Category: Class 11 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Developments Leading to the Bohr’s Model of Atom”.

1. What’s the wavelength for the visible region in electromagnetic radiation?
a) 400 – 750 nm
b) 400 – 750 mm
c) 400 – 750 μm
d) 400 – 750 pm
Answer: a
Clarification: Electromagnetic spectrum is made up of various electromagnetic radiation. They are radio waves, X-rays, gamma rays, UV rays, the visible region, IR waves, and microwaves. Visible rays are the only ones which a human eye can see. They range from 450 – 750 nm.

2. What is the wavenumber of violet color?
a) 25 x 10³ mm^-1
b) 25 x 10³ m^-1
c) 25 x 10³ cm^-1
d) 25 x 10³ nm^-1
Answer: c
Clarification: The wavenumber is the reciprocal or the inverse of wavelength. Wavenumber = 1/Wavelength. Its unit is cm^-1. The wavelength of violet color is 400nm as seen in the electromagnetic spectrum. So wavenumber = 1/400nm = 25 x 10³ cm^-1.

3. Calculate the frequency of the wave whose wavelength is 10nm.
a) 2 Hz
b) 3 Hz
c) 1 Hz
d) 4 Hz
Answer: b
Clarification: The relation between wavelength(λ) and frequency(v) of a wave is given by λ = c/v where c is the speed of light of the light. v = c/λ Frequency of the given wave = (3 x 10⁸m/s)/(10 x 10^-9m) = 3 Hz.

4. If Energy = 4.5 KJ; calculate the wavelength.
a) 4.42 x 10^-29 m
b) 4.42 x 10^-39 m
c) 4.42 x 10^-25 m
d) 4.42 x 10^-22 m
Answer: a
Clarification: We know E = hv through Planck’s Quantum Theory, where E is energy, h is Planck’s constant and v is the frequency. 4.5 KJ = (6.626×10^–34 Js)(3 x 10⁸m/s)/(wavelength). wavelength = 4.42 x 10^-29 m.

5. ________ frequency, is the minimum frequency required to eject an electron when photons hit the metal surface.
a) Required
b) Activated
c) Threshold
d) Limiting
Answer: c
Clarification: In the photoelectric effect, when photons strike on a metal surface, it emits electrons. Thus for emitting an electron, it requires a minimum amount of energy. This is threshold energy acquired through threshold frequency.

6. A metal’s work function is 3.8KJ. Photons strike metal’s surface with an energy of 5.2 KJ. what’s the kinetic energy of the emitted electrons?
a) 3.8 KJ
b) 5.2 KJ
c) 9 KJ
d) 1.4 KJ
Answer: d
Clarification: As per the formula of the photoelectric effect, we have E = K.E. + W_o. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and W_o is the work function. K.E. = 5.2 KJ – 3.8 KJ = 1.4 KJ.

7. When an electron jumps from 3^rd orbit to 2^nd orbit, which series of spectral lines are obtained?
a) Balmer
b) Lyman
c) Paschen
d) Brackett
Answer: a
Clarification: As per the spectral lines of the hydrogen, when an electron jumps from n_th orbit to 2^nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in visible region.

8. Find out the wavenumber, when an electron jumps from 2^nd orbit to 1^st.
a) 82357.75 cm^-1
b) 105,677 cm^-1
c) 82257.75 cm^-1
d) 109,677 cm^-1
Answer: c
Clarification: The Swedish spectroscopist, Johannes Rydberg gave a formula; Wavenumber = R_H[(1/n₁)²-(1/n₂)²]. Here R_H is the Rydberg constant and is equal to 109,677 cm^-1. Wavenumber = 109,677(3/4) = 82257.75 cm^-1.

9. The ultraviolet spectral region is obtained in Balmer series.
a) True
b) False
Answer: b
Clarification: When an electron jumps from n^th orbit to 1st orbit, provided that n = 1, 2, 3, etc, it emits radion in the ultraviolet region. As per the spectral lines of the hydrogen, when an electron jumps from n^th orbit to 2^nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in the visible region.

10. During the photoelectric effect, when photons strike with 5.1eV, electrons emitted from which metal have higher kinetic energy?

a) Na
b) Ag
c) Equal
d) Neither
Answer: a
Clarification: As per the formula of the photoelectric effect, we have E = K.E. + W_o. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and W_o is the work function. K.E. of Na and Ag are 2.8 eV and 0.8 eV.

Chemistry Multiple Choice Questions on “States of Matter – Kinetic Energy and Molecular Speeds”.

1. Calculate the root mean square speed of hydrogen in m/s at 27°C?
a) 2835.43 m/s
b) 2635.43 m/s
c) 2735.43 m/s
d) 2731.43 m/s
Answer: c
Clarification: The formula of root mean square speed is given by u_rms = (sqrt{3RT/M}). We have R = 8.314 kgm²/s², M = 10^-3 kg/mol and T = 300 k. So by substituting the formula we get, u_rms = (sqrt{3×8.314×300/10^{-3}}) = 2735.43 m/s.

