NCERT Solutions for Class 11 Science Chemistry Chapter 9 - Hydrogen

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Answer:

Hydrogen has electronic configuration 1s¹ and has a tendency to gain a single electron in its valence shell to attain stable electronic configuration like halogens with ns²np⁵ configuration belonging to the seventeenth group of the periodic table.

Hence, the correct answer is option (ii).

Answer:

Loss of the electron from hydrogen atom results in nucleus (H⁺) of ~1.5×10^–3 pm size. This is extremely small as compared to normal atomic and ionic sizes of 50 to 200pm. As a consequence, H⁺ ion does not exist freely and is always associated with other atoms or molecules.

Hence, the correct answer is option (iv).

Answer:

No ionic compound is hundred percent ionic and no covalent compound is a hundred percent covalent. So if any compound is having more ionic character then there will be some amount of covalent character as well. To decide the covalent character we use Fajan's rule. In the given compounds metal is different while the surrounding atom is hydrogen, so if the metal is having small size then that metal will have more covalent character due to more polarising power. Order of ionic size in these metals will follow order: Li⁺< Na⁺< K⁺< Rb⁺< Cs⁺.
Theefore, the order of covalent character is: LiH > NaH > KH > RbH > CsH
While the ionic character is: LiH < NaH < KH < RbH < CsH

Hence, the correct answer is option (ii).
 

Answer:

CH₄ is an electron precise hydride because there are 4 electrons present in the valence shell of carbon atom due to which 4 more electrons are required to complete the octet and it forms 4 covalent bonds with 4 hydrogen atoms to do so.

Hence, the correct answer is option (iv).

Answer:

Hydrogen has three isotopes: protium, deuterium and tritium. Of these isotopes, only tritium is radioactive and emits low energy β^– particles (t_½, 12.33 years).

Hence, the correct answer is option (iii).
 

Answer:

(A) H₂O₂ + 2HI → I₂ + 2H₂O 
The oxidation number of oxygen is decreased from –1 (H₂O₂) to –2 (H₂O), therefore, it is reduced and acts as an oxidizing agent.

(B) HOCl + H₂O₂ → H₃O⁺ + Cl^– + O₂
The oxidation number of oxygen is increased from –1 (H₂O₂) to 0 (O₂), therefore, it is oxidized and acts as a reducing agent.

Hence, the correct answer is option (ii).

Answer:

Oxide (BaO₂ .8H₂O + O₂) which contain peroxide linkage (–O–O–) on treatment with dil. H₂SO₄ gives H₂O₂.


Hence, the correct answer is option (ii).

Answer:

2I^– + 2H⁺ + H₂O₂ → I₂ + 2H₂O
In the above reaction, I^– ions are oxidized to I₂ (increase in oxidation number from –1 to 0). Therefore, H₂O₂ acts as an oxidizing agent.

Hence, the correct answer is option (iii).

Answer:

I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂
In the above reaction, I₂ is reduced to I^– (oxidation number decreases from 0 to –1), thus, H₂O₂ acts as a reducing agent.

Hence, the correct answer is option (ii).

Answer:

Hydrogen peroxide acts as an oxidising as well as reducing agent in both acidic and alkaline media.

Hence, the correct answer is option (iii).

Answer:

The gaseous mixture of carbon monoxide and hydrogen is called synthesis gas, which is used in the preparation of methanol. Pure H₂ from water gas cannot be obtained easily because it is difficult to remove CO. Therefore, to increase the production of H₂ from water gas, CO is oxidized to CO₂ by mixing it with more steam and passing the mixture over heated FeCrO₄ catalyst at 673 K.
CO(g) + H₂O(g)$\underset{{\mathrm{FeCrO}}_4}{\overset{673 \mathrm{K}}{\to }}$ CO₂(g) + H₂(g)
This is called water-gas shift reaction. Carbon dioxide is removed by scrubbing with sodium arsenite solution.

Hence, the correct answer is option (iii).

Answer:

When sodium peroxide is treated with dilute sulphuric acid, we get sodium sulphate and hydrogen peroxide which can be understood by the following reaction:

   ${\mathrm{Na}}_2{\mathrm{O}}_2 +{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4 + {\mathrm{H}}_2{\mathrm{O}}_2$

Hence, the correct answer is option (iv).
 

Answer:

H₂O₂ is obtained by the electrolysis of sulphuric acid. The electrolytic oxidation of acidified sulphate solution gives peroxodisulphate, which on hydrolysis gives hydrogen peroxide.

