NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 1)

Question:1

Hydrogen resembles halogens in many respects, for which several factors are responsible. Of the following factors, which one is most important in this respect?
(i) Its tendency to lose an electron to form a cation.
(ii) Its tendency to gain a single electron in its valence shell to attain a stable electronic configuration.
(iii) Its low negative electron gain enthalpy value.
(iv) Its small size.

Answer:

The answer is the option (ii) Its tendency to gain a single electron in its valence shell to attain a stable electronic configuration.
Hydrogen resembles halogens as halogens with the configuration of $n{s}^{2}n{p}^5$in the seventeenth group, also harbouring the ability to accept one electron and configure as an inert gas. Similarly, Hydrogen can also accept an electron to configure itself like helium.

$H\to He$
$1s^1$ $1s^2$

Question:2

Why does $H^{+}$ ion always get associated with other atoms or molecules?
(i) The ionisation enthalpy of Hydrogen resembles that of alkali metals.
(ii) Its reactivity is similar to halogens.
(iii) It resembles both alkali metals and halogens.
(iv) Loss of an electron from a hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to its small size, it cannot exist freely.
Answer:

The answer is the option (iv) Loss of an electron from a hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to its small size, it cannot exist freely.
A positive hydrogen ion is extremely small in size and cannot exist as a single atom but instead only exists as an association with other elements.

$H\to H^{+}+e^{-}$

Question:3

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is
$(i)LiH>NaH>CsH>KH>RbH$
$(ii)LiH<NaH<KH<RbH<CsH$
$(iii)RbH>CsH>NaH>KH>LiH$
$(iv)NaH>CsH>RbH>LiH>KH$
Answer:

The answer is the option $(ii)LiH<NaH<KH<RbH<CsH$
This happens due to the reason that as the atomic size increases, the ionic, electropositive character of the metal hybrides also increases.

Question:4

Which of the following hydrides is an electron-precise hydride?
$(i)B_2{H}_6$
$(ii)N{H}_3$
$(iii)H_{2}O$
$(iv)C{H}_4$
Answer:

The answer is the option $(iv)C{H}_4$
$C{H}_4$ has the precise number of electrons to form normal covalent bonds and is a precise electron donor, in order to figure out their own Lewis structures.

Question:5

Radioactive elements emit $\alpha ,\beta and\gamma$ rays and are characterised by their half-lives. The radioactive isotope of Hydrogen is
(i) Protium
(ii) Deuterium
(iii) Tritium
(iv) Hydronium
Answer:

The answer is the option (iii) Tritium
Tritium is radioactive as the neutron-proton ratio is more than 1.5, and tritium has n=2 and p=1.

Question:6

Consider the reactions

(A) $H_2{O}_2+2HI\to I_2+2H_{2}O$ (B) $HOCl+H_2{O}_2\to H_3{O}^{+}+C{l}^{-}+O_2$
Which of the following statements is correct about $H_2{O}_2$ with reference to these reactions? Hydrogen peroxide is ________.
(i) an oxidising agent in both (A) and (B)
(ii) an oxidising agent in (A) and a reducing agent in (B)
(iii) a reducing agent in (A) and an oxidising agent in (B)
(iv) a reducing agent in both (A) and (B)
Answer:

The answer is the option (ii) an oxidizing agent in (A) and a reducing agent in (B)

1. $H_2{O}_2+2HI\to I_2+2H_{2}O$

2. $HOCl+H_2{O}_2\to H_3{O}^{+}+C{l}^{-}+O_2$

Hydrogen peroxide in (A) is an oxidising agent as the oxygen in equation (a) gets reduced from $H_2{O}_{2}to{H}_{2}O$

$H_2{O}_2+2HI\to I_2+2H_{2}O$
+1 -1 +1 -1 0 +1 -2
Hydrogen peroxide in (B) is a reducing agent as oxygen increases from $H_2{O}_{2}to{O}_2$

Question:7

The oxide that gives $H_2{O}_2$ on treatment with dilute $H_{2}S{O}_4$ is —
$(i)Pb{O}_2$
$(ii)Ba{O}_2.8H_{2}O+O_2$
$(iii)Mn{O}_2$
$(iv)Ti{O}_2$
Answer:

The answer is the option $(ii)Ba{O}_2.8H_{2}O+O_2$
This oxide has peroxide linkage $(O_2^{2-})$ when reacted with dilute $H_{2}S{O}_4$, it produces $H_{2}O$, however, dioxides do not produce the same products and the metal atom doesn’t give out water on treatment with dilute $H_{2}S{O}_4$.

Question:8

Which of the following equations depicts the oxidising nature of $H_2{O}_2$?

