RBSE Solutions Class 11 Chemistry Chapter 10 s-Block Elements

Rbse Class 11 Chemistry Chapter 10 Multiple Choice Questions

Question 1. Which of the following alkaline earth metal is most stable to heat ?
(a) MgCOg
(b) CaCOg
(c) SrCOg
(d) BaCOg
Answer: (d) BaCOg
In simple words: Among the alkaline earth metal carbonates, barium carbonate (BaCO3) is the most stable when heated. This stability increases as you go down the group.

🎯 Exam Tip: Remember that thermal stability of carbonates generally increases down an alkaline earth metal group due to increasing ionic character and decreasing polarizing power.

Question 3. Which of the alkali metals halides have lowest lattice energy ?
(a) LiF
(b) NaCl
(c) KBr
(d) CsI
Answer: (d) CsI
In simple words: The compound CsI (Cesium Iodide) has the lowest lattice energy among the options listed. Lattice energy is lowest when the ions are largest.

🎯 Exam Tip: Lattice energy decreases as the size of the ions (both cation and anion) increases because the electrostatic attraction between their centers becomes weaker due to greater distance.

Question 4. Which of the following do not give flame test ?
(a) Be
(b) K
(c) Sr
(d) Na
Answer: (a) Be
In simple words: Beryllium (Be) does not produce a color in a flame test. This is because its electrons are held very tightly and don't get excited enough to emit visible light.

🎯 Exam Tip: Remember that beryllium (Be) and magnesium (Mg) do not show a flame test because their small size and high ionization enthalpy mean their electrons are too strongly bound to be excited by the flame's energy.

Question 5. Which of the following metal has lowest melting point ?
(a) Na
(b) K
(c) Rb
(d) Cs
Answer: (d) Cs
In simple words: Cesium (Cs) has the lowest melting point among these metals. As you go down the alkali metal group, the melting point decreases.

🎯 Exam Tip: For alkali metals, melting points decrease down the group. This is because metallic bonding weakens as the atomic size increases and the valence electrons are further from the nucleus.

Rbse Class 11 Chemistry Chapter 10 Very Short Answer Type Questions

Question 7. In a period, melting points of alkali metals are lower than alkaline eath metals, Why ?
Answer: Alkali metals have lower melting points than alkaline earth metals when compared in the same period. This is because alkali metals have only one valence electron contributing to metallic bonding, while alkaline earth metals have two valence electrons. The stronger metallic bond in alkaline earth metals requires more energy to break, leading to higher melting points.
In simple words: Alkali metals melt at lower temperatures because they have weaker bonds holding their atoms together compared to alkaline earth metals. Alkaline earth metals have two electrons helping to form bonds, while alkali metals only have one.

🎯 Exam Tip: Focus on the number of valence electrons and the strength of metallic bonding when comparing melting points of elements in the same period.

Question 8. Alkali metals are strong electropositive in nature, Why ?
Answer: Alkali metals are considered strongly electropositive because their atoms have low ionization energies. This means they can very easily lose their single outermost valence electron. When an atom readily loses an electron, it forms a positive ion, which is the characteristic of an electropositive element.
In simple words: Alkali metals easily lose an electron and become positive. This is because it doesn't take much energy to remove their outermost electron.

🎯 Exam Tip: To explain electropositivity, always link it to low ionization energy and the ease with which valence electrons are lost.

Question 9. Which metal is responsible for clotting of blood in our body?
Answer: Calcium ions (\( \text{Ca}^{2+} \)) are responsible for the clotting of blood in our body. They play a crucial role in activating various factors in the complex process of blood coagulation, helping to form a stable blood clot.
In simple words: Calcium helps our blood to clot. Without calcium ions, blood would not be able to form scabs and stop bleeding.

🎯 Exam Tip: Calcium ions are a key cofactor in the blood coagulation cascade; remember their role in forming blood clots.

Question 10. Write the electronic configuration of alkali metals.
Answer: The general electronic configuration for all alkali metals is \( \text{ns}^1 \). Here, 'n' represents the principal quantum number (the period number) of the outermost shell, and '1' indicates that there is one electron in the s-orbital of that shell.
In simple words: All alkali metals have one electron in their outermost shell, in the 's' orbital. So their electron setup ends with \( \text{ns}^1 \), where 'n' is the shell number.

🎯 Exam Tip: Remember that the \( \text{ns}^1 \) configuration is what defines alkali metals and explains their characteristic reactivity.

Question 11. Calculate the oxidation state of Na in Na2O2.
Answer: To calculate the oxidation state of Na in \( \text{Na}_2\text{O}_2 \):
Let the oxidation state of Na be \( \text{x} \).
We know that in peroxides, the oxidation state of oxygen is \( -1 \).
There are two Na atoms and two O atoms.
So, \( 2(\text{x}) + 2(-1) = 0 \)
\( 2\text{x} - 2 = 0 \)
\( 2\text{x} = 2 \)
\( \text{x} = 1 \)
Therefore, the oxidation state of Na in \( \text{Na}_2\text{O}_2 \) is \( +1 \).
In simple words: In the compound \( \text{Na}_2\text{O}_2 \), oxygen has an oxidation state of -1 because it's a peroxide. Since the whole molecule is neutral, the two sodium atoms must balance out the two -1 oxygen atoms, meaning each sodium atom has a +1 oxidation state.

