Answer

Hydrogen is the first element of the periodic table. Its electronic configuration is $[1 s^{1}]$. Due to the presence of only one electron in its $1 s$ shell, hydrogen exhibits a dual behaviour, i.e., it resembles both alkali metals and halogens.

Resemblance with alkali metals:

(i) Like alkali metals, hydrogen contains one valence electron in its valency shell.

$H: 1 s^{1}$

$Li:[He] 2 s^{1}$

$ Na : [Ne]3s^1 $

Hence, it can lose one electron to form a unipositive ion.

(ii) Like alkali metals, hydrogen combines with electronegative elements to form oxides, halides, and sulphides.

Resemblance with halogens:

(i) Both hydrogen and halogens require one electron to complete their octets.

$H: 1 s^{1}$

$F: 1 s^{2} 2 s^{2} 2 p^{5}$

$Cl: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$

Hence, hydrogen can gain one electron to form a uninegative ion.

(ii) Like halogens, it forms a diatomic molecule and several covalent compounds.

Though hydrogen shows some similarity with both alkali metals and halogens, it differs from them on some grounds. Unlike alkali metals, hydrogen does not possess metallic characteristics. On the other hand, it possesses a high ionization enthalpy. Also, it is less reactive than halogens.

Owing to these reasons, hydrogen cannot be placed with alkali metals (group I) or with halogens (group VII). In addition, it was also established that $H^{+}$ ions cannot exist freely as they are extremely small. $H^{+}$ ions are always associated with other atoms or molecules. Hence, hydrogen is best placed separately in the periodic table.

Answer

Hydrogen has three isotopes. They are:

1. Protium, ${ }^{1}_1 H$,
2. Deuterium, ${ }^{2}_1 H$ or $D$, and
3. Tritium, $ _1^{3} H$ or $T$

The mass ratio of protium, deuterium and tritium is 1:2:3.

Answer

The ionization enthalpy of hydrogen atom is very high $(1312 kJ mol^{-1})$. Hence, it is very hard to remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic $(H_2)$ molecule.

Answer

Dihydrogenis produced by coal gasification method as:

$ \underset{(coal)}{C_{(s)}}+H_2 O _{(g)} \xrightarrow{1270 K} CO _{(g)}+H _{2(g)} $

The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.

$ CO_{(g)}+H_2 O _{(g)} \xrightarrow[\text{ Catalyst }]{673 K} CO _{2(g)}+H _{2(g)} $

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.

Answer

Dihydrogen is prepared by the electrolysis of acidified or alkaline water using platinum electrodes. Generally, 15-$20 %$ of an acid $(H_2 SO_4)$ or a base $(NaOH)$ is used.

Reduction of water occurs at the cathode as:

$ 2 H_2 O+2 e^{-} \longrightarrow 2 H_2+2 OH^{-} $

At the anode, oxidation of $ OH^- $ ions takes places as:

$ 2 OH^{-} \longrightarrow H_2 O+\frac{1}{2} O_2+2 e^{-} $

$\therefore$ Net reaction can be represented as:

$ H_2 O _{(l)} \longrightarrow H _{2(g)}+\frac{1}{2} O _{2(g)} $

Electrical conductivity of pure water is very low owing to the absence of ions in it. Therefore, electrolysis of pure water also takes place at a low rate. If an electrolyte such as an acid or a base is added to the process, the rate of electrolysis increases. The addition of the electrolyte makes the ions available in the process for the conduction of electricity and for electrolysis to take place.

Answer

(i)

$H _{2}(g)+Mn O {(s)} \xrightarrow{\Delta} Mn {(s)}+H_2 O {(l)}$

(ii)

$CO {(g)}+2 H _{2}(g) \xrightarrow[\text{ catalyst }]{\Delta} CH_3 OH {(l)}$

(iii)

$ C_3 H _{8}(g)+3 H_2 O {(g)} \xrightarrow[\text{ catalyst }]{\Delta} 3 CO {(g)}+7 H _{2} (g) $

(iv)

$Zn {(s)}+2 NaOH {(aq)} \xrightarrow{\text{ heat }} \underset{\text{(Sodium zincate)}}{Na_2 ZnO _{2}(aq)}+H _{2}(g)$

Answer

The ionization enthalpy of $H-H$ bond is very high $(1312 kJ mol^{-1})$. This indicates that hydrogen has a low tendency to form $H^{+}$ ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules $(H_2)$, hydrides with elements, and a large number of covalent bonds.

Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.

Answer

Molecular hydrides are classified on the basis of the presence of the total number of electrons and bonds in their Lewis structures as:

(a) Electron -deficient hydrides

(b) Electron-precise hydrides

(c) Electron-rich hydrides

(a) An electron-deficient hydride has very few electrons, less than that required for representing its conventional Lewis structure e.g. diborane $(B_2 H_6)$.

In $B_2 H_6$, there are six bonds in all, out of which only four bonds are regular two centered-two electron bonds. The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

(b) An electron-precise hydride has a sufficient number of electrons to be represented by its conventional Lewis structure e.g. $CH_4$. The Lewis structure can be written as:

Four regular bonds are formed where two electrons are shared by two atoms.

(c) An electron-rich hydride contains excess electrons as lone pairs e.g. $NH_3$

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Answer

Electron-deficient hydrides are compounds that do not have enough electrons to form normal covalent bonds. This lack of electrons makes them behave differently compared to electron-rich compounds.

A prime example of an electron-deficient hydride is diborane ( B 2 H 6 ). In diborane, boron has only three valence electrons and forms bonds that do not satisfy the octet rule.

The structure of diborane is non-planar and features unique bonding.

It has a three-center, two-electron bond where two boron atoms share a pair of electrons with a bridging hydrogen atom. This results in a structure that is not typical for most hydrides.

Electron-deficient hydrides act as Lewis acids. They can accept electron pairs from Lewis bases. For instance, when diborane reacts with a Lewis base like trimethylamine $(NMe_3)$, it can accept electron pairs, making it more stable.

$ B_2 H_6+2 NMe_3 \longrightarrow 2 BH_3 \cdot NMe_3 \\ $

Due to their electron deficiency, these hydrides are reactive and can participate in various chemical reactions. They can form adducts with Lewis bases and can also engage in reactions with other electron-rich species.

Electron-deficient hydrides can temporarily become electron-rich when they interact with electron donors. This temporary state allows them to stabilize and participate in chemical reactions.

Hence, electron-deficient hydrides like diborane exhibit unique structural characteristics, such as threecenter, two-electron bonds, and they behave as Lewis acids in chemical reactions. Their reactivity is driven by their electron deficiency, allowing them to interact with electron-rich species to stabilize their structure.

Answer

For carbon hydrides of type $C_n H _{2 n+2} $, the following hydrides are possible for

$ \begin{gathered} n=1 \Rightarrow CH_4 \\ n=2 \Rightarrow C_2 H_6 \\ n=3 \Rightarrow C_3 H_8 \\ \end{gathered} $

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.

Taking $C_2 H_6$ as an example, the total number of electrons are 14 and the total covalent bonds are seven. Hence, the bonds are regular $2 e^{-}-2c$ bonds.

Hence, hydride $C_2 H_6$ has sufficient electrons to be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets. Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.

Answer

Non-stoichiometric hydrides, also known as interstitial hydrides or metallic hydrides, are compounds formed by hydrogen and metals, where the ratio of metal to hydrogen is not fixed. This means that the composition can vary, and the hydrides can exist in forms that do not adhere to the law of constant composition.

Formation of Non-Stoichiometric Hydrides:

These hydrides are typically formed by transition metals, lanthanides, and actinides. In these cases, hydrogen atoms occupy the interstitial sites within the metal lattice structure. The general formula for these hydrides can be represented as $M_x H_x$, where $M$ is the metal and $x$ can be a fraction or decimal, indicating the variable ratio of metal to hydrogen.