$(i)2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$
$(ii)2F{e}^{3+}+2H^{+}+H_2{O}_2\to 2F{e}^{2+}+2H_{2}O+O_2$
$(iii)2I^{-}+2H^{+}+H_2{O}_2\to I_2+2H_{2}O$
$(iv)KI{O}_4+H_2{O}_2\to KI{O}_3+H_{2}O+O_2$

Answer:

The answer is the option $(iii)2I^{-}+2H^{+}+H_2{O}_2\to I_2+2H_{2}O$
Iodine ions have a negative charge and are oxidised to form $I_2$. Therefore, $H_2{O}_2$ is the oxidizing agent.

Question:9

Which of the following equation depicts reducing nature of $H_2{O}_2$ ?
$(i)2{[Fe(CN{)}_6]}^{4-}+2H^{+}+H_2{O}_2\to 2{\left[Fe{\left(CN\right)}_6\right]}^{3-}+2H_{2}O$
$(ii)I_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2$
$(iii)M{n}^{2+}+H_2{O}_2\to M{n}^{4+}+2O{H}^{-}$
$(iv)PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$

Answer:

The answer is the option $(ii)I_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2$
Iodine in the form of $I_2$ is reduced to $I^{-}$ and thus $H_2{O}_2$ acts as the reducing agent.

Question:10

Hydrogen peroxide is _________.
(i) an oxidising agent
(ii) a reducing agent
(iii) both an oxidising and a reducing agent
(iv) neither oxidising nor reducing agent
Answer:

The answer is the option (iii) both an oxidizing and a reducing agent
We have seen examples of $H_2{O}_2$ acting as both the reducing and oxidising agent.

Question:11

Which of the following reactions increases the production of dihydrogen from synthesis gas?
$(i)C{H}_4(g)+H_{2}O(g)\underset{Ni}{\overset{1270K}{\to }}CO(g)+3H_2(g)$
$(ii)C(s)+H_{2}O(g)\stackrel{1270K}{\to }CO(g)+H_2(g)$
$(iii)CO(g)+H_{2}O(g)\underset{catalyst}{\overset{673K}{\to }}C{O}_2(g)+H_2(g)$
$(iv)C_2{H}_6+2H_{2}O\underset{Ni}{\overset{1270K}{\to }}2CO+5H_2$
Answer:

The answer is the option $(iii)CO(g)+H_{2}O(g)\underset{catalyst}{\overset{673K}{\to }}C{O}_2(g)+H_2(g)$
Carbon monoxide can be reacted with syngas mixtures using steam and iron chromate as enhancers to produce extra dihydrogen.

Question:12

When sodium peroxide is treated with dilute sulphuric acid, we get ______.
(i) sodium sulphate and water
(ii) sodium sulphate and oxygen
(iii) sodium sulphate, Hydrogen and oxygen
(iv) sodium sulphate and hydrogen peroxide
Answer:

The answer is the option (iv), sodium sulphate and hydrogen peroxide.
$N{a}_2{O}_2+H_{2}S{O}_4\to N{a}_{2}S{O}_4+H_2{O}_2$

Question:13

Hydrogen peroxide is obtained by the electrolysis of ______.
(i) water
(ii) sulphuric acid
(iii) hydrochloric acid
(iv) fused sodium peroxide
Answer:

The answer is the option (ii), sulphuric acid
$H_{2}S{O}_4\to H^{+}+HS{O}^{-4}$
$2HSO-4\to onelectrolysisH{O}_{3}SOOS{O}_3�H+2e^{-}$
$H{O}_{3}SOOS{O}_3�H+2H_{2}O(hydrolysis)\to 2H_{2}S{O}_4+H_2{O}_2$

Question:14

Which of the following reactions is an example of the use of water gas in the synthesis of other compounds?

$(i)C{H}_4(g)+H_{2}O(g)\underset{Ni}{\overset{1270K}{\to }}CO(g)+H_2(g)$
$(ii)CO(g)+H_{2}O(g)\underset{catalyst}{\overset{673K}{\to }}C{O}_2(g)+H_2(g)$
$(iii)C_n{H}_{2n+n}+n{H}_{2}O(g)\underset{Ni}{\overset{1270K}{\to }}nCO+(2n+1)H_2$
$(iv)CO(g)+2H_2(g)\underset{catalyst}{\overset{cobalt}{\to }}C{H}_{3}OH(l)$
Answer:

The answer is the option $(iv)CO(g)+2H_2(g)\underset{catalyst}{\overset{cobalt}{\to }}C{H}_{3}OH(l)$
Methanol used water and gas for the synthesis of other compounds.

Question:15

Which of the following ions will cause hardness in the water sample?
$(i)C{a}^{2+}$
$(ii)N{a}^{+}$
$(iii)C{l}^{-}$
$(iv)K^{+}$
Answer:

The answer is the option $(i)C{a}^{2+}$
These ions, often in the common forms of $Ca(HC{O}_3)$ or $CaC{l}_2$ will have an effect on the water by making it harsh and hard as the salts of calcium are soluble in water, making it hard.