🎯 Exam Tip: Always remember that oxygen has an oxidation state of \( -1 \) in peroxides (like \( \text{Na}_2\text{O}_2 \)) and \( -\frac{1}{2} \) in superoxides (like \( \text{KO}_2 \)), which is different from its usual \( -2 \) state.

Question 13. Why alkali metals and alkaline earth metals are not obtained by chemical reduction method ?
Answer: Alkali metals and alkaline earth metals are not typically obtained through chemical reduction methods. This is because they are very strong reducing agents themselves and are highly reactive metals. It is extremely difficult to find another substance that is a stronger reducing agent than them to reduce their compounds. Therefore, these metals are usually extracted using electrolytic reduction of their fused salts.
In simple words: We can't get alkali and alkaline earth metals by chemical reduction because they are already very good at reducing things. It's hard to find a chemical that can reduce them further.

🎯 Exam Tip: Recognize that highly reactive metals like alkali and alkaline earth metals require electrolysis for their extraction, as chemical reduction is not effective.

Question 14. Potassium carbonate cannot be prepared by Solvay process. Why ?
Answer: Potassium carbonate cannot be prepared using the Solvay process. This is because, unlike sodium bicarbonate (which is sparingly soluble and precipitates out in the Solvay process), potassium bicarbonate is quite soluble in water. Since potassium bicarbonate does not precipitate out easily, it cannot be effectively separated from the solution to proceed with the Solvay process steps for potassium carbonate.
In simple words: The Solvay process doesn't work for making potassium carbonate because potassium bicarbonate stays dissolved in water and doesn't form a solid that can be easily separated.

🎯 Exam Tip: The key limitation of the Solvay process for potassium carbonate is the high solubility of potassium bicarbonate, preventing its precipitation and separation.

Question 15. What reaction takes place when quick lime is heated with silica ?
Answer: When quick lime (calcium oxide, \( \text{CaO} \)) is heated with silica (\( \text{SiO}_2 \)), it undergoes a chemical reaction to form calcium silicate. This is a solid-state reaction that occurs at high temperatures.
The reaction is:
\( \text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \)
In simple words: When quick lime and sand (silica) are heated together, they combine to make a new substance called calcium silicate.

🎯 Exam Tip: Remember that quick lime (CaO) is a basic oxide and reacts with acidic oxides like silica (SiO2) to form silicates, a common reaction in metallurgy and cement production.

Question 16. Why second ionization energy of alkali metals are higher than first ionization energy?
Answer: The second ionization energy of alkali metals is significantly higher than their first ionization energy due to the following reason:
After losing one electron, an alkali metal atom forms a monovalent cation \( (\text{M}^+) \). This ion now has a stable noble gas electronic configuration, typically \( \text{ns}^2 \text{np}^6 \). Removing a second electron from this stable, fully filled electron shell requires a very large amount of energy. The nuclear charge remains the same, but the electron is now being removed from a more stable and tightly held configuration, closer to the nucleus.
In simple words: Alkali metals become very stable after losing their first electron, as they get a full outer shell like a noble gas. So, it takes a lot more energy to pull away a second electron from this very stable setup.

🎯 Exam Tip: The unusually high second ionization energy of alkali metals is a direct consequence of achieving a stable noble gas configuration after losing the first electron.

Rbse Class 11 Chemistry Chapter 10 Short Answer Type Questions

Question 18. How is lithium aluminum hydride formed ?
Answer: Lithium aluminum hydride (\( \text{LiAlH}_4 \)) is formed by the chemical reaction between lithium hydride (\( \text{LiH} \)) and aluminum chloride (\( \text{AlCl}_3 \)). This reaction typically occurs in an ether solvent.
The balanced chemical equation for its formation is:
\( 4\text{LiH} + \text{AlCl}_3 \rightarrow \text{LiAlH}_4 + 3\text{LiCl} \)
In simple words: Lithium aluminum hydride is made by mixing lithium hydride with aluminum chloride. It's a key compound used in many chemical reactions.

🎯 Exam Tip: Remember lithium aluminum hydride as a powerful reducing agent, and its synthesis involves lithium hydride reacting with a suitable aluminum halide.

Question 19. What is Hydrolith? How it reacts with water?
Answer: Hydrolith is the common name for calcium hydride (\( \text{CaH}_2 \)). It is a white solid that is commonly used as a drying agent and as a source of hydrogen gas.
When hydrolith reacts with water, it produces calcium hydroxide (\( \text{Ca(OH)}_2 \)) and hydrogen gas (\( \text{H}_2 \)). This reaction is quite vigorous and releases a significant amount of heat.
The balanced chemical equation is:
\( \text{CaH}_2 + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + 2\text{H}_2 \)
In simple words: Hydrolith is calcium hydride. When it touches water, it creates calcium hydroxide and hydrogen gas, often quite strongly.

🎯 Exam Tip: Calcium hydride (Hydrolith) is an excellent source of hydrogen gas and acts as a strong reducing agent due to the presence of hydride ions.

Question 20. Which of the following NaOH or Mg (OH)2 is strong base ?
Answer: Sodium hydroxide (\( \text{NaOH} \)) is a much stronger base than magnesium hydroxide (\( \text{Mg(OH)}_2 \)).
\( \text{NaOH} \) is a strong base because it is an alkali metal hydroxide, which completely ionizes (or hydrolyzes) in solution to release a high concentration of hydroxide ions (\( \text{OH}^- \)). On the other hand, \( \text{Mg(OH)}_2 \) is an alkaline earth metal hydroxide and is only sparingly soluble in water, meaning it releases fewer \( \text{OH}^- \) ions and is considered a weak base (though it is used as an antacid).
In simple words: Sodium hydroxide is a strong base because it fully breaks apart in water to give many hydroxide ions. Magnesium hydroxide is a weaker base because it doesn't dissolve much in water.

