250+ TOP MCQs on Bonding in Some Homonuclear Diatomic Molecules and Answers

Chemistry Multiple Choice Questions on “Bonding in Some Homonuclear Diatomic Molecules”.

1. What is the electronic configuration of hydrogen molecule?
a) σ1s2
b) σ1s1
c) σ1s
d) σ*1s
Answer: a
Clarification: In a hydrogen atom, each hydrogen shares an electron from 1s orbital, so in total there are two electrons present in 1s bonding orbital of the molecule. So the electronic configuration of the hydrogen molecule is σ1s2.

2. What is the condition, for a molecule do not exist?
a) NA = NB
b) NA > NB
c) NA < NB
d) NA >/< NB
Answer: a
Clarification: When a molecule’s bond order is equal to zero, the molecule cannot exist. The formula for finding out bond order is (frac{1}{2})(NB – NA). So NA – NB = 0, that is NA = NB. Where NA as the number of anti-bonding molecular orbitals and NB as the number of bonding molecular orbitals.

3. Which of the following molecule doesn’t exist?
a) O2
b) H2
c) He2
d) N2
Answer: b
Clarification: The electronic configuration of He2 is σ1s2σ*1s2. Here NA = NB (where NA as the number of anti-bonding molecular orbitals and NB as the number of bonding molecular orbitals). So the molecule He2 doesn’t exist.

4. The number of electrons in bonding orbital and anti-bonding orbital of Lithium molecule are ______ and _______ respectively.
a) 4, 4
b) 2, 2
c) 2, 4
d) 4, 2
Answer: d
Clarification: The lithium molecule is denoted by Li2. When the lithium molecule is formed, each of the lithium shares an electron in the 2s orbital. Its electronic configuration is σ1s2σ*1s2σ2s2. So the number of electrons in bonding orbital and anti-bonding orbital of Lithium molecule are 4 and 2 respectively.

5. What is the electronic configuration of the carbon atom?
a) 1s22s22p2
b) σ1s2σ*1s2σ2s2σ*2s2π2px2π2py2
c) 1s22s22p1
d) σ1s2σ*1s σ2s2π2px2π2py2
Answer: a
Clarification: The given question electronic configuration of carbon “atom”, but not carbon “molecule”. So by following Aufbau’s principle and noting that carbon contains 6 electrons, the answer is given as 1s22s22p2.

6. O2 is paramagnetic in nature.
a) True
b) False
Answer: a
Clarification: σ1s2σ*1s2σ2s2σ*2s22pz2π2px2π2y2π*2px1π*2py1 is the electronic configuration of O2 molecule. According to this, it has two unpaired electrons, making it paramagnetic and is attracted by the magnetic field.

7. The electronic configurations of molecules change when the number of electrons is ________
a) 10
b) 20
c) 17
d) 14
Answer: d
Clarification: The electronic configuration below number of electrons = 14, i.e. number of electrons = 12 is σ1s2σ*1s2σ2s2 σ*2s2π2px2π2y2but after 14, take number of electrons = 16 is σ1s2σ*1s2σ2s2σ*2s22pz2π2px2π2y2π*2px1π*2py1 So the change comes near 2pz orbital’s place.

8. Which of the following is true regarding nitrogen molecule.
a) Diamagnetic
b) Paramagnetic
c) Bond order is 2
d) Total number of electrons in the molecule is 13
Answer: a
Clarification: The Nitrogen molecule N2 is diamagnetic in nature as it contains zero single electrons, all the electrons are paired. Its bond order is calculated as 3 and the total number of electrons the nitrogen molecule are 14.

9. H2, N2, O2 and Li2 are ______________
a) heteronuclear diatomic molecules
b) heteronuclear triatomic molecules
c) homonuclear diatomic molecules
d) homonuclear triatomic molecules
Answer: c
Clarification: The molecules that have the same atoms in them are called as homonuclear molecules. The molecules that have only two atoms are called diatomic molecules. So H2, N2, O2, and Li2 are called homonuclear diatomic molecules.

10. What is the total number of electrons in the Chlorine molecule?
a) 17
b) 34
c) 18
d) 16
Answer: b
Clarification: A Chlorine atom is made up of 17 electrons. When two Chlorine atoms share an electron covalently, the Chlorine molecule is formed. So the Chlorine molecule Cl2 has a total of 17 + 17 atoms = 34 atoms.

