HBSE Class 11th Chemistry Solved Question Paper 2022

Q1. Number of moles in 54g of Water are :

(A) 2.0

(B) 2.5

(C) 3.0

(D) 3.5 

Ans. (C) 3.0

18g = 1 mole

No. of moles = 54/18 = 3 moles

Q2. Find number of unpaired electrons in Oxygen atom :

(A) 1

(B) 2

(C) 3

(D) 4

Ans. (B) 2

Q3. Arrange B, Al, Mg and K in correct order of their metallic character :

(A) B > Al > Mg > K

(B) Al > Mg > B > K

(C) Mg > Al > K > B

(D) K > Mg > Al > B

Ans. (D) K > Mg > Al > B

Q4. How many vertical columns in Modern Periodic table ?

(A) 2

(B) 8

(C) 32

(D) 18

Ans. (D) 18

Q5. Which compound do not show zero dipole moment ?

(A) CO2

(B) BF3

(C) H2O

(D) CCl4

Ans. (C) H2O

Q6. Which type of intermolecular force exists in Cl2 and CCl4 ?

(A) dipole-dipole forces 

(B) dipole-induced dipole forces

(C) dispersion forces

(D) hydrogen bonds

Ans. (C) dispersion forces

Q7. The temperature at which the volume of gas is zero ?

(A) 0°C

(B) O K

(C) O°F

(D) None of these

Ans. (B) O K

Q8. For the process to occur under adiabetic conditions, the correct condition is :

(A) ∆T = 0

(B) ∆P = 0

(C) ∆q = 0     

(D) w = 0 

Ans. (C) ∆q = 0  

∆q = 0 means there is no exchange of heat.

Q9. The pH of Human Saliva is :

(A) 7.9

(B) 9.2

(C) 7.8

(D) 6.8

Ans. (D) 6.8 

pH of human saliva = 6.2 to 7.6 

Q10. The oxidation number of Cl in KClO3 is :

(A) +5

(B) +3

(C) +1

(D) -1 

Ans. (A) +5

+1 + x +3(-2) = 0

1 + x – 6 = 0

x = +5 

Q11. The oxidation number of Sulphur in SO4²- is :

(A) -2

(B) +6

(C) +3

(D) +4 

Ans. (B) +6

x + 4(-2) = -2 

x = -2 + 8 = +6 

Q12. How many isotopes of Hydrogen ?

(A) 2

(B) 3

(C) 5

(D) 4

Ans. (B) 3

The hydrogen element has three isotopes : hydrogen, deuterium, and tritium.

Q13. Standard electrode potential of Hydrogen is :

(A) 0.0 V

(B) 1.0 V

(C) -1.0 V

(D) 2.0 V

Ans. (A) 0.0 V

Q14. What is formula of Caustic Soda ?

(A) Ca(OH)2

(B) NaOH

(C) KOH

(D) Na2CO3

Ans. (B) NaOH

Q15. Which alkali metal having least melting point ?

(A) Na

(B) K

(C) Rb 

(D) Cs 

Ans. (D) Cs

Caesium (Cs) has large size and weaker metallic bond. 

Q16. Thermodynamically the most stable form of Carbon is :

(A) Diamond

(B) Graphite

(C) Fullerenes

(D) Coal

Ans. (B) Graphite

Q17. CnH2n-2 is general formula of : 

(A) Alkanes 

(B) Arenes

(C) Alkenes

(D) Alkynes

Ans. (D) Alkynes

Q18. Heating a mixture of sodium acetate with soda-lime gives :

(A) Propane

(B) Ethane

(C) Methane 

(D) Butane

Ans. (C) Methane

Q19. Round up 34.216 upto three significant figures.

Ans. 34.2 

Q20. Which of the following species will have largest size ?

Mg, Mg²+, Al, Al³+ 

Ans. Mg 

Mg > Al > Mg²+ > Al³+ 

Q21. Write general electronic configuration of p-block elements.

Ans. ns² np¹-⁶ 

Q22. What is Boyle’s Law ?

Ans. The pressure and volume of a gas are inversely proportional to each other as long as the temperature and the quantity of gas are kept constant. 

P ∝ 1/V  (Temperature = Constant) 

P1/P2 = V2/V1 = constant 

Q23. What is constant in Charle’s Law ?

Ans. Charles’s law states that if a given quantity of gas is held at a constant pressure, its volume is directly proportional to the absolute temperature. 

V ∝ T  (Pressure = Constant) 

Q24. What is oxidation process ? 

Ans. Oxidation is a process which involves the addition of oxygen or any electronegative element or the removal of hydrogen or any electropositive element. 

(Removal of hydrogen or addition of oxygen)

Q25. Which is reducing agent in the following reaction ?

Zn + CuSO4 → ZnSO4 + Cu 

Ans. Zinc(Zn) is reducing agent. 

Q26. Why is Na less reactive than K ?

Ans. Because Na has small size and large ionisation energy than K. 

Q27. 1 kilometer = ……… millimeter or …….. millimeter 

Ans. 1000000 mm or 10⁶ mm

Q28. Binary compound formed by Li and O is ………… 

Ans. Li2O

Q29. Each line of V Vs. P graph at constant temperature is called ………… 

Ans. isotherm or isothermal

Q30. In a reaction acceptor of electrons called ………… 

Ans. oxidising agent

Q31. ………. number of hydrogen bonded water molecules in CuSO4.5H2O. 

Ans. One

Q32. Group 13 hydrides called ………….. 