🎯 Exam Tip: Remember that alkali metal hydroxides are generally strong bases, while alkaline earth metal hydroxides are weaker bases due to lower solubility, except for Ba(OH)2 which is relatively strong.

Question 21. Compare the alkali metals and alkaline earth metals with respect to
• ionisation enthalpy
• atomic and ionic radius
Answer: Let's compare alkali metals (Group 1) and alkaline earth metals (Group 2) based on the given properties:

1. Ionization Enthalpy:
• Alkali metals have very low first ionization enthalpies because they have only one valence electron and losing it results in a stable noble gas configuration. This energy decreases down the group.
• Alkaline earth metals have higher first ionization enthalpies compared to alkali metals in the same period. This is because they have two valence electrons and a higher nuclear charge. However, their second ionization enthalpies are also relatively low (compared to the second IE of alkali metals) as losing the second electron also results in a stable noble gas configuration. The first ionization enthalpy generally decreases down the group.

2. Atomic and Ionic Radius:
• Alkali metals have the largest atomic and ionic radii in their respective periods. As we move down the group, both atomic and ionic radii increase due to the addition of new electron shells.
• Alkaline earth metals have smaller atomic and ionic radii than alkali metals in the same period. This is because alkaline earth metals have a higher effective nuclear charge, which pulls the electrons closer to the nucleus. Both atomic and ionic radii also increase down the group for alkaline earth metals, similar to alkali metals, but they are always smaller than the corresponding alkali metals.
In simple words: Alkali metals lose electrons easily and are big. Alkaline earth metals need more energy to lose their first electron and are slightly smaller than alkali metals in the same row. Both types of metals get bigger as you go down their groups.

🎯 Exam Tip: When comparing periodic trends, always consider the effective nuclear charge and the number of electron shells. For ionization enthalpy, the stability of the resulting ion is also crucial.

Question 22. Why potassium and calcium are used in photoelectric all in place of Lithium?
Answer: Potassium and calcium are often preferred over lithium in photoelectric cells. This is because potassium and calcium have much lower ionization energies compared to lithium. Photoelectric cells work by electrons being ejected from a metal surface when light hits it. Metals with lower ionization energies require less energy to release their electrons, meaning they can respond to light with lower energy (like visible light) more effectively. Lithium's higher ionization energy means it requires higher energy light (like UV) to eject electrons, making it less suitable for common photoelectric applications.
In simple words: Potassium and calcium are better for photoelectric cells than lithium. This is because their electrons are easier to remove with light, making them more sensitive to everyday light.

🎯 Exam Tip: The efficiency of a photoelectric cell is directly related to the ionization energy of the metal used; lower ionization energy means greater sensitivity to lower energy light.

Question 23. Starting with sodium chloride how would proceed to prepare-
(i) sodium hydroxide
(ii) sodium carbonate ?
Answer: Here's how to prepare sodium hydroxide and sodium carbonate starting from sodium chloride:

(i) Preparation of Sodium Hydroxide (\( \text{NaOH} \)):
Sodium hydroxide is prepared by the electrolysis of an aqueous solution of sodium chloride (brine) using the Castner-Kellner cell. In this cell, a mercury cathode and a carbon (graphite) anode are used.

At the anode (oxidation):
\( 2\text{Cl}^- (\text{aq}) \rightarrow \text{Cl}_2 (\text{g}) + 2\text{e}^- \)

At the cathode (reduction):
\( \text{Na}^+ (\text{aq}) + \text{e}^- + \text{Hg} (\text{l}) \rightarrow \text{Na-Hg} (\text{amalgam}) \)
(Sodium ions combine with mercury to form sodium amalgam)

The sodium amalgam then reacts with water to produce sodium hydroxide, hydrogen gas, and mercury:
\( 2\text{Na-Hg} (\text{amalgam}) + 2\text{H}_2\text{O} (\text{l}) \rightarrow 2\text{NaOH} (\text{aq}) + \text{H}_2 (\text{g}) + 2\text{Hg} (\text{l}) \)
Overall reaction:
\( 2\text{NaCl} (\text{aq}) + 2\text{H}_2\text{O} (\text{l}) \rightarrow 2\text{NaOH} (\text{aq}) + \text{Cl}_2 (\text{g}) + \text{H}_2 (\text{g}) \)

(ii) Preparation of Sodium Carbonate (\( \text{Na}_2\text{CO}_3 \)):
Sodium carbonate is prepared using the Solvay process (also known as the ammonia-soda process), which uses sodium chloride, limestone (\( \text{CaCO}_3 \)), and ammonia as raw materials.

