Answer

(b) Hydrogen resembles halogens in many respects for which several factors are responsible. The most important is hydrogen like halogens accept an electron readily to achieve nearest inert gas configuration.

Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals and halogens.

• (a) Its tendency to lose an electron to form a cation: This is incorrect because hydrogen’s tendency to lose an electron to form a cation (H⁺) is more characteristic of alkali metals rather than halogens. Halogens typically gain an electron to form anions (X⁻).

• (c) Its low negative electron enthalpy value: This is incorrect because halogens have high negative electron enthalpy values, meaning they release a significant amount of energy when they gain an electron. Hydrogen’s electron enthalpy is not as negative as that of halogens.

• (d) Its small size: This is incorrect because while hydrogen does have a small atomic size, this property alone does not make it resemble halogens. The resemblance to halogens is more about its electronic configuration and tendency to gain an electron rather than its size.

Answer

(d) $\mathrm{H}^{+}$ion always get associated with other atoms or molecules. The reason is that loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to small size it cannot exist free.

• (a) Ionisation enthalpy of hydrogen resembles that of alkali metals: This is incorrect because the ionisation enthalpy of hydrogen is actually quite different from that of alkali metals. Alkali metals have much lower ionisation enthalpies compared to hydrogen, making this option irrelevant to the association of $\mathrm{H}^{+}$ ions with other atoms or molecules.

• (b) Its reactivity is similar to halogens: This is incorrect because hydrogen’s reactivity is not similar to halogens. Halogens are highly electronegative and tend to gain electrons to form negative ions, whereas hydrogen tends to lose its single electron to form a positive ion ($\mathrm{H}^{+}$).

• (c) It resembles both alkali metals and halogens: This is incorrect because while hydrogen shares some properties with both alkali metals and halogens, this resemblance does not explain why $\mathrm{H}^{+}$ ions always get associated with other atoms or molecules. The key reason is the extremely small size of the $\mathrm{H}^{+}$ ion, which makes it highly reactive and unable to exist freely.

Answer

(b) Metal hydrides are ionic, covalent or molecular in nature. lonic character increases as the size of the atom increases or the electronegativity of the atom decreases. The correct order of increasing ionic character is

$$ \mathrm{LiH}<\mathrm{NaH}<\mathrm{KH}<\mathrm{RbH}<\mathrm{CsH} $$

• Option (a) $\mathrm{LiH}>\mathrm{NaH}>\mathrm{CsH}>\mathrm{KH}>\mathrm{RbH}$: This option is incorrect because it suggests that $\mathrm{LiH}$ has the highest ionic character, which contradicts the trend that ionic character increases with the size of the metal cation. Lithium is the smallest cation among the given options, so $\mathrm{LiH}$ should have the least ionic character.

• Option (c) $\mathrm{RbH}>\mathrm{CsH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{LiH}$: This option is incorrect because it places $\mathrm{RbH}$ and $\mathrm{CsH}$ at the beginning of the sequence, suggesting that $\mathrm{RbH}$ has a higher ionic character than $\mathrm{CsH}$. However, cesium is larger than rubidium, so $\mathrm{CsH}$ should have a higher ionic character than $\mathrm{RbH}$.

• Option (d) $\mathrm{NaH}>\mathrm{CsH}>\mathrm{RbH}>\mathrm{LiH}>\mathrm{KH}$: This option is incorrect because it places $\mathrm{NaH}$ at the beginning of the sequence, suggesting that $\mathrm{NaH}$ has the highest ionic character. Sodium is smaller than potassium, rubidium, and cesium, so $\mathrm{NaH}$ should not have the highest ionic character. Additionally, it incorrectly places $\mathrm{LiH}$ before $\mathrm{KH}$, which contradicts the trend that $\mathrm{LiH}$ should have the least ionic character.

Answer

(d) Electron-precise hydrides contain exact number of electrons to form normal covalent bonds. e.g., $-\mathrm{CH}_{4}$ which has tetrahedral in geometry.

• (a) $B_{2}H_{6}$: This is an electron-deficient hydride. Diborane ($B_{2}H_{6}$) has fewer electrons than needed for conventional 2-center 2-electron bonds, leading to the formation of multi-center bonds (3-center 2-electron bonds).

• (b) $\mathrm{NH}_{3}$: Ammonia ($\mathrm{NH}_{3}$) is an electron-rich hydride. It has a lone pair of electrons on the nitrogen atom, which makes it have more electrons than needed for just forming normal covalent bonds.

• (c) $\mathrm{H}_{2}O$: Water ($\mathrm{H}_{2}O$) is also an electron-rich hydride. It has two lone pairs of electrons on the oxygen atom, resulting in more electrons than required for forming normal covalent bonds.

Thinking Process

To solve this problem, the point kept in mind that nucleids with $n / p$ (neutron/proton) ratio $>1.5$ are usually radioactive.

Answer

(c) The radioactive isotope of hydrogen is tritium. For tritium $(n=3, p=1)$, therefore $n / p$ ratio is 3.

• Protium (a): Protium is the most common isotope of hydrogen with one proton and no neutrons. It is stable and does not exhibit radioactivity.

• Deuterium (b): Deuterium is an isotope of hydrogen with one proton and one neutron. It is also stable and does not emit $\alpha, \beta$, or $\gamma$ rays.

• Hydronium (d): Hydronium is not an isotope of hydrogen; it is a positively charged ion (H₃O⁺) formed when a proton (H⁺) associates with a water molecule (H₂O). It is not radioactive.

Thinking Process

(i) Reducing agents are those substance (atoms, ions or molecules) which can readily lose electrons to other substance.

(ii) Oxidising agents are those substance (atoms, ions or molecules) which can readily accept electrons from other substance.

Answer

(b)

(i)

Thus, here $H_{2} O_{2}$ oxidises $HI$ into $I_{2}$ hence, it behaves as oxidising agent.

• Option (a) is incorrect because in reaction (ii), $H_{2}O_{2}$ is not acting as an oxidizing agent. Instead, it is acting as a reducing agent by reducing $HOCl$ to $Cl^{-}$.

• Option (c) is incorrect because in reaction (i), $H_{2}O_{2}$ is not acting as a reducing agent. Instead, it is acting as an oxidizing agent by oxidizing $HI$ to $I_{2}$.

• Option (d) is incorrect because in both reactions, $H_{2}O_{2}$ is not acting as a reducing agent. In reaction (i), it acts as an oxidizing agent, and in reaction (ii), it acts as a reducing agent.

Answer

(b) Oxides such as $BaO_{2}, Na_{2} O_{2}$ etc; which contain peroxide linkage (i.e., $-O-O$ or $O_{2}^{2-}$ ) on treatment with dilute $H_{2} SO_{4}$ give $H_{2} O_{2}$ but dioxides $(O=M=O$, where $M$ is the metal atom) such as $PbO_{2}, MnO_{2}, TiO_{2}$ do not give $H_{2} O_{2}$ on treatment with dilute $H_{2} SO_{4}$.

$$ \underset{\text{peroxide}}{\underset{Hydrated barium}{BaO_2 . 8H_2O(s)}} + H_2SO_4 (aq) \longrightarrow BASO_4 (s + )\underset{\text{peroxide}}{\underset{Hydrated barium}{ H_2O_2(aq)}} + 8H_2O (l) $$

• (a) $PbO_{2}$: Lead dioxide ($PbO_{2}$) is a dioxide, not a peroxide. It does not contain the peroxide linkage ($-O-O$ or $O_{2}^{2-}$) necessary to produce hydrogen peroxide ($H_{2}O_{2}$) when treated with dilute sulfuric acid ($H_{2}SO_{4}$).