Question:16

Which of the following compounds is used for water softening?
$(i)C{a}_3(P{O}_4{)}_2$
$(ii)N{a}_{3}P{O}_4$
$(iii)N{a}_6{P}_6{O}_{18}$
$(iv)N{a}_{2}HP{O}_4$
Answer:

The answer is the option $(iii)N{a}_6{P}_6{O}_{18}$
Commercially known as Calgon, sodium hexametaphosphate is used to treat water and make it soft.

$2CaC{l}_2+N{a}_2[N{a}_4(P{0}_3{)}_6]\to N{a}_2[C{a}_2(P{0}_3{)}_6]+4NaCl$

Question:17

Elements of which of the following group(s) of periodic table do not form hydrides.
(i) Groups 7, 8, 9
(ii) Group 13
(iii) Groups 15, 16, 17
(iv) Group 14
Answer:

The answer is the option (i) Group 7, 8, 9
These elements are unable to form hydrides.

Question:18

Only one element of ________ forms hydride.
(i) group 6
(ii) group 7
(iii) group 8
(iv) group 9

Answer:

The answer is the option (i) Group 6
Only Chromium (Cr) is the element in group 6, capable of forming a hydride.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 2)

Question:19

Which of the following statements are not true for Hydrogen?
(i) It exists as a diatomic molecule.
(ii) It has one electron in the outermost shell.
(iii) It can lose an electron to form a cation, which can freely exist
(iv) It forms a large number of ionic compounds by losing an electron.
Answer:

The answer is the option (iii) & (iv), i.e. it can lose an electron to form a cation which can freely exist, and it forms a large number of ionic compounds by losing an electron.
Hydrogen cannot form ionic compounds by giving an electron but instead forms many covalent bonds by the sharing of electrons.

Question:20

Dihydrogen can be prepared on a commercial scale by different methods. In its preparation by the action of steam on hydrocarbons, a mixture of CO and H₂ gas is formed. It is known as ____________.
(i) Water gas
(ii) Syngas
(iii) Producer gas
(iv) Industrial gas
Answer:

The answer is the option (i) and (ii), i.e. water gas and syngas.
Synthesis gas is a combination of CO and H₂, also known as water gas.

Question:21

Which of the following statement(s) is/are correct in the case of heavy water?
(i) Heavy water is used as a moderator in a nuclear reactor.
(ii) Heavy water is more effective as a solvent than ordinary water.
(iii) Heavy water is more associated than ordinary water.
(iv) Heavy water has a lower boiling point than ordinary water.
Answer:

The answer is the option (i) and (iii), i.e. Heavy water is used as a moderator in a nuclear reactor, and Heavy water is more associated than ordinary water.
It is associated with water as it has a higher mass as well as facilitates as a moderator in exchange reactions.

Question:22

Which of the following statements about Hydrogen are correct?
(i) Hydrogen has three isotopes, of which protium is the most common.
(ii) Hydrogen never acts as a cation in ionic salts.
(iii) Hydrogen ion, $H^{+}$, exists freely in solution.
(iv) Dihydrogen does not act as a reducing agent.
Answer:

The answer is the option (i) and (ii), i.e. Hydrogen has three isotopes of which protium is the most common, and Hydrogen never acts as a cation in ionic salts.
Protium, one of Hydrogen’s 3 isotopes, is the most common as well and due to its small atomic size, it does not act as a cation but is associated with other molecules and compounds.

Question:23

Some of the properties of water are described below. Which of them is/are not correct?
(i) Water is known to be a universal solvent.
(ii) Hydrogen bonding is present to a large extent in liquid water.
(iii) There is no hydrogen bonding in the frozen state of water.
(iv) Frozen water is heavier than liquid water.
Answer:

The answer is the option (iii) and (iv). There is no hydrogen bonding in the frozen state of water, and Frozen water is heavier than liquid water.
Water exhibits different properties in different states due to hydrogen bonding, which exists in a water molecule. Ice is lighter than water due to the fact that there are empty spaces in the tetrahedra of the hydrogen bonds.

Question:24

Hardness of water may be temporary or permanent. Permanent hardness is due to the presence of
(i) Chlorides of Ca and Mg in water
(ii) Sulphates of Ca and Mg in water
(iii) Hydrogen carbonates of Ca and Mg in water
(iv) Carbonates of alkali metals in water
Answer:

The answer is the option (i) and (ii): Chlorides of Ca and Mg in water and sulphate of Ca and Mg in water.
Two salts of calcium and magnesium, when found as compounds of carbonate, chloride, and sulfate, dissolve in the water and make it hard.