Examples of Non-Stoichiometric Hydrides:

For example, lanthanum hydride (LaH) can have a composition of $ \mathrm{LaH}{2 \cdot 87}$, and titanium hydride (TiH) can vary between $ \mathrm{TiH}{1.5}$ and $ \mathrm{TiH}_{1.8}$. This variability in the metal to hydrogen ratio is what characterizes non-stoichiometric hydrides.

Alkali metals, such as lithium, sodium, and potassium, do not form non-stoichiometric hydrides. Instead, they primarily form ionic hydrides. This is because alkali metals have low electronegativity, which leads to the formation of ionic bonds with hydrogen, resulting in a fixed stoichiometric ratio.

The reason alkali metals do not form non-stoichiometric hydrides is that they typically form ionic hydrides, which have a defined stoichiometry. The ionic hydrides consist of a metal cation and a hydride anion $\left(\mathrm{H}^{-}\right)$, leading to a stable ionic lattice structure, unlike the variable compositions seen in non-stoichiometric hydrides.

So, non-stoichiometric hydrides are variable composition hydrides formed by transition metals, lanthanides, and actinides, where hydrogen occupies interstitial sites in the metal lattice. Alkali metals do not form these types of hydrides; instead, they form ionic hydrides with fixed stoichiometry due to their low electronegativity.

Answer

Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of $Ni, Pd, Ce$, and $Ac$, hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Answer

Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy ( $435.88 kJ mol^{-1}$ ). This energy can be used to generate a temperature of $4000 K$, which is ideal for welding and cutting metals. Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Answer

Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom (like $ \mathrm{F}, \mathrm{O}$, or N ) and is attracted to another electronegative atom. The strength of hydrogen bonding is influenced by the polarity of the bond.

The electronegativities of the relevant atoms are as follows:

Fluorine (F) - 4.0

Oxygen (O) - 3.5

Nitrogen (N) - 3.0

Since fluorine is the most electronegative, it will create the most polar bond when bonded to hydrogen.

HF (Hydrogen Fluoride):

The bond between H and F is highly polar due to the high electronegativity of F . This results in a significant partial positive charge on H and a significant partial negative charge on F , leading to strong hydrogen bonding.

$ \mathrm{H}_2 \mathrm{O}$ (Water):

Water has two hydrogen atoms bonded to one oxygen atom. The O-H bond is also polar, but the overall hydrogen bonding strength is less than that of HF due to the lower electronegativity of O compared to F .

$ \mathrm{NH}_3$ (Ammonia):

Ammonia has one nitrogen atom bonded to three hydrogen atoms. While $N$ is electronegative, it is less so than $F$, resulting in weaker hydrogen bonds compared to HF and $ \mathrm{H}_2 \mathrm{O}$.

Among $ \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$, and HF , HF has the highest magnitude of hydrogen bonding due to the high polarity of the $ \mathrm{H}-\mathrm{F}$ bond resulting from the high electronegativity of fluorine.

So, HF (Hydrogen Fluoride) has the highest magnitude of hydrogen bonding due to its maximum polarity resulting from the high electronegativity of fluorine.

Answer

Saline hydrides such as $ \mathrm{NaH}, \mathrm{CaH}_2$ violently react with water and form metal hydroxide and hydrogen gas.

$ \begin{aligned} & \mathrm{NaH}(\mathrm{~s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g}) \\ & \mathrm{CaH}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{aq})+2 \mathrm{H}_2(\mathrm{~g}) \end{aligned} $

During these reactions, large amount of energy is released so that hydrogen catches fire. $ \mathrm{CO}_2$ cannot extinguish this fire as it is reduced by hot metal hydride.

$ \mathrm{NaH}+\mathrm{CO}_2 \rightarrow \mathrm{HCOONa} $

To extinguish this fire, sand (a highly stable solid) can be used.

Answer

(i) Ionic compounds conduct electricity whereas covalent compounds does not.

$ \mathrm{BeH}_2$ is a covalent hydride and does not conduct electricity. $ \mathrm{CaH}_2$ is an ionic hydride and conducts electricity in molten state. $ \mathrm{TiH}_2$ is metallic in nature and conducts electricity at room temperature.