250+ TOP MCQs on States of Matter – Kinetic Molecular Theory of Gases and Answers

Chemistry Multiple Choice Questions on “States of Matter – Kinetic Molecular Theory of Gases”.

1. Which of the following assumption explains great compressibility of gases?
a) the actual volume of the gas molecules is negligible
b) there is no force of attraction
c) particles are always in random motion
d) different particles have different speeds
Answer: a
Clarification: Gas molecules are considered as point masses because the actual volume of gas molecules is negligible when compared to the space between them, so this assumption explains the greater compressibility of gases.

2. Gases ________ and occupy all the space that is available to them.
a) contract
b) compress
c) expand
d) shrink
Answer: c
Clarification: At normal temperature and pressure, there is no force of attraction between the gas particles. So they expand and occupy the space that is available. This statement supports the assumption.

3. Gases do not have a fixed shape.
a) true
b) false
Answer: a
Clarification: Particles do not have a fixed shape because they are always in random motion and particles never occupy fixed position the keep on moving and never occupy a particular shape this is the reason to prove that the above statement is true.

4. Why do you think the pressure is exerted by the gas on the walls of the container in all directions?
a) they consist of identical particles
b) collide with each other during random motion
c) lack of definite shape
d) more forces of attraction
Answer: b
Clarification: As the particles of gas travel in straight lines and move in a random motion and collide with each other and also collide with the walls of the container, the pressure is exerted on the walls of the container in all directions.

5. Collisions of gas molecules are ___________
a) perfectly elastic
b) inelastic
c) always occur in a proper and predicted motion
d) not conserved
Answer: a
Clarification: The total energy of molecules before the collision is equal to the total energy after the collision. That means there may be an exchange of energy between the molecules but the total energy does not change, so the collisions of gas molecules are elastic.

6. At an instance different particles have ________ speeds.
a) same
b) different
c) opposite
d) similar
Answer: b
Clarification: According to the kinetic molecular theory of gases, the assumption that different particles have different speeds at a time is because particles do not always have the same speeds and move in different directions but we can assume that the distribution of speeds is constant at a particular temperature.

7. There is an increase in temperature of an object then the kinetic energy of the object __________
a) decreases
b) increases
c) remains the same
d) it is not related to the temperature
Answer: b
Clarification: From the kinetic molecular theory of gases we can know that the kinetic energy of an object is proportional directly to the Absolute Temperature of that object so when the temperature of an object increases obviously the kinetic energy also increases.

8. The temperature of a gas is 100 K it is heated until it is 200 k then, what do you understand regarding kinetic energy in this process?
a) halved
b) tripled
c) quadrupled
d) doubled
Answer: d
Clarification: According to the kinetic molecular theory of gases, if the Absolute Temperature of a gas is doubled then the kinetic energy is also doubled because they are directly proportional to each other so here the answer is doubled. If the temperature doubles from 100 k to 200 k, kinetic energy also doubles.

9. Which of the following is not a postulate of Kinetic molecular theory of gases?
a) the actual volume of gas molecules is negligible
b) there are high forces of attraction between the gas molecules
c) collisions are elastic in gas molecules
d) kinetic energy of gas molecules is directly proportional to the absolute temperature
Answer: b
Clarification: According to Kinetic molecular theory of gases, there are no forces of attraction between the particles of gas at normal temperature and pressure. This is because gases expand and occupy all the space available.

10. Which of the following statements do you think is the correct one?
a) in gases there is a predominance of intermolecular energy
b) all molecules have the same speed at different temperatures
c) collisions of gas molecules are elastic
d) do not exert the same pressure in all directions
Answer: c
Clarification: Collisions are elastic as the total amount of energy before and after the collision is the same. The correct statements of the incorrect ones are: in gases, there is a predominance of thermal energy, the distribution of speeds remains constant at a particular temperature and the gases exert the same pressure in all directions of the container.

250+ TOP MCQs on Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G and Answers

Chemistry Multiple Choice Questions and Answers for Class 11 on “Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G”.

