Hydrogen and It's Compounds

The problem asks for the approximate difference between the temperature of maximum density of heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$ and ordinary water $\left(\mathrm{H}_2 \mathrm{O}\right)$.

The temperature of maximum density of $\mathrm{H}_2 \mathrm{O}$ is approximately $3.98^{\circ} \mathrm{C}$.

The temperature of maximum density of $\mathrm{D}_2 \mathrm{O}$ is approximately $11.2^{\circ} \mathrm{C}$.

The difference in temperature in Celsius is

$ \Delta T_C=11.2^{\circ} \mathrm{C}-3.98^{\circ} \mathrm{C}=7.22^{\circ} \mathrm{C} $

The problem specifies the temperatures in Kelvin, with $y$ being the temperature for $\mathrm{H}_2 \mathrm{O}$ and $x$ for $\mathrm{D}_2 \mathrm{O}$. The difference is $(x-y)$ in Kelvin.

The magnitude of a temperature difference is the same whether measured in Celsius or Kelvin.

$ \Delta T_K=\Delta T_C $

So, the difference in Kelvin is 7.22 K . The value 7.22 K is closest to the option 7.0 K .

$\mathrm{LiAlH}_4=$ Total charge $=0$

Li oxidation number $=+1$

Al oxidation number $=+3$

$ \begin{aligned} 1+3+x \times 4 & =0 \\ x & =-1 \\ \mathrm{H}_2 \mathrm{O}=2 \times x-2=0, x & =+1 \\ \text { Hence, } X=\mathrm{LiAlH}_4, Y & =\mathrm{H}_2 \mathrm{O} \end{aligned} $

Among the given options (d) does not match correctly.

Electron precise hydrides are compounds of hydrogen with other elements that have the exact number of valence electron needed to form normal covalent bond. e.g. $\mathrm{CH}_4$.

All reactions are as follows,

I. $2 \mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow \mathrm{~B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2 \uparrow$

II. $\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{~N}\left(\mathrm{CH}_3\right)_3 \longrightarrow 2 \mathrm{BH}_3 \cdot \mathrm{~N}\left(\mathrm{CH}_3\right)_3$

III. $2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaAlO}_2 +3 \mathrm{H}_2 \uparrow$

IV. $2 \mathrm{BF}_3+6 \mathrm{NaOH} \longrightarrow \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}$

Thus, reactions given in I and III only gives $\mathrm{H}_2$ as one of the products.

Statement I is correct, but statement II is not correct. The correct form of statement II is, 10 volume $\mathrm{H}_2 \mathrm{O}_2$ contains $3 \%(w / v) \mathrm{H}_2 \mathrm{O}_2$.

The dielectric constant of $\mathrm{D}_2 \mathrm{O} \,\,{\text {is }}$ less than $\mathrm{H}_2 \mathrm{O}, \mathrm{D}_2 \mathrm{O}$ is heavier. This difference in mass affects the vibrational frequencies of the molecule and their ability to align with an electric field.

$\mathrm{C}, \mathrm{Si}, \mathrm{Ge}$ form electron precise hydrides. Electron precise hydrides are compounds where total number of valence electrons is exactly enough to

form covalent bonds between all the atoms in the molecules resulting in forming a complete octet.

The decomposition of $\mathrm{H}_2 \mathrm{O}_2$ is

$ 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 $

4 mL of $\mathrm{H}_2 \mathrm{O}_2$ solution produces 80 mL of $\mathrm{O}_2$

$ \begin{aligned} 1 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 & =\frac{80}{4} \mathrm{~mL} \text { of } \mathrm{O}_2 \\ & =20 \mathrm{~mL} \text { of oxygen } \end{aligned} $

Sodium hydroxide $(\mathrm{NaOH})$ is commercially prepared by electrolysis of brine solution among the given options. Antichlor is not a use of sodium hydroxide.

Antichlor is a substance used to remove residual chlorine from material that has been bleached.

Among the given options statement given in option (c) is incorrect regarding covalent hydrides. Covalent hydrides can be volatile or non volatile.

$ \mathrm{BeH}_{2} $ and $ \mathrm{MgH}_{2} $ have polymeric structure. Polymeric structure are those materials/compounds that are made up of long chain repeating molecules.

Same structure is for $ \mathrm{MgH}_{2} $.

Hydrogen peroxide $ \left(\mathrm{H}_{2} \mathrm{O}_{2}\right) $ is decomposed by the rough surface of glass, alkali oxide present in it and light. Therefore, to prevent its decomposition $ \mathrm{H}_{2} \mathrm{O}_{2} $ is usually stored in waxed lined plastic bottle and kept in dark.

The complete reaction is as follows.

$ 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \longrightarrow \underset{\substack{\text { (Product } X \text { ) } \\\\ \text { Diborane }}}{2 \mathrm{~B}_2 \mathrm{H}_6 }+3 \mathrm{AlF}_3+3 \mathrm{LiF} $

$\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaH} \longrightarrow \underset{\substack{\text { (Product } Y \text { ) } \\\\ \text { Sodium tetrahydroborate }}}{2 \mathrm{NaBH}_4}$

It is a good reducing agent.