Hence, the increasing order of electrical conductivity is:

$ \mathrm{BeH}_2<\mathrm{CaH}_2<\mathrm{TiH}_2$.

(ii) The ionic character of bond depends on electronegativity difference between two atoms. When the electronegativity difference is larger, the ionic character is smaller.

On moving down the group of alkali metals, the electronegativity decreases from Li to Cs.

Hence the ionic character increases in the order is:

$ \mathrm{LiH}<\mathrm{NaH}<\mathrm{CsH}$.

(iii) Bond dissociation energy depends on bond strength. Bond strength depends on the attractive and repulsion forces present in a molecule. Due to higher nuclear mass of $ \mathrm{D}_2$, the attraction between nucleus and bond pair in D-D is stronger than in $ \mathrm{H}-\mathrm{H}$. This results in greater bond strength and higher bond dissociation enthalpy. Thus, the bond dissociation enthalpy of $ \mathrm{D}-\mathrm{D}$ is higher than that of $ \mathrm{H}-\mathrm{H}$.

The bond dissociation enthalpy of F-F is minimum as the repulsion between the bond pair and lone pairs of F is strong.

Hence, the increasing order of bond dissociation enthalpy is:

$ \mathrm{F}-\mathrm{F}<\mathrm{H}-\mathrm{H}<\mathrm{D}-\mathrm{D}$.

(iv) NaH is an ionic hydride and can easily donate its electrons. It is most reducing. $ \mathrm{MgH}_2$ and $ \mathrm{H}_2 \mathrm{O}$ are covalent hydrides. $ \mathrm{H}_2 \mathrm{O}$ has lower reducing power than $ \mathrm{MgH}_2$ as its bond dissociation energy is higher.

Hence, the increasing order of the reducing property is:

$ \mathrm{H}_2 \mathrm{O}<\mathrm{MgH}_2<\mathrm{NaH}$.

Answer

In water, O is $s p^3$-hydridized. Due to stronger lone pair-lone pair repulsions than bond pair-bond pair repulsions, the HOH bond angle decreases from $109.5^{\circ}$ to $104.5^{\circ}$.

Thus, water is a bent molecule.

$ \mathrm{H}_2 \mathrm{O}_2$ has a non-planar structure.

The two oxygen atoms are linked to each other by a single covalent bond (i.e., peroxide bond) and each oxygen is further linked to a hydrogen atom by a single covalent bond.The two $ \mathrm{O}-\mathrm{H}$ bonds are, however, in different planes. The dihedral angle between the two planes being $111.5^{\circ}$ in the gas phase. Thus, the structure of $ \mathrm{H}_2 \mathrm{O}_2$ is like that of an open book.

Answer

The reaction between two water molecules to produce hydronium ion and hydroxide ion is known as autoprotolysis of water. This is self ionization of water.

$ 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-} $

This reaction indicates the amphoteric nature of water. It can act as an acid as well as a base. One water molecule donates electron whereas other water molecule accepts electron.

Answer

The reaction between of a redox reaction as water is gettting oxidized to fluroine is being reduced to fluoride ion. The oxdation number of various species can be represented as:

Fluorine is redcued from zero to ( -1 ) oxidation state. A decrease in oxidation state indicates the reduction of fluorine.

Water is oxidised from (-2) to zero oxidation state. An increase in oxidation state indicates oxidation of water.

Answer

(i)

Lead sulfide $(\mathrm{PbS})$ reacts with hydrogen peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ to form lead sulfate $\left(\mathrm{PbSO}_4\right)$ and water $\left(\mathrm{H}_2 \mathrm{O}\right)$.

$ \mathrm{PbS}(s)+\mathrm{H}_2 \mathrm{O}_2(a q) \rightarrow \mathrm{PbSO}_4(s)+\mathrm{H}_2 \mathrm{O}(l) $

This is a redox reaction because the oxidation states of lead and oxygen change.