Answer:

Water-gas is the mixture of CO and H₂ which is used for the synthesis of methanol.

$\mathrm{CO}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\underset{\mathrm{Catalyst}}{\overset{\mathrm{Cobalt}}{\to }}{\mathrm{CH}}_3\mathrm{OH} \left(l\right)$

Hence, the correct answer is option (iv).
 

Answer:

Calcium and magnesium salts in the form of hydrogen carbonate, chloride and sulphate in water make water hard.

Hence, the correct answer is option (i).

Answer:

Na₆P₆O₁₈ (Sodium hexametaphosphate) commercially known as Calgon, is used for water softening.
CaCl₂ + Na₂[Na₄(PO₃)₆] →Na₂[Na₂Ca(PO₃)₆] + 2NaCl  

Hence, the correct answer is option (iii).

Answer:

Elements of groups 7, 8, 9 of the periodic table do not form hydrides.

Hence, the correct answer is option (i).

Answer:

Chromium is the only element of group 6 that forms hydride, CrH.

Hence, the correct answer is option (i).

Answer:

Hydrogen loses an electron to form a cation which cannot exist freely. It generally does not form ionic compounds by losing an electron but forms large number of covalent compounds by sharing of electrons.

Hence, the correct answers are options (iii) and (iv).

Answer:

The mixture of CO and H₂ is known as water gas. Since this mixture of CO and H₂ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or 'syngas'.

Hence, the correct answers are options (i) and (ii).

Answer:

Heavy water is used as a moderator in nuclear reactor. It is a less effective solvent than water and has a higher boiling point due to greater molar mass. Also, heavy water is more associated than ordinary water.

Hence, the correct answers are option (i) and (iii).

Answer:

Hydrogen has three isotopes: protium, deuterium and tritium. Protium is the most common. Hydrogen acts as an anion H⁻ in ionic salts. Dihydrogen acts as a reducing agent and H⁺ ion cannot exist freely in the solution.

Hence, the correct answers are option (i) and (ii).

Answer:

There is extensive hydrogen bonding in the frozen state of water i.e., ice. Ice has a network structure and is lighter than water due to the empty spaces present in tetrahedrons formed by hydrogen bonds.

Hence, the correct answers are option (iii) and (iv).

Answer:

Permanent hardness of water is due to the chlorides and sulphates of Ca and Mg in water.

Hence, the correct answers are options (i) and (ii).

Answer:

All the elements of group 14 form electron precise hydrides, i.e., these hydrides have the required number of electrons to write their conventional Lewis structures and have tetrahedral geometries. For example: CH₄ is a group 14 hydride having a tetrahedral geometry.

Hence, the correct answers are options (ii) and (iii).

Answer:

Hydrides of group 13 are electron deficient, i.e., they have too few electrons for writing their conventional Lewis structure and act as Lewis acids, i.e., electron acceptors. Whereas, hydrides of group 15 are electron rich, i.e., they have excess electrons which are present as lone pairs. and act as Lewis bases.

Hence, the correct answers are options (i) and (iv).

Answer:

Metallic hydrides are formed by many d-block and f-block elements and are non-stoichiometric hydrides, being deficient in hydrogen. They conduct heat and electricity. Ionic hydrides conduct electricity only in molten or aqueous state.

​Hence, the correct answers are options (i), (ii) and (iii).

Answer:

Water gas is produced when steam is passed over red hot coke or coal at 1270 K in the presence of nickel as catalyst.
C(s) + H₂O(g) $\stackrel{1270 \mathrm{K}}{\to }$ CO(g) + H₂(g)
                               water gas

Pure H₂ from water gas cannot be obtained easily because it is difficult to remove CO. Therefore, to increase the production of H₂ from water gas, CO is oxidized to CO₂ by mixing it with more steam and passing the mixture over heated FeCrO₄ catalyst at 673 K.
CO(g) + H₂O(g)$\underset{{\mathrm{FeCrO}}_4}{\overset{673 \mathrm{K}}{\to }}$ CO₂(g) + H₂(g)
This is called water-gas shift reaction. Carbon dioxide is removed by scrubbing with sodium arsenite solution.

Answer:

Metallic hydrides/interstitial hydrides are those in which hydrogen occupies the interstitial site in the some metal lattices without changing the type of lattice. They are are formed by d- and f-block elements and their hydrides conduct heat and electricity. They are almost always nonstoichiometric, being deficient in hydrogen.
 