• (c) $MnO_{2}$: Manganese dioxide ($MnO_{2}$) is also a dioxide and lacks the peroxide linkage. Therefore, it does not produce hydrogen peroxide ($H_{2}O_{2}$) upon treatment with dilute sulfuric acid ($H_{2}SO_{4}$).

• (d) $TiO_{2}$: Titanium dioxide ($TiO_{2}$) is another example of a dioxide, which does not contain the peroxide linkage. As a result, it does not yield hydrogen peroxide ($H_{2}O_{2}$) when treated with dilute sulfuric acid ($H_{2}SO_{4}$).

Answer

(c) The reaction in which $H_{2} O_{2}$ is reduced i.e., oxidation state of oxygen decreases from -1 to -2 depicts the oxidising nature of $H_{2} O_{2}$. e.g.,

• Option (a): In this reaction, $H_{2}O_{2}$ is acting as a reducing agent because it is being oxidized to $O_{2}$. The oxidation state of oxygen in $H_{2}O_{2}$ increases from -1 to 0.

• Option (b): In this reaction, $H_{2}O_{2}$ is also acting as a reducing agent because it is being oxidized to $O_{2}$. The oxidation state of oxygen in $H_{2}O_{2}$ increases from -1 to 0.

• Option (d): In this reaction, $H_{2}O_{2}$ is acting as a reducing agent because it is being oxidized to $O_{2}$. The oxidation state of oxygen in $H_{2}O_{2}$ increases from -1 to 0.

Answer

(b) $H_{2} O_{2}$ acts as an oxidising as well as reducing agent in alkaline media. The given below reaction show the reducing action in basic medium

$$ \begin{aligned} I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\ 2 MnO_{4}^{-}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-} \end{aligned} $$

• Option (a): In this reaction, $H_2O_2$ is acting as an oxidizing agent because it is causing the oxidation of $[Fe(CN)_6]^{4-}$ to $[Fe(CN)_6]^{3-}$.

• Option (c): In this reaction, $H_2O_2$ is acting as an oxidizing agent because it is causing the oxidation of $Mn^{2+}$ to $Mn^{4+}$.

• Option (d): In this reaction, $H_2O_2$ is acting as an oxidizing agent because it is causing the oxidation of $PbS$ to $PbSO_4$.

Answer

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

• (a) Hydrogen peroxide is not solely an oxidising agent because it can also act as a reducing agent in certain reactions.
• (b) Hydrogen peroxide is not solely a reducing agent because it can also act as an oxidising agent in certain reactions.
• (d) Hydrogen peroxide is not neither an oxidising nor reducing agent because it can act as both an oxidising and reducing agent depending on the reaction conditions.

Answer

(c) The process of producing syn gas or synthesis gas from coal is called ‘coal gasification’.

$\underset{\text { Coal }}{\mathrm{C}(\mathrm{s})}+\underset{\text { Steam }}{\mathrm{H} _2 \mathrm{O}(g)} \xrightarrow[\mathrm{Ni}]{1270 \mathrm{~K}} \underbrace{\mathrm{CO}(g)+\mathrm{H} _2(g)} _{\text {Syn gas }}$

The production of hydrogen can be increased by reacting carbon monoxide of the syn gas with steam in the presence of iron chromate as a catalyst at $673 \mathrm{~K}$.

$\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \xrightarrow{\mathrm{FeCrO}_4, 673 \mathrm{~K}} \mathrm{CO}_2(g)+\mathrm{H}_2(g)$

$\mathrm{CO}_{2}$ is removed by scrubbing with a solution of sodium arsenite.

• Option (a) is incorrect because it describes the steam reforming of methane, which produces synthesis gas (CO and H₂) but does not specifically increase the production of dihydrogen from synthesis gas.

• Option (b) is incorrect because it describes the reaction of carbon with steam to produce synthesis gas (CO and H₂), but it does not involve a subsequent step to increase the production of dihydrogen from the synthesis gas.

• Option (d) is incorrect because it describes the steam reforming of ethane, which produces synthesis gas (CO and H₂) but does not specifically increase the production of dihydrogen from synthesis gas.

Answer

(d) When sodium peroxide is treated with dilute sulphuric acid, we get sodium sulphate and hydrogen peroxide

$$ Na_{2} O_{2}+\text { dil. } H_{2} SO_{4} \longrightarrow Na_{2} SO_{4}+H_{2} O_{2} $$

• Option (a): Sodium peroxide reacts with dilute sulfuric acid to produce sodium sulfate and hydrogen peroxide, not water. The reaction specifically forms hydrogen peroxide (H₂O₂) rather than water (H₂O).

• Option (b): The reaction between sodium peroxide and dilute sulfuric acid does not produce oxygen gas. Instead, it produces hydrogen peroxide (H₂O₂) along with sodium sulfate.

• Option (c): The reaction does not produce hydrogen gas. The correct products are sodium sulfate and hydrogen peroxide, not hydrogen gas and oxygen.

Answer

(b) Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.

$$ \begin{aligned} 2H_2SO_4 \xrightarrow{} 2H^{+} + 2HSO_4^{-} \end{aligned} $$ $$ 2H^{+} + 2HSO _4^{-} \xrightarrow {electrolysis} HO_3SOOSO _3H \xrightarrow {Hydrolysis} 2H_2SO_4 + H_2O_2 $$

• (a) Water: Electrolysis of water primarily produces hydrogen and oxygen gases, not hydrogen peroxide. The process involves the splitting of water molecules (H₂O) into hydrogen (H₂) and oxygen (O₂) at the electrodes.

• (c) Hydrochloric acid: Electrolysis of hydrochloric acid (HCl) results in the production of chlorine gas (Cl₂) at the anode and hydrogen gas (H₂) at the cathode. Hydrogen peroxide is not formed in this process.

• (d) Fused sodium peroxide: Sodium peroxide (Na₂O₂) does not undergo electrolysis to produce hydrogen peroxide. Instead, it reacts with water to form sodium hydroxide (NaOH) and hydrogen peroxide (H₂O₂), but this is a chemical reaction, not an electrolytic process.

Answer

(d) The water gas is the combination of carbon monoxide and hydrogen.

$$ CO(g)+2 H_{2}(g) \xrightarrow[\text { Catalyst }]{\text { Cobalt }} CH_{3} OH(l) $$

It is an example of water gas used in the synthesis of methanol.

• Option (a): This reaction is not an example of the use of water gas in the synthesis of other compounds. Instead, it is a reaction where methane reacts with water to produce carbon monoxide and hydrogen, which is a process known as steam reforming.

• Option (b): This reaction is not an example of the use of water gas in the synthesis of other compounds. It is the water-gas shift reaction, where carbon monoxide reacts with water to produce carbon dioxide and hydrogen.