**Step I: Preparation of Ammoniated Brine and Reaction with \( \text{CO}_2 \)**
Ammonia (\( \text{NH}_3 \)) is first dissolved in a concentrated sodium chloride solution (brine) to form ammoniated brine. Then, carbon dioxide (\( \text{CO}_2 \)) is passed through this solution. The \( \text{CO}_2 \) is obtained by heating limestone (\( \text{CaCO}_3 \)). This leads to the formation of ammonium bicarbonate, which then reacts with sodium chloride to produce sodium bicarbonate.
\( \text{NH}_3 (\text{g}) + \text{H}_2\text{O} (\text{l}) + \text{CO}_2 (\text{g}) \rightarrow \text{NH}_4\text{HCO}_3 (\text{aq}) \)
\( \text{NH}_4\text{HCO}_3 (\text{aq}) + \text{NaCl} (\text{aq}) \rightarrow \text{NaHCO}_3 (\text{s}) + \text{NH}_4\text{Cl} (\text{aq}) \)
Sodium bicarbonate is less soluble and precipitates out, which can be filtered.

**Step II: Calcination of Sodium Bicarbonate**
The precipitated sodium bicarbonate is then heated (calcined) at 160-230\( ^\circ \text{C} \) to convert it into sodium carbonate, releasing carbon dioxide and water.
\( 2\text{NaHCO}_3 (\text{s}) \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 (\text{s}) + \text{H}_2\text{O} (\text{g}) + \text{CO}_2 (\text{g}) \)

**Step III: Ammonia Recovery**
Ammonia is recovered from the ammonium chloride solution (formed in Step I) by treating it with quick lime (\( \text{CaO} \)), which is also obtained from heating limestone.
\( \text{CaCO}_3 (\text{s}) \xrightarrow{\text{Heat}} \text{CaO} (\text{s}) + \text{CO}_2 (\text{g}) \)
\( \text{CaO} (\text{s}) + \text{H}_2\text{O} (\text{l}) \rightarrow \text{Ca(OH)}_2 (\text{aq}) \)
\( 2\text{NH}_4\text{Cl} (\text{aq}) + \text{Ca(OH)}_2 (\text{aq}) \rightarrow 2\text{NH}_3 (\text{g}) + \text{CaCl}_2 (\text{aq}) + 2\text{H}_2\text{O} (\text{l}) \)
The recovered ammonia is reused in the process, making it economical. Calcium chloride (\( \text{CaCl}_2 \)) is obtained as a by-product.
In simple words: To make sodium hydroxide, we pass electricity through salt water. To make sodium carbonate, we use salt water, ammonia, and carbon dioxide. Sodium bicarbonate first forms and then is heated to get sodium carbonate.

🎯 Exam Tip: Understand the key reactions and steps for both the Castner-Kellner process (for NaOH) and the Solvay process (for Na2CO3), including the roles of each reactant and byproduct.

Question 24. Explain the structure of the following -
(a) BeCl2 (vapour state)
(b) BeCl2 (solid state)
Answer: Let's explain the structure of Beryllium Chloride (\( \text{BeCl}_2 \)) in both its vapour and solid states.

(a) \( \text{BeCl}_2 \) in Vapour State:
In the vapour state, \( \text{BeCl}_2 \) exists primarily as a linear monomer at very high temperatures (around 1200 K). However, at slightly lower temperatures, it tends to form a chlorine-bridged dimer structure. In this dimer, two \( \text{BeCl}_2 \) units are linked by two chlorine atoms, forming a four-membered ring. Each beryllium atom still completes its octet by sharing electrons with the bridging chlorine atoms.

(b) \( \text{BeCl}_2 \) in Solid State:
In the solid state, \( \text{BeCl}_2 \) forms a polymeric, chain-like structure. Each beryllium atom is tetrahedrally coordinated, being surrounded by four chlorine atoms. Two of these chlorine atoms are attached by covalent bonds, and the other two are attached by coordinate covalent bonds, forming bridges to adjacent beryllium atoms. This creates an extended chain where the \( \text{BeCl}_2 \) units are linked head-to-tail through bridging chlorine atoms.
In simple words: In the gas phase, BeCl2 can be a single molecule or two molecules joined by chlorine bridges. In the solid state, it forms a long chain where each beryllium atom is linked to other beryllium atoms through chlorine atoms.

🎯 Exam Tip: Remember that beryllium compounds often show covalent character and polymeric structures due to beryllium's small size and high polarizing power, which contrasts with other group 2 elements.

Question 25. Why alkali metals given blue colour with liquid ammonia ?
Answer: When alkali metals dissolve in liquid ammonia, they give a deep blue solution. This blue color is due to the formation of ammoniated electrons. The alkali metal atoms readily lose their valence electrons, and these electrons, along with the metal ions, become solvated (surrounded) by ammonia molecules. These "ammoniated electrons" absorb energy from the visible region of light. When white light passes through the solution, the absorbed wavelengths cause the remaining transmitted light to appear blue.
The reactions are:
\( \text{M} (\text{s}) + (\text{x} + \text{y})\text{NH}_3 (\text{l}) \rightarrow [\text{M}(\text{NH}_3)_\text{x}]^+ (\text{ammoniated ion}) + [\text{e}(\text{NH}_3)_\text{y}]^- (\text{ammoniated electron}) \)
In simple words: Alkali metals turn liquid ammonia blue because their electrons get surrounded by ammonia molecules. These "ammoniated electrons" soak up certain light colors, making the solution look blue.

🎯 Exam Tip: The unique blue color of alkali metal solutions in liquid ammonia is a result of ammoniated electrons, which are responsible for both the color and the reducing properties of these solutions.

Question 26. H2 molecule exists but He2 molecule does not. Why?
Answer: The existence or non-existence of a molecule can be explained by its bond order, which is calculated using molecular orbital theory.

