$A{\rm{and}}B$ are two salts. $A$ with dilute ${\rm{HCl}}$ and $B$ with conc. ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ react to give reddish brown vapours, hence $A$ and $B$ respectively are:

${\rm{NaN}}{{\rm{O}}_2},{\rm{NaBr}}$

${\rm{NaBr}},{\rm{NaN}}{{\rm{O}}_3}$

${\rm{NaN}}{{\rm{O}}_3},{\rm{NaBr}}$

${\rm{NaBr}},{\rm{NaN}}{{\rm{O}}_2}$

Which reagent would enable you to remove ${\rm{SO}}_4^{2 - }$ ions from solution containing both ${\rm{SO}}_4^{2 - }$ and ${\rm{C}}{{\rm{l}}^ - }$ ions?

${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2}$

An alloy of ${\rm{Na}} + {\rm{K}}$ is:

Used in specially designed thermometers

${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ is reduced by ${{\rm{I}}_2}$ to

${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_4}{{\rm{O}}_6}$

${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$

S-BLOCK ELEMENTS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Medium

1 of 111Question permalink

Understanding +4 | -1

Which of the following exists in polymeric form?

View answer and explanation

Correct option: B

\mathrm{BeC}{\mathrm{l}_2}

Option B is correct because

BeCl_2

exists in a polymeric form in the solid state, with chlorine atoms bridging between beryllium atoms.

Solution approach

Correct Answer + Core Reason: $\mathrm{BeC}{\mathrm{l}_2}$ . Beryllium chloride exists as a polymer in the solid state.

Step-by-Step Solution:

Step 1: Consider the structure of each compound.

Step 2: $\mathrm{AlC}{\mathrm{l}_3}$ exists as a dimer ( $Al_2Cl_6$ ) in the vapor phase but not as a polymer in the solid state.

Step 3: $\mathrm{BeC}{\mathrm{l}_2}$ exists as a polymeric chain structure in the solid state, with chlorine atoms bridging between beryllium atoms.

Step 4: ${\mathrm{B}_2}{\mathrm{H}_6}$ is a dimer but not a polymer.

Step 5: $\mathrm{SiC}$ is a giant covalent network solid, but not a polymer in the same sense as $BeCl_2$ .

Why other options are wrong:
- Option A: $AlCl_3$ exists as a dimer ( $Al_2Cl_6$ ) but not a polymer in the solid state.
- Option C: $B_2H_6$ is a dimer, not a polymer.
- Option D: $SiC$ is a network solid, not a polymer in the same way as $BeCl_2$ .

Key Concepts:
- Structure of inorganic compounds
- Polymer formation
- Dimer formation

Memory Aids:
- $BeCl_2$ = Polymeric chain structure

Quick Tips:
- Remember the structures of common inorganic compounds.

2 Easy

2 of 111Question permalink

Application +4 | -1

If 5 moles of

\mathrm{BaCl}_{2}

is mixed with 2 moles of

\mathrm{Na}_{3} \mathrm{PO}_{4}

, the maximum number of moles of

\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}

formed is ________ (Nearest integer)

View answer and explanation

Correct option: A

1

Option A (1) is correct because the reaction stoichiometry and limiting reactant determine the maximum moles of Ba3(PO4)2 formed.

Solution approach

Correct Answer + Core Reason: 1. The limiting reactant determines the maximum amount of product formed.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$3 \mathrm{BaCl}_{2}(aq) + 2 \mathrm{Na}_{3} \mathrm{PO}_{4}(aq) \rightarrow \mathrm{Ba}_{3}(\mathrm{PO}_{4})_{2}(s) + 6 \mathrm{NaCl}(aq)$

Step 2: Determine the mole ratio of reactants:\nThe mole ratio of $BaCl_2$ to $Na_3PO_4$ is 3:2.

Step 3: Calculate the required moles of $BaCl_2$ to react with 2 moles of $Na_3PO_4$ :\nMoles of $BaCl_2$ required = $(3/2) \times 2 = 3$ moles.

Step 4: Identify the limiting reactant:\nSince we have 5 moles of $BaCl_2$ , which is more than the 3 moles required, $Na_3PO_4$ is the limiting reactant.

Step 5: Calculate the moles of $Ba_3(PO_4)_2$ formed:\nFrom the balanced equation, 2 moles of $Na_3PO_4$ produce 1 mole of $Ba_3(PO_4)_2$ .\nTherefore, moles of $Ba_3(PO_4)_2$ formed = $(1/2) \times 2 = 1$ mole.

Why other options are wrong:
- Option B (0): This is incorrect because the reaction proceeds to form a product, and the limiting reactant is present.
- Option C (0): This is incorrect because the reaction proceeds to form a product, and the limiting reactant is present.
- Option D (0): This is incorrect because the reaction proceeds to form a product, and the limiting reactant is present.

Key Concepts:
- Stoichiometry
- Limiting Reactant
- Mole Ratio

Memory Aids:
- Remember to balance the chemical equation before performing stoichiometric calculations.

Quick Tips:
- Identify the limiting reactant first to determine the maximum amount of product formed.

3 Hard

3 of 111Question permalink

Remembering +4 | -1

Alkali metal superoxides are obtained by the reaction of

View answer and explanation

Correct option: A

Oxygen with alkali metal in liquid ammonia

Option A (Oxygen with alkali metal in liquid ammonia) is correct because alkali metal superoxides are obtained by reacting alkali metals with excess oxygen in liquid ammonia.

Solution approach

Correct Answer + Core Reason: Oxygen with alkali metal in liquid ammonia. This reaction yields alkali metal superoxides.

Step-by-Step Solution:

Step 1: Understand the formation of superoxides.

Step 2: Alkali metals react with oxygen to form oxides, peroxides, or superoxides, depending on the metal and the reaction conditions.

Step 3: Superoxides are formed when alkali metals react with excess oxygen in liquid ammonia.

Why other options are wrong:
- Option B: Water with alkali metal in liquid ammonia: Alkali metals react violently with water.
- Option C: Hydrogen peroxide with alkali metals: This reaction is not typically used to prepare superoxides.
- Option D: Hydrogen peroxide with alkali metals in liquid ammonia: This reaction is not typically used to prepare superoxides.

Key Concepts:
- Alkali metal oxides, peroxides, and superoxides
- Reaction of alkali metals with oxygen

Memory Aids:
- Alkali Metal + Excess Oxygen in Liquid Ammonia = Superoxide

Quick Tips:
- Remember the specific conditions required for the formation of superoxides.

4 Hard

4 of 111Question permalink

Remembering +4 | -1

Pearl ash and caustic potash are chemically:

View answer and explanation

Correct option: A

{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}\mathrm{and KOH}

Option A (

{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}\mathrm{and KOH}

) is correct because pearl ash is potassium carbonate (

{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}

) and caustic potash is potassium hydroxide (KOH).

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}\mathrm{and KOH}$ . Pearl ash is potassium carbonate and caustic potash is potassium hydroxide.

Step-by-Step Solution:

Step 1: Identify the chemical names for pearl ash and caustic potash.

Step 2: Pearl ash is the common name for potassium carbonate ( $\mathrm{K}_2\mathrm{CO}_3$ ).