Question:25

Which of the following statements is correct?
(i) Elements of group 15 form electron-deficient hydrides.
(ii) All elements of group 14 form electron-precise hydrides.
(iii) Electron-precise hydrides have tetrahedral geometries.
(iv) Electron-rich hydrides can act as Lewis acids.
Answer:

The answer is the option (ii) and (iii). All elements of group 14 form electron-pair hydrides, and electron-pair hydrides have tetrahedral geometries.
All group 14 elements are tetrahedral in terms of geometry and form their own Lewis structures. They are electron-precise hydrides that have enough electrons.

Question:26

Which of the following statements is correct?
(i) Hydrides of group 13 act as Lewis acids.
(ii) Hydrides of group 14 are electron deficient hydrides.
(iii) Hydrides of group 14 act as Lewis acids.
(iv) Hydrides of group 15 act as Lewis bases.
Answer:

The answer is the option (i) and (iv), i.e. Hydrides of group 13 act as Lewis acids and Hydrides of group 15 act as Lewis bases.
All the elements of group 13 are hydrides, which act as Lewis acids since they form electron-deficient compounds. The group 14 elements are known as electron-rich hydrides and thus have extra electrons, which are present in the form of lone pairs. Groups 15 to 17 have elements that form compounds that have 1 to 3 lone pairs and therefore act as Lewis bases.

Question:27

learn.careers360.com/ncert/question-which-of-the-following-statements-is-correct-i-metallic-hydrides-are-deficient-of-hydrogen/

(i) Metallic hydrides are deficient of Hydrogen.
(ii) Metallic hydrides conduct heat and electricity.
(iii) Ionic hydrides do not conduct electricity in solid state.
(iv) Ionic hydrides are very good conductors of electricity in solid state.

Answer:

The answer is the option (i), (ii), and (iii). Metallic hydrides are deficient of Hydrogen, Metallic hydrides conduct heat and electricity, and Ionic hydrides do not conduct electricity in the solid state.
Hydride and not volatile or conductive in the solid state, but are crystalline instead. While metallic hydrides are non-stoichiometric hydrides. They only conduct electricity in their molten state.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Short Answer Type

Question:28

How can the production of Hydrogen from water gas be increased by using a water gas shift reaction?
Answer:

The production of water gas is a chemical reaction. It is produced by superheated steam is passed over coal, nickel acts as the catalyst. The by-products are carbon dioxide and Hydrogen.
$CO+H_{2}O+H_2\underset{catalyst}{\overset{673K}{\to }}C{O}_2+2H_2$

Question:29

What are metallic/interstitial hydrides? How do they differ from molecular hydrides?
Answer:

Metallic hydrides are usually formed by d and f block elements. These hydrides are good conductors of heat and electricity; they lack Hydrogen, which makes them non-stoichiometric. These differ from molecular hydrides as they are formed by s, p block elements while the metals of group 7, 8, 9 do not form hydrides. Molecular hydrides are not good conductors of electricity or heat, unlike the metallic hydrides. They are also volatile compounds which have low melting and boiling points, but metallic hydrides, on the other hand, are hard in texture and have a certain metallic lustre.

Question:30

Name the classes of hydrides to which$H_{2}O$,$B_2{H}_6$ and $NaH$ belong.
Answer:

$H_{2}O$ – Is a molecular hydride which is covalent and electron-rich.
$B_2{H}_6$- Is a molecular hydride which is deficient in electrons.
$NaH$– Ionic hydride

Question:31

If the same mass of liquid water and a piece of ice is taken, then why is the density of ice less than that of liquid water?
Answer:

Ice and water have different structures and since water expands on freezing the volume of the same amount to ice is more than that of liquid water. Therefore, the density of water is much higher than that of ice and therefore, ice will float on water.

Question:32

Complete the following equations:
$(i)PbS(s)+H_2{O}_2(aq)\to$
$(ii)CO(g)+2H_2(g)\underset{catalyst}{\overset{cobalt}{\to }}$
Answer:

$(i)PbS(s)+H_2{O}_2(aq)\to PbS{O}_4+2H_{2}O$
$(ii)CO(g)+2H_2(g)\underset{catalyst}{\overset{cobalt}{\to }}C{H}_{3}OH$

Question:33

Give reasons:
(i) Lakes freeze from the top towards the bottom.
(ii) Ice floats on water.

Answer:

1. The lake freezes from top to bottom because the temperature during winter keeps on decreasing, and the movement of water happens in such a way that the cold water is heavier and so it sinks to the bottom. While warm water replaces it by coming on the surface. The process repeats until the temperature decreases below 4 degrees and the lake keeps on freezing from top to bottom.