2. What is the ratio of u_rms to u_mp in oxygen gas at 298k?
a) 1.124
b) 1.224
c) 1.228
d) 1.128
Answer: b
Clarification: The ratio of root mean square speed, represented as u_rms to the most probable speed, represented as u_mp is always the same for identical conditions and same gas. It is (sqrt{3RT/M}) divided by (sqrt{8RT/πM}) = 1.224.

3. The speed of three particles is recorded as 3 m/s, 4 m/s, and 5 m/s. What is a root mean square speed of these particles?
a) 4.082 m/s
b) 2.07 m/s
c) 3.87 m/s
d) 3.082 m/s
Answer: a
Clarification: The root means square speed of particles is nothing but the square root over the sum of squares of the particle’s speeds by a total number of particles. So by substituting, √3² + 4² + 5²/3 = 4.082 m/s.

4. What is the ratio of root mean square speed of 16 grams of Oxygen to 4 grams of hydrogen?
a) 2
b) 3
c) 4
d) 1
Answer: a
Clarification: The formula of root mean square speed of particles is given as (sqrt{3RT/M}). We know that the velocity of gas molecules is inversely proportional to the root over the mass of the gas here the mass of oxygen to the mass of hydrogen ratio is the answer. So (sqrt{16/4}) = 2.

5. Which of the following is greater for identical conditions and the same gas?
a) most probable speed
b) average speed
c) root mean square speed
d) most probable and average speed have the same value
Answer: c
Clarification: According to the formula, the root mean square speed is greater than the average speed and the average speed is greater than the most probable speed at given identical conditions and for the same gas.

6. The root mean square speed of a gas at a certain condition is 1.128 times greater than the most probable speed.
a) true
b) false
Answer: b
Clarification: The ratio of root mean square speed to the mean probable speed is 1.224. So the above statement is considered to be wrong. The ratio between the main probable speed and the average speed and root mean square speed is 1: 1.128: 1.224.

7. What is the most probable speed of oxygen gas with the mass of 32 grams at 27-degree centigrade?
a) 33.74 m/s
b) 44.78 m/s
c) 57.94 m/s
d) 549.14 m/s
Answer: b
Clarification: The formula for the most probable speed of a gas is given as (sqrt{8RT/πM}). Here R is a universal gas constant which is always equal to 8.314 kgm²s^-2, T=300 Kelvin and M = 0.032 kg So by substituting, we get an answer as 44.78 m/s.

8. Which among the following options do you think has the highest average speed?
a) chlorine
b) hydrogen
c) neon
d) oxygen
Answer: b
Clarification: The formula of average speed is given by (sqrt{2RT/M}), where R is universal gas constant, T is a temperature in Kelvin and M is the mass in kilograms. From the formula, we understand that the average speed is inversely proportional to the root over the mass. As hydrogen has the least mass among the options it has the highest average speed.

9. What is the ratio of the velocities of 2 moles of hydrogen to five moles of helium?
a) (sqrt{14})
b) (sqrt{10})
c) (sqrt{20})
d) (sqrt{50})
Answer: b
Clarification: The formula of average speed is given by (sqrt{2RT/M}). We know that the velocity of gas molecules is inversely proportional to the root over the mass of the gas. So here the ratio of velocities is (sqrt{5×4/2×1}) = (sqrt{10}).

10. What is the mean velocity of one Mole neon gas at a temperature of 400 Kelvin?
a) 11.533 m/s
b) 357.578 m/s
c) 367.79 m/s
d) 34 m/s
Answer: a
Clarification: The formula for mean velocity of a gas is given by the expression (sqrt{2RT/M}). for one mole of neon gas M is taken as 0.02 kg, temperature as 400 k, R as 8.314 kgm²s^-2, so by substituting we get an answer as 11.533 m/s.

Chemistry Multiple Choice Questions on “Applications of Equilibrium Constants”.

1. If K_C > 10x, the products predominate over reactants. Then what is the value of x?
a) 2
b) 3
c) 4
d) 1
Answer: b
Clarification: If K_C > 10x, the products are predominant, over the reactants in a chemical reaction because if K_C is very large the reaction proceeds nearly to completion an example for this is at 300K, H_2(g) + Cl_2(g) (rightleftharpoons) 2HCl_(g), K_C = 4 x 10³¹.

2. What will happen if K_C > 10^-3 in a chemical reaction?
a) products are predominant
b) reactants are predominant
c) equilibrium
d) dynamic equilibrium
Answer: d
Clarification: If K_C > 10^-3, the reactants are predominant over the products in a chemical reaction and the products are predominant, over the reactants in a chemical reaction, if K_C > 10³, where K_C is the equilibrium constant.

