CH3CHO :
(a) Acetaldehyde
(b) Methanal
(c) Formyl chloride
(d) Ethanal
Answer – (d) Ethanal

13. The correct value of Avogadro’s constant will be :
(a) 6.022 × 1023
(b) 6.022 × 1026
(c) 6.022 × 1025
(d) 6.022 × 1022
Answer – (a) 6.022 × 1023

14. How many molecules of water of crystalization in Potash Alum?
(a) 12
(b) 24
(c) 6
(d) 3
Answer – (b) 24

15. Assertion (A) : In a reaction
Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn is a reductant but itself get oxidized.
Reason (R) : In a Redox reaction oxidant is reduced by accepting electrons and Reductant is oxidized by losing electrons.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Answer – (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

16. Assertion (A) : Graphite is an Element.
Reason (R) : Element is the pure form of substance containing the same kind of atoms.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Answer – (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

17. Assertion (A) : Boiling point of alkanes increase with increase in molecular weight.
Reason (R) : Vander Waals’s force increase with increase in molecular weight.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Answer – (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

18. Assertion (A) : Saturated hydrocarbons are chemically less reactive.
Reason (R) : All Isomeric paraffins have the same parent name.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Answer – (c) (A) is true but (R) is false.

Section – B (2 Marks)

19. (a) How many total electrons are present in the Nitrate Ion?
Answer : 32 electrons
NO3 = 1×7 + 3×8 + 1 = 7 + 24 + 1 = 32 e

(b) Write Electronic Configuration of chromium atom.
Answer : 1s2 2s2 2p6 3s2 3p6 3d5 4s

20. BF3 does not have proton but still acts as an acid and reacts with NH3, why is it so? What type of bond is formed between two?
Answer – BF3 is electron deficient and hence acts as Lewis acid. NH3 has one lone pair which it can donate to BF3 and form a coordinate bond.

21. Balance the following redox reaction by lon Electron method :
Cr2O72–(aq) + SO2(g) Cr3+(aq) + SO42–(aq) (Acidic)
Answer : Cr2O72–(aq) + 3SO2(g) + 2H+ 2Cr3+(aq) + 3SO42–(aq) + H2O(l)

22. Explain why (CH3)3C+ more stable than CH3CH2+ and CH3+ is the least stable cation?
Answer – (CH3)3C+ is the most stable because it is stabilized by both strong inductive effects and extensive hyperconjugation from three methyl groups. CH3CH2+ is less stable because it has only one methyl group providing inductive stabilization and hyperconjugation. CH3+ is the least stable because it has no alkyl groups to provide inductive stabilization or hyperconjugation.

23. (a) If the pH of a solution is 7, calculate its pOH value.
Answer : pO + pOH = 14
pOH = 14 – 7 = 7

(b) What happens to the pH, if a few drops of acid are added to CH3COONH4 solution?
Answer – The pH will remain almost constant (being a buffer solution)

24. What is Electrophilic substitution of Benzene? Show its mechanism.
Answer – Electrophilic substitution of benzene is a chemical reaction in which an electrophile replaces a hydrogen atom in a benzene ring. Benzene is particularly stable due to its delocalized π electrons, but it can still undergo substitution reactions where one of these hydrogen atoms is replaced by an electrophile.

OR

Distinguish between Carbocation and Carbanion.
Answer –

Carbocation Carbanion
1. The ion which contains the positive charge on Carbon atom is called Carbocation. 1. The ion which contains the negative charge on the Carbon atom is called Carbanion.
2. The hybridization of carbocation is sp2. 2. The hybridization of carbanion is sp3.
3. The geometry of Carbon in carbocation is trigonal planar. 3. The geometry of Carbon in carbanion is pyramidal.
4. In any chemical reaction, they act as electrophiles. 4. In any chemical reaction, they act as nucleophile.
5. Example, CH3+ 5. Example, CH5

 

25. (a) What is the basic difference in Mendeleev’s Periodic Law and Modern Periodic Law?
Answer – Mendeleev’s Periodic Law states that the physical and chemical properties of elements are periodic functions of their Atomic Weights.
On the other hand, the Modern periodic Law states that the physical and chemical properties of elements are periodic functions of their Atomic Numbers.

(b) What is the reason that the first lonization energy of Nitrogen is higher while that of oxygen is lower?
Answer – Because nitrogen is more stable due to its half-filled electronic configuration.