Step 3: Caustic potash is the common name for potassium hydroxide (KOH).

Why other options are wrong:
- Option B: $\mathrm{KOH and}{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}$ : This is incorrect because the order is reversed.
- Option C: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}\mathrm{and KOH}$ : This is incorrect because pearl ash is potassium carbonate, not sodium carbonate.
- Option D: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}\mathrm{and NaOH}$ : This is incorrect because neither pearl ash nor caustic potash are sodium compounds.

Key Concepts:
- Common names of chemical compounds
- Potassium carbonate
- Potassium hydroxide

Memory Aids:
- Pearl ash = Potassium Carbonate
- Caustic Potash = Potassium Hydroxide

Quick Tips:
- Memorize the common names of important chemical compounds.

5 Easy

5 of 111Question permalink

Understanding +4 | -1

When sodium chloride solution is electrolysed, the gas that is liberated at the cathode is

View answer and explanation

Correct option: C

Hydrogen

Option C is correct because during the electrolysis of aqueous sodium chloride, water is reduced at the cathode to produce hydrogen gas and hydroxide ions.

Solution approach

Correct Answer + Core Reason: C. Hydrogen. During electrolysis of aqueous NaCl, hydrogen gas is produced at the cathode due to the reduction of water.

Step-by-Step Solution:

Step 1: Electrolysis of aqueous NaCl involves the following ions: $\mathrm{Na}^+, \mathrm{Cl}^-, \mathrm{H}^+, \mathrm{OH}^-$ .

Step 2: At the cathode (reduction): Both $\mathrm{Na}^+$ and $\mathrm{H}^+$ can be reduced. However, $\mathrm{H}^+$ is more easily reduced than $\mathrm{Na}^+$ . The reaction is: $2\mathrm{H}^+(aq) + 2e^- \rightarrow \mathrm{H}_2(g)$ . Since water is the source of H+, we can write: $2\mathrm{H}_2\mathrm{O}(l) + 2e^- \rightarrow \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)$

Step 3: At the anode (oxidation): Both $\mathrm{Cl}^-$ and $\mathrm{OH}^-$ can be oxidized. However, $\mathrm{Cl}^-$ is more easily oxidized than $\mathrm{OH}^-$ . The reaction is: $2\mathrm{Cl}^-(aq) \rightarrow \mathrm{Cl}_2(g) + 2e^-$ .

Step 4: Therefore, hydrogen gas is liberated at the cathode.

Why other options are wrong:
- Option A: Oxygen is produced at the anode under different electrolytic conditions (e.g., electrolysis of water with a suitable electrolyte).
- Option B: Chlorine is produced at the anode, not the cathode.
- Option D: Air is a mixture of gases and is not produced during electrolysis.

Key Concepts:
- Electrolysis
- Reduction at cathode
- Oxidation at anode
- Standard reduction potentials

Memory Aids:
- AN OX (Anode Oxidation), RED CAT (Reduction Cathode)

Quick Tips:
- Consider the standard reduction potentials of the ions present in the electrolytic solution.

6 Medium

6 of 111Question permalink

Application +4 | -1

What is product of reaction :

\underset {\text{(dilute)}}{Ba(OH)_2}+H_2O_2+ClO_2 \longrightarrow

View answer and explanation

Correct option: D

Ba\left(ClO_2\right)_2

Option D (

Ba(ClO_2)_2

) is correct because the reaction of barium hydroxide, hydrogen peroxide, and chlorine dioxide produces barium chlorite and other products.

Solution approach

Correct Answer + Core Reason: $Ba\left(ClO_2\right)_2$ . The reaction produces barium chlorite.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$Ba(OH)_2 + 2ClO_2 + H_2O_2 \rightarrow Ba(ClO_2)_2 + 2H_2O + O_2$

Step 2: Identify the product:\nThe product of the reaction is barium chlorite, $Ba(ClO_2)_2$ .

Why other options are wrong:
- **Option A ( $HOCl$ ): Hypochlorous acid is not the main barium-containing product in this reaction.
- Option B ( $Ba(OCl)_2$ ): Barium hypochlorite is not the main barium-containing product in this reaction.
- Option C ( $Ba(ClO_3)_2$ ): Barium chlorate is not the main barium-containing product in this reaction.

Key Concepts:
- Chemical Reactions
- Redox Reactions

Quick Tips:**
- Understanding the reactivity of chlorine dioxide is crucial for predicting the products.

7 Easy

7 of 111Question permalink

Understanding +4 | -1

Which among the following elements has the tendency to form covalent compounds.

View answer and explanation

Correct option: B

Be

The correct answer is B (

Be

) because Beryllium, due to its small size and high polarizing power, has a greater tendency to form covalent compounds compared to other alkaline earth metals.

Solution approach

Correct Answer + Core Reason: $Be$ . Beryllium has a greater tendency to form covalent compounds due to its small size and high polarizing power.

Step-by-Step Solution:

Step 1: Recall the factors affecting the formation of covalent compounds.

Step 2: Small size and high polarizing power favor covalent bond formation.

Step 3: Beryllium has the smallest size and highest polarizing power among the given alkaline earth metals.

Step 4: Therefore, Beryllium has the greatest tendency to form covalent compounds.

Why other options are wrong:
- Option A: $Ba$ is a large atom with low polarizing power and tends to form ionic compounds.
- Option C: $Mg$ is larger than Be and has a lower polarizing power, so it tends to form ionic compounds more readily than Be.
- Option D: $Ca$ is larger than Be and has a lower polarizing power, so it tends to form ionic compounds more readily than Be.

Key Concepts:
- Fajan's rules and polarizing power.
- Covalent character of ionic compounds.

Memory Aids:
- Small size and high charge favor covalent character.

Quick Tips:
- Remember Fajan's rules when predicting the covalent character of ionic compounds.

8 Easy

8 of 111Question permalink

Remembering +4 | -1

The non-metal which is not affected by

\mathrm{NaOH}

:

View answer and explanation

Correct option: A

\mathrm{C}

Option A (

\mathrm{C}

) is correct because carbon is generally unreactive with NaOH under normal conditions.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}$ . Carbon is generally unreactive with NaOH.

Step-by-Step Solution:

Step 1: Consider the reactivity of the given non-metals with NaOH.

Step 2: Analyze the given options:
- A. $\mathrm{C}$ (Carbon) is generally unreactive with NaOH under normal conditions.
- B. $\mathrm{Si}$ (Silicon) reacts with NaOH to form sodium silicate and hydrogen gas.
- C. $\mathrm{P}$ (Phosphorus) reacts with NaOH to form phosphine and sodium hypophosphite.
- D. $\mathrm{S}$ (Sulfur) reacts with NaOH to form sodium sulfide and sodium thiosulfate.

Step 3: Therefore, carbon is the non-metal that is not affected by NaOH.

Why other options are wrong:
- Option B: Silicon reacts with NaOH.
- Option C: Phosphorus reacts with NaOH.
- Option D: Sulfur reacts with NaOH.

Key Concepts:
- Reactivity of Non-metals
- Reactions with NaOH

Memory Aids:
- Remember the general reactivity trends of non-metals.