2. The density of ice is less than water due to its structure, which forms empty spaces between the water molecules. 4 atoms of Hydrogen surround 1 of oxygen, and therefore make ice float on water

Question:34

What do you understand by the term ‘auto protolysis' of water? What is its significance?
Answer:

Autoprotolysis is the process when two similar molecules react with each other to produce products that are called ions with Proton transfer. Autoprotolysis of water means the transfer of one Proton from a certain molecule to another. This explains the ability of water act as both acid and base. Therefore, making it amphoteric in nature.
$H_{2}O(l)+H_{2}O(l)\stackrel{reversible}{\to }{H}_3{O}^{+}(aq)+O{H}^{-}(aq)$

Question:35

Discuss briefly de-mineralisation of water by ion exchange resin.
Answer:

Demineralization of water means that all the soluble salts present in water are removed through cation and anion exchange. In the cation exchange process, cations of sodium, calcium and magnesium replace those of Hydrogen. While the anion exchange process exchanges OH. These both combine to produce water.
$H^{+}+O{H}^{-}\to H_{2}O$

Question:36

Molecular hydrides are classified as electron-deficient, electron precise and electron-rich compounds. Explain each type with two examples.
Answer:

There are three major types of molecular hydrides.
Electron-precise hydrides are the ones that have just enough exact number of electrons to facilitate normal covalent bonds. These are the hydrides which are primarily comprised of group 14 elements: $C{H}_4$.
Electron-deficient hydrides are the ones which do not possess enough electrons to facilitate normal covalent bonds. Hydrides of group 13 a prime examples; $B_2{H}_6$.
Electron-rich hydrides are ones which have an excess number of electrons remaining after normal covalent bonds. Elements of group 15 16 17 are prime examples; $N{H}_3$.

Question:37

How is heavy water prepared? Compare its physical properties with those of ordinary water.
Answer:

Heavy water is prepared by the exhaustive electrolysis of water.

Question:38

Write one chemical reaction for the preparation of $D_2{O}_2$.
Answer:

$D_{2}S{O}_4$ , when reacted in water over $Ba{O}_2$ produces $D_2{O}_2$.
$Ba{O}_2+D_{2}S{O}_4\to BaS{O}_4+D_2{O}_2$

Question:39

Calculate the strength of 5 volume $H_2{O}_2$ solution.
Answer:

5 volume $H_2{O}_2$solution signifies that 1 L of 5 volume $H_2{O}_2$ when decomposed, gives 5L of $O_2$ at NTP.
$H_2{O}_2\to 2H_{2}O+O_2$

$H_2{O}_2$- 68g
$2H_{2}O+O_2$– 22.4l at NTP
22.4L of $O_2$ at NTP is gotten from $H_2{O}_2$ is 68g

5L of $O_2$ at NTP is produced from $H_2{O}_2$ = $\frac{68}{22.4}\times 5=15.18g$
But 5L of $O_2$ at NTP is from 5 volume of 1L $H_2{O}_2$

Strength of $H_2{O}_2$ in 5 volume $H_2{O}_2$ = 15.18L
% strength of $H_2{O}_2$ sol = $\frac{15.18}{1000}\times 100=1.518$%

Question:40

(i) Draw the gas phase and solid phase structure of$H_2{O}_2$.
(ii) $H_2{O}_2$ is a better oxidising agent than water. Explain.
Answer:

(i) The structure of differ$H_2{O}_2$s according to its physical state, as explained below: -

ii) $H_2{O}_2$ is a better oxidising agent than water as it works in both acidic and alkaline media. Water as an oxidising agent is reduced to $H_2$ and reacts with only active metals with other conditions such as the electrode potential being less than -0.83V.

Question:41

Melting point, enthalpy of vapourisation and viscosity data of $H_{2}O$ and $D_{2}O$are given below:

On the basis of this data, explain in which of these liquids intermolecular forces are stronger?

Answer:

The given properties, such as boiling point, enthalpy of vaporization, and viscosity,y are all dependent on the intermolecular forces of the liquids. Water has a lower intermolecular force of attraction as compared to $D_{2}O$ , and thus the values are lower for water.

Question:42

Dihydrogen reacts with dioxygen $(O_2)$ to form water. Write the name and formula of the product when the isotope of Hydrogen, which has one proton and one neutron in its nucleus is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen? Justify your answer.
Answer:

Deuterium (D) is the isotope of Hydrogen that contains one proton and one neutron. Thus, on the reaction of water with deuterium, heavy water (deuterium oxide) is produced. Deuterium’s bond is stronger than the normal hydrogen bond, and thus $H_2$ reacts more with deuterium than oxygen.

$2D_2+O_2\to (heat)2D_{2}O$

Question:43

Explain why HCl is a gas and HF is a liquid.
Answer:

F is a stronger electronegative than chlorine, and so forms a stronger bond with Hydrogen than chlorine. In order to break HF bonds, more energy is required as compared to break HCl bonds. This is the reason why HF has a higher boiling point than HCl and is a liquid at room temperature.