Section – C (3 Marks)

26. A measured temperature on the Farenheit scale is 200°F. What will this reading be on Celsius?
Answer – Relationship between Fahrenheit scale & Celsius scale is given by :
F = 9/5 C + 32
C = 5/9 × (F–32)
C = 5/9 × (200–32)
C = 5/9 × 168 = 93.3
Thus, the temperature is 93.3°C

(b) What will be the molarity of a solution which contains 5.85 grams of NaCl (s) per 500 ml ?
Answer – Mass of solute (NaCl) = 5.85 g
Molar mass of solute (NaCl) = 58.5 g/mol
Number of moles of solute (NaCl) = (Mass of solute) ÷ (Molar mass of solute NaCl) = 5.85/58.5 = 0.1 mol
Molarity = (Number of moles of solute) ÷ (Volume of solution in Litre) = 0.1/500 × 1000 = 0.2 mol L–1

(c) What is the relation between Molecular formula and empirical formula?
Answer : Molecular Formula = n × Empirical Formula
MF = n × EF
Where n = Molar mass / Empirical formula mass

27. An element “X” has atomic number 13 :
(a) Write electronic configuration of “X”.
Answer – The element with atomic number 13 is aluminum (Al). Electronic configuration is : 1s2 2s2 2p6 3s2 3p1

(b) State the group to which “X” belongs?
Answer – Group no. 13

(c) Is “X” a metal or non-metal?
Answer – Metal

28. (a) In a process, 701 Joule of Heat is absorbed by a system and 394 J of work is done by the system. What is the change in Internal energy for the process?
Answer – According to the first law of thermodynamics,
∆U = q + w = (+701) + (–394) = 701 – 394 = 307 Joule
Hence, the change in internal energy for the given process is 307 J.

(b) What will be the enthalpy of all elements in their standard state?
Answer – Zero

OR

Expansion of the gas in the vacuum is called free expansion, calculate the work done as well as the change in the Internal energy if 1 litre of the ideal gas expands isothermally into a vacuum until the total volume is 5 litres?
Answer – Work done (w) = 0
Change in the Internal energy (∆U) = 0

29. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen, calculate the masses of CO2 and H2O produced when 0.20 gram of this compound is subjected to complete combustion.
Answer – Calculation of mass of CO2 produced :
Mass of compound = 0.20 g
Percentage of carbon = 69%
Percentage of carbon = 12/44 × (Mass of CO2 / Mass of compound) × 100
60 = 12/44 × (Mass of CO2 / 0.20 g) × 100
Mass of CO2 = (69×44×0.20) ÷ (12×100) = 0.506 g
Calculation of mass of H2O produced :
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8%
Percentage of hydrogen = 2/18 × (Mass of H2O / Mass of compound) × 100
4.8 = 2/18 × (Mass of H2O / 0.20 g) × 100
Mass of H2O = (4.8×18×0.20) ÷ (2×100) = 0.0864 g
Hence, the mass of CO2 is 0.506 g and mass of H2O is 0.864 g.

OR

Explain the term Inductive effect and Electromeric effect with suitable examples.
Answer : Inductive Effect – The permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect. Inductive effect could be +I effect or –I effect. Example: CH3–CH2–Cl
• Electromeric Effect – It involves the complete transfer of the shared pair of π electrons to either of the two atoms linked by multiple bonds in the presence of an attacking reagent. Electromeric effect could be +E effect or –E effect. Example: CH2=CH2

30. (a) What is Kc for the reaction in state of Equilibrium?
2SO2(g) + O2(g) 2SO3(g)
[SO2] = 0.6 M, [O2] = 0.82 and [SO3] = 1.90 M
Answer : Kc = [SO3]2/[SO2]2[O2] = (1.90)²/(0.6)²(0.82) = 12.22
Kc = 12. 22 M–1 (approx.)

(b) At a certain temperature and total pressure of 105 Pa. Iodine contains 40% by volume of I atoms. I2(g) 2I(g), calculate Kp for the equilibrium.
Answer – Partial pressure of I atoms,
PI = 40/100 × 105 = 4 × 104 Pa
Partial pressure of I2 atoms,
PI2 = 60/100 × 105 = 6 × 104 Pa
Kp = (PI)2/(PI2) = (4×104)2/(6×104) = 2.67 × 104 Pa

Section – D (4 Marks)

31. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to surroundings is taken into account. It states that energy supplied to the system goes in partly to increase the internal energy of the system and the rest in work on the environment. Mathematically ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in Internal energy of the system. ∆Q and ∆W depends on the path taken to go from initial to final state, but the combination ∆Q – W path independent.
Questions :
(a) First law of thermodynamics is concerned with the conservation of ……………
Answer – Energy

(b) Which is not path function ∆Q, ∆W, ∆Q – ∆W ?
Answer : ∆Q – ∆W

(c) What is system?
Answer – A system is defined as any portion of matter under thermodynamic consideration, which is separated from the rest of the universe by real or imaginary boundaries.

(d) Thermodynamics is the study of …………….
Answer – the relationships between heat, work, and energy.

32. Three students, Vishal, Vaibhav and Vikas were determining the element present in a given organic compound by the Lassaigne Test. Vaibhav added FeSO4 and dil. H2SO4 to Na – extract and observed Prussian blue colour. Vishal was keeping some pieces of sodium metal in his pocket, Vikas asked him not to do so.
Questions :
(a) Name the element present in organic compound.
Answer – Nitrogen

(b) Write the test for chlorine.
Answer – Lassaigne Test

(c) Write the value associated.
Answer – The value associated with Vishal being asked not to keep pieces of sodium metal in his pocket is safety. Sodium metal reacts vigorously with moisture in the air and can cause burns or even ignite, so it is important to handle it with care and avoid keeping it in pockets or exposed to moisture.