1. In a reaction, at 300k, KC is given as 2 x 1013, then what is the value of ΔG?
a) – 7.64 × 104 J
b) – 7.64 × 104 J mol–1
c) – 7.64 × 10 J mol–1
d) – 7.64 × 104 mol–1
Answer: b
Clarification: We know that ΔG0 = – RT lnKc, where ΔG0 is the standard Gibbs free energy, R is universal gas constant, T is the temperature and KC is equilibrium constant; substituting KC as 2 x 1013, ΔG0 = (– 8.314J mol–1K–1 × 300K) × ln(2×1013); ΔG0 = – 7.64 × 104 J mol–1.

2. For a chemical reaction, the value of ΔG0 is -831.4 J/mol. Then what is the value of KC at 100 k?
a) 1.0077
b) 1.077
c) 1.007
d) 2.7
Answer: d
Clarification: We know that ΔG0 = – RT lnKc, where ΔG0 is the standard Gibbs free energy, R is universal gas constant, T is the temperature and KC is equilibrium constant; substituting ΔG0 as -831.4 J/mol, we get lnk = -831.4J/mol divided by -8.314J mol–1K–1 × 100K = 1; lnk = 1; K = e1 = 2.7

3. If the value of ΔG0 is -2502 J/mol and K is 2, what is the temperature of the reaction that is occurring?
a) 200 k
b) 101 k
c) 100 k
d) 300 k
Answer: c
Clarification: We know that ΔG0 = – RT lnKc, where ΔG0 is the standard Gibbs free energy, R is universal gas constant, T is the temperature and KC is equilibrium constant; substituting ΔG0 as -2502 J/mol, we get -2502 J/mol = -8.314J mol–1K–1 × T ln2 = 2502 J/mol = T = 2502/2.502 = 100 k.

4. In a reaction, if the value of Gibbs free energy is greater than zero what does it infer?
a) K is greater than 1
b) K is less than 1
c) K is equal to 1
d) Cannot deduce K from Gibbs free energy
Answer: b
Clarification: If the value of Gibbs free energy is greater than the zero that means, -ΔG0/RT is negative and that e-ΔG0/RT is greater than 1, so the K is greater than 1 this means the reaction is nonspontaneous and proceeds in the forward direction.

5. When is a reaction nonspontaneous?
a) Gibbs free energy is positive
b) Gibbs free energy is negative
c) Gibbs free energy is zero
d) Does not depend on Gibbs free energy
Answer: a
Clarification: When Gibbs free energy is positive, the reaction that occurs is nonspontaneous and a reaction occurs backward that is the products are converted into reactants. Simply the reverse reaction could occur.

6. What did the Q depicted in the equation; ΔG = ΔG0 + RT lnQ?
a) reaction coefficient
b) reaction quotient
c) equilibrium constant
d) free energy
Answer: b
Clarification: In the equation ΔG = ΔG0 + RT lnQ, ΔG is the Gibbs free energy, R is the universal gas constant, T is the temperature and Q is the reaction quotient. When Gibbs free energy is zero, the reaction quotient becomes equilibrium constant.

7. If the value of Gibbs free energy for a reaction is 20J/mol, the reaction is ___________
a) spontaneous
b) nonspontaneous
c) may be spontaneous
d) may not be spontaneous
Answer: b
Clarification: Is the value of Gibbs free energy for a reaction is 20 J/mol, that means it is positive, -ΔG0/RT is negative and that e-ΔG0/RT is greater than 1, so the K is greater than 1 this means the reaction is nonspontaneous.

8. For a reaction, Kc = 3.81 × 10–3 and ΔG0 = 13.8 kJ/mol. Then what is the value of R?
a) -8.314J mol–1K–1
b) 8.314J mol–1K–1
c) cannot say as the temperature is not given
d) -8.314J mol–1
Answer: b
Clarification: Though the temperature is not given, universal gas constant value always remains the same, whatever may be the other values. So the universal gas constant is given by a constant value that is -8.314J mol–1K–1.

9. What happens when reaction quotient is equal to the equilibrium constant?
a) the reaction proceeds in the forward direction
b) the reaction proceeds in the backward direction
c) the reaction reaches equilibrium
d) cannot predict
Answer: c
Clarification: When Gibbs free energy is zero, the reaction reaches equilibrium and at equilibrium, the reaction quotient is replaced by the equilibrium constant, as both the values are equal. That is when reaction quotient is equal to the equilibrium constant reaction reaches equilibrium.