• Option (c): This reaction is not an example of the use of water gas in the synthesis of other compounds. It describes the steam reforming of hydrocarbons to produce carbon monoxide and hydrogen, which is a method to generate synthesis gas (syngas), not a direct use of water gas.

Answer

(a) Bicarbonates, chlorides and sulphates of $\mathrm{Ca}$ and $\mathrm{Mg}$ are responsible for the hardness of water.

Note Hard water forms scum/precipitate with soap. Soap containing sodium stearate $\left(C_{17} H_{35} COONa\right)$ reacts with hard water to precipitate out $Ca/ Mg$ stearate.

$$2 C_{17} H_{35} COONa(aq)+M^{2+}(aq) \longrightarrow\left(C_{17} H_{35} COO\right)_{2} M \downarrow +2 \mathrm{Na}^{+}(\mathrm{aq})(\text { where, } \mathrm{M}=\mathrm{Ca} / \mathrm{Mg}) $$

It is unsuitable for laundry and boilers.

• (b) $\mathrm{Na}^{+}$: Sodium ions do not cause hardness in water. Hardness is primarily due to the presence of calcium and magnesium ions, which form insoluble compounds with soap, leading to scum formation. Sodium ions, on the other hand, do not form such insoluble compounds and therefore do not contribute to water hardness.

• (c) $\mathrm{Cl}^{-}$: Chloride ions do not cause hardness in water. While chloride ions can be present in hard water, they do not form insoluble compounds with soap. Hardness is specifically due to calcium and magnesium ions, which react with soap to form precipitates.

• (d) $\mathrm{K}^{+}$: Potassium ions do not cause hardness in water. Similar to sodium ions, potassium ions do not form insoluble compounds with soap. The hardness of water is due to the presence of calcium and magnesium ions, which react with soap to form insoluble precipitates.

Answer

(c) For water softening, sodium hexametaphosphate is used. The chemical formula is $Na_2[Na_4(PO_3)_6]$ . It is also known as calgon .

$ \underset{\text{(From hard water)}}{2CaCl_2} + \underset{\text{Sodium hexametaphosphate}}{Na_2[Na_4(PO_3)_6]} \longrightarrow \underset{\text{Complex salt (soluble)}}{Na_2[Ca_2(PO_3)_6]} + 4NaCl $

• (a) $Ca_{3}\left(PO_{4}\right)_{2}$: This compound, calcium phosphate, is not used for water softening. Instead, it is often found in hard water as a precipitate and does not help in removing hardness.

• (b) $Na_{3} PO_{4}$: Sodium phosphate can be used for water softening, but it is not as effective as sodium hexametaphosphate. It can precipitate calcium and magnesium ions but is not commonly used for this purpose.

• (d) $Na_{2} HPO_{4}$: Disodium hydrogen phosphate is not typically used for water softening. It is more commonly used as a buffering agent and does not effectively remove calcium and magnesium ions from hard water.

Answer

(a) Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are $CH_{4}, NH_{3}, H_{2} O$ and $HF$. For convenience hydrogen compounds of non-metals have also been considered as hydrides.

• (b) Group 13: Elements in Group 13, such as boron, aluminum, and gallium, do form hydrides. For example, boron forms borane ($BH_3$), and aluminum forms alane ($AlH_3$).

• (c) Groups 15, 16, 17: Elements in these groups also form hydrides. For instance, nitrogen in Group 15 forms ammonia ($NH_3$), oxygen in Group 16 forms water ($H_2O$), and fluorine in Group 17 forms hydrogen fluoride ($HF$).

• (d) Group 14: Elements in Group 14, such as carbon and silicon, form hydrides as well. Carbon forms methane ($CH_4$) and silicon forms silane ($SiH_4$).

Answer

(a) Only one element of group 6, i.e., Cr forms hydride.

Note Metallic (or interstitial) hydrides are formed by many d-block and f-block elements. However, the metals of group 7,8 and 9 do not form hydride. Even from group 6, only chromium forms $\mathrm{CrH}$. These hydrides conduct heat and electricity though not as efficiently as their parent metals do.

• Group 7: The metals of group 7 do not form hydrides.
• Group 8: The metals of group 8 do not form hydrides.
• Group 9: The metals of group 9 do not form hydrides.

Answer

$(c, d)$

$\mathrm{H}^{+}$does not exist freely and is always associated with other atoms or molecules.

Like alkali metals, hydrogen forms oxides, halides and sulphides. However, unlike alkali metals, it has a very high ionisation enthalpy and does not possess metallic characteristics under normal conditions.

• (a) This statement is true. Hydrogen naturally exists as a diatomic molecule (H₂).
• (b) This statement is true. Hydrogen has one electron in its outermost shell.

Answer

$(a, b)$

Dihydrogen can be prepared on commercial scale by different methods. Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yield hydrogen.

$$ C_nH_{2n+2} + nH_2O +2 \xrightarrow[Ni]{1270 K} nCO_2 + (2n+1)H_2$$

e.g. $$ CH_2(g) + H_2O(g) \xrightarrow[Ni]{1270 k} CO(g) + 3H_2(g) $$

The mixture of $CO$ and $H_{2}$ is called water gas. As this mixture of $CO$ and $H_{2}$ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or ‘syn gas’.

• Producer gas: Producer gas is a mixture of carbon monoxide (CO) and nitrogen (N₂) obtained by passing air over red-hot coke. It does not contain hydrogen (H₂) and is different from the mixture of CO and H₂ formed by the action of steam on hydrocarbons.

• Industrial gas: The term “industrial gas” is a broad category that refers to gases used in various industrial processes, such as oxygen, nitrogen, argon, and others. It is not a specific term for the mixture of CO and H₂ formed by the action of steam on hydrocarbons.

Answer

$(a, c)$

Heavy water is used as moderator in nuclear reactor. Boiling point of heavy water is higher than ordinary water and it is not as effective in the form of solvent as water due to its low dielectric constant.

• Heavy water is not more effective as a solvent than ordinary water because it has a lower dielectric constant.
• Heavy water does not have a lower boiling point than ordinary water; in fact, it has a higher boiling point.

Answer

$(a, b)$

Among the three isotopes of hydrogen, protonium is the most common. In ionic salts, hydrogen exists as hydride $\left(\mathrm{H}^{-}\right)$.

• Option (c) is incorrect because the hydrogen ion, $\mathrm{H}^{+}$, does not exist freely in solution. Instead, it associates with water molecules to form hydronium ions, $\mathrm{H}_3\mathrm{O}^{+}$.

• Option (d) is incorrect because dihydrogen ($\mathrm{H}_2$) does act as a reducing agent. It can donate electrons to other substances, thereby reducing them.

Answer

(c, $d)$

There is $\mathrm{H}$-bonding even in frozen state of water, i.e., ice is lighter than liquid water.

The crystalline form of water is ice. At atmospheric pressure, ice crystallises in the hexagonal form, but at very low temperatures it condenses to cubic form. Density of ice is less than that of water. Therefore, an ice cube floats on water.

• (a) Water is known to be a universal solvent: This statement is correct. Water is often referred to as a universal solvent because it can dissolve a wide variety of substances due to its polar nature and ability to form hydrogen bonds.