For \( \text{H}_2 \) molecule:
Each H atom has 1 electron, so \( \text{H}_2 \) has 2 electrons.
Electronic configuration of H: \( 1\text{s}^1 \).
Molecular orbital configuration of \( \text{H}_2 \): \( (\sigma 1\text{s})^2 \)
Number of electrons in bonding molecular orbitals \( (\text{n}_\text{b}) = 2 \)
Number of electrons in antibonding molecular orbitals \( (\text{n}_\text{a}) = 0 \)
Bond order \( = \frac{\text{n}_\text{b} - \text{n}_\text{a}}{2} = \frac{2 - 0}{2} = 1 \)
Since the bond order is 1, \( \text{H}_2 \) exists and is a stable molecule.

For \( \text{He}_2 \) molecule:
Each He atom has 2 electrons, so \( \text{He}_2 \) would have 4 electrons.
Electronic configuration of He: \( 1\text{s}^2 \).
Molecular orbital configuration of \( \text{He}_2 \): \( (\sigma 1\text{s})^2 (\sigma^* 1\text{s})^2 \)
Number of electrons in bonding molecular orbitals \( (\text{n}_\text{b}) = 2 \)
Number of electrons in antibonding molecular orbitals \( (\text{n}_\text{a}) = 2 \)
Bond order \( = \frac{\text{n}_\text{b} - \text{n}_\text{a}}{2} = \frac{2 - 2}{2} = 0 \)
Since the bond order is 0, \( \text{He}_2 \) does not exist as a stable molecule.
In simple words: The \( \text{H}_2 \) molecule exists because it has a bond order of 1, meaning its atoms are strongly linked. But the \( \text{He}_2 \) molecule does not exist because it has a bond order of 0, meaning there's no net bond holding the helium atoms together.

🎯 Exam Tip: Always use molecular orbital theory and bond order calculations to determine the stability and existence of diatomic molecules. A bond order of zero implies the molecule is unstable and does not exist.

Question 28. Dehydrated calcium chloride is used as a dehydration agent. Why?
Answer: Dehydrated calcium chloride (\( \text{CaCl}_2 \)) is an effective dehydration (drying) agent because it has a very high solvation energy for the calcium ion. This means that \( \text{Ca}^{2+} \) ions strongly attract water molecules and can form stable hydrates (compounds that contain water molecules). This strong tendency to absorb and bind water molecules makes it useful for removing moisture from gases and organic liquids.
When \( \text{CaCl}_2 \) reacts with water, it typically forms hydrated forms like \( \text{CaCl}_2 \cdot 2\text{H}_2\text{O} \), \( \text{CaCl}_2 \cdot 4\text{H}_2\text{O} \), or \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \). The general reaction can be shown as:
\( \text{CaCl}_2 + \text{nH}_2\text{O} \rightarrow \text{CaCl}_2 \cdot \text{nH}_2\text{O} \)
In simple words: Calcium chloride is good at drying things because its calcium ions strongly pull in and hold onto water molecules, forming special water-containing compounds.

🎯 Exam Tip: Remember that compounds with high solvation energy or strong tendencies to form stable hydrates are often effective drying agents due to their ability to absorb water from their surroundings.

Question 29. LiF is almost insoluble in water where as Licl is soluble only in water but also in acetone. Explain the reason.
Answer: Lithium fluoride (\( \text{LiF} \)) is almost insoluble in water, while lithium chloride (\( \text{LiCl} \)) is soluble in both water and acetone. This difference is mainly due to the balance between lattice energy and hydration energy.

\( \text{LiF} \) has an exceptionally high lattice energy. This is because both lithium ion (\( \text{Li}^+ \)) and fluoride ion (\( \text{F}^- \)) are very small, leading to very strong electrostatic attraction between them. The hydration energy released when \( \text{LiF} \) dissolves in water is not enough to overcome this very high lattice energy, so it remains largely insoluble.

On the other hand, \( \text{LiCl} \) has a lower lattice energy compared to \( \text{LiF} \). The chloride ion (\( \text{Cl}^- \)) is larger than \( \text{F}^- \), which reduces the electrostatic attraction in the crystal lattice. The hydration energy released when \( \text{LiCl} \) dissolves in water is sufficient to overcome its lattice energy, making it soluble. Additionally, \( \text{LiCl} \) exhibits significant covalent character due to the high polarizing power of the small \( \text{Li}^+ \) ion. This covalent nature allows it to dissolve in organic solvents like acetone, which are non-polar or weakly polar, by forming strong ion-dipole interactions.
In simple words: Lithium fluoride (LiF) doesn't dissolve well in water because its ions stick together too strongly. Lithium chloride (LiCl) dissolves well in water because its ions don't stick as strongly and it can also dissolve in acetone because it has some covalent properties.

🎯 Exam Tip: Solubility is a balance between lattice energy and solvation energy. For compounds with small cations, covalent character can also play a role in solubility in organic solvents.

