Category: Class 11 Chemistry Objective Questions

Chemistry Assessment Questions for NEET Exam on “Strategies to Control Environmental Pollution”.

1. Which of the following is a major cause of environmental degradation?
a) Sewage treatment
b) Improper waste disposal
c) Microwave-assisted reactions
d) Bioamplification
Answer: b
Clarification: Disposing of waste has huge environmental impacts as a certain type of wastecan be hazardous and can contaminate the environment if not handled properly. Some waste generates methane gas, which contributes to the greenhouse effect.

2. Identify the type of waste which can be degraded by composting, vermicomposting and landfills method.
a) Plastic
b) Tins and metals
c) Biodegradable
d) Non-biodegradable
Answer: c
Clarification: Biodegradable wastes are deposited in landfills and are converted into compost. These wastes mix with the soil and produce manure. Examples include paper, leaves and vegetable peels.

3. Dumping of sewage sludge into the sea is preferred and beneficial over dumping them into the land. State true or false.
a) True
b) False
Answer: b
Clarification: Dumping of sewer sludge into land is preferred. This is because it contains compounds of nitrogen and phosphorus which act as a good fertilizer for the soil.

4. Which method is best suitable for disposing of plastic wastes and polythene bags?
a) Burning and incineration
b) Digesting
c) Dumping
d) Recycling
Answer: d
Clarification: Recycling is the most useful method for plastic waste disposal as several waste materials can be used as raw materials to manufacture useful products again.

5. Which among the following is a biodegradable waste?
a) Polythene
b) Polystyrene
c) Cellulose
d) Butadiene
Answer: c
Clarification: Cellulose is an organic material derived from plants. Hence they are degraded in nature by microorganisms. They are deposited in landfills and are converted into compost.

To practice Chemistry Assessment Questions for NEET Exam,

Chemistry Objective Questions & Answers on “Towards Quantum Mechanical Model of the Atom”.

1. Who found out about dual behavior of a matter?
a) De Broglie
b) Bohr
c) Rutherford
d) Thomson
Answer: a
Clarification: A French physicist named Louis de Broglie proposed that matter exhibits both particle and wave like nature. This means that like photons, electrons also should have both momentum and wavelength.

2. A ball of mass 0.5kg is moving with velocity 6.626 m/s. What’s the wavelength of that ball?
a) 1 x 10^-34 m
b) 2 x 10^-34 m
c) 2 x 10^-32 m
d) 2 x 10^-3 m
Answer: b
Clarification: Louis de Brogie gave the realation between momentum and wavelength as λ = h/p. Here h is Planck’s constant, whose value is 6.626 x 10^-34 J/s. Wavelength = h/mv = 2 x 10^-34 m (momentum p = mass m x velocity v).

3. Mass of a photon is given by 3.313 x 10^-34 kg. Find it’s wavelength.
a) 0.67A°
b) 0.67nm
c) 0.37A°
d) 1.67A°
Answer: a
Clarification: Louis de Brogie gave the realation between momentum and wavelength as λ = h/p. Here h is Planck’s constant, whose value is 6.626 x 10^-34 J/s. Wavelength = h/mc = 6.626 x 10^-34 Js/(3.313 x 10^-34 kg x 3 x 10⁸ m/s) = 0.67A°.

4. Determining the exact position and velocity of an electron is impossible at a time.
a) True
b) False
Answer: a
Clarification: A German physicist, Werner Heisenberg stated Heisenberg’s principle of uncertainty, that states that determining the exact position and velocity of an electron is impossible at a time, as a result of dual nature of matter and radiation.

5. As per Heisenberg’s principle of uncertainty, the relation between relative momentum and relative position is __________
a) independent
b) equal
c) directly proportional
d) inversely proportional
Answer: d
Clarification: Heisenberg’s principle of uncertainty states that the product of relative momentum and velocities is equal to greater than the h/4π, where is “h” is the Planck’s constant and is equal to 6.626 x 10^-34 Js.

6. The uncertainty of a ball is given by 0.5A°. Then calculate the uncertainty in momentum.
a) 2.055 x 10^-24 kgm/s
b) 1.015 x 10^-24 kgm/s
c) 1.055 x 10^-24 kgm/s
d) 1.095 x 10^-24 kgm/s
Answer: c
Clarification: Heisenberg’s principle of uncertainty states that Δx. Δp ≥ h/4π, x is position, p is momentum and “h” is the Planck’s constant and is equal to 6.626 x 10^-34 Js. Relative momentum Δp = h/4πΔx = 1.055 x 10^-24 kgm/s.

