Class 11 Chemistry NCERT Chapter s-Block: Higher Order Thinking Skills (HOTS) Questions
Some NCERT Exemplar class 11 chemistry chapter 10 s-block elements questions and answers are given below that will help you tackle complex problems. The questions below will help you evaluate your understanding of the concepts.
Question 1: Complete the following two reactions.
(i) $4 \mathrm{LiNO}_3 \longrightarrow x+\mathrm{O}_2$
(ii) $2 \mathrm{NaNO}_3 \longrightarrow \mathrm{y}+\mathrm{O}_2$
(1). $\mathrm{x}=\mathrm{LiNO}_2 ; y=\mathrm{NaNO}_2$
(2). $\mathrm{x}=\mathrm{LiNO}_2 ; y=\mathrm{NaNO}_2$
(3). $\mathrm{x}=\mathrm{LiO}+\mathrm{NO}_2 ; y=\mathrm{Na}_2 \mathrm{O}+\mathrm{NO}_2$
(4). $\mathrm{x}=\mathrm{Li}_2 \mathrm{O}+\mathrm{NO}_2 ; y=\mathrm{NaNO}_2$
Answer:
$
\begin{aligned}
& 4 \mathrm{LiNO}_3 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{NO}_2+\mathrm{O}_2 \\
& 2 \mathrm{NaNO}_3 \longrightarrow 2 \mathrm{NaNO}_2+\mathrm{O}_2
\end{aligned}
$
Hence, the answer is option (3).
Question 2: Suppose metal reacts with the oxygen to form oxide, then aqueous solution of this oxide, when added to a solution of HI, turn yellowish brown in colour. This compound is-
(1) $\mathrm{Na}_2 \mathrm{O}$
(2) $\mathrm{Li}_2 \mathrm{O}$
(3) NaOH
(4) $\mathrm{Na}_2 \mathrm{O}_2$
Answer:
$\mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}+\mathrm{H}_2 \mathrm{O}_2$ aqueous solution of $\mathrm{Na}_2 \mathrm{O}_2$ form $\mathrm{H}_2 \mathrm{O}_2$ which acts as an oxidizing agent, hence converting lodide to lodine $\left(\mathrm{I}_2\right)$.
Hence, the answer is option (4).
Question 3. Given below are two statements:
Statement I : On passing $\mathrm{HCl}_{(\mathrm{g})}$ through a saturated solution of $\mathrm{BaCl}_2$, at room temperature white turbidity appears.
Statement II : When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is correct but Statement II is incorrect
(2) Both Statement I and Statement II are incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are correct
Answer:
$\mathrm{BaCl}_2, \mathrm{NaCl}$ are soluble but on adding $\mathrm{HCl}(\mathrm{g})$ to $\mathrm{BaCl}_2, \mathrm{NaCl}$ solutions, Sodium or Barium chlorides may precipitate out, as a consequence of the law of mass action
Hence, the answer is the option (1).
Question 4: Which one the following compounds will readily react with dilute $\mathrm{NaOH}$ ?
(1) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$
(2) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
(3) $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}$
(4) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$
Answer:
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Stronger ACID than $\mathrm{H}_2 \mathrm{O}$
Hence, the answer is the option (4)
Question 5: In Solvay ammonia process, sodium bicarbonate is precipitated due to-
(1) Presence of $\mathrm{NH}_3$
(2) Reaction with $\mathrm{CO}_2$
(3) Reaction with brine solution
(4) Reaction with NaOH
Answer:
$
\mathrm{NH}_4 \mathrm{HCO}_3+\mathrm{NaCl} \longrightarrow \mathrm{NaHCO}_3 \downarrow+\mathrm{NH}_4 \mathrm{Cl}
$
Hence, the answer is the option (3).
Question 6: A: Lithium carbonate is not so stable to heat
R: Lithium being very small in size, polarizes large $\mathrm{CO}_3^{2-}$ ions, leading to the formation of more stable $\mathrm{Li}_2 \mathrm{O}$ and $\mathrm{CO}_2$
(1) A is correct, and R is the correct explanation.
(2) A is correct, and R is also correct but not the correct explanation.
(3) A is correct, and R is incorrect.
(4) A is incorrect, and R is incorrect.
Answer:
Lithium carbonate is unstable to heat; lithium being very small in size, polarises a large $\mathrm{CO}_3^{2-}$ ion leading to the formation of more stable $\mathrm{Li}_2 \mathrm{O}$ and $\mathrm{CO}_2$
Hence, the answer is the option (1).
Question 7: Li is strongest reducing agent among alkali metals due to which of the following factor?
(1) Ionisation Enthalpy
(2) Electron Affinity
(3) Hydration energy
(4) Lattice energy.
Answer:
Li due to the highest hydration energy among the alkali metals, which is the strongest reducing agent.
Hence, the answer is the option (3).
Approach to Solve Questions of Chapter 10 s-block
To solve class 11 chemistry NCERT Exemplar chapter 10 s-block elements questions, it is important to follow a systematic approach. It is recommended to strategies your study plan to solve the questions of this chapter. The following are the points that will help you build a good approach.
1. While solvinge questions from s-block elements, it's important to understand the key concepts like electronic configuration, atomic and ionic radii, ionisation enthalpy, physical and chemical properties, anomalous behaviour of lithium and beryllium, diagonal relationships, etc.
2. Many questions revolve around periodic trends. Students need to understand how and why these trends occur across and down the group. The Key Properties students should focus on are:
- Atomic and ionic radii
- Ionisation enthalpy
- Hydration enthalpy
- Reactivity with air and water
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3. Some anomalies and exceptions exist from the generally observed properties. So, it's important to remember those exceptions, as questions are asked frequently about these topics.
- Anomalous behaviour of lithium and beryllium
- Diagonal relationship Li resembles Mg, and Be resembles Al.
4. Students must remember the basic chemical reactions of the NCERT Exemplar class 11 chemistry chapter 10 s-block elements like
- Reactions of s-block elements with oxygen, water, and hydrogen.
- Reactions of s-block elements with Halogens
5. Questions asked from this chapter are mostly based on trends and comparisons. A Proper understanding of basic concepts and practice helps students clear their doubts. Students can refer NCERT Exemplar textbooks for practice. For revision students can follow Class 11 chemistry chapter 10 s-block elements Notes.
General Trends in s-block Elements
Atomic trends play an important role in predicting the physical and chemical properties of s-block elements. Some of the important trends are given below.
1. The Atomic Radius of s-block elements increases down the group because of the addition of electron shells.
2. Ionisation enthalpy decreases down the group because it becomes easier to remove the outermost electron due to increased atomic size and shielding effect.
3. Electronegativity decreases down the group because the tendency to attract electrons decreases down the group.
4. Metallic character increases down the group because elements become more metallic as they lose electrons more readily.
5. Melting and Boiling points decrease down the group because of weaker metallic bonding.
6. Density generally increases down the group, but group 1 shows irregular trends.
7. Reactivity increases down the group, both alkali and alkaline earth metals become more reactive, especially with water and oxygen.
8. Hydration energy decreases as we move down the group because the interaction between the ion and water molecules becomes weaker.