Quick Tips:
- Recall the reactions of common non-metals with NaOH.

9 Hard

9 of 111Question permalink

Application +4 | -1

Reaction of

\mathrm{BeO}

with ammonia and hydrogen fluoride gives

A

which on thermal decomposition gives

\mathrm{BeF}_{2}

and

\mathrm{NH}_{4} \mathrm{F}

. What is '

{A}

'?

View answer and explanation

Correct option: D

\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4}

Option D ((

\mathrm{NH}_{4}

)

_{2} \mathrm{BeF}_{4}

) is correct because the reaction sequence and products indicate the formation of ammonium tetrafluoroberyllate.

Solution approach

Correct Answer + Core Reason: $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4}$ . The reaction forms ammonium tetrafluoroberyllate.

Step-by-Step Solution:

Step 1: Write the reaction sequence:
$BeO + NH_3 + HF \rightarrow A \xrightarrow{\Delta} BeF_2 + NH_4F$

Step 2: Deduce the structure of A:\nSince A decomposes to $BeF_2$ and $NH_4F$ , A must contain both Be and $NH_4$ and F.\nThe most likely formula for A is $(NH_4)_2BeF_4$ .

Step 3: Write the reaction:
$BeO + 2NH_4HF_2 \rightarrow (NH_4)_2BeF_4 + H_2O$

Why other options are wrong:
- **Option A ( $\mathrm{H}_{3} \mathrm{NBeF}_{3}$ ):** This formula does not account for the formation of $NH_4F$ upon decomposition.
- **Option B ( $\left(\mathrm{NH}_{4}\right) \mathrm{BeF}_{3}$ ):** This formula does not balance the charges and does not account for the formation of $NH_4F$ upon decomposition.
- **Option C ( $\left(\mathrm{NH}_{4}\right) \mathrm{Be}_{2} \mathrm{F}_{5}$ ):** This formula does not account for the formation of $NH_4F$ upon decomposition and is less likely due to the stoichiometry.

Key Concepts:
- Complex Formation
- Thermal Decomposition

Quick Tips:
- Consider the decomposition products to deduce the structure of the intermediate.

10 Medium

10 of 111Question permalink

Understanding +4 | -1

Solution of sodium metal in liquid

\mathrm{N}{\mathrm{H}_3}

is strongly reducing due to the presence of :

View answer and explanation

Correct option: B

Solvated electrons

Option B is correct because the blue color and reducing properties of the solution are due to the presence of solvated electrons, which are free electrons surrounded by ammonia molecules.

Solution approach

Correct Answer + Core Reason: Solvated electrons. The reducing property is due to the presence of solvated electrons in liquid ammonia.

Step-by-Step Solution:

Step 1: Understand the behavior of alkali metals in liquid ammonia.

Step 2: When sodium metal dissolves in liquid ammonia, it forms a solution containing solvated sodium ions ( $Na^+$ ) and solvated electrons ( $e^-$ ).

Step 3: $Na(s) + (x+y)NH_3(l) \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$

Step 4: The solvated electrons are responsible for the strong reducing properties and the characteristic blue color of the solution.

Why other options are wrong:
- Option A: Sodium atoms themselves are not the primary reason for the reducing nature in the solution; it's the solvated electrons.
- Option C: $NaOH$ is not formed in this process. Sodium reacts with ammonia to form sodium amide ( $NaNH_2$ ) under different conditions.
- Option D: Sodium amide ( $NaNH_2$ ) can be formed, but it's not the primary reason for the reducing nature of the solution.

Key Concepts:
- Alkali metals in liquid ammonia
- Solvated electrons
- Reducing properties

Memory Aids:
- Solvated electrons = Reducing power + Blue color

Quick Tips:
- Remember that solvated electrons are key to understanding the properties of alkali metal solutions in liquid ammonia.

11 Medium

11 of 111Question permalink

Application +4 | -1

A doctor by mistake administers a

\mathrm{Ba}\left(\mathrm{NO}_3\right)_2

solution to a patient for radiography investigations. Which of the following should be given as the best to prevent the absorption of soluble barium?

View answer and explanation

Correct option: B

\mathrm{Na}_2 \mathrm{SO}_4

Option B (Na2SO4) is correct because it will react with Ba(NO3)2 to form insoluble BaSO4, preventing the absorption of soluble barium.

Solution approach

Correct Answer + Core Reason: $\mathrm{Na}_2 \mathrm{SO}_4$ . It precipitates $Ba^{2+}$ as insoluble $BaSO_4$ .

Step-by-Step Solution:

Step 1: Understand the problem:\nWe need to find a substance that will react with $Ba(NO_3)_2$ to form an insoluble barium salt, preventing its absorption.

Step 2: Consider the options:
- NaCl: Forms $BaCl_2$ , which is soluble.
- $Na_2SO_4$ : Forms $BaSO_4$ , which is insoluble.
- KCl: Forms $BaCl_2$ , which is soluble.
- $NH_4Cl$ : Forms $BaCl_2$ , which is soluble.

Step 3: Write the reaction:
$Ba(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaNO_3(aq)$

Step 4: Conclusion:
$Na_2SO_4$ is the best option because it forms insoluble $BaSO_4$ .

Why other options are wrong:
- **Option A ( $\mathrm{NaCl}$ ): NaCl would form soluble barium chloride, which would not prevent the absorption of barium.
- Option C ( $\mathrm{KCl}$ ): KCl would also form soluble barium chloride, which would not prevent the absorption of barium.
- Option D ( $\mathrm{NH}_4 \mathrm{Cl}$ ):** $NH_4Cl$ would also form soluble barium chloride, which would not prevent the absorption of barium.

Key Concepts:
- Solubility Rules
- Precipitation Reactions

Memory Aids:
- Remember that barium sulfate ( $BaSO_4$ ) is insoluble.

Quick Tips:
- Know the common solubility rules to predict precipitate formation.

12 Medium

12 of 111Question permalink

Analyzing +4 | -1

A\mathrm{and}B

are two salts.

A

with dilute

\mathrm{HCl}

and

B

with conc.

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

react to give reddish brown vapours, hence

A

and

B

respectively are:

View answer and explanation

Correct option: D

\mathrm{NaN}{\mathrm{O}_2},\mathrm{NaBr}

Option D (

\mathrm{NaN}{\mathrm{O}_2},\mathrm{NaBr}

) is correct because

\mathrm{NaBr}

with conc.

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

and

\mathrm{NaN}{\mathrm{O}_2}

with dilute

\mathrm{HCl}

produce reddish-brown vapors.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaN}{\mathrm{O}_2},\mathrm{NaBr}$ . $\mathrm{NaN}{\mathrm{O}_2}$ with dilute $\mathrm{HCl}$ and $\mathrm{NaBr}$ with conc. ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ produce reddish-brown vapors.

Step-by-Step Solution:

Step 1: Identify the reactions that produce reddish-brown vapors.