Answer:

H₂O is an electron rich covalent or molecular hydride.
B₂H₆ is an electron deficient molecular hydride.
NaH is an ionic hydride.

Answer:

The mass per unit volume is called density. Since water expands on freezing, it crystallises into an open hexagonal form. Hydrogen bonding gives ice a rather open type structure with wide holes. Therefore, volume of ice for the same mass of water is more than liquid water. Hence, density of ice is lower than liquid water and ice floats on water.

Answer:

(i) PbS (s) + 4H₂O₂(aq) → PbSO₄ (s) + 4H₂O(l)
(ii) $\mathrm{CO}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\underset{\mathrm{Catalyst}}{\overset{\mathrm{Cobalt}}{\to }}{\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{l}\right)$

Answer:

(i) 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 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 keep on freezing from top to bottom.
(ii) Density of ice is less than water due to presence of empty spaces created because of H-bonding between H₂O molecules. Therefore, ice floats on water.

Answer:

Autoprotolysis is a reaction in which two same molecules react to give ions with proton transfer. Water undergoes autoprotolysis, i.e., a proton from one molecule is transferred to another molecule.


Significance: Autopyrolysis of water shows that water is amphoteric in nature, i.e., has the ability to act as an acid as well as a base.

Answer:

The demineralised water is the one which is free from all soluble mineral salts and is obtained by passing water successively through a cation exchange and an anion exchange resin. In cation exchange process, H⁺ exchanges for Na⁺, Ca²⁺, Mg²⁺ and in the anion exchange process, OH⁻ exchanges for anions like Cl⁻, HCO₃⁻, SO₄²⁻ etc.
The H⁺ and OH^– released in the above two processes neutralize each other to form water.
H⁺ + OH^– → H₂O

Answer:

(i) Electron deficient hydrides are the hydrides that do not have sufficient number of electrons to form normal covalent bonds. For example: the hydrides of group 13 such as B₂H₆, (AlH₃)_n etc.
(ii) Electron precise hydrides are those which have exact number of electrons to form normal covalent bonds. Examples are the hydrides of group 14 such as CH₄, SiH₄, etc.
(iii) Electron rich hydrides are those which have more number of electrons than normal covalent bonds. The excess electrons are present in the form of lone pairs. Examples are the hydrides of group 15, 16 and 17 such as NH₃, H₂O etc.

Answer:

Heavy water is prepared by exhaustive electrolysis of water.
The comparison between the physical properties of water and heavy water is given below.

Answer:

D₂O₂ can be prepared by the following reaction:

Answer:

5 volume solution of H₂O₂ means that 1L of this H₂O₂ solution will give 5 L of oxygen at STP
2H₂O₂(l) → O₂(g) + 2H₂O(l)
It shows that 68 g H₂O₂ gives 22.7 L of O₂ at STP,
So 5 L oxygen will be obtained from $\frac{68}{22.7}\times 5=15 \mathrm{g} {\mathrm{H}}_2{\mathrm{O}}_2$
∴ Strength of H₂O₂ in 5 volume H₂O₂ solution = 15 g L⁻¹ = 1.5% H₂O₂ solution

Answer:

(i) The structure of H₂O₂ in gas phase is shown below.

Dihedral angle is 111.5°.

The structure of H₂O₂ in solid phase at 110K is shown below.

Dihedral angle is 90.2°.

(ii) H₂O₂ decomposes readily to yield oxygen because of its weak peroxide (−O−O−) linkage. Therefore, it is a better oxidising agent than water. 
2H₂O₂ → O₂ + H₂O

Answer:

The intermolecular forces are stronger in D₂O than H₂O because it has higher melting point, enthalpy of vaporisation and viscosity.

Answer:

Deuterium is the isotope of hydrogen which has one proton and one neutron in its nucleus. It is represented by D or ${}_1{}^2\mathrm{H}$.
It reacts with oxygen to form Deuterium oxide or Heavy water (D₂O).
The reactivity of both the isotopes will be different towards oxygen.
Deuterium will be less reactive than hydrogen because of greater D−D bond dissociation enthalpy.
 

Answer:

Fluorine is more electronegative than chlorine and therefore, forms stronger hydrogen bonds. As a result, more energy is required to break the H−F bond than H−Cl bond. Hence, HF is a liquid while HCl is a gas.