• (b) Hydrogen bonding is present to a large extent in liquid water: This statement is correct. Hydrogen bonding is indeed present to a large extent in liquid water, which is responsible for many of its unique properties, such as high boiling point, high specific heat, and surface tension.

Answer

$(a, b)$

Permanent hardness is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. Permanent hardness is not removed by boiling.

• (c) hydrogen carbonates of $\mathrm{Ca}$ and $\mathrm{Mg}$ in water: Hydrogen carbonates of calcium and magnesium cause temporary hardness, not permanent hardness. Temporary hardness can be removed by boiling the water, which precipitates the carbonates.

• (d) carbonates of alkali metals in water: Carbonates of alkali metals do not cause hardness in water. Alkali metal carbonates are generally soluble in water and do not contribute to either temporary or permanent hardness.

Answer

$(b, c)$

Electron precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 forms electron-precise compounds (e.g., $\mathrm{CH}_{4}$ ) which are tetrahedral in geometry.

• (a) Elements of group 15 form electron deficient hydrides: This statement is incorrect because elements of group 15 typically form electron-rich hydrides, not electron-deficient ones. For example, ammonia (NH₃) has a lone pair of electrons, making it electron-rich.

• (d) Electron rich hydrides can act as Lewis acids: This statement is incorrect because electron-rich hydrides have excess electrons and typically act as Lewis bases, not Lewis acids. Lewis acids are electron pair acceptors, whereas electron-rich hydrides are more likely to donate electron pairs.

Answer

$(a, d)$

All elements of group 13 will form electron deficient compounds which acts as Lewis acids.

All elements of group 14 will form electron precise compounds.

Electron rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 forms such compounds. $NH_{3}$ has 1-lone pair, $H_{2} O$ has $2$ and $HF$ has $3$ lone pairs act as Lewis bases.

• (b) Hydrides of group 14 are electron deficient hydrides: This statement is incorrect because hydrides of group 14 are electron-precise compounds, meaning they have a complete octet and are neither electron-deficient nor electron-rich.

• (c) Hydrides of group 14 act as Lewis acids: This statement is incorrect because hydrides of group 14 do not act as Lewis acids. Instead, they are electron-precise and do not have the ability to accept electron pairs.

Answer

$(a, b, c)$

The ionic hydrides are crystalline, non-volatile and non-conducting in solid state. However, their molten state conduct electricity.

• Option (d) is incorrect because ionic hydrides are not good conductors of electricity in the solid state. They are crystalline and non-conducting in their solid form. They only conduct electricity when they are in a molten state or dissolved in water.

Answer

Water gas is produced when superheated steam is passed over red hot coke or coal at $1270 \mathrm{~K}$ in presence of nickel as catalyst.

$ \underset{\text{Coke}}{C(s)} + \underset{\text{Steam}}{H_2O(g)} +121.3 KJ \xrightarrow[Nickel]{1270 k} \underbrace{CO(g)+ H_2(g)}_{\text{Water gas}}$

It is inconvinient to obtain pure $H_{2}$ from water gas as $CO$ is difficult to remove. Hence, to increase the production of $H_{2}$ from water gas, $CO$ is oxidised to $CO_{2}$ by mixing it with more steam and passing the mixture over $FeCrO_{4}$ catalyst at $673 ~K$.

$ \underbrace{CO(g)+ H_2(g)}_{\text{Water gas}} + \underset{\text{Steam}}{H_2O(g)} \xrightarrow[FeCrO_4]{673 K} CO_2(g) + 2H_2(g) $

This is called water-gas shiff reaction. Carbon dioxide is removed by scrubbing with mixture of sodium arsenite solution or by passing the mixture through water under $30 ~atm$ pressure when $CO_{2}$ dissolves leaving behind $H_{2}$ which is collected.

Answer

Metallic/interstitial hydrides are formed by many $d$-block and $f$-block elements. These hydrides conduct heat and electricity.

Unlike saline hydride, they are almost always non-stoichiometric, being deficient in hydrogen. e.g., $LaH_{2.87}, YbH_{2.55}, TiH_{1.5-1.8}, ZrH_{1.3-1.75}, VH_{0.56}, NiH_{0.6-0.7}, PdH_{0.6-0.8}$ etc. In such hydrides, the law of constant composition does not hold good.

Comparision between molecular and metallic hydrides

Answer

$H_{2} O$ - Covalent or molecular hydride (electron rich hydride).

$B_{2} H_{6}$ - Covalent or molecular hydride (electron deficient hydride).

$NaH$ - Ionic or saline hydride.

Note Molecular hydrides are further classified according to the relative number of electrons and bonds in their Lewis structures.

(i) Electron deficient hydride has too few electrons for writing its conventional Lewis structure.

(ii) Electron precise compounds have the required number of electrons to write their conventional Lewis structures.

(iii) Electron rich hydrides have excess electrons which are present as lone pairs.

Answer

In ice, molecules of $\mathrm{H}_{2} \mathrm{O}$ are not packed so closely as in liquid water. There exists vacant spaces in the crystal lattice. This results in larger volume and lower density (density $=$ mass/volume).

In other words, density of ice is lower than liquid water and hence ice floats on water.

Hexagonal honey comb structure of ice

Answer

(i) When $\mathrm{PbS}$ react with hydrogen peroxide, then $\mathrm{PbSO}_{4}$ and water are formed.

$$ PbS(s)+4 H_{2} O_{2}(aq) \longrightarrow PbSO_{4}+4 H_{2} O $$

(ii) When carbon mono-oxide reacts with hydrogen in the presence of cobalt catalyst, then methanol is formed.

$$ CO(g)+2 H_{2}(g) \xrightarrow[\text { catalyst }]{\text { Cobalt }} CH_{3} OH(l) $$

Answer

(i) Density of ice is less than that of liquid water. During severe winter, the temperature of lake water keeps on decreasing. Since, cold water is heavier, therefore, it moves towards bottom of the lake and warm water from the bottom moves towards surface. This process continues. The density of water is maximum at $277 \mathrm{~K}$.

Therefore, any further decrease in temperature of the surface water will decrease in density. The temperature of surface water keeps on decreasing and ultimately it freezes.

Thus, the ice layer at lower temperature floats over the water below it. Due to this, freezing of water into ice takes place continuously from top towards bottom.

(ii) Density of ice is less than that of liquid water, so it floats over water.

Answer

Auto-protolysis means self ionisation of water.

$ \underset{\text{Acid }_1}{H _2O(l)} + \underset{Base _2}{H _2O(l)} \leftrightharpoons \underset{Acids_2}{H_3O^+(aq)} + \underset{Base_1}{OH^-(aq)} $

Due to auto-protolysis, water is amphoteric in nature. It reacts with both acids and bases.

e.g., $$ \underset{Acid_1}{H_2O(l)} + \underset{Base_2}{NH_3(aq)} \longrightarrow \underset{Acid_2}{NH_4^+(aq)} + \underset{Base_1}{OH^-(aq)} $$

$$ \underset{Base_1}{H_2O(l)} + \underset{Acid_2}{H_2S(aq)} \longrightarrow \underset{Acid_1}{H_3O^+(aq)} + \underset{Base_2}{HS^-(aq)} $$

Answer

Water which is free from all soluble minerals salts is called demineralised water. Demineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In cation exchanger, $Ca^{2+}, Mg^{2+}, Na^{+}$and other cations present in water are removed by exchanging them with $H^{+}$ions while in anion exchanger, $Cl^{-}, HCO_{3}^{-}, SO_{4}^{2-}$, etc., present in water are removed by exchanging them with $OH^{-}$ions.

$$ \underset{\text { (Released in cation exchanger) }}{\mathrm{H}^{+}} \underset{\text { (Released in anion exchanger) }}{\mathrm{OH}^{-}} \longrightarrow \mathrm{H}_{2} \mathrm{O} $$

Synthetic ion exchange resins are of two types.