3. Reaction quotient is depicted by this symbol __________
a) K
b) Q_C
c) R
d) q
Answer: b
Clarification: For any reversible reaction at any stage other than equilibrium, the ratio of the molar concentrations of the products to that of the reactants, where is concentration term is raised to the power equal to the stoichiometric efficient to the substance concerned is called the reaction quotient, Q_C.

4. For a reaction aA + bB → cC + dD, which is not in equilibrium the Q_C is given as __________
a) [A]^a[B]^b/[C]^c[D]^d
b) [C]^c[D]^d/[A]^a[B]^b
c) [A][B]/[C][D]
d) [C][D]/[A][B]
Answer: b
Clarification: A very basic reaction like aA + bB → cC + dD, where the capital letters represent the compounds or molecules and the small letters are the coefficients of them the reaction quotient Q_C, is given by [C]^c[D]^d/[A]^a[B]^b.

5. What will happen If Q_C > K_C?
a) Q_C decreases till equilibrium
b) Q_C increases till equilibrium
c) Q_C remains constant
d) cannot say
Answer: a
Clarification: If Q_C > K_C, the value of Q_C will tend to decrease to reach the value of equilibrium constant (that is towards equilibrium) and the reaction will continue in the opposite direction, where Q_C is reaction quotient and K_C is the equilibrium constant.

6. At equilibrium, K_C is _______________
a) greater than reaction quotient
b) equal to the reaction quotient
c) less than the reaction question
d) independent of reaction question
Answer: b
Clarification: At equilibrium, the equilibrium constant and the reaction quotient is equal. The equilibrium constant is depicted by the symbol K_C and the reaction quotient is represented by the symbol Q_C.

7. What do you think will happen if reaction quotient is smaller than the equilibrium constant?
a) equilibrium constant will change
b) reaction quotient remains constant
c) reaction quotient increases continuously
d) reaction quotient increases till K_C
Answer: d
Clarification: If the reaction quotient is less than the equilibrium constant K_C, the reaction quotient will tend to increase and the reaction will proceed in the forward direction, till it reaches the value of the equilibrium constant.

8. If [H₂]t=0.10 M, [I₂]t = 0.20 M and [HI]t = 0.40 M, in the reaction H_2(g) + I_2(g) (rightleftharpoons) 2HI_(g), what is the value of Q_C?
a) 8
b) 4
c) 2
d) 1
Answer: a
Clarification: We know that reaction quotient Q_c = [HI]t² / [H₂]t[I₂]t and given that [H₂]t=0.10 M, [I₂]t = 0.20 M and [HI]t = 0.40 M, So by substituting, we get Q_c = (0.40 M x 0.40 M)/(0.20 M x 0.10 M) = 8.0, 8 is the answer.

9. What do you understand from the reaction if reaction quotient is 2 and the equilibrium constant is 3?
a) the equilibrium constant increases
b) the equilibrium constant decreases
c) the equilibrium constant remains the same
d) reaction quotient increases
Answer: d
Clarification: In a reaction, if reaction quotient is less than the equilibrium constant, the reaction quotient will tend to increase and the reaction will proceed in the forward direction till it reaches equilibrium.

10. Equilibrium constant depends on the temperature.
a) true
b) false
Answer: a
Clarification: Yes, the equilibrium constant depends on the temperature and its unique for a chemical reaction at a given temperature. So the above-given statement about the equilibrium constant is considered to be true.

Chemistry Written Test Questions and Answers for Class 11 on “Dihydrogen as a Fuel”.

1. Absorption of hydrogen on a metal surface is known as __________
a) adsorption
b) occlusion
c) absorption
d) emission
Answer: b
Clarification: Occlusion is nothing but the absorption of hydrogen at a metal surface which is a primary cause of reduction, hydrogenation, etc. As the temperature increases, occlusion decreases i.e. they are dependent on each other.

2. In a hydrogen molecule, when the spins are in the same direction. What is it known as?
a) adsorbent hydrogen
b) nascent hydrogen
c) parahydrogen
d) ortho hydrogen
Answer: d
Clarification: When in the hydrogen molecule, the nucleus spins are in the same direction, it is known as orthohydrogen. When the spins are in the opposite direction, it is known as parahydrogen. At room temperature hydrogen consists of 75% ortho and 25% para hydrogens.

3. What is freshly prepared hydrogen known as?
a) nascent hydrogen
b) atomic hydrogen
c) parahydrogen
d) absorbed hydrogen
Answer: a
Clarification: Nascent hydrogen is the hydrogen is freshly obtained from a chemical reaction and is more reactive than normal hydrogen comparatively as it is not stable. The activity of nascent hydrogen depends upon the reaction by which, it is obtained.