7. If the uncertainties in position and momentum are equal, then the uncertainty in position is given by ____
a) √h/4π
b) √h4π
c) √h/4
d) √h/π
Answer: a
Clarification: As we know, Heisenberg’s principle of uncertainty states that Δx. Δp ≥ h/4π, x is position, p is momentum and “h” is the Planck’s constant. Δx = Δp; Δx. Δx = h/4π; Δx = √h/4π

8. If the kinetic energy of an electron is 5J. Find out its wavelength.
a) 0.313 x 10¹⁵ m/s
b) 3.013 x 10¹⁵ m/s
c) 3.310 x 10¹⁵ m/s
d) 3.313 x 10¹⁵ m/s
Answer: d
Clarification: We know that the mass of an electron is 9.1 x 10^-31 kg. Given that the kinetic energy of an electron is 5J. K.E = mv²/2 and by substituting we get v = √1.098 x 10³¹ m/s = 3.313 x 10¹⁵m/s.

9. An object has a mass of 6 kg and velocity of 10 m/s. The speed is measured with 5% accuracy, then find out Δx in m.
a) 0.12676 x 10^-34
b) 0.1566 x 10^-34
c) 0.176 x 10^-34
d) 0.276 x 10^-34
Answer: c
Clarification: Speed’s uncertainty is 10 x 5/100 = 0.5 m/s. We have Heisenberg’s principle of uncertainty i.e. Δx. Δp ≥ h/4π. Δx = h/2mπ. Therefore uncertainty in position = 6.626 x 10^-34 Js/12 x 3.1416 = 0.176 x 10^-34.

10. Δx. Δp ≥ h/4π.
a) True
b) False
Answer: a
Clarification: Heisenberg’s principle of uncertainty states that the product of relative momentum and velocities is equal to greater than the h/4π, where is “h” is the Planck’s constant and is equal to 6.626 x 10^-34 Js. Hence the above statement is true.

Chemistry Multiple Choice Questions on “Hydrogen Bonding”.

1. Nitrogen, fluorine and oxygen are ___________ in nature.
a) electronegative
b) electropositive
c) metallic
d) semi-metallic
Answer: a
Clarification: Electronegativity is the tendency of a neutrally isolated gaseous atom to attract an electron. This is high in the case of nitrogen, oxygen, and fluorine. So the negative charge in the hydrogen bond is towards them.

2. Which bond acts like a bridge two molecules formed by a covalent bond?
a) Covalent bond
b) Ionic bond
c) Hydrogen bond
d) Metallic bond
Answer: c
Clarification: For example, consider a molecule that’s formed due to a covalent between hydrogen and fluorine. Here hydrogen acquires a positive charge. Hydrogen bond formation occurs between hydrogen in a molecule and fluorine of the other molecule.

3. A molecule named o-nitrophenol consists of ______________hydrogen bond/s.
a) intermolecular
b) intramolecular
c) both intermolecular and intramolecular
d) neither intermolecular nor intramolecular
Answer: b
Clarification: The presence of hydrogen between molecules is intermolecular hydrogen bond and presence of hydrogen bond in the molecule itself is intramolecular hydrogen bond. In o-nitrophenol, there is a hydrogen bond between the hydrogen of the hydroxide group and oxygen.

4. In a hydrogen bond, hydrogen has a positive charge.
a) True
b) False
Answer: a
Clarification: IN the hydrogen bond, the other element is highly electronegative, So the hydrogen becomes electropositive comparatively in this case and there is a displacement of electrons towards the electronegative side. Hence hydrogen has a positive charge.

5. Water molecules contain _____________ hydrogen bond/s.
a) intermolecular
b) intramolecular
c) both intermolecular and intramolecular
d) neither intermolecular nor intramolecular
Answer: a
Clarification: In a water molecule, the hydrogen bonds are formed between other molecules but not within the same molecule. The presence of hydrogen between molecules is intermolecular hydrogen bond and presence of hydrogen bond in the molecule itself is intramolecular hydrogen bond.

6. The magnitude of the H-bonding depends on the ___________ of the compound.
a) surroundings
b) system
c) atmosphere
d) physical state
Answer: d
Clarification: The magnitude of the H-bonding depends on the physical state of the compound. In the gaseous state, it is minimum and in the solid state, it is maximum. There is a great influence on a compound’s structure and properties.