Step 2: Consider the reaction of sodium nitrite ( $\mathrm{NaN}{\mathrm{O}_2}$ ) with dilute hydrochloric acid ( $\mathrm{HCl}$ ). This reaction produces nitrous acid ( $\mathrm{HN}{\mathrm{O}_2}$ ), which decomposes to nitrogen monoxide ( $\mathrm{NO}$ ), which further reacts with oxygen to form nitrogen dioxide ( $\mathrm{NO_2}$ ), a reddish-brown gas.\nReaction: $2\mathrm{NaN}{\mathrm{O}_2} + 2\mathrm{HCl} \to 2\mathrm{NaCl} + 2\mathrm{HN}{\mathrm{O}_2$ ; $2\mathrm{HN}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O} + 2\mathrm{NO} + {1 \over 2}{\mathrm{O}_2} \to \mathrm{NO}_2$ (reddish-brown).

Step 3: Consider the reaction of sodium bromide ( $\mathrm{NaBr}$ ) with concentrated sulfuric acid ( ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ ). This reaction produces hydrogen bromide ( $\mathrm{HBr}$ ), which is then oxidized by the sulfuric acid to bromine ( $\mathrm{Br_2}$ ), a reddish-brown gas.\nReaction: $2\mathrm{NaBr} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4} + 2\mathrm{HBr}$ ; $2\mathrm{HBr} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{B}{\mathrm{r}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ (reddish-brown).

Step 4: Therefore, A is $\mathrm{NaN}{\mathrm{O}_2}$ and B is $\mathrm{NaBr}$ .

Why other options are wrong:
- Option A: $\mathrm{NaBr},\mathrm{NaN}{\mathrm{O}_3}$ : $\mathrm{NaN}{\mathrm{O}_3}$ with dilute $\mathrm{HCl}$ does not produce reddish-brown fumes.
- Option B: $\mathrm{NaN}{\mathrm{O}_3},\mathrm{NaBr}$ : $\mathrm{NaN}{\mathrm{O}_3}$ with dilute $\mathrm{HCl}$ does not produce reddish-brown fumes.
- Option C: $\mathrm{NaBr},\mathrm{NaN}{\mathrm{O}_2}$ : $\mathrm{NaBr}$ should react with conc. ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ and $\mathrm{NaN}{\mathrm{O}_2}$ with dilute $\mathrm{HCl}$ . The order is reversed.

Key Concepts:
- Reactions of salts with acids
- Formation of reddish-brown vapors ( $\mathrm{NO_2}$ and $\mathrm{Br_2}$ )

Memory Aids:
- Remember that bromides with conc. ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ and nitrites with dilute acids can produce reddish-brown fumes.

Quick Tips:
- Know the characteristic reactions of halides and nitrites with acids.

13 Easy

13 of 111Question permalink

Application +4 | -1

Which reagent would enable you to remove

\mathrm{SO}_4^{2 - }

ions from solution containing both

\mathrm{SO}_4^{2 - }

and

\mathrm{C}{\mathrm{l}^ - }

ions?

View answer and explanation

Correct option: C

\mathrm{Ba}{\left( {\mathrm{OH}} \right)_2}

Option C is correct because

Ba(OH)_2

will precipitate

SO_4^{2-}

ions as

BaSO_4

, which is insoluble, while not reacting with

Cl^-

ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ba}{\left( {\mathrm{OH}} \right)_2}$ . Barium hydroxide precipitates sulfate ions as barium sulfate.

Step-by-Step Solution:

Step 1: Consider the reactions of each reagent with $SO_4^{2-}$ and $Cl^-$ ions.

Step 2: $\mathrm{NaOH}$ will not precipitate $SO_4^{2-}$ ions.

Step 3: $\mathrm{P}{\mathrm{b}^{2 + }}$ will precipitate both $SO_4^{2-}$ and $Cl^-$ ions as $PbSO_4$ and $PbCl_2$ , respectively.

Step 4: $\mathrm{Ba}{\left( {\mathrm{OH}} \right)_2}$ will precipitate $SO_4^{2-}$ ions as $BaSO_4$ , which is insoluble, while barium chloride ( $BaCl_2$ ) is soluble and remains in solution.

Step 5: $\mathrm{BaS}{\mathrm{O}_4}$ is already a precipitate and will not remove $SO_4^{2-}$ ions from solution.

Why other options are wrong:
- Option A: NaOH will not precipitate sulfate ions.
- Option B: $Pb^{2+}$ will precipitate both sulfate and chloride ions.
- Option D: $BaSO_4$ is a product, not a reagent to remove sulfate ions.

Key Concepts:
- Precipitation reactions
- Solubility rules
- Selective precipitation

Memory Aids:
- $Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4$ (Insoluble)

Quick Tips:
- Remember the solubility rules for common ions.

14 Easy

14 of 111Question permalink

Remembering +4 | -1

An alloy of

\mathrm{Na} + \mathrm{K}

is:

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Zero, 1) is correct because electrolysis of

H_2SO_5

gives

H_2O_2

at the anode, and

H_2SO_5

has zero peroxy bonds while

H_2O_2

has one.

Solution approach

Correct Answer + Core Reason: Zero, 1. $X$ is $H_2SO_5$ (Caro's acid) which has zero peroxy bonds, and $Y$ is $H_2O_2$ which has one peroxy bond.

Step-by-Step Solution:

Step 1: Identify the compounds $X$ and $Y$ .
$X$ is Caro's acid ( $H_2SO_5$ ), which undergoes electrolysis to produce $Y$ , which is hydrogen peroxide ( $H_2O_2$ ).

Step 2: Determine the number of peroxy bonds in $H_2SO_5$ and $H_2O_2$ .
$H_2SO_5$ has the structure $(HO)(O)S(O)_2(OH)$ , which contains no O-O bonds (peroxy bonds).
$H_2O_2$ has the structure $H-O-O-H$ , which contains one O-O bond (peroxy bond).

Why other options are wrong:
- Option A: Incorrect, $H_2SO_5$ has zero peroxy bonds.
- Option B: Incorrect, $H_2SO_5$ has zero peroxy bonds and $H_2O_2$ has one.
- Option D: Incorrect, $H_2O_2$ has one peroxy bond.

Key Concepts:
- Electrolysis
- Peroxy Bonds
- Structure of Caro's Acid and Hydrogen Peroxide

Memory Aids:
- Remember the structures of Caro's acid and hydrogen peroxide.

Quick Tips:
- Draw the structures of the compounds to determine the number of peroxy bonds.

15 Easy

15 of 111Question permalink

Analyzing +4 | -1

Which possesses highest lattice energy?

View answer and explanation

Correct option: B

\mathrm{LiF}

Option B (LiF) is correct because lattice energy is inversely proportional to the size of the ions and directly proportional to the charge. LiF has the smallest ions among the given options, leading to the highest lattice energy.

Solution approach

Correct Answer + Core Reason: $\mathrm{LiF}$ . It has the smallest cation and anion, leading to the highest lattice energy.

Step-by-Step Solution:

Step 1: Recall the factors affecting lattice energy.\nLattice energy is proportional to $\frac{{q_1 q_2}}{{r_1 + r_2}}$ , where $q_1$ and $q_2$ are the charges of the ions, and $r_1$ and $r_2$ are their radii.