Cation exchange resins contain large organic molecule with $\mathrm{SO}_{3} \mathrm{H}$ group and are water soluble. It is changed to $R \mathrm{Na}$ by treating it with $\mathrm{NaCl}$. The resin $R \mathrm{Na}$ exchanges $\mathrm{Mg}^{2+}$ and $\mathrm{Ca}^{2+}$ ions present in hard water to make the water soft.

$$ 2 \mathrm{RNa}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{R}_{2} \mathrm{M}(\mathrm{s})+2 \mathrm{Na}^{+}(\mathrm{aq}) \quad\left(M=\mathrm{Ca}^{2+} \text { or } \mathrm{Mg}^{2+}\right) $$

The resin can be regenerated by passing $\mathrm{NaCl}$ (aqueous solution) in it.

Pure demineralised (deionised) water is obtained by passing water successively through a cation exchange and anion exchange resins. In the cation exchange process,

$$ 2 R \mathrm{H}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \rightleftharpoons \underset{\begin{array}{l} \text { (Cation exchange } \\ \text { resin in the } \mathrm{H}^{+} \text {form) } \end{array}}{M R_{2}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})} $$

$\mathrm{H}^{+}$exchanges for $\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$ and other cations present in water.

This process results in proton release and thus, makes the water acidic. In the anion exchange process

$$ R NH_{2}(s)+H_{2} O(l) \rightleftharpoons R \stackrel{+}{N} H_{3} \cdot OH^{-}(s) $$

$R \stackrel{+}{N} H_{3} \cdot OH^{-}$is substituted ammonium hydroxide anion exchange resin.

$$ R \stackrel{+}{N} H_{3} \cdot OH^{-}(s)+X^{-}(aq) \rightleftharpoons R \stackrel{+}{N} H_{3} \cdot X^{-}(s)+OH^{-}(aq) $$

Answer

Molecular hydrides are classified according to the relative numbers of electrons and bonds in Lewis structure as follow

(i) Electron deficient hydrides These type of hydrides contain central atom with incomplete octet. These are formed by 13 group elements, e.g., $BH_{3}$, $AlH_{3}$, etc. To complele their octet they generally exist in polymeric forms such as $B_{2} H_{6}, ~B_{4} H_{10},\left(AlH_{3}\right)_{n}$ etc. These hydrides act as Lewis acids.

(ii) Electron precise hydrides These hydrides have exact number of electrons required to form normal covalent bonds. These are formed by 14 group elements, e.g., $CH_{4}, SiH_{4}$, etc. These are tetrahedral in shape.

(iii) Electron rich hydrides These hydrides contain central atom with excess electrons, which are present as ione pairs.

These are formed by 15,16 and 17 group elements, e.g., $NH_{3}, H_{2} O, HF$, etc. These hydrides act as Lewis bases.

Answer

Heavy water is prepared by prolonged electrolysis of water. Comparison of physical properties of heavy water with those of ordinary water is as follows

Answer

The one chemical reaction for the preparation of $D_{2} O_{2}$ is by the action of $D_{2} SO_{4}$ dissolved in water over $BaO_{2}$.

$$ BaO_{2}+D_{2} SO_{4} \longrightarrow BaSO_{4}+D_{2} O_{2} $$

Answer

By definition, 5 volumes $H_{2} O_{2}$ solution means that $1 ~L$ of this $H_{2} O_{2}$ solution on decomposition produces $5 ~L$ of $O_{2}$ at STP.

$$ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} $$

$$ 2 \times 34 g \longrightarrow 22.7L \quad at \quad STP $$

If $22.7 LO_{2}$ at STP will be obtained from $H_{2} O_{2}=68 ~g$

$\therefore 5 ~L$ of $O_{2}$ at STP will be obtained from $H_{2} O_{2}=\frac{68 \times 5}{22.7} ~g=14.98=15 ~g$

$\therefore$ Strength of $H_{2} O_{2}$ in 5 volume $H_{2} O_{2}$ solution $=15 ~g ~L^{-1}$.

$\Rightarrow \quad$ Percentage strength of $H_{2} O_{2}$ solution $=\frac{15}{1000} \times 100=1.5 $%

Therefore, strength of $H_{2} O_{2}$ in 5 volume $H_{2} O_{2}$ solution $=15 ~g / L=1.5 $% $H_{2} O_{2}$ solution.

Answer

(i) $H_{2} O_{2}$ has a non-planar structure. The molecular dimensions in the gas phase and solid phase are given below

(a) gas phase

(b) solid phase

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~k $, dihedral angle is $90.2^{\circ}$.

(ii) $H_{2} O_{2}$ is better oxidising agent than water as discussed below

(a) $H_{2} O_{2}$ oxidises an acidified solution of $KI$ to give $I_{2}$ which gives blue colour with starch solution but $H_{2} O$ does not.

$$ \begin{gathered} 2 KI +H_{2} SO_{4}+H_{2} O_{2} \\ ~K_{2} SO_{4}+2 H_{2} O+I_{2} \end{gathered} $$

(b) $H_{2} O_{2}$ turns black PbS to white $PbSO_{4}$ but $H_{2} O$ does not.

$$ PdS+4 H_{2} O_{2} \rightarrow PbSO_{4}+4 H_{2} O $$

Thinking Process

The data given in the question shows that melting point, enthalpy of vaporisation and viscosity of $D_{2} O$ is more than that of $H_{2} O$. The intermolecular force is directly proportional to these three parameters.

Answer

Given that,

From this data, it is concluded that the values of melting point, enthalpy of vaporisation and viscosity depend upon the intermolecular forces of attraction.

Since, their values are higher for $D_{2} O$ as compared to those of $H_{2} O$, therefore, intermolecular forces of attraction are stronger in $D_{2} O$ than in $H_{2} O$.

Answer

The isotope of hydrogen which contains one proton and one neutron is deuterium (D). Therefore, when dideuterium reacts with dioxygen, heavy water $\left(\mathrm{D}_{2} \mathrm{O}\right)$ is produced.

$$ \underset{\text { Dideuterium }}{2 D_{2}(g)}+\underset{\text { Dioxygen }}{O_{2}(g)} \xrightarrow{\text { Heat }} \underset{\begin{array}{c} \text { Deuterium oxide } \\ \text { (Heavy water) } \end{array}}{2 D_{2} O} $$

The reactivity of $H_{2}$ and $D_{2}$ towards oxygen will be different. Since, the $D-D$ bond is stronger than $H-H$ bond, therefore, $H_{2}$ is more reactive than $D_{2}$.