7. Which of the following molecule can form a hydrogen bond with hydrogen?
a) Sodium
b) Oxygen
c) Aluminum
d) Rubidium
Answer: b
Clarification: Most Electronegative elements can only form hydrogen bonds with hydrogen, Among Sodium, Oxygen, Aluminium, and Rubidium, Oxygen is the electronegative element. So only oxygen can form a hydrogen bond.

8. Which of the following molecules doesn’t involve hydrogen bond formation?
a) H₂O
b) O-nitrophenol
c) NaCl
d) HF
Answer: c
Clarification: Water, Hydrogen fluoride include intermolecular hydrogen while o-nitrophenol has intramolecular hydrogen bonding. Sodium Chloride NaCl has an ionic bond that is much stronger than hydrogen bond.

9. In a hydrogen bond, the electron pair that is shared moves away from hydrogen.
a) True
b) False
Answer: a
Clarification: Yes, in the formation of a hydrogen bond, the electron pair moves away from hydrogen. This occurs due to the high electronegativity of other atoms participating in a hydrogen bond. So naturally, hydrogen has positive charges and electrons move away.

10. Alcohol and HF molecule contains __________ & _____________ hydrogen bonds.
a) intramolecular, intermolecular
b) intermolecular, intermolecular
c) intermolecular, intramolecular
d) intramolecular, intramolecular
Answer: b
Clarification: The presence of hydrogen between molecules is intermolecular hydrogen bond and the presence of hydrogen bond in the molecule itself is intramolecular hydrogen bond. So it’s intermolecular in case of alcohol and water.

Chemistry Multiple Choice Questions on “States of Matter – Behaviour of Real Gases: Deviation from Ideal Gas Behaviour”.

1. The plot PV vs v at constant temperature is a straight line for real gases.
a) true
b) false
Answer: b
Clarification: The plot of PV vs P is not a straight line for real gases because they deviate from Ideal behaviour. are there are two types of deviations one is a positive deviation and the other is a negative deviation.

2. PV/nRT is known as ____________
a) compressibility factor
b) volume factor
c) pressure factor
d) temperature factor
Answer: a
Clarification: PV/nRT is known as compressibility factor and is represented by the letter Z. It is a ratio of PV and nRT; where p is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is temperature.

3. Which of the following conditions do you think a real gas behaves as an ideal gas?
a) high pressure
b) low pressure
c) intermediate pressure
d) at any pressure
Answer: b
Clarification: At low-pressure conditions, Z = 1 handset behaves as an ideal gas but at high-pressure Z is greater than 1 and for intermediate pressure that is less than 1. So at low-pressure condition, a real gas behaves as an ideal gas.

4. What is the temperature known as where a real gas obeys Boyle’s law or as an ideal gas?
a) Boyle temperature
b) Charge temperature
c) Critical temperature
d) Absolute Temperature
Answer: a
Clarification: The temperature at which a real gas obeys Boyle’s law and other ideal gas law at a certain range of pressure is called Boyle temperature or Boyle point. It is unique for every gas and depends upon its nature.

5. Compressibility can be expressed as _______
a) real volume divided by the ideal volume
b) real universal gas constant by ideal universal gas constant
c) real temperature by ideal temperature
d) real volume divided by real pressure
Answer: a
Clarification: The deviation of real gas behaviour from ideal gas behaviour is known from the compressibility factor. This compressibility factor can also be measured as the ratio of real volume to ideal volume.

6. Above Boyle temperature real gases show __________ deviation from ideal gases.
a) positive
b) negative
c) no
d) both positive and negative
Answer: a
Clarification: Above Boyle temperature, the value of the compressibility factor is greater than 1. So the gases show positive deviation from ideal gases as the forces of attraction between the gas molecules are very low.

7. Which of the following is a corrected equation of ideal gas equation?
a) (P – an²V²)(V – nb) = nRT
b) (P – an²/V²)(V + nb) = nRT
c) (P + an²/V²)(V – nb) = nRT
d) (P – an²/V²)(V – nb) = nRT
Answer: d
Clarification: (P – an²/V²)(V – nb) = nRT; where p is pressure, a is the magnitude of intermolecular attractive forces within a gas, n is the number of moles, v is volume, b is a van der Waal constant, R is the universal gas constant and T is temperature.

8. The value of a in van der Waal equation is _____________ /dependent on ___________
a) pressure
b) temperature
c) pressure and temperature
d) independent of pressure and temperature
Answer: d
Clarification: Value of an in van der Waal equation represents a measure of the magnitude of intermolecular attractive forces within the gas and it is also independent of temperature and pressure. The van der Waal’s equation is given by (P – an²/V²)(V – nb) = nRT.