Question 30. The solution of sodium carbonate is alkaline. Why?
Answer: An aqueous solution of sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) is alkaline (basic) in nature because of hydrolysis. Sodium carbonate is a salt formed from a strong base (\( \text{NaOH} \)) and a weak acid (\( \text{H}_2\text{CO}_3 \)). When \( \text{Na}_2\text{CO}_3 \) dissolves in water, the carbonate ion (\( \text{CO}_3^{2-} \)) reacts with water to produce bicarbonate ions (\( \text{HCO}_3^- \)) and hydroxide ions (\( \text{OH}^- \)). The production of hydroxide ions makes the solution alkaline.
The hydrolysis reaction is:
\( \text{CO}_3^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \rightleftharpoons \text{HCO}_3^- (\text{aq}) + \text{OH}^- (\text{aq}) \)
The sodium ions (\( \text{Na}^+ \)) do not hydrolyze as they are from a strong base.
In simple words: Sodium carbonate solution is basic because when it dissolves in water, the carbonate part reacts with water to make hydroxide ions, which makes the solution alkaline.

🎯 Exam Tip: Remember that salts formed from a strong base and a weak acid undergo anionic hydrolysis, releasing hydroxide ions and making the solution alkaline.

Question 31. Write the uses of following-
(a) Lime stone
(b) Uses of Sodium Carbonate
Answer: Here are the uses for limestone and sodium carbonate:

(a) Uses of Limestone (\( \text{CaCO}_3 \)):
• It is a crucial raw material in the manufacture of cement, where it is heated with clay.
• It is used in the production of quicklime (\( \text{CaO} \)) and slaked lime (\( \text{Ca(OH)}_2 \)).
• It acts as a flux in metallurgy (e.g., in the extraction of iron) to remove impurities.
• It is used as an antacid to neutralize stomach acid.
• It is used as an abrasive, for example, in toothpaste.
• It is utilized in agriculture to reduce soil acidity.

(b) Uses of Sodium Carbonate (\( \text{Na}_2\text{CO}_3 \)):
• It is a primary ingredient in the manufacture of glass, soap, and borax.
• It is used as a washing soda for laundry and cleaning purposes.
• It is employed in the paper, paint, and textile industries.
• It serves as an important laboratory reagent for both qualitative and quantitative analysis.
• It is used in water softening to remove permanent hardness.
In simple words: Limestone is used to make cement, lime, in farming, and as an antacid. Sodium carbonate is used to make glass, soap, and borax, for washing, and to soften water.

🎯 Exam Tip: When listing uses, aim for 3-5 distinct applications for each substance, covering different industries or household functions.

Question 32. Explain the following :
BeO is almost insoluble but BeSO4 is soluble in water.
Answer: Beryllium oxide (\( \text{BeO} \)) is almost insoluble in water, whereas beryllium sulfate (\( \text{BeSO}_4 \)) is soluble. This difference in solubility can be explained by considering their lattice energies and hydration energies.

\( \text{BeO} \) has a very high lattice energy due to the small size and high charge of both \( \text{Be}^{2+} \) and \( \text{O}^{2-} \) ions. The strong electrostatic forces hold the ions tightly in the crystal lattice. The hydration energy of the ions when placed in water is not enough to overcome this very high lattice energy, making \( \text{BeO} \) insoluble.

In contrast, \( \text{BeSO}_4 \) is soluble in water. Although it also has a relatively high lattice energy, the hydration energy of the \( \text{Be}^{2+} \) ion is exceptionally high. The small size of the \( \text{Be}^{2+} \) ion allows it to be extensively hydrated by water molecules, releasing a large amount of hydration energy. This high hydration energy of \( \text{Be}^{2+} \) effectively overcomes the lattice energy of \( \text{BeSO}_4 \), allowing it to dissolve in water.
In simple words: Beryllium oxide doesn't dissolve much in water because its ions are packed very tightly with strong bonds. Beryllium sulfate dissolves in water because, even with strong bonds, the small beryllium ion gets heavily surrounded by water molecules, releasing enough energy to break the bonds and dissolve.

🎯 Exam Tip: Remember that for small ions like Be2+, hydration energy can be exceptionally high, and it's the balance between lattice energy and hydration energy that dictates solubility.

Question 33. Write balanced equation for relations between
(a) Be2 C and water
(b) KO2 and water
(c) Lithium and nitrogen
Answer: Here are the balanced chemical equations for the given reactions:

(a) Reaction between \( \text{Be}_2\text{C} \) and water:
When beryllium carbide (\( \text{Be}_2\text{C} \)) reacts with water, it produces beryllium hydroxide (\( \text{Be(OH)}_2 \)) and methane gas (\( \text{CH}_4 \)).
\( \text{Be}_2\text{C} (\text{s}) + 4\text{H}_2\text{O} (\text{l}) \rightarrow 2\text{Be(OH)}_2 (\text{s}) + \text{CH}_4 (\text{g}) \)

(b) Reaction between \( \text{KO}_2 \) and water:
Potassium superoxide (\( \text{KO}_2 \)) reacts vigorously with water to produce potassium hydroxide (\( \text{KOH} \)), oxygen gas (\( \text{O}_2 \)), and hydrogen peroxide (\( \text{H}_2\text{O}_2 \)).
\( 4\text{KO}_2 (\text{s}) + 2\text{H}_2\text{O} (\text{l}) \rightarrow 4\text{KOH} (\text{aq}) + 3\text{O}_2 (\text{g}) \)

(c) Reaction between lithium and nitrogen:
Lithium is the only alkali metal that directly reacts with nitrogen gas at room temperature to form lithium nitride (\( \text{Li}_3\text{N} \)).
\( 6\text{Li} (\text{s}) + \text{N}_2 (\text{g}) \rightarrow 2\text{Li}_3\text{N} (\text{s}) \)
In simple words: When beryllium carbide mixes with water, it makes beryllium hydroxide and methane. Potassium superoxide with water makes potassium hydroxide, oxygen, and hydrogen peroxide. Lithium and nitrogen react to form lithium nitride.