Answer

$F$ is smaller and more electronegative than $Cl$, so it forms stronger $H$-bonds as compared to $Cl$. As the consequence, more energy is needed to break the $H$-bonds in $HF$ than $HCl$ and hence the boiling point of $HF$ is higher than that of $HCl$.

That’s why $\mathrm{HF}$ is liquid and $\mathrm{HCl}$ is a gas.

Answer

The first element of the periodic table is hydrogen and its molecular form is dihydrogen $\left(H_{2}\right)$. When $H_{2}$ reacts with $O_{2}$, water is formed. Water is a liquid at room temperature. When liquid water freezes, it expands to form ice.

Density of ice is lower than that of liquid water and hence ice floats over water. Water is amphoteric in nature. It acts as a base in presence of strong acids and as an acid in presence of strong bases.

$$ \underset{Base_1}{H_2O(l)} + \underset{Acid_2}{H_2S(aq)} \longrightarrow \underset{Acid_1}{H_3O^+(aq)} + \underset{Base_2}{HS^-(aq)} $$

$$ \underset{Acid_1}{H_2O(l)} + \underset{Base_2}{NH_3(aq)} \longrightarrow \underset{Acid_2}{NH_4^+(aq)} + \underset{Base_1}{OH^-(aq)} $$

Due to amphoteric character, water undergoes self ionisation as shown below

$ \underset{\text{Acid}_1}{H _2O(l)} + \underset{Base _2}{H _2O(l)} \leftrightharpoons \underset{\substack{Acids_2 \\ \text{(Conjugate} \\ \text{acid)}}}{H_3O^+(aq)} + \underset{\substack{Base_1 \\ \text{(Conjugate} \\ \text{base)}}}{OH^-(aq)}$

This self ionisation of water is called auto-protolysis or autoionisation.

Answer

(i) The name of the compound is hydrogen peroxide, $H_{2} O_{2}$. It acts as an oxidising agent as well as reducing agent in both acidic and basic medium.

(ii) $H_{2} O_{2}$ decomposes slowly on exposure to light and dust particles. In the presence of metal surfaces or traces of alkali present in glass containers, the decomposition of $H_{2} O_{2}$ is catalysed.

It is, therefore, stored in wax lined glass or plastic vessels in dark. Urea is added as a negative catalyst or stabiliser to check its decomposition.

$$ 2 H_{2} O_{2}(l) \stackrel{h v}{\longrightarrow} 2 H_{2} O(l)+O_{2}(g) $$

Answer

Hydrogen resembles alkali metals, i.e., $\mathrm{Li}, \mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}$ and $\mathrm{Fr}$ of group I of the periodic table in the following respects

(i) Like alkali metals, hydrogen also contain one electron in its outermost (valence) shell and exhibit +1 oxidation state.

(ii) Like alkali metals, hydrogen also loses its only electron to form hydrogen ion, i.e., $\mathrm{H}^{+}$ (proton).

(iii) Like alkali metals, hydrogen combines with electronegative elements (non-metals) such as oxygen, halogens and sulphur forming their oxides, halides and sulphides respectively.

(iv) Like alkali metals, hydrogen also acts as a strong reducing agent.

Answer

Hydrogen has one electron which it can either lose or gain or share to acquire noble gas, i.e., helium gas configuration.

Therefore, in principle, it can form either ionic or covalent bonds. But the ionisation enthalpy of hydrogen is very high ( $\left.1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ and its electron gain enthalpy is only slightly negative $\left(-73 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$.

From this consequence, it does not have a high tendency to form ionic bonds but rather prefers to form only covalent bonds.

Answer

The ionisation enthalpy of hydrogen higher than that of sodium. Both hydrogen and sodium have one electron in the valence shell. But the size of hydrogen is much smaller than that of sodium and hence, the ionisation enthalpy of hydrogen is much higher $\left(1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ than that of sodium $\left(496 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$.

Answer

Basic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen. Hydrogen is a gas at room temperature.

However, by cooling and applying high pressure, gaseous $H_{2}$ can be converted into liquid $H_{2}$ which has much smaller volume and hence can be transported easily. Thus, the basic property of hydrogen which is useful for hydrogen economy is that it can be converted into a liquid by cooling under high pressure.

Answer

Following are the importance of heavy water

(i) It is extensively used as a moderator in nuclear reactors.

(ii) It is used as a tracer compound in the study of reaction mechanism.

(iii) It is used for the preparation of other deuterium compounds such as $CD_{4}, D_{2} SO_{4}$, etc.

Answer

The Lewis structure of hydrogen peroxide is

Answer

Following are the chemical equation of $H_{2} O_{2}$ in which it behaves as an oxidising as well as reducing agent

(i) $H_{2} O_{2}$ oxidises acidified $KI$ to iodine.

$$ 2 KI+H_{2} O_{2}+H_{2} SO_{4} \longrightarrow I_{2}+K_{2} SO_{4}+2 H_{2} O $$

(ii) $H_{2} O_{2}$ reduces $KMnO_{4}$ to $MnO_{2}$ in alkaline medium.

$$ 2 KMnO_{4}+3 H_{2} O_{2} \longrightarrow 2 MnO_{2}+2 KOH+3 O_{2}+2 H_{2} O $$

Answer

The bleaching action of hydrogen peroxide is due to the nascent oxygen which it liberates on decomposition.

$$ H_{2} O_{2} \longrightarrow H_{2} O+[O] $$

The nascent oxygen combines with colouring matter, in turn, gets oxidised. Thus, the bleaching action of $H_{2} O_{2}$ is due to the oxidation of colouring matter by nascent oxygen. It is used for the bleaching of delicate materials like ivory, feathers, silk, wool etc.

Colouring matter $+[\mathrm{O}] \longrightarrow$ Colourless matter

Answer

Oxygen is more electronegative $(\mathrm{E.N.}=3.5)$ than hydrogen $(\mathrm{E.N.}=2.1)$ hence, $\mathrm{O}-\mathrm{H}$ bond is polar. In the water molecule, two polar $\mathrm{O}-\mathrm{H}$ bonds are present which are held together at an angle of $104.5^{\circ}$. Due to the resultant of these two dipoles, water molecule is polar and has an dipole moment of 1.84 Debye.

Answer

Water show high boiling point as compared to hydrogen sulphide due to high electronegativity of oxygen ( $\mathrm{E.N.}=3.5$ ). Water undergoes extensive $\mathrm{H}$ - bonding as a result of which water exists as associated molecule.

For breaking these hydrogen bond, a large amount of energy is needed and hence the boiling point of $\mathrm{H}_{2} \mathrm{O}$ is high.

In other words, due to lower electronegativity of $S(E.N.=2.5)$, hydrogen sulphide do not undergo $H$-bonding. Consequently, $H_{2} ~S$ exists as discrete molecule and hence its boiling point is much lower than that of $H_{2} O$. Thats why $H_{2} ~S$ is a gas at room temperature.

Answer

Dilute solutions of $H_{2} O_{2}$ cannot be concentrated by heating because it decomposes much below its boiling point.

$$ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} $$

1 % $H_{2} O_{2}$ is extracted with water and concentrated to $\sim 30 $% (by mass) by distillation under reduced pressure. It can be further concentrated to $\sim 85 $% by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $H_{2} O_{2}$.

