Chemistry of Calcium (Ca) - Practice Questions | Calcium (Ca) Chemistry Guide

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements about Calcium is INCORRECT?

A) It reacts moderately with cold water, evolving hydrogen gas. B) It burns in air to form a mixture of oxide and nitride. C) Its compounds are generally ionic in nature. D) It is a strong oxidizing agent.

Solution: D

Explanation: Calcium is an alkaline earth metal (Group 2 element). Metals, especially s-block elements, tend to lose electrons readily to form positive ions. This property makes them strong reducing agents, as they themselves get oxidized while reducing other species. Therefore, calcium is a strong reducing agent, not an oxidizing agent.

A) Calcium reacts with cold water, though less vigorously than Group 1 metals, producing Ca(OH)₂ and H₂ gas.
B) When burned in air, calcium forms calcium oxide (CaO) and calcium nitride (Ca₃N₂).
C) Due to the significant electronegativity difference between calcium and non-metals, its compounds (like CaO, CaCl₂, CaCO₃) are predominantly ionic.

Question 2

Plaster of Paris is chemically:

A) Calcium sulphate dihydrate B) Calcium sulphate hemihydrate C) Calcium carbonate D) Calcium chloride

Solution: B

Explanation: Plaster of Paris is calcium sulphate hemihydrate, with the chemical formula CaSO₄·½H₂O. It is prepared by heating gypsum (CaSO₄·2H₂O) to about 373 K (100°C), which causes it to lose three-fourths of its water of crystallization.

Calcium sulphate dihydrate is gypsum (CaSO₄·2H₂O).
Calcium carbonate is chalk, limestone, or marble (CaCO₃).
Calcium chloride (CaCl₂) is a deliquescent salt often used as a desiccant.

Question 3

Temporary hardness of water is primarily due to the presence of:

A) Calcium sulphate and magnesium sulphate B) Calcium chloride and magnesium chloride C) Calcium bicarbonate and magnesium bicarbonate D) Calcium carbonate and magnesium carbonate

Solution: C

Explanation: Temporary hardness in water is caused by the dissolved bicarbonates of calcium [Ca(HCO₃)₂] and magnesium [Mg(HCO₃)₂]. This type of hardness is called “temporary” because it can be removed by simple boiling, which converts the soluble bicarbonates into insoluble carbonates (or hydroxides in the case of magnesium), which then precipitate out.

A) Sulphates of calcium and magnesium cause permanent hardness.
B) Chlorides of calcium and magnesium cause permanent hardness.
D) Carbonates of calcium and magnesium are generally insoluble; their dissolved bicarbonates are responsible for temporary hardness.

Assertion-Reason Questions

Question 4

Assertion (A): Calcium hydroxide, Ca(OH)₂, is a stronger base than magnesium hydroxide, Mg(OH)₂. Reason (R): Down the group, the ionic character of the M-OH bond increases due to decreasing electronegativity and increasing atomic size, facilitating the release of OH⁻ ions.

A) Both A and R are true, and R is the correct explanation of A. B) Both A and R are true, but R is not the correct explanation of A. C) A is true, but R is false. D) A is false, but R is true.

Solution: A

Explanation: Assertion (A) is true. As we move down Group 2, the basicity of hydroxides increases. Thus, Ca(OH)₂ is a stronger base than Mg(OH)₂ because Ca(OH)₂ is more soluble in water and dissociates to a greater extent, releasing more OH⁻ ions.

Reason (R) is also true and correctly explains A. Moving down the group from Mg to Ca, the atomic size increases, and electronegativity decreases. This leads to an increase in the ionic character of the M-OH bond (Ca-OH bond is more ionic than Mg-OH bond). A more ionic M-OH bond means the bond between the metal and oxygen is weaker, making it easier for the hydroxide ion (OH⁻) to dissociate from the metal, thereby increasing the basic strength.

Question 5

Assertion (A): Calcium reacts less vigorously with water than sodium. Reason (R): The molar mass of calcium is higher than sodium.

A) Both A and R are true, and R is the correct explanation of A. B) Both A and R are true, but R is not the correct explanation of A. C) A is true, but R is false. D) A is false, but R is true.

Solution: C

Explanation: Assertion (A) is true. Sodium (Group 1) reacts explosively with cold water due to its high reactivity and the immediate ignition of evolved hydrogen. Calcium (Group 2) reacts with cold water, but the reaction is less vigorous. The Ca(OH)₂ formed is sparingly soluble and tends to coat the surface of the metal, slowing down the reaction.