🎯 Exam Tip: Pay close attention to the products of reactions involving metal carbides with water (producing methane or acetylene), superoxides with water (producing O2), and the unique reaction of lithium with nitrogen.

Rbse Class 11 Chemistry Chapter 10 Long Answer Type Questions

Question 35. Lithium does not show similarity with other elements of the group. Explain the reasons.
Answer: Lithium (Li) is the first element of Group 1 (alkali metals) and shows several properties that are different from the other elements in its group. This is often referred to as the "anomalous behavior of lithium," and it's primarily due to the following reasons:

• **Exceptionally Small Size:** Lithium has an exceptionally small atomic and ionic size compared to other alkali metals. This small size leads to a very high charge density.
• **High Polarizing Power:** Due to its small size and high charge, \( \text{Li}^+ \) has a very high polarizing power. This means it can distort the electron clouds of neighboring anions to a significant extent, leading to greater covalent character in its compounds (e.g., \( \text{LiCl} \) is more covalent than \( \text{NaCl} \)).
• **High Ionization Enthalpy:** Lithium has the highest ionization enthalpy among the alkali metals, meaning it requires more energy to remove an electron compared to other group members.
• **Absence of d-orbitals:** Lithium lacks d-orbitals in its valence shell, which limits its covalency to a maximum of four. Other elements in the group have vacant d-orbitals, which can be used for bond formation.

These factors contribute to differences such as lithium reacting less vigorously with water than other alkali metals, forming stable covalent compounds, and exhibiting a diagonal relationship with magnesium.
In simple words: Lithium acts differently from other alkali metals because it is very small, can pull electron clouds strongly, needs more energy to lose an electron, and doesn't have d-orbitals.

🎯 Exam Tip: When discussing anomalous behavior, always link it back to the small size, high polarizing power, and absence of d-orbitals of the first element in a group.

Question 36. Write the properties of hydrides of Group I and Group II elements and draw explain the structure of Beryllium hydride.
Answer: Let's discuss the properties of hydrides of Group 1 and Group 2 elements, and then the structure of beryllium hydride.

**Properties of Hydrides of Group 1 (Alkali Metals):**
• Alkali metal hydrides (\( \text{MH} \)) are typically ionic solids with high melting points. The ionic character increases down the group from \( \text{LiH} \) to \( \text{CsH} \).
• They are powerful reducing agents. Their reducing character increases as you move down the group, meaning \( \text{CsH} \) is a stronger reducing agent than \( \text{LiH} \).
• As atomic size increases down the group, the M-H bond strength in these hydrides decreases, making them less stable thermally.
• They react with water to produce hydrogen gas and the corresponding metal hydroxide.

**Properties of Hydrides of Group 2 (Alkaline Earth Metals):**
• Beryllium hydride (\( \text{BeH}_2 \)) is covalent and polymeric. Magnesium hydride (\( \text{MgH}_2 \)) has some covalent character. The other alkaline earth metal hydrides (\( \text{CaH}_2 \), \( \text{SrH}_2 \), \( \text{BaH}_2 \)) are ionic and crystalline solids, known as hydroliths.
• Ionic alkaline earth hydrides are strong reducing agents.
• They react with water to liberate hydrogen gas, making them useful as sources of hydrogen and as reducing agents.

**Structure of Beryllium Hydride (\( \text{BeH}_2 \)):**
Beryllium hydride (\( \text{BeH}_2 \)) is an electron-deficient covalent compound that exists as a polymeric structure. In this polymer, each beryllium atom is connected to two hydrogen atoms, and each hydrogen atom also bridges between two beryllium atoms. This forms a chain-like structure with three-center two-electron bonds. Each beryllium atom effectively achieves a stable octet (or close to it) through these bridging bonds, where a single electron pair is shared among three atoms (Be-H-Be).

In simple words: Hydrides of alkali metals are usually ionic, have high melting points, and are strong reducing agents. Hydrides of alkaline earth metals can be ionic or covalent; beryllium hydride is covalent and forms long chains by sharing hydrogen atoms between beryllium atoms.

🎯 Exam Tip: Differentiate between the ionic and covalent nature of hydrides across Group 1 and Group 2, particularly highlighting beryllium's unique polymeric structure due to its small size and tendency to form covalent bonds.

Question 37. Explain the importance of Na, K, Mg and Ca in biological fluids.
Answer: Sodium (Na), Potassium (K), Magnesium (Mg), and Calcium (Ca) are essential elements that play vital roles in the biological fluids and overall functioning of the human body.