Oxidation state of H in $ \mathrm{NaBH}_{4} $ is -1 .

[ $ \mathrm{Na}=+1, \mathrm{~B}=+3 $ oxidation state ]

$ [1+3+4 x=0 $ then $ x=-1] $

Hence, option (a) is incorrect about compound $ Y $,

$\mathrm{H}_2 \mathrm{O}_2$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

$\mathrm{H}_2 \mathrm{O}_2$ structure in solid phase, dihedral angle is $90.2^{\circ}$.

The reaction and their products are

$ \begin{aligned} & 2 \mathrm{NaBH}_4+\mathrm{I}_2 \xrightarrow{\Delta} \mathrm{~B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2 \uparrow \\ & 2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{\Delta} \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NaF} \\ & 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \xrightarrow{ } 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{Li}\left[\mathrm{AlF}_4\right] \\ & 3 \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NH}_3 \xrightarrow{\Delta} 2 \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6+12 \mathrm{H}_2 \uparrow \end{aligned} $

Hence, reaction given in (I) and (IV) give hydrogen as one product.

Let’s go through each option step by step:

Option A: Saline hydrides (ionic hydrides like NaH or CaH₂) contain the hydride ion ($\mathrm{H}^-$). During electrolysis, the hydride ions lose electrons (oxidation) at the anode:

$2\,\mathrm{H}^- \rightarrow \mathrm{H}_2 + 2\,e^-$

This makes Option A correct.

Option B: Methane ($\mathrm{CH}_4$) is described as an electron precise hydride because all of its bonds are localized two-electron bonds, satisfying the octet rule for the carbon atom. Therefore, this statement is also correct.

Option C: Chromium hydride falls under the category of metallic (or interstitial) hydrides. Such hydrides typically have a metallic lattice structure and, as a result, conduct heat and electricity. So, Option C is correct.

Option D: Hydrides of group 15 elements (like $\mathrm{NH}_3$, $\mathrm{PH}_3$, etc.) possess a lone pair on the central element. This lone pair usually makes them behave as Lewis bases (electron pair donors), not Lewis acids (electron pair acceptors). Hence, Option D is incorrect.

The incorrect statement is Option D.

Water gas is produced by alternately passing steam and air over a bed of red-hot coke or coal that is kept at a high temperature. It consists of a mixture of carbon monoxide (CO) and hydrogen (H₂).

Electron precise hydrides are those in which every bond is a localized two‐electron, two‐center bond—that is, the bonding can be described fully by standard Lewis structures without invoking multi‐center bonding. Let’s break down the options:

Group 13 elements (like boron) form hydrides such as BH₃. However, BH₃ is electron deficient (having only 6 valence electrons around boron) and typically exists as diborane (B₂H₆) where bridging hydrogen atoms create three‐center two‐electron bonds. This makes them non–electron precise.

Group 14 elements (like carbon) form hydrides such as methane (CH₄). In CH₄, carbon has four valence electrons and forms four bonds with hydrogen atoms. Each CH bond is a two‐electron bond with the electron count exactly sufficient to satisfy the octet rule—this is the hallmark of an electron precise hydride.

Group 15 hydrides (e.g., ammonia, NH₃) and Group 16 hydrides (e.g., water, H₂O) also obey the octet rule and have conventional two‐electron bonds. They are electron precise, but the term is most typically used in discussions contrasting them with electron deficient species (like those of Group 13).

Since the question asks for an example, the most classic and illustrative case is found in the hydrides of Group 14 elements.

Thus, the correct answer is:

Option A: Group 14 elements.

Let's analyze each statement one by one:

Statement (i): Saline hydrides produce $\mathrm{H_2}$ gas when reacted with water.

Saline (or ionic) hydrides (such as $\mathrm{NaH}$ or $\mathrm{LiH}$) react with water to form the corresponding hydroxide and hydrogen gas. For example:

$\mathrm{NaH} + \mathrm{H_2O} \rightarrow \mathrm{NaOH} + \mathrm{H_2}$

This statement is correct.

Statement (ii): Presently $\sim 77\%$ of the industrial dihydrogen is produced from coal.

Most industrial hydrogen is produced by the steam reforming of natural gas (methane), not from coal gasification. Globally, the majority of hydrogen comes from hydrocarbon sources like natural gas.

Therefore, this statement is not correct.

Statement (iii): Commercially marketed $\mathrm{H_2O_2}$ contains $3\%$ $\mathrm{H_2O_2}$.

Over-the-counter hydrogen peroxide solutions are commonly sold as a 3% solution. This is routinely labeled and used for antiseptic purposes.

This statement is correct.

Given the above analysis, only statements (i) and (iii) are correct.

Thus, the correct option is:

Option B: (i), (iii) only.

Statement I is correct, but Statement II is incorrect.