Step 2: Compare the ionic sizes.\nLi+ is the smallest cation, and F- is the smallest anion among the given options.

Step 3: Determine the compound with the highest lattice energy.\nLiF has the smallest cation and anion, so it has the highest lattice energy.

Why other options are wrong:
- Option A: $\mathrm{NaCl}$ has larger ions than LiF.
- Option C: $\mathrm{Csl}$ has the largest ions among the given options.
- Option D: $\mathrm{KF}$ has a larger cation than LiF.

Key Concepts:
- Lattice energy
- Factors affecting lattice energy (charge and size)

Memory Aids:
- Smaller ions and higher charges lead to higher lattice energy.

Quick Tips:
- Consider the sizes and charges of the ions when comparing lattice energies.

16 Hard

16 of 111Question permalink

Understanding +4 | -1

Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?

View answer and explanation

Correct option: D

In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group.

Option D is correct because the reactivity of alkali metals increases down the group due to decreasing ionization energy, while the reactivity of halogens decreases down the group due to decreasing electronegativity and electron affinity.

Solution approach

Correct Answer + Core Reason: In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group. This is due to ionization energy and electronegativity trends.

Step-by-Step Solution:

Step 1: Recall the periodic trends of reactivity for alkali metals and halogens.

Step 2: Alkali metals: Reactivity increases down the group because ionization energy decreases, making it easier to lose an electron.

Step 3: Halogens: Reactivity decreases down the group because electronegativity and electron affinity decrease, making it harder to gain an electron.

Why other options are wrong:
- Option A: The reactivity decreases in the alkali metals is incorrect.
- Option B: In both the alkali metals and the halogens the chemical reactivity decreases is incorrect for alkali metals.
- Option C: Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens is incorrect for halogens.

Key Concepts:
- Periodic trends
- Reactivity of alkali metals
- Reactivity of halogens
- Ionization energy
- Electronegativity
- Electron affinity

Memory Aids:
- Alkali Metals: Down the group, reactivity INCREASES (easier to lose electron)
- Halogens: Down the group, reactivity DECREASES (harder to gain electron)

Quick Tips:
- Remember the key factors influencing reactivity: ionization energy for metals and electronegativity/electron affinity for nonmetals.

17 Medium

17 of 111Question permalink

Understanding +4 | -1

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

is reduced by

{\mathrm{I}_2}

to

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_4}{\mathrm{O}_6}

Option D is correct because sodium thiosulphate (

\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3

) reacts with iodine (

{\mathrm{I}_2}

) to form sodium tetrathionate (

\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6

) and sodium iodide.

Solution approach

Correct Answer + Core Reason: D. $\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_4}{\mathrm{O}_6}$ . $\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}$ is oxidized by ${\mathrm{I}_2}$ to $\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_4}{\mathrm{O}_6}$ .

Step-by-Step Solution:

Step 1: The reaction between sodium thiosulphate and iodine is a redox reaction.

Step 2: The balanced chemical equation is: $2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3(aq) + {\mathrm{I}_2}(aq) \rightarrow \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6(aq) + 2\mathrm{NaI}(aq)$

Step 3: In this reaction, iodine ( ${\mathrm{I}_2}$ ) is reduced to iodide ( $\mathrm{I}^-$ ), and thiosulphate ( $\mathrm{S}_2\mathrm{O}_3^{2-}$ ) is oxidized to tetrathionate ( $\mathrm{S}_4\mathrm{O}_6^{2-}$ ).

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{a}_2}\mathrm{S}$ is not formed in this reaction.
- Option B: $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}$ is not formed in this reaction.
- Option C: $\mathrm{NaHS}{\mathrm{O}_3}$ is not formed in this reaction.

Key Concepts:
- Redox reactions
- Sodium thiosulphate
- Iodine
- Sodium tetrathionate

Memory Aids:
- Thiosulphate + Iodine = Tetrathionate + Iodide

Quick Tips:
- Balance the redox reaction to determine the products.

18 Hard

18 of 111Question permalink

Analyzing +4 | -1

Two metals ‘x’ and ‘y’ belong to the same group of S – block. ‘x’ forms a soluble sulphate but insoluble oxide, ‘y’ forms soluble oxide but insoluble sulphate x and y are respectively

View answer and explanation

Correct option: A

Mg and Ba

Option A (Mg and Ba) is correct because Magnesium forms a soluble sulfate but an insoluble oxide, while Barium forms a soluble oxide but an insoluble sulfate.

Solution approach

Correct Answer + Core Reason: Mg and Ba. Magnesium forms a soluble sulphate but insoluble oxide, while Barium forms a soluble oxide but insoluble sulphate.

Step-by-Step Solution:

Step 1: Analyze the properties of Group 2 elements.\nGroup 2 elements (alkaline earth metals) form sulfates and oxides.

Step 2: Consider the solubility of sulfates.\nSolubility of sulfates decreases down the group: $\mathrm{MgSO}_4$ is soluble, $\mathrm{CaSO}_4$ is sparingly soluble, $\mathrm{SrSO}_4$ and $\mathrm{BaSO}_4$ are insoluble.

Step 3: Consider the solubility of oxides.\nSolubility of oxides increases down the group: $\mathrm{MgO}$ is insoluble, $\mathrm{CaO}$ , $\mathrm{SrO}$ , and $\mathrm{BaO}$ are soluble.

Step 4: Identify the metals x and y.
- x forms a soluble sulfate but insoluble oxide: This is Magnesium (Mg).
- y forms a soluble oxide but insoluble sulfate: This is Barium (Ba).

Why other options are wrong:
- Option B: Ba and Mg is incorrect because the properties are reversed.
- Option C: Be and Mg is incorrect because Be doesn't form a soluble sulphate.
- Option D: Na and K are Group 1 elements, not Group 2.

Key Concepts:
- Solubility trends in Group 2
- Properties of alkaline earth metals

Memory Aids:
- Sulfates decrease in solubility down Group 2.
- Oxides increase in solubility down Group 2.

Quick Tips:
- Remember the solubility trends for Group 2 compounds.

19 Easy

19 of 111Question permalink

Remembering +4 | -1

On heating washing soda, we get:

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{O}\left( v \right)

Option D (

{\mathrm{H}_2}\mathrm{O}\left( v \right)

) is correct because washing soda (Na2CO3.10H2O) loses water molecules upon heating.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{O}\left( v \right)$ . Washing soda loses water of crystallization on heating.

Step-by-Step Solution:

Step 1: Washing soda is $Na_2CO_3.10H_2O$ .

Step 2: On heating, it loses its water of crystallization.
$Na_2CO_3.10H_2O(s) \rightarrow Na_2CO_3(s) + 10H_2O(g)$

Step 3: The product is water in the vapor phase.

Why other options are wrong:
- Option A: $\mathrm{CO}$ - Carbon monoxide is not produced during the heating of washing soda.
- Option B: $\mathrm{CO} + \mathrm{C}{\mathrm{O}_2}$ - Carbon monoxide and carbon dioxide are not produced during the heating of washing soda.
- Option C: $\mathrm{C}{\mathrm{O}_2}$ - Carbon dioxide is not produced during the heating of washing soda.