Question:44

When the first element of the periodic table is treated with dioxygen, it gives a compound whose solid state floats on its liquid state. This compound has an ability to act as an acid as well as a base. What products will be formed when this compound undergoes autoionisation?
Answer:

The first element is Hydrogen, and its molecular form is dihydrogen, so when it reacts with oxygen, it forms water, which is physical solid-state is ice and has a density lower than water, so it floats on water. Water has an amphoteric nature which acts as a base went around acids and vice versa.
$H_{2}O(asanacid)+N{H}_3\to NH{4}^{+}+O{H}^{-}(aq)$
$H_{2}O(asabase)+H_{2}S\to H_3{O}^{+}+H{S}^{-}$

Question:45

Rohan heard that instructions were given to the laboratory attendant to store a particular chemical, i.e., keep it in the darkroom, add some urea in it, and keep it away from dust. This chemical acts as an oxidising as well as a reducing agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.
(i) Write the name of this compound.
(ii) Explain why such precautions are taken for storing this chemical.

Answer:

1. $H_2{O}_2$

2. The compound is used in many industries such as textile and paper, bleaching agent and is composed of light and dust particles. It is widely employed to curb pollution by reducing it through the oxidizing action of harmful cyanides and reducing the effluents. It can function as both an oxidising and reducing agent and thus is stored in the dark, away from dust, as these can cause harmful chemical reactions.

Question:46

Give reasons why Hydrogen resembles alkali metals?
Answer:

Hydrogen has a specific, unique electronic configuration which is similar to alkali metals but belongs to group 1 in the periodic table. Its strong nature,e like the alkali metals, due to its configuration, makes it give one electron to combine to form unipositive ions.

Question:47

Hydrogen generally forms covalent compounds. Give reason.
Answer:

Due to its configuration, Hydrogen also resembles certain halogens, given its ionization enthalpy. Therefore, when acting as a halogen, it forms hydrides by combining with the elements to form a diatomic molecule, along with covalent compounds.

Question:48

Why is the ionisation enthalpy of Hydrogen higher than that of sodium?
Answer:

Hydrogen has a higher ionization enthalpy than sodium because sodium has a different configuration that is in the last shell, the electron is in 3s, which, if lost, the configuration resembles neon, which is a noble gas. However, with Hydrogen, the case is different, and the electron is in the 1s orbital, which,h on losing, does not amount to attainment of noble gas configuration.

Question:49

The basic principle of a hydrogen economy is transportation and storage of energy in the form of liquid or gaseous Hydrogen. Which property of Hydrogen may be useful for this purpose? Support your answer with the chemical equation if required.
Answer:

Hydrogen is an excellent source of energy. It can be alternate only used as fuel to empower automobiles and generate nuclear energy. It is looked at as a possible option and an economy run using Hydrogen as a fuel can be called as a hydrogen economy.
Hydrogen is freely abundant and replaceable, and so is regarded as a natural source since it also occurs as water, therefore can be largely utilised.
If Hydrogen is combusted, then it produces water and is not polluting for the environment.
A cell containing Hydrogen will give more power than the normal ones and as is looked at as the better alternative.
Hydrogen also acts as a great reducing agent and can be a substitute for the industries in place of carbon.

Question:50

What is the importance of heavy water?
Answer:

Heavy water is composed by electrolysis of water or through the fertilizer industries. It is a the great catalyst in nuclear reactors and can be used to study reactions.

Question:51

Write the Lewis structure of hydrogen peroxide.
Answer:

Question:52

An acidic solution of hydrogen peroxide behaves as an oxidising as well as the reducing agent. Illustrate it with the help of a chemical equation.
Answer:

1. Reactions which justify oxidising nature are-
H₂O₂ + 2KI $\to$ 2KOH + I₂
2. Reactions which justify reducing nature are-
H₂O₂ + Cl₂$\to$ 2HCl + O₂

Question:53

With the help of suitable examples, explain the property of $H_2{O}_2$ that is responsible for its bleaching action?
Answer:

$H_2{O}_2$ serves as a great bleaching agent and disinfectant, commercially sold as perhydrol and disintegrates on exposure to direct light.
The bleaching action of H2O2 is due to labile oxygen, which it releases on decomposition.
H₂O₂ → H₂O+[O]
The nascent oxygen combines with coloured material and makes it colourless. It can bleach feathers, silk, wool, paper, etc.

Question:54

Why is water molecule polar?
Answer:

The polar nature of the water molecule is due to the structure of the bent molecule and the bond length of 95.7pm and the angle of 104.5.

Question:55

Why does the water show a high boiling point as compared to hydrogen sulfide? Give reasons for your answer.
Answer:

Water has a higher boiling point than hydrogen sulfide because the hydrogen bonding in water is quite different, and hydrogen sulfide does not have hydrogen bonding at all.

