Category: Class 11 Chemistry Objective Questions

This set of Chemistry Multiple Choice Questions on “Electronic Configurations and Types of Elements: s, p, d and f Blocks”.

1. Name the element that belongs to s-block but is placed in the p-block.
a) Hydrogen
b) Helium
c) Argon
d) Aluminum
Answer: b
Clarification: Helium’s electronic configuration is 1s². As the last electron is filled in s-orbital, it belongs to s-block. Since the 1^st shell cannot accommodate any orbital than s, 1s² is completely filled orbital, hence making it, a noble gas. Noble gases are placed in p-block.

2. __________ has both the characteristics of Alkali metals and halogens.
a) Helium
b) Chlorine
c) Sodium
d) Hydrogen
Answer: d
Clarification: As per the outer shell configuration of hydrogen (that is 1s¹), it has only one electron in s-orbital making it eligible as an Alkali metal. It requires only 1 electron to obtain a noble gas configuration, which is a characteristic of halogen.

3. ns¹ and ns² are the outer shell configurations of elements in s-block.
a) True
b) False
Answer: a
Clarification: Yes, ns¹ and ns² are the outer shell configurations of elements in s-block. The reason behind this is that they are ready to lose 1 electron or 2 electrons depending on the group number and have low ionization enthalpies.

4. The p-block elements along with s-block elements are called as ________ elements.
a) Inner transition
b) Representative
c) Radioactive
d) Transition
Answer: b
Clarification: The p-block elements comprise of elements from group-13 to group-18 while s-block elements are 1^st and 2nd groups. They two together form “Representative elements” or “Main group elements”.

5. What happens to the non-metallic nature as we move from left to right in groups?
a) Increases
b) Decreases
c) Remains constant
d) Irregular
Answer: a
Clarification: Non-metallic nature is defined as a tendency to gain electrons thus having high negative electron gain enthalpies, but the 18th group has no reactivity. So the non-metallic nature increases from left to right in groups leaving noble gases.

6. The outer shell electronic configuration of transition block elements is given by ________
a) (n-1)d^1-10ns²
b) (n-1)d^1-10(n-1)s^0-2
c) (n-1)d^1-10ns^0-2
d) nd^1-10ns^0-2
Answer: c
Clarification: Transition block elemts are d-block elements. The electrons fill in the d-orbital, and s-orbital orbital gets varied for maintaing stability of atoms. Therefore the outer shell electronic configuration of transition block elements is given by (n-1)d^1-10ns^0-2.

7. Which of the following is not true about transuranium elements?
a) Atomic number > 92
b) Elements after Uranium
c) Decay radioactively as they are unstable
d) Example is Thorium
Answer: d
Clarification: The elements after Uranium (Z = 92) are known as transuranium elements, they are unstable and decay radioactively into other elements. But the atomic number of Thorium is 90, hence it is not a transuranium element.

8. Chalcogens are the elements of Group ___________
a) 18
b) 16
c) 12
d) 2
Answer: b
Clarification: Group 16 consists of elements that belong to chalcogens also known as the oxygen family. The elements of this group are oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and Polonium (Po). They are called so because they are mostly found in earth’s crust.

9. Which of the following is not a Metalloid?
a) Germanium
b) Silicon
c) Aluminum
d) Tellurium
Answer: c
Clarification: Metalloids are those with both the properties of metals and non-metals. They are also known as Semi-metals. They are 8 metalloids known till date, namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At).

10. Metals consist of ______% of the elements known.
a) 78
b) 32
c) 22
d) 68
Answer: a
Clarification: Elements are classified into metals, non-metals, and metalloids based on their properties. While 78% of the elements that are known today are metals. Metallic nature is prominent on the left side of the periodic table.

Chemistry Question Paper on “Thermodynamics – Enthalpy Change, ∆rH of a Reaction – Reaction Enthalpy”.

1. The standard state of a substance is considered when the temperature is 298 k and the pressure is ____________
a) 1 ATM
b) 1 bar
c) 1 Pascal
d) 760 mm HG
Answer: b
Clarification: All the standard states of a substance are considered when the temperature is 298 Kelvin and the pressure is 1 bar. 1 bar = 0.987 atmospheric pressure = 10000 Pascal = 750.0617 mm of Mercury.

2. All the enthalpies of fusion are positive.
a) true
b) false
Answer: a
Clarification: Fusion is a process of conversion of liquid to solid the enthalpy is energy that is required for a process. As the melting of a solid is endothermic, the enthalpies of fusion are positive so the above statement is true.