10. Is a relationship between reaction quotient and Gibbs free energy at a temperature T?
a) ΔG = ΔG0 + RT lnQ
b) ΔG = ΔG0 + RT lnk
c) ΔG = ΔG0 + R lnQ
d) ΔD = ΔG0 + RT lnQ
Answer: a
Clarification: The relationship between reaction quotient and Gibbs free energy at temperature t is given as as ΔG = ΔG0 + RT lnQ, where ΔG is the Gibbs free energy, R is the universal gas constant, T is the temperature and Q is the reaction quotient.

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250+ TOP MCQs on s-Block Elements – Group 1 Elements: Alkali Metals and Answers

Chemistry Multiple Choice Questions on “s-Block Elements – Group 1 Elements: Alkali Metals”.

1. Which of the following metal is not an alkali metal?
a) magnesium
b) rubidium
c) sodium
d) caesium
Answer: a
Clarification: Alkali metals are the elements of group 1. The outer shell configuration of group 1 elements is ns1, where n is the number of it’s period. Magnesium is not an alkali metal because it’s outer shell configuration is ns2.

2. Alkali metals have the biggest atomic radii.
a) true
b) false
Answer: a
Clarification: The alkali metals have the biggest atomic radii in their respective periods, atomic radii increases as we go down the group due to the addition of a new shell in each subsequent step. So the above statement is true.

3. The melting point of alkali metal is _____________
a) depends on the atmosphere
b) low
c) high
d) zero
Answer: b
Clarification: The melting and boiling points of alkali metals are quite low and they decrease down the group due to weakening of their metallic bonds. Francium is the only element in this group which is a liquid at room temperature.

4. Is there removal of second electron difficult in alkali metals?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: a
Clarification: The first ionization enthalpy of alkali metals is the lowest among the elements in their respective periods and increases on moving down the Group. The second ionization enthalpies of the alkali metals are very high because by releasing an electron, ions require noble gas configuration, so removal of the second electron is difficult.

5. Alkali metals are strongly _____________
a) neutral
b) electropositive
c) electronegative
d) non-metallic
Answer: b
Clarification: Due to low ionization enthalpies, alkali metals are strongly electropositive or metallic in nature and electropositive nature increases from Lithium to caesium due to decrease in ionization enthalpy.

6. Alkaline earth metals show +1 Oxidation state and their atomic volume _____________ down the group.
a) is irregular
b) increase
c) decrease
d) do not change
Answer: c
Clarification: The alkali metal atom show only +1 Oxidation State, because of their unipositive Ion at the time the stable noble gas configuration. The atomic volume of alkali metals is the highest in its period and goes on increasing down the group from top to bottom.

7. Does the degree of hydration depend upon the size of the cation?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: a
Clarification: The degree of hydration depends upon the size of the cation, smaller the size of the cation, greater is its hydration enthalpy. The relative degree of hydration in an increasing order is Li+ > Na+ > K+ > Rb+ > Cs+.

8. The flame of caesium is in the colour _____________
a) white
b) red violet
c) yellow
d) blue
Answer: d
Clarification: Alkali metals and their salt impart characteristic colours to the flame because the outer electrons get excited to higher energy levels. When the electrons return to the original state, it releases visible light of a characteristic wavelength which provides colour to the flame. The colour of the Flame of the caesium is blue.

9. Caesium has the highest electrical conductivity in its group.
a) true
b) false
Answer: a
Clarification: Due to the presence of loosely held Valence Electrons which are free to move throughout the metal structure, the alkali metals are good conductors of heat and electricity. Electrical conductivity increases from top to bottom in the order, so caesium has the highest electrical conductivity in its group.

10. All alkali metals are good dash agents?
a) oxidizing
b) reducing
c) both oxidising and reducing
d) neither oxidizing not reducing
Answer: b
Clarification: All the alkali metals are good reducing agents due to their low ionization energies. The reducing character of group 1 elements follows the increasing order of Sodium, Potassium, rubidium, Caesium and lithium.

11. What happens when alkali metals are exposed to moist air?
a) formation of nitrates
b) formation of oxides
c) formation of chlorides
d) formation of sulphates
Answer: b
Clarification: On exposure to moist air, the surface gets tarnished due to the formation of oxides, hydroxide and carbonates. Few examples are sodium hydroxide, sodium carbonate, potassium hydroxide etc.