(ii)

In acidic medium, permanganate ion $\left(\mathrm{MnO}_4^{-}\right)$ reacts with hydrogen peroxide to produce manganese ions $\left(\mathrm{Mn}^{2+}\right)$, water, and oxygen gas.

$ \begin{gathered} \mathrm{2MnO}_4^{-}(a q)+\mathrm{5H}_2 \mathrm{O}_2(a q) \rightarrow 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l)+5 \mathrm{O}_2(g) \end{gathered} $

This is also a redox reaction as the oxidation states of manganese and oxygen change.

(iii)

Calcium oxide $(\mathrm{CaO})$ reacts with water vapor to form calcium hydroxide $\left(\mathrm{Ca}(\mathrm{OH})_2\right)$.

$ \mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(s) $

This is a hydrolysis reaction because a compound reacts with water to produce another compound.

(iv)

Aluminum chloride $\left(\mathrm{AlCl}_3\right)$ reacts with water to produce aluminum oxide $\left(\mathrm{Al}_2 \mathrm{O}_3\right)$ and hydrochloric acid $(HCl)$ .

$ 2 \mathrm{AlCl}_3(g)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Al}_2 \mathrm{O}_3(s)+6 \mathrm{HCl}(a q) $

This is a hydrolysis reaction.

(v)

Calcium nitride $\left(\mathrm{Ca}_3 \mathrm{~N}_2\right)$ reacts with water to form calcium hydroxide $\left(\mathrm{Ca}(\mathrm{OH})_2\right)$ and ammonia $\left(\mathrm{NH}_3\right)$.

$ \mathrm{Ca}_3 \mathrm{~N}_2(s)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2(s)+2 \mathrm{NH}_3(g) $

This is a hydrolysis reaction.

Answer

Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.

The three-dimensional structure of ice is represented as:

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of $276 pm$. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

Answer

The presence of calcium and magnesium bicarbonates $ \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2$ and $ \mathrm{Mg}\left(\mathrm{HCO}_3\right)_2$ causes temporary hardness in water.

The presence of soluble salts of calcium and magnesium, i.e., sulphates and chlorides of calcium and magnesium cause permanent hardness in water.

Answer

The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., $ Na^+ $ , $ Ca^{2+} $ , $ Mg^{2+} $ etc) and anions ( e.g., $ Cl^- $ , $ SO_4^{2-} $ , $ HCO_3^- $ etc ) present in water by $ H^+ $ and $ OH^- $ ions respectively.

Synthetic resins are of two types:

1. Cation exchange resins

2. Anion exchange resins

Cation exchange resins are large organic molecules that contain the $-SO_3 H$ group. The resin is firstly changed to $RNa$ (from $RSO_3 H$ ) by treating it with $NaCl$. This resin then exchanges $Na^{+}$ ions with $Ca^{2+}$ and $Mg^{2+}$ ions, thereby making the water soft.

$ 2 RNa + M^{2+}(aq) \longrightarrow R_2 M {(s)}+2 Na^{+}{(a q)} $

There are cation exchange resins in $H^{+}$ form. The resins exchange $H^{+}$ ions for $Na^{+}, Ca^{2+}$, and $Mg^{2+}$ ions.

$ 2 RH+M^{2+}{(a q)} \leftarrow MR _{2}(s)+2 H^{+}{(a q)} $

Anion exchange resins exchange $OH^{{-}}$ ions for anions like $Cl^{^{-}}, HCO_3^{-}$, and $SO_4^{2-}$ present in water.

$ RNH_{2}(s)+H_2O{(l)} \leftrightharpoons RNH_3^{+} \cdot OH^{-}(s) $

$ \hspace{40mm} \downarrow + X_{(a q)}^{-} $

$ \hspace{40mm} RNH_3^{+} \cdot X^{-}{(s)}+OH ^{-}{(aq)}$

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.

This acidic water is then passed through the anion exchange process where $OH^{-}$ ions neutralize the $H^{+}$ ions and deionize the water obtained.