3. Consider that, a ball is immersed in water at room temperature and then taken out having 18 grams of water on it, how much amount of energy is required to dry that water at room temperature?
a) 41.43 KJ/mol
b) 49.53 KJ/mol
c) 41.3 KJ/mol
d) 41.53 KJ/mol
Answer: d
Clarification: Heat required to eliminate water : n x Δ_vapH^– = (1 mol) × (44.01 kJ mol^–1) = 44.01 kJ mol^-1. Δ_vapor = Δ_vapH – ΔnRT = 44.01 kJ mol^-1 – 1×8.314 J/K-mol x 298 k x 10^-3 = 41.53 KJ/mol. So the amount of energy needed is 41.53 KJ/mol.

4. Calculate the internal energy change when 2 moles of water at 0 degrees converts into ice at 0-degree centigrade?
a) 12 KJ per mole
b) 6 KJ per mole
c) 1 KJ per mole
d) 102 KJ per mole
Answer: a
Clarification: Energy change when 1 mol of water at 0-degree centigrade changes into ice at 0 degrees in centigrade is 6 kJ/mol, So the internal energy change when 2 moles of water at 0 degrees converts into ice at 0 degrees is 12 kJ/mole.

5. What is a change in energy if 18 grams of water is heated from room temperature to 20 degrees above it?
a) 1.50 KJ
b) 0.506 KJ
c) 1.06 KJ
d) 1.506 KJ
Answer: d
Clarification: We know that Q = msΔT, where Q is the energy,m is the mass of water, s is the specific heat of water and T is the temperature. So the change in energy required here = 18 g x 4.184 J/g-K x 20K = 1.506KJ.

6. When a chemical reaction is reversed the value of enthalpy is reversed in sign.
a) true
b) false
Answer: a
Clarification: For example, the formation of ammonia has an enthalpy of -91.8 KJ per mole and the decomposition of ammonia has an enthalpy of + 91.8 KJ per Mol. So the above statement that when a chemical reaction is reversed the value of enthalpy is reversed in the sign is true.

7. Consider the equation 2 H₂ + O₂ → 2 H₂O, what does the 2 in the coefficient of H₂O molecule represent?
a) number of particles
b) the number of molecules
c) number of moles
d) number of atoms
Answer: c
Clarification: In a balanced thermochemical equation, the coefficients always refer to the number of the moles (but never molecules) of reactants and products involved in a reaction so 2 in the coefficient of H₂O refers to the number of the moles of water.

8. What is the unit of standard enthalpy of fusion or molar enthalpy of fusion?
a) KJ Mol
b) KJ per Mol
c) Mol per KJ
d) 1/KJ Mol
Answer: b
Clarification: The enthalpy change that occurs during melting of one mole of a solid substance in the standard state is called standard enthalpy of fusion or molar enthalpy of fusion, it is represented by the symbol Δ_fusH^–, the units of this are KJ per Mol.

9. Which of the following is not an application of Hess’s law?
a) determination of heat of formation
b) determination of heat of transition
c) determination of Gibb’s energy
d) determination of heat of hydration
Answer: c
Clarification: The following are the applications of Hess’s law; determination of heat of formation, determination of heat of transition and determination of heat of hydration, also to calculate bond energies.

10. ΔH_r = Σ ΔH_f[products] – Σ ΔH_f[reactants].
a) true
b) false
Answer: a
Clarification: The equation ΔH_r = Σ ΔH_f[products] – Σ ΔH_f[reactants] says that the enthalpy of a reaction is the difference between the enthalpy of products and enthalpy of reactants. The above statement regarding enthalpy is true.

To practice all questions papers on Chemistry,

Chemistry Multiple Choice Questions on “Classical Idea of Redox Reactions-Oxidation and Reduction Reactions”.

1. In the reaction of formation of magnesium oxide magnesium undergoes ________
a) reduction
b) oxidation
c) hydrogenation
d) decomposition
Answer: b
Clarification: Oxidation is a process that involves the addition of oxygen to an element or compound or the removal of hydrogen from a compound, an example is of this is the formation of magnesium oxide from magnesium.

2. A zinc ion is formed due to oxidation.
a) true
b) false
Answer: a
Clarification: Oxidation is the loss of electrons by an atom, ion or molecule. it is also known as de^–electronation. Zn → Zn²⁺ + 2e^–, the loss of electrons from zinc i.e. formation of zinc is an example of oxidation.