12. Sodium Peroxide is _____________ in colour and potassium superoxide is used as a source of _____________
a) blue, yellow
b) yellow, hydrogen
c) blue, oxygen
d) yellow, oxygen
Answer: d
Clarification: Sodium Peroxide acquires yellow colour due to the presence of superoxide as an impurity. Potassium superoxide is used as a source of oxygen in submarines, space shuttles and an emergency breathing apparatus such as oxygen masks.

13. Which of the following is true regarding the reactivity order of alkali metals towards hydrogen?
a) Li < Na < K > Rb
b) Lithium < Na < K < Rb < Cs
c) Li > Na < Cs
d) Li < Rb > Cs
Answer: b
Clarification: Two moles of alkali metal reacts with one mole of hydrogen molecule in order to form 2 moles of alkali metal hydride. The correct order of reactivity of alkali metals towards hydrogen is Li < Na < K < Rb < Cs.

14. Lithium fluoride is _____________ in water.
a) completely soluble
b) soluble
c) insoluble
d) cannot say
Answer: c
Clarification: All alkali halides except Lithium fluoride are soluble in water, this is because Lithium fluoride is soluble in nonpolar solvents as it has a strong covalent bond. Lithium fluoride is represented by the formula LiF.

15. Alkali metals dissolving in Ammonia liquid give the blue solution, this is due to the formation of ammoniated _____________
a) ions
b) metal cations only
c) metal cations and electrons
d) electrons only
Answer: c
Clarification: They give deep blue solution due to the formation of ammonia the metal cations and electrons, the blue colour is due to the oxidation of ammonia electron to higher energy levels and the absorption of photons occurs in the red region of the spectrum.

250+ TOP MCQs on Structural Representations of Organic Compounds and Answers

Chemistry Multiple Choice Questions on “Structural Representations of Organic Compounds”.

1. Identify the condensed formula of ethane from the following.
a) CH3-CH3
b) HC-H2-C-H3
c) CH2=CH2
d) CH3CH3
Answer: d
Clarification: In condensed formula, all the atoms are represented but single bonds are not shown. Only double and triple bonds will be represented. CH2=CH2 is not ethane; it is ethene.

2. Which is not a heteroatom?
a) Oxygen
b) Nitrogen
c) Carbon
d) Sulphur
Answer: c
Clarification: A heteroatom is any atom other than carbon or hydrogen atom. All the others apart from carbon mentioned above are heteroatoms. In other words, a heteroatom is a non-carbon atom present in a carbon structure.

3. No atoms are shown while representing structures in bond-line formula.
a) True
b) False
Answer: b
Clarification: Only carbon and hydrogen atoms are not shown in bond-line representation. But other atoms like oxygen, nitrogen, chlorine, etc. are shown. Bond-line representation of molecules is a skeletal representation in which the bonds are shown as single lines according to the bond order, i.e. one line for a single bond, two lines for a double bond and so on.

4. Identify the condensed formula of the given compound from the following.
ClCH2CH2CH2CH2CHCH2CH2CH2CHCH2CH3
a) Cl (CH2)2 (CH2)2CH (CH2)2CH2CH2CH3
b) Cl (CH2)3CH2CHCH2CH2CH2CH (CH3) 2
c) Cl (CH2)4CH (CH2)3CH (CH2) CH3
d) Cl (CH2)3CH2CH2CH (CH3)2CH3CHCH2CH3
Answer: c
Clarification: For determining the condensed formula, combine all the CH2 terms together and assign the appropriate number to them, each time checking that carbon only forms four bonds with the other atoms. Here, in the first part, i.e. ClCH2CH2CH2CH2CHCH2CH2CH2CHCH2CH3, there are four CH2 terms; so we group all these terms and represent them as (CH2)4. Similarly, combine the CH2 terms in the other parts and eventually we get the condensed formula of the above mentioned molecule.

5. Determine the bond-line formula of CH3CH2COCH2CH3.
a)
b)
c)
d)
Answer: c
Clarification: In CH3CH2COCH2CH3, the ketone group(C=O) is attached to the third carbon and therefore the double bond should be represented at the end of the second line, i.e. at the position of the third carbon. In all the other options, the ketone group is not attached to the third carbon, thereby, making them incorrect.

250+ TOP MCQs on Strategies to Control Environmental Pollution and Answers

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.

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