Question:56

Why can dilute solutions of hydrogen peroxide not be concentrated by heating? How can a concentrated solution of hydrogen peroxide be obtained?
Answer:

Hydrogen peroxide causes burns when it is heated, so it should not be concentrated. You should use other methods other than heating to make it concentrated, such as distillation with water under pressure. Further distillation under pressure and careful extraction can give you pure hydrogen peroxide.

Question:57

Why is hydrogen peroxide stored in wax-lined bottles?
Answer:

Hydrogen peroxide is highly reactive, and that is stored in dark coloured bottles because it disintegrates on exposure to direct light.
$2H_2{O}_2\to 2H_{2}O+O_2$

Question:58

Why does hard water not form lather with soap?
Answer:

Hard water contains soluble salts which make it rough and form a precipitate when used along with soap. Due to the harsh chemicals dissolved in it, hard water is unsuitable for washing and laundry. The accumulation of salts makes it unsuitable for boilers.

Question:59

Phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why?
Answer:

Phosphoric acid is preferred as it does not dissolve, but sulphuric acid can react with hydrogen peroxide to disintegrate and decompose.

Question:60

How will you account for ${104.5}^{\circ }$ bond angle in the water?
Answer:

$S{p}^3$ hybridisation is present in oxygen along with a 109-degree angle, but the repulsive bond decreases the angle to a 104 degrees.

Question:61

Write redox reaction between fluorine and water?
Answer:

$2F_2(g)+2H_{2}O\to 4H^{+}(aq)+4F^{-}(aq)+O_2$

Question:62

Write two reactions to explain the amphoteric nature of water?

Answer:

$H_{2}O(l)+N{H}_3\to O{H}^{-}(aq)+N{H}_4(aq)$
$H_{2}O(l)+H_{2}S(aq)\to H_2{O}^{+}+H{S}^{-}$

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Matching Type

Question:63

Correlate the items listed in Column I with those listed in Column II. Find out as many correlations as you can.

Answer:

(i) → (i), (j)
(ii) → (d), (e), (g), (h), (i)
(iii) → (f)
(iv) → (a), (c)
(v) → (b), (d)
(vi) → (e)

Question:64

Match Column I with Column II for the given properties/applications mentioned therein.

Answer:

(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question:65

Match the terms in Column I with the relevant item in Column II.

Answer:
(i) → (e) (ii) → (d) (iii) → (b) (iv) → (a) (v) → (c)

Question:66

Match the items in Column I with the relevant item in Column II.

Answer:

i. → (b), (d)
ii. → (c)
iii. → (a), (c)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Assertion and Reason Type

Question:67

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the options given below each question.

Assertion (A): Permanent hardness of water is removed by treatment with washing soda.
Reason (R): Washing soda reacts with soluble magnesium and calcium sulphate to form insoluble carbonates.

(i) Statements A and R both are correct, and R is the correct explanation of A.
(ii) A is correct, but R is not correct.
(iii) A and R both are correct, but R is not the correct explanation of A.
(iv) A and R both are false.
Answer:

The answer is the option (i) Statements A and R both are correct, and R is the correct explanation of A.

Question:68

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the options given below each question.
Assertion (A): Some metals like platinum and palladium can be used as storage media for Hydrogen.
Reason (R): Platinum and palladium can absorb large volumes of Hydrogen.
(i) Statements A and R both are correct, and R is the correct explanation of A.
(ii) A is correct, but R is not correct.
(iii) A and R both are correct, but R is not the correct explanation of A.
(iv) A and R both are false.

Answer:

The answer is the option (i) statements A and R both are correct, and R is the correct explanation of A.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Long Answer Type

Question:69

Atomic Hydrogen combines with almost all elements, but molecular Hydrogen does not. Explain.
Answer:

Molecular Hydrogen is very stable, but atomic Hydrogen is very reactive, and the chemical behaviour of any molecule is determined by the bond dissociation enthalpy. In dihydrogen, the hydrogen bond dissociation enthalpy is very high, and therefore, it only reacts with a certain number of elements.

Question:70

How can $D_{2}O$ be prepared from water? Mention the physical properties in which $D_{2}O$ differs from $H_{2}O$. Give at least three reactions of $D_{2}O$showing the exchange of Hydrogen with deuterium.

Answer:

Extended electrolysis of water can produce D2O; it differs from water as it has a high molecular mass.