Answer

Amphoteric nature of water is represented by following chemical reactions.

(1) Water as a base:

$ \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HS}^{-}(\mathrm{aq}) $

(2) Water as an acid:

$ \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{NH}_3(\mathrm{aq}) \rightleftharpoons \mathrm{OH}^{-}+\mathrm{NH}_4^{+} $

(3) Self ionization of water in which water simultaneous acts as acid and base.

$ 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-} $

Answer

To demonstrate that hydrogen peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ can act as both an oxidizing agent and a reducing agent, we can write down specific chemical reactions for each case.

Hydrogen Peroxide as an Oxidizing Agent

Reaction 1:

When hydrogen peroxide reacts with iron(II) ions in an acidic medium, it oxidizes $ \mathrm{Fe}^{2{+}}$ to $ \mathrm{Fe}^{3{+}}$.

$ 2 \mathrm{Fe}^{2+}+2 \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O} $

Reaction 2:

In another reaction, hydrogen peroxide oxidizes manganese(II) ions to manganese(IV) ions in a basic medium.

$ \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{MnO}_2+2 \mathrm{OH}^{-} $

Reaction 3:

Hydrogen peroxide can also oxidize lead sulfide $(\mathrm{PbS})$ to lead sulfate $\left(\mathrm{PbSO}_4\right)$.

$ \mathrm{PbS}+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{PbSO}_4+4 \mathrm{H}_2 \mathrm{O} $

Hydrogen Peroxide as a Reducing Agent

Reaction 1:

In acidic conditions, hydrogen peroxide can reduce permanganate ions $\left(\mathrm{MnO}_4^{-}\right)$ to manganese(II) ions ( $ \mathrm{Mn}^{2+}$ ).

$ 2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{O}_2+8 \mathrm{H}_2 \mathrm{O} $

Reaction 2:

Hydrogen peroxide can also reduce iodine $\left(I_2\right)$ to iodide ions $\left(l^{-}\right)$.

$ \mathrm{I}_2+\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{I}^{-}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 $

Reaction 3:

Additionally, hydrogen peroxide can reduce hypochlorous acid ( HOCl ) to chloride ions $\left(\mathrm{Cl}^{-}\right)$.

$ \mathrm{HOCl}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{O}_2 $

These reactions illustrate that hydrogen peroxide can act as an oxidizing agent by accepting electrons and oxidizing other substances, as well as a reducing agent by donating electrons and reducing other substances.

Answer

Demineralised water is free from all soluble mineral salts. It does not contain any anions or cations.

Demineralised water is obtained by passing water successively through a cation exchange (in the $H^{+}$ form) and an anion exchange (in the $OH^{-}$ form) resin.

During the cation exchange process, $H^{+}$ exchanges for $Na^{+}, Mg^{2+}, Ca^{2+}$, and other cations present in water.

$ 2 RH {(s)}+M^{2+} {(a q)} \leftrightharpoons MR _{2}(s)+2 H^{+} {(a q)} \ldots \ldots{(1)} $

In the anion exchange process, $OH^{-}$ exchanges for anions such as $CO_3^{2-}, SO_4^{2-}, Cl^{-}, HCO_3^{-}$ etc. present in water.

$RNH _{2}(s)+H_2 O {(l)} \leftrightharpoons RNH_3^{+} \cdot OH ^{-} {(s)} $

$ RNH_3^{+} OH ^{-}{(s)}+X ^{-} {(a q)} \leftrightharpoons RNH_3^{+} \cdot X ^{-}{(s)}+OH ^{-} {(a q)} \ldots \ldots{(2)}$

$OH^{-}$ ions liberated in reaction (2) neutralize $H^{+}$ ions liberated in reaction (1), thereby forming water.

$ H^{+}{(a q)} + OH^{-}{(a q)} \longrightarrow H_2 O {(l)} $

Answer

Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants, and animals for survival. Demineralised water is free of all soluble minerals. Hence, it is not fit for drinking. It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.