Answer

Hydrogen peroxide is decomposed by rough surfaces of glass, alkali oxides present in it and light to form $H_{2} O$ and $O_{2}$.

$$ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} $$

To prevent this decomposition, hydrogen peroxide is usually stored in paraffin wax coated plastic or teflon bottles.

Answer

Hard water contains salts of calcium and magnesium ions. Hard water does not give lather with soap and forms scum/precipitate with soap. Soap containing sodium stearate $\left(C_{17} H_{35} COONa\right)$ reacts with hard water to precipitate out as $Ca / Mg$ stearate.

$$ 2 C_{17} H_{35} COONa(aq)+M^{2+}(aq) \longrightarrow\left(C_{17} H_{35} COO\right)_{2} M \downarrow+2 Na^{+}(aq) $$

(where, $M$ is $Ca / Mg$ )

It is therefore, unsuitable for laundry.

Answer

$H_{2} SO_{4}$ acts as a catalyst for decomposition of $H_{2} O_{2}$. Therefore, some weaker acids such as $H_{3} PO_{4}, H_{2} CO_{3}$ is preferred over $H_{2} SO_{4}$ for preparing $H_{2} O_{2}$ from peroxides.

$3 \mathrm{BaO}_2+2 \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow \underset{\substack{\text { (Insoluble) }}}{\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2}+3 \mathrm{H}_2 \mathrm{O}_2$

this method has the advantage over $BaO_{2}$ − $H_{2}SO_{4}$ method since almost all the heavy metal (e.g.,Pb etc.) impurities present in $Ba O_{2}$ and which catalyse the decomposition of $H_{2} O_{2}$ are removed as insoluble phosphates. As a result , the resulting solution of $H _{2} O _{2}$ has good keeping properties.

Answer

In water, oxygen has $s p^{3}$-hybridisation and the bond angle of $\mathrm{HOH}$ should have been $109^{\circ} 28^{\prime}$. In $\mathrm{H}_{2} \mathrm{O}$, the oxygen atom is surrounded by two shared pairs and two lone pairs of electrons. From VSEPR theory, lone pair - lone pair repulsions are stronger than bond pair-bond pair repulsions.

As a result, the bond angle of $\mathrm{HOH}$ in water slightly decreases from the regular tetrahedral angle of $109^{\circ} .28^{\prime}$ to $104.5^{\circ}$.

Answer

Fluorine is a strong oxidising agent, it oxidises $H_{2} O$ to $O_{2}$ or $O_{3}$. The reactions are as follows

$$ \begin{aligned} 2 ~F_{2}(g)+2 H_{2} O(l) \longrightarrow O_{2}(g)+4 H^{+}(aq)+4 ~F^{-}(aq) \\ 3 ~F_{2}(g)+3 H_{2} O(l) \longrightarrow O_{3}(g)+6 H^{+}(aq)+6 ~F^{-}(aq) \end{aligned} $$

Answer

Water has the ability to act as an acid as well as base, i.e., it behaves as an amphoteric substance. From the Bronsted Lowry theory, it acts as an acid with $NH_{3}$ and a base with $H_{2} ~S$.

$$ \begin{gathered} H_{2} O(l)+NH_{3}(aq) \longrightarrow OH^{-}(aq)+NH_{4}^{+}(aq) \\ H_{2} O(l)+H_{2} ~S(aq) \longrightarrow H_{3} O^{+}(aq)+HS^{-}(aq) \end{gathered} $$

The auto - protolysis (self-ionisation) of water takes place. The reaction are as follows

$ \underset{Acid}{H_2O(l)}+ \underset{Base}{H_2O(l)} \longrightarrow \underset{\text{Conjugate acid}}{H_3O^+}(aq) + \underset{\text{Conjugate base}}{OH^-(aq)}$

Answer

A. $\rightarrow(9,10)$

B. $\rightarrow(4,5,7,8,9)$

C. $\rightarrow(6)$

D. $\rightarrow(1,3)$

E. $\rightarrow(2,4)$

F. $\rightarrow(5)$

A. Synthesis gas - Synthesis of methanol,Mixture of $\mathrm{Co}$ and $\mathrm{H}_{2}$

B. Dihydrogen — Reducing agent ,Stoichiometric compounds of s-block elements ,$Zn+NaOH$ ,$Zn+$ dil. $H_{2} SO_{4}$ ,Synthesis of methanol

C. Heavy water - Prolonged electrolysis of water

D. Calgon $-Na_{2}\left[Na_{4}\left(PO_{3}\right)_{6}\right]$ ,Softening of water

E. Hydrogen peroxide - Oxidising agent ,Reducing agent

F. Salt like hydrides - Stoichiometric compounds of s-block elements.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(4)$

C. $\rightarrow$ (2)

D. $\rightarrow(1)$

E. $\rightarrow$ (3)

A. Electrolysis of water produce hydrogen and oxygen.

B. Lithium aluminium hydride is used as reducing agent.

C. Hydrogen chloride is a polar molecule.

D. Heavy water is used in atomic reactor as moderator.

E. Atomic hydrogen recombines on metal surface to generate high temperature.

Answer

A. $\rightarrow(2,4)$

B. $\rightarrow$ (3)

C. $\rightarrow(1,3)$

A. Hydrogen peroxide is used as a perhydrol and propellant.

B. Sodium hexametaphosphate is used in Calgon method.

C. Permanent hardness of hard water is removed by zeolite and sodium hexametaphosphate.

Answer

(a) Statements of assertion and reason both are correct and reason is the correct explanation of assertion.

$ Na_2CO_3 + \underset{\substack{\text{or }\\ \text{ CaSO}_4 \\ \text{( from hard } \\ \text{ water )}}}{MgSO_4} \longrightarrow Na_2SO_4 +\underset{\substack{\text{or}\\ \text{CaCO}_3 \\ \text{(insoluble)}}}{MgCO_3} $

Answer

(a) Statements of assertion and reason both are correct and reason is the correct explanation of assertion. Since, metals like Pd and Pt adsorbs a large volume of hydrogen, hence, these are used as a storage media for it.

Answer

Atomic hydrogen is highly unstable. Since, the electronic configuration of atomic hydrogen is $1 s^{1}$, it needs one more electron to complete its configuration and gain stability. Therefore, atomic hydrogen is very reactive and combines with almost all the elements.

It, however, reacts in three different ways i.e.,

(i) by loss of its single electron to form $\mathrm{H}^{+}$,

(ii) by gain of one electron to form $\mathrm{H}^{-}$and

(iii) by sharing its electron with other atoms to form single covalent bonds. In contrast, the bond dissociation energy form $\mathrm{H}-\mathrm{H}$ bond is very high ( $435.88 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ). As a result, molecular hydrogen is almost inert at room temperature and hence reacts only with a few elements.

Answer

(i) $\mathrm{D}_{2} \mathrm{O}$ can be prepared by prolonged electrolysis of water.