(d) What is the nature of Lassaigne extract?
Answer – Alkaline

OR

Give the IUPAC names of the following compounds :

Answer – (a) 3-chloro-propanal
(b) 2,5-dimethylheptane
(c) 2,2-dicholoroethan-1-ol
(d) Butan-1-al

Section – E (5 Marks)

33. (a) Calculate the energy of each photons which :
(i) Correspond to light of frequency 3 × 1015 Hz.
Answer – Energy of a photon is given by the expression, E = h𝛎
h = Planck’s constant = 6.626 × 10–34 Js
𝛎 = frequency of light = 3 × 1015 Hz
E = h𝛎 = (6.626 × 10–34) × (3 × 1015) = 1.988 × 10–18 J
E = 1.988 × 10–18 J

(ii) Have wavelength of 0.50 Å.
Answer – Energy of a photon having wavelength (λ) is given by the expression, E = hc/λ
h = Planck’s constant = 6.626 × 10–34 Js
s = speed of light = 3 × 108 m/s
λ = wavelength = 0.50 Å = 0.50 × 10–10 m
E = hc/λ = (6.626×10–34×3×108)/(0.50×10–10) = 3.98 × 10–15 J
E = 3.98 × 10–15 J

(b) What is the number of photons of light with wavelength 4000 Pm which provide 1 Joule of energy?
Answer – Energy of a photon having wavelength (λ) is given by the expression, E = hc/λ
h = Planck’s constant = 6.626 × 10–34 Js
s = speed of light = 3 × 10⁸ m/s
λ = wavelength = 4000 Pm = 4000 × 10–12 m = 4 × 10–9 m
E = hc/λ = (6.626×10–34×3×108)/(4×10–9) = 4.97 × 10–17 J
Number of photons providing 1 joule of energy,
n = 1/E = 1/(4.97×10–17) = 2.01 × 1016 photons

OR

(a) Total number of orbitals associated with third shell will be …………..
Answer – Total number of orbitals = n2 = 32 = 9

(b) Orbital angular momentum depends on …………..
Answer – azimuthal quantum number (l)

(c) Number of angular nodes for 4d orbital is ……………
Answer – Two

(d) The number of radial nodes for 3p orbital is ……………
Answer – One

(e) Which of the orbital has lowest energy among 4d, 4f, 5s, 5p ?
Answer – Lowest energy for 5s

34. (a) Which is more polar CO2 or NO2? Give reason.
Answer – N2O is more polar than CO2. This is because CO2 is a linear and symmetrical molecule so the net dipole moment is zero. Hence, it is non polar. Now, N2O is also linear and symmetrical but it is considered as the resonance hybrid of two structures due to which its net dipole moment is 0.116D. Hence, it is polar.

(b) Why ethyl alcohol completely miscible with water?
Answer – Ethyl alcohol is miscible with water because both water and ethanol are polar in nature and ethanol has −OH groups capable of informing hydrogen bond with water molecule.

(c) What are the main postulates of valence shell electron pair repulsion (VSEPR) theory?
Answer – The total number of valence shell electron pairs determines the form of the molecule. The shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom. Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.

OR

(a) Compare relative stability of the following species and indicate their magnetic properties on the basis of Molecular Orbital Theory : O2, O2+, O2
Answer – Bond order of O2+ is 2.5, bond order of O2 is 2, bond order of O2 is 1.5.
Order of the stability is O2+ > O2 > O2

(b) Although CO2 and H2O are triatomic molecule the shape of H2O molecule is bent while that of CO2 is linear, explain on the basis of dipole moment.
Answer – The difference in dipole moments between H2O and CO2 is responsible for their different molecular shapes. The presence of non-bonding electron pairs in H2O leads to an uneven distribution of charge, resulting in a bent shape with a net dipole moment. Meanwhile, the symmetrical arrangement of polar bonds in CO2 leads to a linear shape with no net dipole moment.

35. Explain in brief :
(a) Wurtz Reaction
Answer – Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.
R–X + 2Na + X–R + dry ether R–R + 2NaX
Example: CH3–Br + 2Na + Br–CH3 CH3–CH3 + 2NaBr

(b) Markovnikoff Rule
Answer – According to this rule, in an unsymmetrical alkene when the addition of halogen acid (HX) takes place, then the negative part of the halogen acid is attached to that carbon atom which is double bonded as well have the least number of hydrogen atoms.
For example: The addition of hydrogen bromide to propene.
CH3–CH=CH2 + HBr CH3–CHBr–CH3
In this reaction, bromine is a negative part of halogen acid which attaches to the double-bonded carbon atom and has the least number of the hydrogen atom.

(c) Friedel Craft Reaction (Alkylation)
Answer – Friedel-Crafts Alkylation refers to the replacement of an aromatic proton with an alkyl group. This is done through an electrophilic attack on the aromatic ring with the help of a carbocation. The Friedel-Crafts alkylation reaction is a method of generating alkylbenzenes by using alkyl halides as reactants.

(d) Suggest a route for the preparation of the Nitrobenzene starting from Acetylene.
Answer –

(e) How will you get Bromobenzene from Benzene?
Answer –

 

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