Answer

Water is essential for all forms of life. It constitutes around 65 percent of the human body and 95 percent of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant.

The high heat of vapourization and heat of capacity of water helps in moderating the climate and body temperature of all living beings.

It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.

Answer

Water is often referred to as the “universal solvent” due to its ability to dissolve a wide range of substances. This property is primarily attributed to its molecular structure and polarity.

Properties of Water:

High Dielectric Constant: Water has a high dielectric constant, which reduces the electrostatic forces between charged particles (ions). This allows ionic compounds to separate into their constituent ions when dissolved in water.

Dipole Moment: Water molecules are polar, meaning they have a positive end (hydrogens) and a negative end (oxygen). This polarity enables water to interact with various solutes effectively.

(i) Dissolving Ionic Compounds:

Ionic compounds dissolve in water due to ion-dipole interactions. The positive and negative ions of the compound are attracted to the oppositely charged ends of the water molecules, leading to the dissociation of the ionic compound into its ions.

Dissolving Covalent Compounds:

Covalent compounds can also dissolve in water, primarily through the formation of hydrogen bonds. When a covalent compound has polar covalent bonds, it can interact with water molecules, allowing it to dissolve.

(ii) Hydrolysis by Water:

Hydrolysis refers to the reaction of water with other substances to form new products. Water can hydrolyze various compounds, including:

Nonmetallic Oxides: These can react with water to form acids or bases e.g.,

${CO}_2+{H}_2 {O} \rightarrow {H}_2{CO}_3$ .

Hydrides: Certain hydrides react with water to release hydrogen gas e.g.,

${NaH}+{H}_2 {O} \rightarrow {NaOH}+{H}_2$ .

Carbides: Some carbides react with water to produce hydrocarbons or other products e.g.,

${CaC}_2+2 {H}_2{O} \rightarrow$ ${Ca}({OH})_2+{C}_2{H}_2$.

Phosphides: These can react with water to produce phosphine and hydroxides e.g.,

${Ca}_3 {P}_2+6 {H}_2{O} \rightarrow$ $3 {Ca}({OH})_2 + 2 {PH}_3$.

Nitrides: Certain nitrides can also hydrolyze to form ammonia e.g.,

${AlN}+3 {H}_2{O} \rightarrow {Al}({OH})_3+{NH}_3$.

Answer

Heavy water is injurious to human beings, plants and animals since it slows down the rates of reactions occurring in them. Thus heavy water does not support life so it is not used for drinking.

Heavy water $\left(D_2 \mathrm{O}\right)$ is a oxide of deuterium, the isotope of Hydrogen. Heavy water is extensively used as a moderator in nuclear reactors and in exchange reactions for the study of reaction mechanisms. It can be prepared by exhaustive electrolysis of water or as a by product in some fertilizer industries.

$ \mathrm{CaC}_2+2 \mathrm{D}_2 \mathrm{O} \rightarrow \mathrm{C}_2 \mathrm{D}_2+\mathrm{Ca}(\mathrm{OD})_2 $

$ \mathrm{SO}_3+\mathrm{D}_2 \mathrm{O} \rightarrow \mathrm{D}_2 \mathrm{SO}_4 $

$ \mathrm{Al}_4 \mathrm{C}_3+12 \mathrm{D}_2 \mathrm{O} \rightarrow 3 \mathrm{CD}_4+4 \mathrm{Al}(\mathrm{OD})_3$

Answer

Hydrolysis

Hydrolysis is a chemical reaction where water is used to break down a compound. It is a type of double decomposition reaction where water acts as a reactant. In hydrolysis, the water molecule splits into hydrogen and hydroxide ions, which then react with the compound to form new products.

Hydrolysis can be classified as a substitution reaction because it involves the replacement of a part of the compound with water.

The result of hydrolysis is often the formation of saturated compounds from unsaturated ones, as the addition of water can lead to the saturation of the compound.

Hydration

Hydration, on the other hand, is a process where water molecules are added to a substance. In this case, water combines with a substrate without breaking it down. Hydration typically involves the addition of water to a compound, resulting in a hydrated form of that compound.