9. What are the units of “b” in van der Waals equation?
a) L/mol
b) L mol
c) 1/L mol
d) L
Answer: a
Clarification: The ideal gas equation is given as (P – an²/V²)(V – nb) = nRT. So by considering the equation, we can understand that the units of the volume are equal to the units of a number of moles X be so the units of b. So b’s units = volume / number of moles so it is L/mol.

10. A gas that is of 2 moles occupies a volume of about 500 ml at 300 Kelvin and 50 atmospheric pressure, calculate the compressibility factor of the gas.
a) 1.863
b) 0.7357
c) 0.5081
d) 1.8754
Answer: c
Clarification: Compressibility factor Z = PV/nRT; Z = 50 atm x (500/1000) ml / 2 x 0.082 x 300 k = 25/6×8.2 = 0.5081. That means Z < 1, so this is a negative deviation from ideal gas behaviour. So the gas is more compressible than expected.

Chemistry Problems on “Factors Affecting Equilibria”.

1. Which of the following factors do you think will not affect the state of the equilibrium?
a) concentration
b) pressure
c) temperature
d) color
Answer: d
Clarification: According to Le chatelier’s principle, if a system at equilibrium is subjected to a change in concentration, pressure or temperature the equilibrium shifts in the direction that tends to nullify the effect of the change.

2. What will happen if at equilibrium the concentration of one of the reactants is increased?
a) equilibrium will shift in the forward direction
b) equilibrium population will not change
c) equilibrium will shift in the backward direction
d) equilibrium will move to and fro
Answer: a
Clarification: As per the Le chatelier’s principle, if the concentration of one of the reactants is increased at equilibrium the equilibrium will shift in the forward direction and vice versa is also possible. Therefore we can say that there is an effect of change of concentration on equilibrium.

3. In the reaction, H_2(g) + Br_2(g) (rightleftharpoons) 2HBr_(g), what will happen if there is a change in pressure?
a) equilibrium moves left
b) equilibrium moves right
c) there is no change in equilibrium
d) we cannot say
Answer: c
Clarification: According to Le chatelier’s principle, there will be no effect of pressure on equilibrium if the chemical equation has the same number of moles of reactants and products hear the moles on the left side is equal to the moles on the right side that is 2.

4. Consider the chemical equation TiCl₄ + 2 H₂O (rightleftharpoons) TiO₂ + 4 HCl if the pressure is increased what will happen to the equilibrium?
a) it moves backward direction
b) moves forward direction
c) remains constant
d) cannot say
Answer: a
Clarification: When there is a change in a number of moles, the equilibrium will shift in the direction having a smaller number of moles when the pressure is increased and vice-versa. Here the equilibrium moves in the backward direction because the reactants have less number of moles than products.

5. If a reaction produces heat during the process then it is ____________
a) exothermic
b) endothermic
c) both exothermic and endothermic
d) neither exothermic nor endothermic
Answer: a
Clarification: If a reaction releases heat then it is exothermic in nature and if the reaction consumes heat it is endothermic in nature. An example of the exothermic reaction is the mixing of washing soda in water and mixing glucose in water is an example of an endothermic reaction.

6. The formation of products is favoured by ____________ temperature in an endothermic reaction.
a) high
b) low
c) moderate
d) 0
Answer: a
Clarification: When the process is exothermic, low temperature of favors Forward reaction and when the process is endothermic high temperature favors the formation of the products. This is according to le chatelier’s principle.

7. At constant pressure, if the inert gases added then the equilibrium will shift in the direction of ___________
a) decrease in the number of moles
b) increase in the number of moles
c) does not depend on the number of moles
d) does not change
Answer: b
Clarification: At constant pressure, if inert gas added, it will increase the volume of the system there for the equilibrium will shift in a direction, in which there is an increase in the number of moles of gases as per le chatelier’s principle.

8. Which side does the catalyst shift the equilibrium position?
a) left side
b) right side
c) may be left or right side
d) does change the position
Answer: d
Clarification: As per Le chatelier’s principle on the effect of catalyst states that the presence of the catalyst does not change the position of the equilibrium it simply fast and the attainment of the equilibrium.

9. The equilibrium position does not change when there is an addition of inert gas at constant volume.
a) true
b) false
Answer: a
Clarification: If keeping the volume of the system constant and inert gases added the relative molar concentration of the substance will not change hence the equilibrium position of the reaction remains unaffected.