Ans. Electron Deficient Hydrides

Q33. ……….. is oxidation state of K in KO2.

Ans. +1 

Q34. Group 1 Li, Na, K, Rb, Cs are called ……….. metals. 

Ans. Alkali 

Q35. The type of hybridization of carbon in diamond is ……….. 

Ans. sp³

Q36. Calculate the molarity of NaOH in the solution prepared by dissolving its 4g NaOH in water to form 250mL of solution. 

Ans. Weight of NaOH (W) = 4g 

Molecular weight of NaOH (MW) = 40g 

Volume of solution (V) = 250mL 

Molarity (M) = W/MW × 1000/V = 4/40 × 1000/250 = 4/10 = 0.4 M 

Q37. What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10ms-¹ ? 

Ans. Wavelength, λ = h/mv = (6.626×10-³⁴) ÷ (0.1×10) = 6.626 × 10-³⁴ metre 

(h = de-broglie constant, m = mass, v = velocity) 

Q38. What is enthalpy of formation ? Give one example. 

Ans. Heat evolved or absorbed when 1 mole of the substance is formed form its elements under given conditions of temperature and pressure. 

e.g.  C (graphite) + 2H2 (g) → CH4 (g)   ΔfH° = – 74.81 kJmol-¹

Q39. What will be the conjugate acids for the Bronsted bases NH3 and HCOO- ? 

Ans. 

Conjugate Acid  → Bronsted Base 

NH3                       → NH2– 

HCOO-                 → HCOOH 

Q40. What are Electrophiles ? Give its types with examples. 

Ans. Electrophiles are electron deficient species and can accept an electron pair from electron rich species. e.g. Br+, Cl+, CH3+ 

Q41. Explain Hund’s rule of maximum multiplicity with example. 

Ans. Hunds Rule of Maximum Multiplicity rule states that for a given electron configuration, the term with maximum multiplicity falls lowest in energy. According to this rule electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied. 

Q42. Define closed system, intensive properties and internal energy. 

Ans. Closed system– The system which allows the exchange of energy not of matter with the surroundings is called a closed system.  

Intensive properties– The properties which do not depend upon the quantity of matter or substance present in the system. 

Internal Energy– Internal energy is the inherent microscopic energy of a system associated with the random motion of its molecules.

Q43. Define acid and base with example in Bronsted-Lowry concept. 

Ans. Bronsted Lowry Concept of acid and Base : As per this concept, acids are proton donor while bases are proton acceptor.

In this example, water is proton donor and hence it is acting like an acid. Ammonia; being proton acceptor, is a base.

Q44. Explain catenation, inert pair effect and allotropy. 

Ans. Catenation– Atoms of carbon and some other elements link with one another through strong covalent bonds to form long chains or branches. This is called catenation. Usually, it is shown by C, S and Si.

Inert pair effect– On moving down the group, the tendency of s electrons to participate in chemical bonding decreases. This effect is called inert pair effect. Due to this, in group 13, the stability of +3 oxidation state decreases and the stability of +1 oxidation state increases.

Allotropy– Allotropy is the phenomenon in which an element exist in more than one form having same chemical properties by different physical properties. For example, diamond, graphite and fullerenes are allotropes of carbon.

Q45. Give IUPAC names of the following :

(a) 3-Bromo 3-Chloroheptane 

(b) 3-Chloropropanal

(c) 2,5-Dimethylheptane 

Q46. What is meant by hybridization of atomic orbitals ? Describe the shapes of sp, sp² and sp³ hybrid orbitals with example. 

Ans. Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes.

For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new sp² hybrid orbitals.

These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule. 

Shape of sp hybrid orbitals– sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals as :

Shape of sp² hybrid orbitals– sp² hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p-orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as:

Shape of sp³ hybrid orbitals– Four sp³ hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. The four sp3 hybrid orbitals are arranged in the form of a tetrahedron as : 

                                               OR

What is meant by the term bond order ? Calculate the bond order of N2, O2, O2+ and O2-. 

Ans. Bond order– The number of chemical bonds present between two atoms of a molecule or one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule. 

Bond order of N2 = (10-4)/2 = 6/2 = 3

Bond order of O2 = (10-6)/2 = 4/2 = 2

Bond order of O2+ = (9-6)/2 = 3/2 = 1.5 

Bond order of O2– = (10-7)/2 = 3/2 = 1.5 

Q47(i) What is Wurtz reaction ? 

Ans. Wurtz’s reaction is an organic chemical coupling reaction wherein sodium metal is reacted with two alkyl halides in the environment provided by a solution of dry ether in order to form a higher alkane along with a compound containing sodium and the halogen.

R–X + 2Na + X–R → R–R + 2Na–X (Basic reaction)

R = alkyl group, X = halogen (F, Cl, Br, I)

(ii) What is Markownikoff rule ? Explain with example. 

Ans. When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent is attached to the unsaturated C atom having less number of hydrogen atoms. Thus, when HBr is added to propene, isopropyl bromide is obtained as major product. 

CH3−CH=CH2 + HBr → CH3−CH(Br)–CH3 

                                           OR

(i) What is anti-Markownikoff rule ? Explain with example. 

Ans. When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent is attached to the unsaturated C atom having less number of hydrogen atoms. Thus, when HBr is added to propene, isopropyl bromide is obtained as major product. 

CH3−CH=CH2 + HBr → CH3−CH(H)–CH2(Br)  

(ii) What is Friedel-Crafts alkylation reaction ? 

Ans. 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.