Hydration is considered an addition reaction since it involves the addition of water to the compound.

In contrast, hydration leads to the formation of hydrated compounds, which are typically the result of adding water to a dehydrated compound.

Answer

Saline hydrides areionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as:

$ AH +H_2 O \longrightarrow AOH +H _{2} $

(where, $A=Na, Ca, \ldots .$. )

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide. The dry organic solvent distills over.

Answer

The elements with atomic numbers $15,19,23$ and 44 are nitrogen, potassium, vanadium and ruthenium respectively.

Ammonia is a covalent compound. It contains lone pair of electrons on N . Hence, it is electron rich hydride.

Potassium hydride is ionic in nature due to high electropositive nature of potassium. It is crystalline and non volatile.

Vanadium and ruthenium hydries are nonstoichiometric metallic hydrides with hydrogen deficiency.

KH violently reacts with water.

$ \mathrm{KH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{KOH}+\mathrm{H}_2 $

Ammonia is a lewis base.

$ \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{NH}_4^{+} $

Vanadium and ruthenium hydrides do not react with water.

The increasing order of reactivity of hydrides towards water is

$ (\mathrm{V}, \mathrm{Ru})<\mathrm{H}<\mathrm{NH}_3<\mathrm{KH} . $

Answer

Potassium Chloride

(i)

Potassium chloride $(KCl)$ is the salt of a strong acid $(HCl)$ and strong base $(KOH)$. Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows:

$ KCl {(s)} \xrightarrow{\text{ water }} K {(a q)}^{+}+Cl {(a q)}^{-} $

(ii,iii)

In acidified and alkaline water, the ions do not react and remain as such.

Aluminium (III) Chloride

(i)

Aluminium (III) chloride is the salt of a strong acid $(HCl)$ and weak base $[Al(OH)_3]$. Hence, it undergoes hydrolysis in normal water.

$ AlCl _{3}(s)+3 H_2 O {(l)} \xrightarrow[\text{ Water }]{\text{ Normal }} Al(OH) _{3}(s)+3 H^{+} {(a q)}+3 Cl^{-} {(a q)} $

(ii)

In acidified water, $H^{+}$ ions react with $Al(OH)_3$ forming water and giving $Al^{3+}$ ions. Hence, in acidified water, $AlCl_3$ will exist as $ \mathrm{Al}^{3+}(a q)$ and $ \mathrm{Cl}^{-}(a q)$ ions.

$ \begin{aligned} & AlCl _{3} (s) \xrightarrow[\text{ Water }]{\text{ Acidified }} Al ^{3+}{(a q)}+3 Cl^{-} (a q) \end{aligned} $

(iii)

In alkaline water, the following reaction takes place:

$ Al(OH) _{3}(s)+\underbrace{OH ^{-} {(a q)}} _{\text{from alkaline water }} \longrightarrow[Al(OH)_4] ^{-} {(a q)}+2 H_2 O {(l)} $

Answer

Hydrogen Peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ acts as bleaching agent due to the release of nascent oxygen.

$ \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}+[\mathrm{O}] $

The nascent oxygen combines with colouring matter which in turn gets oxidized. Thus, the bleaching action of hydrogen peroxide is due to the oxidation of colouring matter by nascent oxygen. It is used for the bleaching of silk, ivory, wool or feather.

Answer

(i) Hydrogen economy

Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas.

Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.

(ii) Hydrogenation

Hydrogenation is the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati, ghee etc.

(iii) Syngas

Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

(iv) Water shift reaction

It is a reaction of carbon monoxide of syngas mixture with steam in the presence of a catalyst as:

This reaction is used to increase the yield of dihydrogen obtained from the coal gasification reaction as:

(v) Fuel cells

Fuel cells are devices for producing electricity from fuel in the presence of an electrolyte. Dihydrogen can be used as a fuel in these cells. It is preferred over other fuels because it is eco-friendly and releases greater energy per unit mass of fuel as compared to gasoline and other fuels.