(ii) Physical properties

(a) $\mathrm{D}_{2} \mathrm{O}$ is colourless, odourless, tasteless liquid. It has maximum density $-1.1073 \mathrm{~g} \mathrm{~mL}^{-1}$ at $11.6^{\circ} \mathrm{C}$ (Maximum density of water at $4^{\circ} \mathrm{C}$ ).

(b) Solubility of salts in heavy water is less than in ordinary water because it is more viscous than ordinary water.

(c) Nearly, all physical constants of $D_{2} O$ are higher than $H_{2} O$. It is due to the greater nuclear mass of deuterium atom than $H$-atom and stronger $H$-bonding in $D_{2} O$ than $H_{2} O$.

(iii) Exchange reactions of hydrogen with deuterium

$$ \begin{aligned} NaOH+D_{2} O & \longrightarrow NaOD+HOD \\ HCl+D_{2} O & \longrightarrow DCl+HOD \\ NH_{4} Cl+D_{2} O & \longrightarrow NH_{3} DCl+HOD \end{aligned} $$

Answer

(i) Industrially, $H_{2} O_{2}$ is prepared by the auto-oxidation of 2-alkylanthraquinols.

$$ \text { 2-ethylanthraquinol } \underset{H_{2} / Pd}{\stackrel{O_{2} / \text { (air) }}{\rightleftarrows}} H_{2} O_{2}+\text { Oxidised product } $$

In this case, 1 % $H_{2} O_{2}$ is formed. It is extracted with water and concentrated to $\sim 30 $% (by mass) by distillation under reduced pressure.

It can be further concentrated to $\sim 85 $% by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $H_{2} O_{2}$.

(ii) $H_{2} O_{2}$ has a non-planar structure.

The molecular dimensions in the gas phase and solid phase are shown below

(a)

(b)

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~K$, dihedral angle is $90.2^{\circ}$

In the gas phase, $\mathrm{H}_{2} \mathrm{O}$ is a bent molecule with a bond angle of $104.5^{\circ}$ and $\mathrm{O}-\mathrm{H}$ bond length of $95.7 \mathrm{pm}$ as shown below

(a)

(b)

(a) The bent structure of water;

(b) The water molecule as a dipole and

(c) The orbital overlop picture in water molecule.

(iii) Following are the three important uses of $H_{2} O_{2}$

(a) In daily life, it is used as a hair bleach and as a mild disinfectant. As an antiseptic it is sold in the market as perhydrol.

(b) It is used in the synthesis of hydroquinone, tartaric acid and certain food products and pharmaceuticals (cephalosporin) etc.

(c) It is employed in the industries as a bleaching agent for textiles, paper pulp, leather, oils, fats etc.

Answer

(i) $H_{2} O_{2}$ is industrially manufactured by the auto-oxidation of 2alkylanthraquinols

2-ethylanthraquinol $ \rightleftharpoons [H_2/Pd]{O_2/(air)} H_2O_2 +$ Oxidised product

In this case, 1 % $ H_{2} O_{2}$ is formed. It is extracted with water and concentrated to $\sim 30 $% (by mass) by distillation under reduced pressure. It can be further concentrated to $\sim 85 $% by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $H_{2} O_{2}$.

(ii) $H_{2} O_{2}$ acts as an oxidising as well as reducing agent in both acidic and alkaline media. Following reactions are described below

(a) Oxidising action in acidic medium

$$ \begin{array}{r} 2 Fe^{2+}(aq)+2 H^{+}(aq)+H_{2} O_{2}(aq) \longrightarrow 2 Fe^{3+}(aq)+2 H_{2} O(l) \\ PbS(s)+4 H_{2} O_{2}(aq) \longrightarrow PbSO_{4}(~s)+4 H_{2} O(l) \end{array} $$

(b) Reducing action in acidic medium

$$ \begin{aligned} 2 MnO_{4}^{-}+6 H^{+}+5 H_{2} O_{2} & \longrightarrow 2 Mn^{2+}+8 H_{2} O+5 O_{2} \\ HOCl+H_{2} O_{2} & \longrightarrow H_{3} O^{+}+Cl^{-}+O_{2} \end{aligned} $$

(c) Oxidising action in basic medium

$$ \begin{aligned} 2 Fe^{2+}+H_{2} O_{2} & \longrightarrow 2 Fe^{3+}+2 OH^{-} \\ Mn^{2+}+H_{2} O_{2} & \longrightarrow Mn^{4+}+2 OH^{-} \end{aligned} $$

(d) Reducing action in basic medium

$$ \begin{aligned} I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\ 2 MnO_{4}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-} \end{aligned} $$

Answer

(i) Molar mass of $H_{2} O_{2}=34 ~g ~mol^{-1}$

$1 ~L$ of $5 M$ solution of $H_{2} O_{2}$ will contain $34 \times 5 ~g H_{2} O_{2}$

$2 ~L$ of $5 M$ solution of $H_{2} O_{2}$ will contain $34 \times 5 \times 2=340 ~g H_{2} O_{2}$

Mass of $H_{2} O_{2}$ present in $2 ~L$ of 5 molar solution $=340 ~g$

(ii) $0.2 ~L$ (or $200 ~mL$ ) of $5 M$ solution will contain

$$ \begin{gathered} \frac{340 \times 0.2}{2}=34 ~g H_{2} O_{2} \\ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} \end{gathered} $$

$68 ~g H_{2} O_{2}$ on decompósitíoñiquil give $32 ~g O_{2} \quad 2 \times 16=32 ~g$

$\therefore 34 ~g H_{2} O_{2}$ on decomposition will give $\frac{32 \times 34}{68}=16 ~g O_{2}$

Answer

Since, a colourless liquid ‘A’ contains only hydrogen and oxygen and decomposes slowly on exposure to light but is stabilised by addition of urea, therefore, liquid $A$ may be hydrogen peroxide.

(i) The structure of $H_{2} O_{2}$ is

(a) Gas phase

5&top_left_x=818)

(b) Solid phase

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~k$, dihedral angle is $90.2^{\circ}$.

(ii) $2 H_{2} O_{2} \xrightarrow[\text { Sunlight }]{\text { hv }} 2 H_{2} O+O_{2}$

Answer

It is LiH because it has significant covalent character due to the smallest alkali metal, Li. $\mathrm{LiH}$ is very stable. It is almost unreactive towards oxygen and chlorine.

It reacts with $Al_{2} Cl_{6}$ to form lithium aluminium hydride.

$$ 8 LiH+Al_{2} Cl_{6} \longrightarrow 2 LiAlH_{4}+6 LiCl $$

Answer

Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid.

$$ 2 \mathrm{Na}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{Na}^{+} \mathrm{H}^{-} $$

It reacts violently with water to produce $\mathrm{H}_{2}$ gas

$$ 2 NaH+2 H_{2} O \longrightarrow 2 NaOH+2 H_{2} $$

In solid state, $\mathrm{NaH}$ does not conduct electricity. On electrolysis, in its molten state it gives $\mathrm{H}_{2}$ at anode and $\mathrm{Na}$ at cathode.

$$ \mathrm{Na}^{+} \mathrm{H}^{-}(l) \xrightarrow{\text { Electrolysis }} \underset{\text { At cathode }}{2 \mathrm{Na}(l)}+\underset{\text { At anode }}{\mathrm{H}_{2}(g)} $$