3. Removal of oxygen from a compound is an example of ________
a) oxidation
b) reduction
c) oxygenation
d) dehydrogenation
Answer: b
Clarification: Reduction is a chemical process, which involves the addition of hydrogen or an element or compound for the removal of oxygen from a compound. An example of reduction is a chemical reaction;
H₂ + F₂ → 2HF.

4. Reduction involves in __________ oxidation number.
a) decrease
b) increase
c) independence
d) remain constant
Answer: a
Clarification: The oxidation number is recharged in which an atom appears to have when all the atoms are removed from it as ions. It may have +ve or -ve sign. Reduction results in a decrease in the oxidation number.

5. Formation of copper from copper iron is an example of a reduction.
a) true
b) false
Answer: a
Clarification: Reduction is the gain of electrons by an atom, ion or molecule this process is known as electronation. Prayer for the formation of copper from copper iron as an example of reduction reaction because the copper gains two electrons in this case.

6. Oxidation is the same as __________
a) addition of hydrogen
b) removal of oxygen
c) addition of oxygen
d) removal of Nitrogen
Answer: c
Clarification: Oxidation is the same as the addition of oxygen or removal of hydrogen where reduction is the addition of hydrogen or removal of oxygen. Oxidation is caused by an oxidising agent and reduction is caused by a reducing agent.

7. Formation of zinc sulphide is an example of ___________
a) reduction
b) oxidation
c) removal of oxygen
d) addition of hydrogen
Answer: b
Clarification: Addition of electronegative element or removal of any other electropositive element is also considered as a process that involves oxidation. Here, sulphur is an electronegative element. So the formation of zinc sulphide is an Oxidation reaction.

8. SnCl₂ + 2FeCl₂ → SnCl₄ + 2FeCl₂ is an example of __________________ reaction.
a) only oxidation
b) only reduction
c) redox
d) neither oxidation nor reduction
Answer: c
Clarification: Chemical reactions which involves both oxygen as well as reduction process simultaneously are known as redox reactions. The above given reaction involves both oxidation and reduction reactions, so it is a Redox reaction.

9. SnCl₂ + 2FeCl₂ → SnCl₄ + 2FeCl₂. Which of the following element undergoes oxidation in the reaction given?
a) iron
b) tin
c) chlorine
d) ferrous
Answer: b
Clarification: Tin changes is oxidation state from + 2 to + 4. Increase in oxidation number indicates Oxidation reaction. Oxidation is the loss of electrons by an atom, ion or molecule, it is also known as the de^–electronation.

10. Which of the following do you think is a correct statement?
a) oxidation is caused by a reducing agent
b) the oxidation reaction is a Redox reaction
c) addition of electropositive element is a type of oxidation
d) reduction is the addition of hydrogen
Answer: d
Clarification: Oxidation is caused by an oxidizing agent, both oxidation and reduction reaction is combined together to form a Redox reaction and the addition of electropositive element is a type of reduction so the correct statement is that reduction as the addition of hydrogen.

Chemistry Multiple Choice Questions on “s-Block Elements – Important Compounds of Calcium”.

1. Which of the following acid chemical formula first slaked lime?
a) Calcium chloride
b) Calcium oxide
c) Calcium hydroxide
d) Calcium carbonate
Answer: c
Clarification: When 1 mole of burnt lime, that is calcium oxide is combined with 1 mole of the hydrogen molecule, a hissing sound appears and slaked lime is produced along with the heat. The chemical formula for slaked lime Ca(OH)₂.

2. Quicklime as same as calcium oxide.
a) True
b) False
Answer: a
Clarification: Yes, calcium oxide chemical formula is CaO and it is also called as quicklime or simply lime. It is prepared by the thermal decomposition of calcium carbonate at 1072-1270 Kelvin, carbon dioxide comes as a byproduct and it is a basic oxide.

3. Which of the following is not a compound of mortar?
a) Phenol
b) Quicklime
c) Sand
d) Water
Answer: a
Clarification: Quicklime or calcium oxide (CaO) is used as a basic flux, for removing the hardness of water, also used in mortar. A mixture of quick lime and sand in the ratio 1:3 with enough water to make a thick place is called motor.

4. What is the enthalpy of heat for dissolving quicklime and water?
a) 55 kJ/mol
b) 63 kJ/mol
c) 75 kJ/mol
d) 78 kJ/mol
Answer: b
Clarification: Calcium hydroxide is prepared by dissolving quicklime and water and the enthalpy change for this reaction is 63 kJ/mole (the sign is negative). Here the chemical formulae for quicklime and Calcium hydroxide are CaO and Ca(OH)₂ has twice respectively.