Key Concepts:
- Properties of washing soda
- Water of crystallization

Memory Aids:
- Washing soda formula: $Na_2CO_3.10H_2O$

Quick Tips:
- Remember that washing soda contains water of crystallization.

20 Medium

20 of 111Question permalink

Remembering +4 | -1

In a sodium chloride crystal, each chloride ion is surrounded by:

View answer and explanation

Correct option: B

6\mathrm{N}{\mathrm{a}^ + }\mathrm{ions}

Option B (

6\mathrm{N}{\mathrm{a}^ + }\mathrm{ions}

) is correct because in the NaCl crystal structure, each chloride ion is surrounded by six sodium ions in an octahedral arrangement.

Solution approach

Correct Answer + Core Reason: $6\mathrm{N}{\mathrm{a}^ + }\mathrm{ions}$ . In NaCl crystal, each chloride ion is surrounded by six sodium ions.

Step-by-Step Solution:

Step 1: Visualize the NaCl crystal structure.

Step 2: The NaCl structure is a face-centered cubic (FCC) lattice.

Step 3: Each chloride ion ( $\mathrm{Cl}^−$ ) is located at the corners and face centers of the cube.

Step 4: Each chloride ion is surrounded by six sodium ions ( $\mathrm{Na}^+$ ) in an octahedral arrangement.

Why other options are wrong:
- Option A: $4\mathrm{N}{\mathrm{a}^ + }\mathrm{ions}$ : This is incorrect. The coordination number is 6.
- Option C: $1\mathrm{N}{\mathrm{a}^ + }\mathrm{ion}$ : This is incorrect. The coordination number is 6.
- Option D: $2\mathrm{N}{\mathrm{a}^ + }\mathrm{ions}$ : This is incorrect. The coordination number is 6.

Key Concepts:
- NaCl crystal structure
- Coordination number
- Face-centered cubic lattice

Memory Aids:
- NaCl - 6:6 coordination

Quick Tips:
- Remember the coordination numbers for common crystal structures.

21 Medium

21 of 111Question permalink

Analyzing +4 | -1

Which order is NOT correct

View answer and explanation

Correct option: C

Melting point :

LiCl > NaCl > KCl > RbCl > CsCl

Option C (Melting point :

LiCl > NaCl > KCl > RbCl > CsCl

) is NOT correct because as the size of the cation increases, the lattice energy decreases, and hence the melting point decreases.

Solution approach

Correct Answer + Core Reason: Melting point : $LiCl > NaCl > KCl > RbCl > CsCl$ . This order is incorrect because melting point decreases as the size of the cation increases.

Step-by-Step Solution:

Step 1: Analyze the given orders.

Step 2: Thermal stability: $BeC{O_3} < MgC{O_3} < CaC{O_3} < {K_2}C{O_3}$ . Thermal stability of carbonates increases down the group due to increasing ionic size and decreasing polarizing power of the cation. This order is correct.

Step 3: Mobility in aqueous solution: $R{b^ + } > {K^ + } > N{a^ + } > L{i^ + }$ . Mobility of ions in aqueous solution depends on the hydration of ions. Smaller ions are more hydrated and hence have lower mobility. This order is correct.

Step 4: Melting point: $LiCl > NaCl > KCl > RbCl > CsCl$ . As the size of the cation increases, the lattice energy decreases, and hence the melting point decreases. This order is correct.

Step 5: Hydration energy: $B{e^{2 + }} > M{g^{2 + }} > C{a^{2 + }} > S{r^{2 + }} > B{a^{2 + }}$ . Hydration energy decreases as the size of the ion increases. This order is correct.

Step 6: Therefore, the melting point order is NOT correct. As the size of the cation increases, the lattice energy decreases, and hence the melting point decreases. The correct order should be $LiCl > NaCl > KCl > RbCl > CsCl$ .

Why other options are wrong:
- Option A: Thermal stability : $BeC{O_3} < MgC{O_3} < CaC{O_3} < {K_2}C{O_3}$ : This order is correct.
- Option B: Mobility in aqueous solution : $R{b^ + } > {K^ + } > N{a^ + } > L{i^ + }$ : This order is correct.
- Option D: Hydration energy : $B{e^{2 + }} > M{g^{2 + }} > C{a^{2 + }} > S{r^{2 + }} > B{a^{2 + }}$ : This order is correct.

Key Concepts:
- Thermal stability of carbonates
- Mobility of ions in aqueous solution
- Melting point of ionic compounds
- Hydration energy of ions

Memory Aids:
- Remember that thermal stability increases down the group for carbonates.
- Remember that mobility decreases with increasing hydration.
- Remember that melting point decreases with increasing cation size.
- Remember that hydration energy decreases with increasing ion size.

Quick Tips:
- Understand the factors affecting thermal stability, mobility, melting point, and hydration energy.

22 Easy

22 of 111Question permalink

Understanding +4 | -1

Gypsum is added to clinker during cement manufacture to:

View answer and explanation

Correct option: A

Decrease the rate of setting of cement

Option A is correct because Gypsum (

CaSO_4 \cdot 2H_2O

) is added to cement to slow down the hydration process of cement compounds, preventing flash setting.

Solution approach

Correct Answer + Core Reason: Decrease the rate of setting of cement. Gypsum acts as a retarder, controlling the hydration rate of cement.

Step-by-Step Solution:

Step 1: Understand the role of gypsum in cement manufacturing.

Step 2: Gypsum ( $CaSO_4 \cdot 2H_2O$ ) reacts with tricalcium aluminate ( $C_3A$ ) to form calcium sulfoaluminate, which prevents rapid hydration of $C_3A$ .

Step 3: This controlled hydration allows for proper setting and hardening of the cement.

Why other options are wrong:
- Option B: Gypsum does not primarily make the cement impervious. Impermeability is related to the cement's composition and curing process.
- Option C: Gypsum does not directly bind the particles of calcium silicate. Calcium silicates hydrate to form C-S-H gel, which is responsible for binding.
- Option D: Gypsum influences the formation of colloidal gel but its primary function is to control the rate of setting, not just facilitate gel formation.

Key Concepts:
- Cement chemistry
- Hydration of cement compounds
- Retarders in cement

Memory Aids:
- Gypsum = Go Slow (retards setting)

Quick Tips:
- Remember that gypsum is a common additive to control the setting time of cement.

23 Easy

23 of 111Question permalink

Analyzing +4 | -1

A sudden large jump between the values of second and third ionization energies of an element would be associated with the electronic configuration:

View answer and explanation

Correct option: D

1{s^2},2{s^2}2{p^6},3{s^2}

Option D (1s2,2s22p6,3s2) is correct because removing the third electron requires significantly more energy as it comes from a stable noble gas core.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2}2{p^6},3{s^2}$ . Removing the third electron requires significantly more energy as it comes from a stable noble gas core.

Step-by-Step Solution:

Step 1: Understand Ionization Energy:\nIonization energy is the energy required to remove an electron from an atom or ion in the gaseous phase.\nA large jump in ionization energy indicates that an electron is being removed from a core electron, which is held more tightly.