Reason (R) is false. While the molar mass of calcium (40.08 g/mol) is indeed higher than that of sodium (22.99 g/mol), molar mass is not the primary factor determining the vigor of reaction with water. The reactivity of metals with water is primarily governed by their standard electrode potential, ionization enthalpy, and hydration enthalpy. Sodium has a lower first ionization enthalpy and higher negative standard electrode potential compared to calcium, making it more reactive. Additionally, calcium needs to lose two electrons, which requires more energy than sodium losing one.

Short Answer Questions

Question 6

Explain what happens chemically when gypsum is heated to 393 K. Write the balanced chemical equation for the reaction.

Model Answer: When gypsum (calcium sulphate dihydrate, CaSO₄·2H₂O) is heated to approximately 393 K (120°C), it partially loses its water of crystallization. It forms calcium sulphate hemihydrate (CaSO₄·½H₂O), commonly known as Plaster of Paris, and releases 1.5 molecules of water.

If gypsum is heated beyond 393 K, it loses all its water of crystallization to form anhydrous calcium sulphate (CaSO₄), also known as ‘dead burnt plaster’, which loses its setting properties.

Chemical Equation: CaSO₄·2H₂O(s) $\xrightarrow{393 K}$ CaSO₄·½H₂O(s) + 1.5H₂O(g)

Question 7

How is temporary hardness of water removed by the boiling method? Write the relevant chemical equations for calcium and magnesium bicarbonates.

Model Answer: Temporary hardness of water is caused by the presence of dissolved bicarbonates of calcium [Ca(HCO₃)₂] and magnesium [Mg(HCO₃)₂]. This hardness can be removed by boiling the water.

Upon heating, the soluble bicarbonates decompose to form insoluble carbonates (in the case of calcium) and insoluble hydroxide (in the case of magnesium), which then precipitate out of the solution. These precipitates can be removed by filtration, thereby softening the water.

Chemical Equations:

For Calcium bicarbonate: Ca(HCO₃)₂(aq) $\xrightarrow{\text{heat}}$ CaCO₃(s)↓ + H₂O(l) + CO₂(g) (Calcium carbonate precipitates out)
For Magnesium bicarbonate: Mg(HCO₃)₂(aq) $\xrightarrow{\text{heat}}$ Mg(OH)₂(s)↓ + 2CO₂(g) (Magnesium hydroxide precipitates out, as MgCO₃ is unstable and hydrolyses to Mg(OH)₂ at elevated temperatures)

High-Order Thinking Skills (HOTS) Question

Question 8

A white solid ‘X’ on heating gives a colourless gas ‘Y’ and a residue ‘Z’. The gas ‘Y’ turns lime water milky. Residue ‘Z’ vigorously reacts with water to form ‘P’, which is extensively used in whitewashing. Identify X, Y, Z, and P. Write balanced chemical equations for all reactions involved.

Model Answer:

Identification of X, Y, Z, and P:

Gas ‘Y’ turns lime water milky: This is a characteristic test for carbon dioxide (CO₂). Thus, Y = CO₂.
White solid ‘X’ on heating gives gas ‘Y’ (CO₂) and residue ‘Z’: This indicates that X is a carbonate that decomposes on heating. Since Z reacts with water to form ‘P’ (used in whitewashing), and whitewashing uses slaked lime [Ca(OH)₂], it implies that Z is quicklime (CaO), which reacts with water to form Ca(OH)₂. Therefore, X must be calcium carbonate (CaCO₃).
Residue ‘Z’ reacts with water to form ‘P’: If X is CaCO₃ and Y is CO₂, then the residue ‘Z’ formed from the decomposition of CaCO₃ must be calcium oxide (quicklime). Thus, Z = CaO.
‘P’ is extensively used in whitewashing: When quicklime (CaO) reacts with water, it forms calcium hydroxide (slaked lime), which is used for whitewashing. Thus, P = Ca(OH)₂.

Summary of Identities:

X = Calcium Carbonate (CaCO₃)
Y = Carbon Dioxide (CO₂)
Z = Calcium Oxide (CaO)
P = Calcium Hydroxide (Ca(OH)₂)

Balanced Chemical Equations:

Decomposition of X (CaCO₃) to Y (CO₂) and Z (CaO): CaCO₃(s) $\xrightarrow{\Delta}$ CaO(s) + CO₂(g)
Test for Y (CO₂) using lime water (Ca(OH)₂ - part of P): CO₂(g) + Ca(OH)₂(aq) $\rightarrow$ CaCO₃(s)↓ + H₂O(l) (The formation of insoluble CaCO₃ causes the milky appearance)
Reaction of Z (CaO) with water to form P (Ca(OH)₂): CaO(s) + H₂O(l) $\rightarrow$ Ca(OH)₂(aq) (This reaction is highly exothermic and is known as slaking of lime)

Chemistry of Calcium (Ca) - Practice Questions