4. What is the problem that arises with hydrogen as a fuel?
a) storage
b) production
c) process
d) availability
Answer: a
Clarification: The production of hydrogen is very easy and its process is also simple because the availability is enormous in the atmosphere as well as in the form of water, but the problem comes in the storage because we need to store hydrogen in a high-pressure tank or in a cryogenic tank.

5. Is liquid hydrogen fuel dangerous?
a) yes
b) no
c) may be
d) may not be
Answer: a
Clarification: Yes, hydrogen fuel is dangerous as it has low ignition energy and high combustion energy. It tends to leak easily from tanks, this is a reason for many explosions at hydrogen fuel manufacturing or storing centers.

6. Can dihydrogen be used as a Fuel?
a) yes
b) no
c) may be
d) cannot say
Answer: a
Clarification: Yes, dihydrogen can be used as a fuel and the process of usage of liquid hydrogen as an alternate source of energy is given by the term hydrogen economy. In the near future, there is a possibility of using liquid hydrogen as a fuel.

7. What do you think the hydrogen economy involves?
a) transfer
b) production
c) implementation
d) process
Answer: b
Clarification: Hydrogen economy is a technology that involves production, transportation, and storage of energy in the form of liquid hydrogen as it is an alternate source of energy, there are many ways in which hydrogen can be used as fuel.

8. Hydrogen is stored in an alloy.
a) true
b) false
Answer: a
Clarification: The storage of hydrogen in liquid form can be done in vacuum insulated cryogenic tanks or in a metal or in an alloy like Iron Titanium alloy such as interstitial hydride. Hence, the above statement is true.

9. What can photo hydrogen obtain?
a) rotational energy
b) wind energy
c) renewable energy
d) water energy
Answer: c
Clarification: Photohydrogen as used to obtain renewable energy from sunlight. This can be possible by using microscopic organisms such as bacteria or algae. This is also a form of obtaining energy through hydrogen.

10. How can we produce a large scale of hydrogen?
a) titration
b) by electrolysis
c) hydrogenation
d) storage
Answer: c
Clarification: Large scale production of hydrogen can be done by electrolysis of water or by thermochemical reaction cycle. Hydrogen fuel has many advantages over conventional fuels that it is nonpolluting and it liberates a large amount of energy on combustion.

To practice Chemistry Written Test Questions and Answers for Class 11,

Chemistry Written Test Questions for NEET Exam on “Tetravalence of Carbon: Shapes of Organic Compounds”.

1. What is the hybridization of CH₃CH₂CH₂CN?
a) sp, sp³
b) sp², sp³
c) sp
d) sp, sp²
Answer: a
Clarification: The complete structure of the compound mentioned is → CH₃-CH₂-CH₂-C≡N. When carbon forms only single bonds, then it is sp³ hybridized. When it forms triple bond with nitrogen, then it is sp hybridized. So CH₃CH₂CH₂CN is sp and sp³ hybridized.

2. What is the valency of carbon?
a) Pentavalent
b) Divalent
c) Trivalent
d) Tetravalent
Answer: d
Clarification: The atomic number of carbon is 6 and thus it has 4 valence electrons. So, it can either accept 4 electrons or donate 4 electrons in order to become stable. But it is difficult for carbon to neither lose 4 electrons due to its strong force of attraction with the nucleus nor gain 4 electrons since the protons in the nucleus are not sufficient to hold 8 electrons in them. Therefore, carbon forms 4 covalent bonds with other atoms, thereby, exhibiting tetravalency.

3. How many σ and π bonds are present in the following molecule?
N≡C-CH-C≡N
a) σ = 5; π = 4
b) σ = 6; π = 3
c) σ = 4; π = 2
d) σ = 3; π = 5
Answer: a
Clarification: Firstly, you have to draw the complete Lewis structure of the molecule. Make sure the valencies of all atoms are satisfied and then you have to count the total number of bonds present. In a triple bond, one of the bonds is a sigma (σ) bond and the other two are π bonds. In a double bond, one is σ bond and the other is a π bond. All the single bonds are σ bonds.

4. Arrange the following in the increasing order of electronegativity.
a) sp² < sp < sp³
b) sp³ < sp² < sp
c) sp < sp² < sp³
d) sp³ < sp < sp²
Answer: b
Clarification: Greater the s-character of the hybrid orbitals, greater is the electronegativity. Sp hybrid orbitals have 50% s-character, thereby, greater electronegativity. Sp³ has only 25% s-character and sp² hybrid orbitals have only 33% s-character, making sp the greatest electronegative orbital among them.

5. σ bonds are stronger than π bonds.
a) True
b) False
Answer: a
Clarification: In a σ bond, linear overlapping takes place whereas in a π bond, parallel overlapping takes place. Linear overlapping results in greater extent of overlapping which makes σ bond stronger than the π bond.