Step 2: Analyze the electronic configurations:
- A: $1s^2 2s^2 2p^6 3s^1$ - Removing the 2nd electron is from a neutral atom, and the 3rd electron is from a stable noble gas core.
- B: $1s^2 2s^2 2p^6 3s^2 3p^1$ - Removing the 2nd and 3rd electrons are from the same shell.
- C: $1s^2 2s^2 2p^6 3s^2 3p^2$ - Removing the 2nd and 3rd electrons are from the same shell.
- D: $1s^2 2s^2 2p^6 3s^2$ - Removing the 2nd electron is from a neutral atom, and the 3rd electron is from a stable noble gas core.

Step 3: Determine the configuration with a large jump between the 2nd and 3rd ionization energies:\nBoth A and D have a large jump between the 2nd and 3rd ionization energies. However, the question asks for a sudden large jump between the values of the *second* and *third* ionization energies. This means that after removing two electrons, the next electron is much harder to remove. This happens when the first two electrons are valence electrons, and the third is a core electron. Therefore, the correct answer is D.

Why other options are wrong:
- **Option A ( $1{s^2},2{s^2}2{p^6},3{s^1}$ ): While there will be a jump, it won't be as significant as in option D because the 2nd electron is easier to remove than the 3rd.
- Option B ( $1{s^2},2{s^2}2{p^6},3{s^2}3{p^1}$ ): The 2nd and 3rd ionization energies will be relatively close as both are valence electrons.
- Option C ( $1{s^2},2{s^2}2{p^6},3{s^2}3{p^2}$ ): The 2nd and 3rd ionization energies will be relatively close as both are valence electrons.

Key Concepts:
- Ionization Energy
- Electronic Configuration
- Core and Valence Electrons

Quick Tips:**
- Look for electronic configurations where removing the third electron requires breaking into a stable noble gas core.

24 Hard

24 of 111Question permalink

Analyzing +4 | -1

A metal

'M'

reacts with

{\mathrm{N}_2}

to give a compound

'A'\left( {{M_3}N} \right).\mathrm{ }'A'

on heating at high temperature gives back

'M'\mathrm{and '}A'

on reacting with

{\mathrm{H}_2}\mathrm{O}

gives a gas

B.\mathrm{ }'B'

turns

\mathrm{CuS}{\mathrm{O}_4}

solution blue on passing through it.

M\mathrm{and}B

can be

View answer and explanation

Correct option: B

\mathrm{Li and N}{\mathrm{H}_3}

Option B (

\mathrm{Li and N}{\mathrm{H}_3}

) is correct because Lithium reacts with nitrogen to form Lithium nitride (

\mathrm{L}{\mathrm{i}_3}\mathrm{N}

), which reacts with water to produce ammonia (

\mathrm{N}{\mathrm{H}_3}

), and ammonia turns

\mathrm{CuS}{\mathrm{O}_4}

solution blue.

Solution approach

Correct Answer + Core Reason: $\mathrm{Li and N}{\mathrm{H}_3}$ . Lithium reacts with nitrogen to form lithium nitride, which reacts with water to produce ammonia, and ammonia turns $\mathrm{CuS}{\mathrm{O}_4}$ solution blue.

Step-by-Step Solution:

Step 1: Identify the metal $'M'$ that reacts with ${\mathrm{N}_2}$ to form $'{A}' ({M_3}N)$ .

Step 2: Consider the reaction of Lithium with Nitrogen:
$6Li + N_2 \to 2Li_3N (A)$

Step 3: $'A'$ on heating gives back $'M'$ :
$2Li_3N \xrightarrow{\Delta} 6Li + N_2$

Step 4: $'A'$ on reacting with ${\mathrm{H}_2}\mathrm{O}$ gives a gas $'B'$ :
$Li_3N + 3H_2O \to 3LiOH + NH_3 (B)$

Step 5: $'B'$ turns $\mathrm{CuS}{\mathrm{O}_4}$ solution blue:
$CuSO_4 + 2NH_4OH \to Cu(OH)_2 + (NH_4)_2SO_4$
$Cu(OH)_2 + 4NH_3 \to [Cu(NH_3)_4]$

25 Medium

25 of 111Question permalink

Analyzing +4 | -1

Which of the following has minimum values of cation-anion size ratio?

View answer and explanation

Correct option: C

\mathrm{MgC}{\mathrm{l}_2}

Option C (

\mathrm{MgC}{\mathrm{l}_2}

) is correct because Mg has the smallest cation size and Cl has a relatively large anion size compared to the other options, resulting in the smallest cation-anion size ratio.

Solution approach

Correct Answer + Core Reason: $\mathrm{MgC}{\mathrm{l}_2}$ . Magnesium has the smallest cation size among the given options, leading to the minimum cation-anion size ratio.

Step-by-Step Solution:

Step 1: Understand the concept of cation-anion size ratio.\nThe cation-anion size ratio affects the structure and properties of ionic compounds. A smaller ratio generally indicates a lower coordination number.

Step 2: Compare the cation sizes.\nThe cations are $\mathrm{Na}^+$ , $\mathrm{K}^+$ , $\mathrm{Mg}^{2+}$ , and $\mathrm{Ca}^{2+}$ .\nIonic radii: $\mathrm{Na}^+ (102 \text{{pm}})$ , $\mathrm{K}^+ (138 \text{{pm}})$ , $\mathrm{Mg}^{2+} (72 \text{{pm}})$ , $\mathrm{Ca}^{2+} (100 \text{{pm}})$ .
$\mathrm{Mg}^{2+}$ has the smallest ionic radius.

Step 3: Compare the anion sizes.\nThe anions are $\mathrm{Cl}^-$ and $\mathrm{F}^-$ .\nIonic radii: $\mathrm{Cl}^- (181 \text{{pm}})$ , $\mathrm{F}^- (133 \text{{pm}})$ .

Step 4: Calculate or estimate the cation-anion size ratios.
- $\mathrm{NaCl}$ : $102/181 \approx 0.56$
- $\mathrm{KCl}$ : $138/181 \approx 0.76$
- $\mathrm{MgC}{\mathrm{l}_2}$ : $72/181 \approx 0.40$
- $\mathrm{Ca}{\mathrm{F}_2}$ : $100/133 \approx 0.75$

Step 5: Determine the minimum value.
$\mathrm{MgC}{\mathrm{l}_2}$ has the smallest cation-anion size ratio.

Why other options are wrong:
- Option A: $\mathrm{NaCl}$ has a larger cation size than $\mathrm{Mg}^{2+}$ , leading to a larger ratio.
- Option B: $\mathrm{KCl}$ has the largest cation size, leading to a larger ratio.
- Option D: $\mathrm{Ca}{\mathrm{F}_2}$ has a larger cation and a smaller anion than $\mathrm{MgC}{\mathrm{l}_2}$ , leading to a larger ratio.

Key Concepts:
- Cation-anion size ratio
- Ionic radii

Memory Aids:
- Smaller cations and larger anions lead to smaller ratios.

Quick Tips:
- Remember the trends in ionic radii across the periodic table.

26 Medium

26 of 111Question permalink

Remembering +4 | -1

Manufacture of

\mathrm{NaOH}

is done by:

View answer and explanation

Correct option: A

Castner- Kellner process

Option A (Castner- Kellner process) is correct because this process is specifically used for the manufacture of NaOH.