10. What will happen to the position of equilibrium if the concentration of one of the products is increased?
a) shifts left
b) shift right
c) does not change
d) main shift left or right
Answer: a
Clarification: According to Le chatelier’s principle if at equilibrium the concentration of one of the products is increased the equilibrium will shift in the backward direction that is towards the left side and vice versa.

Chemistry Problems,

Chemistry Multiple Choice Questions on “s-Block Elements – General Characteristics of the Compounds of the Alkali Metals”.

1. Which of the following mixture is known as fusion mixture?
a) Sodium carbonate and potassium chloride
b) Sodium carbonate and potassium carbonate
c) Sodium bicarbonate and sodium carbonate
d) Potassium bicarbonate and sodium carbonate
Answer: b
Clarification: The mixture of sodium carbonate and potassium carbonate together is known as a fusion mixture. Only Potassium carbonate is as known as pearl ash. It is sometimes used in quantitative analysis.

2. The solubility of carbonates _____________ down the group.
a) is irregular
b) remains the same
c) decreases
d) increases
Answer: d
Clarification: The solubility of carbonates and bicarbonates of alkali metals increases on moving down the group, this is due to the increase in lattice enthalpies. The order is as follows in an increasing way of lithium bicarbonate, sodium bicarbonate, potassium bicarbonate, rubidium bicarbonate, and cesium bicarbonate.

3. On heating Lithium nitrate which of the following compound is not formed?
a) Hydrogen
b) Lithium oxide
c) Nitrous oxide
d) Oxygen
Answer: a
Clarification: On heating, lithium nitrate decomposes to give nitrous oxide, oxygen, and lithium oxide whereas the nitrates of other alkali metals decompose on heating and give nitrites and oxygen. Therefore hydrogen isn’t formed.

4. Which of the following is called as Chile saltpeter?
a) Rubidium nitrate
b) Sodium nitrate
c) Lithium nitrate
d) Potassium Nitrate
Answer: b
Clarification: Sodium nitrate (NaNo₃) is called Chile saltpeter and potassium nitrate (KNO₃) is called Indian saltpeter. Chile saltpeter which is Sodium nitrate is commonly and naturally found in Chile and Peru.

5. Which of the following as a chemical formula of Glauber’s salt?
a) NaO₄.10H₂O
b) NaSO₄.1H₂O
c) NaSO₄.10H₂O
d) NaSO.10H₂O
Answer: c
Clarification: Glauber’s salt contains two atoms of sodium, one atom of sulfur, 4 atoms of oxygen, and 10 moles of water. It is the sulfate of sodium and is represented chemically as NaSO₄.10H₂O. It is soluble in water.

6. Which of the following alkali metal cannot form superoxide?
a) Potassium
b) Lithium
c) Sodium
d) Cesium
Answer: b
Clarification: All the alkali metals when heated with oxygen form different types of oxides. Lithium forms lithium oxide and some amount of lithium peroxide, while Sodium, Potassium, rubidium, and cesium can form superoxide.

7. Which of the following is true regarding the basic strength?
a) Potassium oxide is more basic than cesium oxide
b) Lithium oxide is more basic than sodium oxide
c) Cesium oxide is more basic than potassium oxide
d) Sodium oxide is more basic than cesium oxide
Answer: c
Clarification: All the oxides, peroxides and superoxides of alkali metals which are formed when heated with oxygen are basic in nature. The basic strength of oxides increases in the order of lithium oxide, sodium oxide, potassium oxide, and cesium oxide.

8. Superoxides are colored and _____________
a) attractive
b) magnetic
c) paramagnetic
d) diamagnetic
Answer: c
Clarification: All the superoxides of alkali metals are colored and paramagnetic, as they possess three electrons bond where one unpaired electron is present. Few examples of superoxide are potassium superoxide, rubidium superoxide, and cesium superoxide.

9. Lithium chloride is more covalent than potassium chloride.
a) True
b) False
Answer: a
Clarification: Lithium chloride is more covalent than Potassium Chloride, due to the smaller size of lithium bigger the onion, larger as its polarizability and the covalent character follow the order of Lithium iodide is greater than Lithium Bromide, greater than lithium chloride, greater than Lithium fluoride.

10. Which of the following is true regarding the reactivity of alkali metals towards a particular halogen?
a) Rubidium is greater than that of sodium
b) Sodium is greater than that of lithium
c) Lithium is greater than that of rubidium
d) Rubidium is greater than that of cesium
Answer: a
Clarification: Alkali metals combine readily with halogens to form ionic halides. The reactivity order of alkali metals towards a particular halogen increase in the order; lithium, sodium, potassium, rubidium, and cesium.