Answer:

The first element of the periodic table is hydrogen. It reacts with dioxygen to from water (H₂O). The autoionisation of water takes place as follows:
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) 
 

Answer:

(i) Hydrogen peroxide

(ii) In the presence of metal surfaces or traces of alkali (present in glass containers), hydrogen peroxide decomposes into oxygen and water rapidly. It is, therefore, stored in wax-lined glass or plastic vessels in dark. Urea can be added as a stabiliser. It is kept away from dust because dust can induce explosive decomposition of the compound.

Answer:

The following reasons make hydrogen resemble alkali metals:

• Hydrogen has electronic configuration 1s¹ which is similar to the outer electronic configuration (ns¹) of alkali metals.
• Hydrogen has resemblance to alkali metals, which lose one electron to form unipositive ions.
• Like alkali metals, hydrogen forms oxides, halides and sulphides.

Answer:

The ionisation enthalpy of hydrogen is 1312 kJ mol^–1 which is quite similar to halogens. Therefore, instead of losing one electron to form a unipositive ion, it forms a diatomic molecule, combines with elements to form hydrides and a large number of covalent compounds

Answer:

The ionisation enthalpy of hydrogen is higher than that of sodium because removal of one electron from hydrogen leads to formation of unstable H⁺ ion while removal of one electron from sodium leads to formation of Na⁺ ion which has stable inert gas configuration.

Answer:

Dihydrogen gas is converted into liquid state by cooling to 20K. Tanks of metal alloy like NaNi₅, Ti–TiH₂, Mg–MgH₂ etc. are in use for storage of dihydrogen in small quantities. Some of the metals (e.g., Pd, Pt) can accommodate a very large volume of hydrogen and, therefore, can be used as its storage media. This property may be useful for hydrogen storage and as a source of energy.

Answer:

Heavy water is extensively used as a moderator in nuclear reactors and in exchange reactions for the study of reaction mechanisms. It is used for the preparation of other deuterium compounds, for example:

Answer:

The  Lewis structure of hydrogen peroxide is drawn below.

Answer:

Hydrogen peroxide as an oxidizing agent:

Hydrogen peroxide as a reducing agent:

Answer:

The oxidizing property of H₂O₂ is responsible for its bleaching action. It releases nascent oxygen [O] which decolorizes colored substances. It is used as a hair bleach and employed in the industries as a bleaching agent for textiles, paper pulp, leather, oils, fats, etc.
 

Answer:

Water molecule is polar because of the difference in electronegativity of oxygen and hydrogen. Therefore, hydrogen acquires a partial positive charge while oxygen acquires partial negative charge.

Answer:

Water shows high boiling point as compared to hydrogen sulphide due to the presence of extensive hydrogen bonding between water molecules.

Answer:

Dilute solutions of hydrogen peroxide not be concentrated by heating because it decomposes below its boiling point.
It can be concentrated by distillation under reduced pressure. 

Answer:

In the presence of metal surfaces or traces of alkali (present in glass containers), hydrogen peroxide decomposes into oxygen and water rapidly. It is, therefore, stored in wax-lined glass or plastic vessels in dark.

Answer:

Presence of calcium and magnesium salts in the form of hydrogencarbonate, chloride and sulphate in water makes water ‘hard’. Hard water forms scum/precipitate with soap. Soap containing sodium stearate (C₁₇H₃₅COONa) reacts with hard water to precipitate out Ca/Mg stearate.

Answer:

Phosphoric acid is a weak acid and does not catalyse the decomposition of hydrogen peroxide. Also, the impurities present in barium peroxide are removed as insoluble phosphates. Therefore, phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides.

Answer:

In case of H₂O molecule, the shape should have been tetrahedral if there were all bond pairs of electrons but two lone pairs of electrons are present so the shape is distorted tetrahedral or angular. The reason is lone pair-lone pair repulsion is more than lone pair-bond pair repulsion which is more than bond pair-bond pair repulsion. Thus, the angle is reduced to 104.5° from 109.5°.

Answer:

Water can be easily oxidised to O₂ with fluorine. The redox reaction between fluorine and water is as follows:
2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^–(aq) + O₂(g)

Answer:

Water is amphoteric in nature (acts both as acid and base). In Bronsted sense, it acts as an acid with NH₃ while as a base with H₂S

Answer:

Answer:

Answer:

Answer:

Answer:

Permanent Hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. It can be removed by treatment with washing soda. Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.