5. Which of the following compound is formed when slaked lime is treated with excess dioxide in the presence of water?
a) Barium sulphate
b) Calcium carbonate
c) Calcium hydroxide
d) Calcium bicarbonate
Answer: d
Clarification: When 1 mole of Calcium Hydroxide is treated with an excess of carbon dioxide in the presence of water, it results in the formation of calcium bicarbonate which is soluble. It is one of the most important properties of slaked lime.

6. Limestone is insoluble in water.
a) True
b) False
Answer: a
Clarification: Calcium carbonate is also known as limestone or marble are chalk, its chemical formula is given by CaCO₃. It is insoluble in water but dissolves in the presence of carbon dioxide due to the formation of calcium bicarbonate.

7. Can limestone be prepared through slaked lime?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: a
Clarification: Yes, limestone can be prepared through slaked lime. Limestone is calcium carbonate while slaked lime is calcium hydroxide. When limestone is passed through carbon dioxide, it results in the formation of calcium carbonate which is a precipitate and water.

8. Which of the following do you think is the correct formula for gypsum?
a) CaSO₄.2H₂O
b) CaSO₄.1/2H₂O
c) CaSO₄.H₂O
d) CaSO₄
Answer: a
Clarification: Gypsum is a compound of calcium. It is chemically calcium sulphate dehydrate and it is also known as alabaster. It is added to cement to slow down at speed of setting. It’s really very essential in our day to day life.

9. Gypsum is the same as plaster of Paris.
a) True
b) False
Answer: b
Clarification: Plaster of Paris is calcium sulphate hemihydrate and gypsum is calcium sulphate dihydrate. On heating gypsum at 390 Kelvin, it gives plaster of Paris. The chemical formula of gypsum and plaster of Paris is CaSO₄.2H₂O and CaSO₄.1/2H₂O respectively.

10. What is dead burnt plaster?
a) Hexahydrate calcium sulphate
b) Hemihydrate calcium sulphate
c) Anhydrous calcium sulphate
d) Decahydrate calcium sulphate
Answer: c
Clarification: When plaster of Paris or calcium sulphate hemihydrate, which is given by the chemical formula CaSO₄.1/2H₂O is heated about 393 Kelvin, no water of crystallization is left and anhydrous calcium sulphate is formed, it is known as dead burnt plaster.

11. Is the plaster of Paris useful in setting a solid mass?
a) Yes
b) No
c) Maybe
d) Cannot say
Answer: a
Clarification: When plaster of Paris or calcium sulphate hemihydrate which is given by the chemical formula CaSO₄.1/2H₂O is mixed with water, it forms of plastic mass which sets into a solid mass with slight expansion due to dehydration and its reverse conversion into gypsum.

12. What is chloride of lime?
a) Calcium sulphate
b) Calcium chloro hypochlorite
c) Calcium chloride
d) Calcium hydroxide
Answer: b
Clarification: Chloride of lime is chemically known as calcium chloro hypochlorite and it is given by the chemical formula CaOCl₂, it is also known as bleaching powder. Bleaching powder is prepared by the fusion of calcium hydroxide and chlorine.

13. What is the average percentage of available chlorine theoretically?
a) 85
b) 49
c) 58
d) 66
Answer: b
Clarification: With an excess of dilute sulfuric acid and carbon dioxide, calcium chloro hypochlorite forms chlorine, which is known as available chlorine. The average percentage of available chlorine is 35 to 40% but theoretically, it should be 49%, which diminishes on keeping the Powder due to a chemical change.

14. Which of the following is not a use of calcium chloro hypochlorite?
a) Sterilization of water
b) Manufacture of chloroform
c) Germicide
d) Painting
Answer: d
Clarification: Calcium chloro hypochlorite or chloride of lime or bleaching powder which is chemically CaOCl₂ is used for bleaching, as a disinfectant and germicide in the sterilization of water, for making wool which is unshrinkable and in the manufacture of chloroform.

15. In the average composition of portland cement, the percentage of magnesium oxide is a ___________
a) 3 to 4%
b) 2 to 3%
c) 4 to 5%
d) 1 to 2%
Answer: b
Clarification: Cement is an important building material. The average composition of portland cement is calcium oxide of 50 to 60%, Silicon dioxide of 20 to 25%, aluminium oxide of 5 to 10%, magnesium oxide of 2 to 3%, ferric oxide of 1 to 2% and sulphur trioxide of 1 to 2%.