Solution approach

Correct Answer + Core Reason: Castner- Kellner process. This process is used for the manufacture of NaOH.

Step-by-Step Solution:

Step 1: The Castner-Kellner process involves the electrolysis of brine (concentrated NaCl solution) using a mercury cathode.

Step 2: Sodium ions are discharged at the mercury cathode to form sodium amalgam.

Step 3: The sodium amalgam is then reacted with water to produce NaOH, hydrogen gas, and mercury.

Why other options are wrong:
- Option B: Solvay process - This process is used for the manufacture of sodium carbonate ( $Na_2CO_3$ ).
- Option C: Brine process - This is a general term for electrolysis of NaCl solution, but Castner-Kellner is the specific process.
- Option D: Mond’s process - This process is used for the purification of nickel.

Key Concepts:
- Manufacture of NaOH
- Castner-Kellner process

Memory Aids:
- Castner-Kellner: NaOH
- Solvay: $Na_2CO_3$

Quick Tips:
- Remember the specific processes for manufacturing different chemicals.

27 Medium

27 of 111Question permalink

Application +4 | -1

Zinc carbonate can be obtained from a solution of zinc chloride by adding:

View answer and explanation

Correct option: A

\mathrm{NaHC}{\mathrm{O}_3}

Option A (

\mathrm{NaHC}{\mathrm{O}_3}

) is correct because sodium bicarbonate reacts with zinc chloride to form zinc carbonate precipitate, sodium chloride, and carbon dioxide.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaHC}{\mathrm{O}_3}$ . Sodium bicarbonate can precipitate zinc carbonate from zinc chloride solution.

Step-by-Step Solution:

Step 1: Write the reaction between zinc chloride and sodium bicarbonate.
$ZnCl_2 + 2NaHCO_3 \to ZnCO_3 + 2NaCl + H_2O + CO_2$

Step 2: Zinc carbonate ( $ZnCO_3$ ) precipitates out of the solution.

Why other options are wrong:
- Option B: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ would also precipitate zinc carbonate, but $NaHCO_3$ is a weaker base and allows for better control of the precipitation.
- Option C: $\mathrm{CaC}{\mathrm{O}_3}$ is insoluble and will not react with zinc chloride in solution.
- Option D: $\mathrm{MgC}{\mathrm{O}_3}$ is also insoluble and will not react with zinc chloride in solution.

Key Concepts:
- Precipitation reactions
- Reaction of zinc chloride with bicarbonates

Memory Aids:
- Remember that bicarbonates can precipitate metal carbonates.

Quick Tips:
- Consider the solubility of the reactants and products.

28 Easy

28 of 111Question permalink

Understanding +4 | -1

Which one of the following on hydrolysis gives the corresponding metallic hydroxide,

{H_2}{O_2}\,\,and\,\,{O_2}

View answer and explanation

Correct option: C

K{O_2}

The correct answer is C (

KO_2

) because potassium superoxide reacts with water to produce potassium hydroxide, hydrogen peroxide, and oxygen.

Solution approach

Correct Answer + Core Reason: $KO_2$ . Potassium superoxide reacts with water to produce the corresponding metallic hydroxide, hydrogen peroxide, and oxygen.

Step-by-Step Solution:

Step 1: Write the hydrolysis reaction of $KO_2$ :
$2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$

Step 2: Analyze the products of the reaction.

Step 3: $KO_2$ produces $KOH$ , $H_2O_2$ , and $O_2$ upon hydrolysis.

Why other options are wrong:
- Option A: $L{i_2}O + H_2O \rightarrow 2LiOH$ . It doesn't produce $H_2O_2$ and $O_2$ .
- Option B: $N{a_2}{O_2} + 2H_2O \rightarrow 2NaOH + H_2O_2$ . It doesn't produce $O_2$ .
- Option D: $BeO$ is amphoteric and does not readily hydrolyze.

Key Concepts:
- Hydrolysis of metal oxides and superoxides.

Memory Aids:
- Superoxides produce $O_2$ upon hydrolysis.

Quick Tips:
- Remember the hydrolysis products of different metal oxides.

29 Easy

29 of 111Question permalink

Remembering +4 | -1

Metallic hydride among the following is

View answer and explanation

Correct option: C

CrH

CrH (Option C) is the correct answer because transition metals form metallic hydrides due to their ability to accommodate hydrogen in their lattice structure.

Solution approach

Correct Answer + Core Reason: CrH. Transition metals form metallic hydrides.

Step-by-Step Solution:

Step 1: Identify the types of hydrides.

Step 2: Metallic hydrides are formed by transition metals.

Step 3: Among the given options, Cr (Chromium) is a transition metal.

Step 4: Therefore, CrH is the metallic hydride.

Why other options are wrong:
- Option A: NaH is an ionic hydride formed by an alkali metal.
- Option B: LiH is an ionic hydride formed by an alkali metal.
- Option D: $MgH_2$ is an ionic hydride formed by an alkaline earth metal.

Key Concepts:
- Types of hydrides: Ionic, Metallic, Covalent
- Transition metals form metallic hydrides

Memory Aids:
- Transition metals = Metallic Hydrides

Quick Tips:
- Remember the position of elements in the periodic table to identify the type of hydride formed.

30 Hard

30 of 111Question permalink

Analyzing +4 | -1

Which of the following acts as reducing agent as well as oxidising agent?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

Option B (

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

) is correct because in

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

, oxygen has an oxidation state of -1, which can be both increased (oxidation) and decreased (reduction).

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ . In $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ , oxygen has an oxidation state of -1, which can be both increased (oxidation) and decreased (reduction).

Step-by-Step Solution:

Step 1: Determine the oxidation state of oxygen in each compound.

Step 2: In $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ , the oxidation state of oxygen is -2. It can only be oxidized.

Step 3: In $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ , the oxidation state of oxygen is -1. It can be both oxidized and reduced.

Step 4: In $\mathrm{NaN}{\mathrm{O}_3}$ , the oxidation state of nitrogen is +5. It can only be reduced.

Step 5: In $\mathrm{KN}{\mathrm{O}_3}$ , the oxidation state of nitrogen is +5. It can only be reduced.

Step 6: Therefore, $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ can act as both a reducing agent and an oxidizing agent.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ : Oxygen has an oxidation state of -2 and can only be oxidized, so it can only act as a reducing agent.
- Option C: $\mathrm{NaN}{\mathrm{O}_3}$ : Nitrogen has an oxidation state of +5 and can only be reduced, so it can only act as an oxidizing agent.
- Option D: $\mathrm{KN}{\mathrm{O}_3}$ : Nitrogen has an oxidation state of +5 and can only be reduced, so it can only act as an oxidizing agent.

Key Concepts:
- Oxidation states
- Oxidizing and reducing agents

Memory Aids:
- Remember that a substance can act as both an oxidizing and reducing agent if the element has an intermediate oxidation state.

Quick Tips:
- Determine the oxidation state of the element in question to determine if it can be oxidized or reduced.

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S-block elements practice · Mixed difficulty

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