Hence, the correct answer is option A.

Answer:

The property of absorption of hydrogen on transition metals is widely used in catalytic reduction / hydrogenation reactions for the preparation of large number of compounds. Platinum and palladium can absorb large volumes of hydrogen and, therefore, can be used as its storage media.

Hence, the correct answer is option A.

Answer:

Atomic hydrogen [H] is produced at a high temperature in an electric arc or under ultraviolet radiations. Since its orbital is incomplete with 1s¹ electronic configuration, it does combine with almost all the elements. It undergoes reactions by (i) loss of the only electron to give H⁺, (ii) gain of an electron to form H^–, and (iii) sharing electrons to form a single covalent bond.
On the hand, molecular hydrogen [H₂] has the highest H–H bond dissociation enthalpy for a single bond between two atoms of any element. It is because of this factor that the dissociation of dihydrogen into its atoms is only ~0.081% around 2000 K which increases to 95.5% at 5000 K. Also, it is relatively inert at room temperature due to the high H–H bond enthalpy.  

Answer:

Heavy water (D₂O) can be prepared by exhaustive electrolysis of water.

The physical properties in which D₂O differs from H₂O are as follows:

Answer:

Hydrogen peroxide (H₂O₂) can be concentrated to ~30% (by mass) by distillation under reduced pressure. It can be further concentrated to ~85% by careful distillation under low pressure.

• The structure of H₂O₂ in gas phase is shown below.

Dihedral angle is 111.5°.

• The structure of H₂O₂ in solid phase at 110K is shown below.

Dihedral angle is 90.2°

• The structure of water is shown below:

Bent molecule with bond angle 104.5° and O − H bond length 95.7 pm.

The important uses of H₂O₂ are as follows:

• It is used as a hair bleach and as a mild disinfectant.
• It is used to manufacture chemicals like sodium perborate and per -carbonate, which are used in high quality detergents.
• It is used in the synthesis of hydroquinone, tartaric acid and certain food products and pharmaceuticals (cephalosporin) etc

Answer:

(i) Electrolytic oxidation of acidified sulphate solution gives peroxodisulphate, which on hydrolysis gives hydrogen peroxide.

(ii) As Oxidizing Agent

• In acidic medium

• In basic medium

As Reducing Agent

• In acidic medium

• In basic medium

Answer:

Molar mass of H₂O₂ = (2$\times$1 + 2$\times$16) = 34 g
Volume of solution = 2 L
Molarity of solution = 5 M
$$\begin{gathered} \mathrm{Molarity}=\frac{{\displaystyle \frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{L})} \\ \mathrm{mass}=\mathrm{Molarity}\times \mathrm{Volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{L})\times \mathrm{molar} \mathrm{mass} \\ \mathrm{mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2=5\times 2\times 34 =340 \mathrm{g} \\ \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{in} 2 \mathrm{L} \mathrm{of} 5 \mathrm{molar} \mathrm{solution} = 340 \mathrm{g} \\ \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{in} 200 \mathrm{mL} \left(0.2 \mathrm{L}\right) \mathrm{of} 5 \mathrm{molar} \mathrm{solution} = 34 \mathrm{g} \end{gathered}$$
The decomposition of H₂O₂ takes place as follows:
2H₂O₂ → O₂ + 2H₂O
The decomposition of 2 moles (68 g) of H₂O₂liberates 1 mole (32 g) of O₂.
So, the decomposition of 34 g of H₂O₂ will liberate 16 g of O₂.

Answer:

(i) The colourless liquid 'A' is H₂O₂.
The structure of H₂O₂ in gas phase is shown below.

Dihedral angle is 111.5°.

The structure of H₂O₂ in solid phase at 110 K is shown below.

Dihedral angle is 90.2°.

Answer:

Lithium hydride is the ionic hydride which has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides.
Its chemical formula is LiH.
The reaction of LiH with Al₂Cl₆ is as follows:
8LiH + Al₂Cl₆ → 2LiAlH₄ + 6LiCl

Answer:

The formula of the crystalline ionic solid given in the question is NaH.

It reacts violently with water as follows:
NaH(s) + H₂O(aq) → NaOH(aq) + H₂(g)

On electrolysis of the melt of thus solid, it liberates H₂ gas at anode.
2H⁻(melt)  H₂(g) + 2e⁻