Category: Class 11 Chemistry Objective Questions

Chemistry MCQs on “Bohr’s Model for Hydrogen Atom”.

1. Angular momentum of an electron is quantized.
a) True
b) False
Answer: a
Clarification: According to Bohr’s postulate, angular momentum is quantized and this is given by the expression m_evr = nh/2π. (n =1, 2, 3….). m_evr is the angular momentum and h is the Planck’s constant. Movement of an electron can only be possible in orbits whose angular momentum is the integral multiple of h/2π

2. The energy of 1^st orbit in a hydrogen atom __________
a) 3.18×10^–12 J
b) –2.18×10^–18 J
c) –3.18×10^–18 J
d) 2.18×10^–18 J
Answer: b
Clarification: The energy of an nth orbit in a hydrogen atom is given by the formula E_n = -R_H/n², where is the energy of nth orbit and R_H is the Rydberg constant. The energy of 1^st orbit in a hydrogen atom = -2.18 x 10^-18 J/1 = -2.18 x 10^-18 J.

3. What is the ratio of the atomic radius of the 5^th orbit in chlorine atom and 3^rd orbit in Helium atom?
a) 153:50
b) 50:153
c) 153:100
d) 100:153
Answer: b
Clarification: The atomic radius of an atom is given by the formula r_n = 52.9n²/Z pm, where r_n is the radius of nth orbit of an atom and Z is the atomic number of that atom. The ratio of the atomic radius of the 5^th orbit in chlorine atom and 3^rd orbit in Helium atom is 25/17:9/2 = 50:153.

4. What’s the radius of 1^st orbit of He⁺ atom?
a) 0.1058 nm
b) 0.2156 nm
c) 0.00529 nm
d) 0.02645 nm
Answer: d
Clarification: The atomic radius of an atom is given by the formula r_n = 52.9n²/Z pm, where r_n is the radius of nth orbit of an atom and Z is the atomic number of that atom. For He⁺, n = 1 and Z =2. Radius = 52.9(1)/2 pm = 0.02645 nm.

5. Bohr’s model couldn’t explain Zeeman and stark effect.
a) False
b) True
Answer: b
Clarification: Yes, it’s a limitation of Bohr’s model that it could not the splitting of spectral lines in the magnetic field that is Zeeman effect and also in the electric field also known as a stark effect. so the above statement is true.

6. Bohr’s model could not explain the ability of atoms to form molecules by ______
a) Attraction
b) Physical bonds
c) Chemical bonds
d) Polarity
Answer: c
Clarification: Though Bohr’s postulates could explain angular momentum, radius, and energy of an orbit, line spectrum of the hydrogen atom, it also had some drawbacks. Among the drawbacks not able to explain the ability of atoms to form molecules by chemical bonds is also one.

7. Calculate the wavelength of a photon that traveled from 5^th orbit to 2^nd orbit.
a) 434 nm
b) 456 nm
c) 863 nm
d) 268 nm
Answer: a
Clarification: The energy of an nth orbit in a hydrogen atom is given by the formula E_n = -R_H/n², where is the energy of nth orbit and R_H is the Rydberg constant. E₅ – E₂ = -4.58 x 10^-19J. λ(wavelength) = c(speed of light)h(Planck’s constant)/E = 434nm.

8. The energy of a hydrogen atom is positive.
a) True
b) False
Answer: b
Clarification: The energy of a hydrogen atom is negative. It means the energy of a hydrogen atom is then that lower than that of a free electron that is at rest. This means the hydrogen atom has negative electronic energy.

9. Which of the following is the value for Rydberg constant?
a) 2.95 x 10^-18 J
b) -2.95 x 10^-18 J
c) -2.18 x 10^-18 J
d) 2.18 x 10^-18 J
Answer: c
Clarification: The energy of an nth orbit in a hydrogen atom is given by the formula E_n = -R_H/n², where is the energy of nth orbit and R_H is the Rydberg constant. When experiments were conducted, the product of the energy of nth orbit to the square of n is constant i.e. Rydberg constant.

10. Iω = _________
a) m_ev²r
b) m_evr²
c) m_er
d) m_evr
Answer: d
Clarification: We know that linear momentum is given by mv and that angular momentun in Iω; m = mass, v = velocity, r = radius, I = inertia of momentum and w = angular velocity. I =m_er² and ω = v/r and Iω = m_er²v/r = m_evr.

To practice MCQs on all areas of Chemistry,

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) N_A = N_B
b) N_A > N_B
c) N_A < N_B
d) N_A >/< N_B
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})(N_B – N_A). So N_A – N_B = 0, that is N_A = N_B. Where N_A as the number of anti-bonding molecular orbitals and N_B 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 N_A = N_B (where N_A as the number of anti-bonding molecular orbitals and N_B 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*2s²π2p_x²π2p_y²
c) 1s²2s²2p¹
d) σ1s²σ*1s σ2s²π2p_x²π2p_y²
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*2s²2p_z²π2p_x²π2_y²π*2p_x¹π*2p_y¹ 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*2s²π2p_x²π2_y²but after 14, take number of electrons = 16 is σ1s²σ*1s²σ2s²σ*2s²2p_z²π2p_x²π2_y²π*2p_x¹π*2p_y¹ So the change comes near 2p_z 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.

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.

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, K_C 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 lnK_c, where ΔG₀ is the standard Gibbs free energy, R is universal gas constant, T is the temperature and K_C is equilibrium constant; substituting K_C 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 K_C at 100 k?
a) 1.0077
b) 1.077
c) 1.007
d) 2.7
Answer: d
Clarification: We know that ΔG₀ = – RT lnK_c, where ΔG₀ is the standard Gibbs free energy, R is universal gas constant, T is the temperature and K_C 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 lnK_c, where ΔG₀ is the standard Gibbs free energy, R is universal gas constant, T is the temperature and K_C 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, K_c = 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.

To practice Chemistry Multiple Choice Questions and Answers for Class 11,

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.

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.