Reactions showing the exchange of Hydrogen with deuterium

• $NaOH+D_{2}O\to NaOD+HOD$

• $HCl+D_{2}O\to DCl+HOD$

• $N{H}_{4}Cl+D_{2}O\to N{H}_{3}DCl+HOD$

Question:71

How will you concentrate $H_2{O}_2$ ? Show differences between structures of $H_2{O}_2$ and $H_{2}O$ by drawing their spatial structures. Also mention three important uses of $H_2{O}_2$.
Answer:

To obtain hydrogen peroxide, you can use evaporation and barium peroxide by removing excess water. To make it more concentrated, use the process of distillation, and low pressure and then pure hydrogen peroxide can be obtained.

The spatial structures of $H_{2}O$ and $H_2{O}_2$ are as follows: -

Three important uses of $H_2{O}_2$ are as follows: -

• Peroxide is used as a disinfectant and antiseptic in the market.

• It is used to produce other chemicals in the industry and is also applied as a commercial bleaching product.

• It finds great application in the textile industry.

Question:72

(i) Give a method for the manufacture of hydrogen peroxide and explain the reactions involved therein.
(ii) Illustrate oxidising, reducing and acidic properties of hydrogen peroxide with equations.
Answer:

Industrial preparation:$H_2{O}_2$ is prepared by the auto-oxidation of 2-alkylanthraquinols $H_{2}O(l)+H_{2}S(aq)\to$
$2F{e}^{2+}+2H^{+}+H_{2}O\to 2F{e}^{3+}+2H_{2}O$
$PbS+H_2{O}_2\to PbS{O}_4+4H_{2}O$
(ii) (a) Reducing action in acidic medium
$2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$
$HOCl+H_2{O}_2\to H_3{O}^{-}+C{l}^{-}+O_2$
(b)Oxidising action in basic medium
$2F{e}^{2+}+H_2{O}_2\to 2F{e}^{3+}+2O{H}^{-}$
(c)
$I_2+H_2{O}_2+OH{�}^{-}\to 2I^{-}+2H_{2}O+O_2$

Question:73

What mass of hydrogen peroxide will be present in 2 litres of a 5 molar solution? Calculate the mass of oxygen that will be liberated by the decomposition of 200 mL of this solution.

Answer:

We know that the molecular mass of $H_2{O}_2$ – 34
Mass of $H_2{O}_2$ present in 1 molar solution – 34 g
Therefore, the mass of $H_2{O}_2$ present in 2 litres of 1 molar solution – $2\times 34=68g$
Hence, the mass of $H_2{O}_2$ that is present in 200ml solution of 1 molar $H_2{O}_2$ will be

$=\frac{34}{5}=6.8g$
$2H_2{O}_2\to 2H_{2}O+O_2$
68g 32g
6.8g 3.2g

Question:74

A colourless liquid ‘A’ contains H and O elements only. It decomposes slowly on exposure to light. It is stabilised by mixing urea to store in the presence of light.
(i) Suggest possible structure of A.
(ii) Write chemical equations for its decomposition reaction in light.
Answer:

(i) A is Hydrogen Peroxide $(H_2{O}_2)$ as it decomposes slowly on exposure to light and is stabilised by mixing urea to store in the presence of light.

Its possible structure is as follows: -


(ii) The chemical equations for its decomposition reaction in light are as follows-
$2H_2{O}_2(l)\to 2H_{2}O(l)+O_2(g)$

Question:75

An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $A{l}_{2}C{l}_6$.
Answer:

Since the ionic hydride of alkali metal has a significant covalent character, therefore, it is LiH. Since LiH is very stable, therefore, it is almost unreactive towards O₂?and Cl₂. It reacts with Al₂Cl₆? form lithium aluminum hydride.
$8\mathrm{LiH}+{\mathrm{Al}}_2{\mathrm{Cl}}_6\to 2{\mathrm{LiAlH}}_4+6\mathrm{LiCl}$

Question:75

An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $A{l}_{2}C{l}_6$.
Answer:

The ionic hydride of an alkali metal, which has a significant covalent character and is almost unreactive towards oxygen and chlorine at moderate temperatures, is lithium hydride is unreactive. That is why it is used for the synthesis of other useful hydrides.

Its reaction to $A{l}_{2}C{l}_6$ can be explained as follows: -

$8LiH+A{l}_{2}C{l}_6\to 2LiAl{H}_4+6LiCl$

Question:76

Sodium forms a crystalline ionic solid with dihydrogen. The solid is non-volatile and non-conducting in nature. It reacts violently with water to produce dihydrogen gas. Write the formula of this compound and its reaction with water. What will happen on the electrolysis of the melt of this solid?
Answer:

Saline hydrides produce dihydrogen gas during a violent reaction with water. The ionic solid crystal that is formed is non-conducting and non-volatile in nature, and reacts aggressively with water to produce dihydrogen gas.
The formula for this compound is NaH; its reaction with water is explained below: -
$NaH(s)+H_{2}O(aq)\to NaOH(aq)+H_2(g)$
Dihydrogen gas at the anode is liberated during electrolysis, which confirms the existence of H-ion