• **Sodium (\( \text{Na}^+ \)):** Sodium ions are primarily found in the extracellular fluid (outside cells), including blood plasma and interstitial fluid. They are crucial for maintaining water balance, regulating blood pressure, transmitting nerve impulses, and muscle contraction. An adult body typically contains about 90 g of sodium.
• **Potassium (\( \text{K}^+ \)):** Potassium ions are the most abundant cations within the intracellular fluid (inside cells). Along with sodium, potassium is vital for nerve signal transmission, muscle contraction, and maintaining the cell membrane's potential. It also helps regulate water flow across cell membranes. An adult body typically contains about 170 g of potassium.
• **Magnesium (\( \text{Mg}^{2+} \)):** Magnesium ions are important cofactors for many enzymatic reactions. They are essential for all enzymes that use ATP in phosphate transfer reactions, crucial for energy production and metabolism. Magnesium is also involved in muscle and nerve function, blood glucose control, and protein synthesis. The main pigment for light absorption in plants, chlorophyll, contains magnesium. An adult body contains about 25 g of magnesium.
• **Calcium (\( \text{Ca}^{2+} \)):** Calcium ions are the most abundant mineral in the body, with about 99% present in bones and teeth, providing structural support. In biological fluids, calcium plays critical roles in blood clotting, neuromuscular function, inter-neuronal transmission, cell membrane integrity, and hormone secretion. An adult body contains about 1200 g of calcium.
In simple words: Sodium helps with water balance and nerve signals outside cells. Potassium helps with nerve signals and muscle function inside cells. Magnesium is key for energy reactions in the body. Calcium builds strong bones and teeth, and helps with blood clotting and nerve function.

🎯 Exam Tip: For each ion, remember its primary location (intra- or extracellular) and at least 2-3 key biological functions to ensure a comprehensive answer.

Question 39. Explain the chemical properties of alkali metals and explain inspite of highest ionisation enthalpy energy in the group, Lithium is a strong reducing agent why ?
Answer: Let's explore the chemical properties of alkali metals and then explain lithium's reducing power.

**Chemical Properties of Alkali Metals:**
Alkali metals (Group 1 elements) are highly reactive due to their large atomic size and low ionization enthalpies, which make them readily lose their single valence electron.

1. **Reaction with Oxygen:** They react readily with oxygen to form different types of oxides, causing them to tarnish in dry air.
• Lithium primarily forms a normal oxide: \( 4\text{Li} (\text{s}) + \text{O}_2 (\text{g}) \rightarrow 2\text{Li}_2\text{O} (\text{s}) \) (Lithium oxide)
• Sodium mainly forms a peroxide: \( 2\text{Na} (\text{s}) + \text{O}_2 (\text{g}) \rightarrow \text{Na}_2\text{O}_2 (\text{s}) \) (Sodium peroxide)
• Potassium, rubidium, and cesium form superoxides: \( \text{K} (\text{s}) + \text{O}_2 (\text{g}) \rightarrow \text{KO}_2 (\text{s}) \) (Potassium superoxide)

2. **Reaction with Water:** They react vigorously with water to form hydroxides and hydrogen gas. The reactivity increases down the group.
\( 2\text{M} (\text{s}) + 2\text{H}_2\text{O} (\text{l}) \rightarrow 2\text{MOH} (\text{aq}) + \text{H}_2 (\text{g}) \)

3. **Reaction with Dihydrogen (Hydrogen Gas):** They react with dihydrogen at high temperatures to form metal hydrides.
\( 2\text{M} (\text{s}) + \text{H}_2 (\text{g}) \xrightarrow{\text{Heat}} 2\text{MH} (\text{s}) \)
(Example: \( 2\text{Na} (\text{s}) + \text{H}_2 (\text{g}) \rightarrow 2\text{NaH} (\text{s}) \))

4. **Reaction with Halogens:** They react readily and vigorously with halogens to form ionic halides.
\( 2\text{M} (\text{s}) + \text{X}_2 (\text{g}) \rightarrow 2\text{MX} (\text{s}) \) (Where \( \text{X} \) is a halogen)

5. **Reaction with Liquid Ammonia:** Alkali metals dissolve in liquid ammonia to give deep blue solutions, forming ammoniated ions and ammoniated electrons, as explained in Question 25.
\( \text{M} (\text{s}) + (\text{x} + \text{y})\text{NH}_3 (\text{l}) \rightarrow [\text{M}(\text{NH}_3)_\text{x}]^+ + [\text{e}(\text{NH}_3)_\text{y}]^- \)

**Lithium as a Strong Reducing Agent:**
Despite having the highest ionization enthalpy among the alkali metals, lithium is the strongest reducing agent in the aqueous solution. This apparent contradiction is explained by the following reasons:
• **Maximum Hydration Energy:** Lithium has an exceptionally small ionic size (\( \text{Li}^+ \)). Because of this, it is highly hydrated in aqueous solutions, meaning many water molecules surround it very strongly. The large amount of energy released during this hydration process (hydration enthalpy) is very high for lithium. This high hydration energy more than compensates for its high ionization enthalpy.
• **Small Size of Lithium Ion:** The small size of the \( \text{Li}^+ \) ion leads to the highest charge density, which facilitates stronger interaction with polar water molecules.

Therefore, the overall energy change for the conversion of \( \text{Li} (\text{s}) \rightarrow \text{Li}^+ (\text{aq}) + \text{e}^- \) is highly negative (favorable) due to the dominant effect of very high hydration energy, making lithium an excellent reducing agent in aqueous solutions.
In simple words: Alkali metals react easily with oxygen, water, hydrogen, and halogens. Lithium is a strong reducing agent even though it's harder to remove its electron than other alkali metals. This is because its tiny ion gets very strongly surrounded by water molecules, releasing a lot of energy that makes it eager to lose an electron.

🎯 Exam Tip: For chemical properties, include typical reactions and any group trends (e.g., reactivity increasing down the group). For lithium's reducing power, emphasize the role of its exceptionally high hydration energy overriding its high ionization enthalpy.