Statement I: In dry cleaning, the solvent $\mathrm{Cl}_2\mathrm{C}=\mathrm{CCl}_2$ (also known as perchloroethylene) was traditionally used, but it has now been replaced by liquefied $\mathrm{CO}_2$ due to environmental concerns and safety issues associated with perchloroethylene.

Statement II: The correct statement is that earlier, chlorine gas was used to bleach paper. Today, this has been largely replaced by $\mathrm{H}_2\mathrm{O}_2$ (hydrogen peroxide).

Hydrogen peroxide is preferred because its decomposition products—water and oxygen—are not harmful to the environment. This makes it a safer and more environmentally friendly option compared to chlorine gas, which can produce toxic by-products.

In option (d), dihydrogen is not released.

$\mathrm{B}_2 \mathrm{H}_6(g)+3 \mathrm{O}_2 \xrightarrow[\text { Heat }]{ } \mathrm{B}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{H}_2 \mathrm{O}$ Dibrone

In all other options, dihydrogen is released.

Oxidation of sodium borohydrate.

$ 2 \mathrm{NaBH}_4(s)+\mathrm{I}_2 \longrightarrow 2 \mathrm{NaI}+\mathrm{B}_2 \mathrm{H}_6+\mathrm{H}_2^{\uparrow} $

Hydrolysis of boranes

$ \begin{aligned} & \text { Hydrolysis of boranes } \\ & \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{H}_2 \uparrow \end{aligned} $

Ammonia with diborane

$ 3 \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NH}_3 \longrightarrow 2 \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6+12 \mathrm{H}_2 $

The type of hydrides that have the exact number of electrons to form a covalent bond is called electron precise. These types of compounds are usually formed by group-14 elements.

Group 13 elements forms electron deficient hydrides.

Group 15 elements formed electron excess.

• In saline hydrides oxidation state of hydrogen is -1 .

So, statement I is incorrect.

• LaH is an interstitial hydride. So, statement II is correct.

• $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ have the tendency to form hydrogen bonds. So, statement III is also correct.

  Electrolysis of molten sodium hydride librates hydrogen gas at anode and sodium is formed at cathode. So, statement IV is incorrect.

  Thus, statements II and III are correct about the hydrides.

Statements II and IV are correct, while statements I and III are incorrect. Here are the corrected versions of statements I and III:

I. Sodium hydride reacts with water to release hydrogen gas:

$ \mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(a q) \longrightarrow \mathrm{NaOH}(a q) + \mathrm{H}_2(g) $

III. Ammonia and water molecules are examples of electron-rich compounds.

The correct match is $\mathrm{A}-\mathrm{III}, \mathrm{B} \cdot \mathrm{IV}$, C-I, D-II..

$ \begin{aligned} & \text { } \mathrm{CaC}_2+2 \mathrm{D}_2 \mathrm{O} \longrightarrow \mathrm{C}_2 \mathrm{D}_2+\mathrm{Ca}(\mathrm{OD})_2 \\ & \therefore P \text { is } \mathrm{C}_2 \mathrm{D}_2 \end{aligned} $

Deionised or demineralised water is obtained by passing hard water through both cation and anion exchanger one after other.

(a) $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ are polymeric hydrides, a large number of $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ units join together by hydrogen bonding to produce a long chain.

(b) LiH is by no means unreactive towards air at moderate temperature because it is a powerful reducing agent.

(c) $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ are covalent in nature due to small size of $\mathrm{Mg}^{2+}$ and $\mathrm{Be}^{2+}$ ions and $\mathrm{H}^{-}$ions is large anion.

(d) The statement (d )is incorrect. The correct stability order of alkali hydrides is

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

Hence, (a), (b), (c) are correct statements.

The extra neutron in the deuterium isotope makes heavy water denser. Also, the freezing point of heavy water is higher than the ordinary water. The freezing point of heavy water at 1 atm pressure is $3.8^{\circ} \mathrm{C}$.

The most effective water softening method is ion exchange process because in this method the dissolved calcium and magnesium ions that cause hardness of water are removed easily with the help of an ion exchange process. Also, in this method relatively large number of calcium and magnesium ions displace smaller sodium ions.

This method is cheaper method and the chemicals used in it are safe to work with and easy to handle.

Consider LiH as the molten ionic hydrides of s-block,

$ \mathrm{LiH} \longrightarrow \mathrm{Li}^{+}+\mathrm{H}^{-} $

Upon passing electric current, the oxidation reaction that take place at anode.

At anode, $2 \mathrm{H}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g})+2 e^{-}$

So, $\mathrm{H}_2$ gas is liberated at anode.

$\mathrm{CaC}_2+2 \mathrm{D}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { Deutero } \\ \text { acetylene }}}{\mathrm{C}_2 \mathrm{D}_2}+\mathrm{Ca}(\mathrm{OD})_2$

The hybridisation of carbon atoms in deutero acetylene is $s p$.

$\mathrm{D}-\mathrm{C} \equiv \mathrm{C}-\mathrm{D}$

Calcium carbide reacts with heavy water to form $\mathrm{C}_2 \mathrm{D}_2$ and $\mathrm{Ca}(\mathrm{OD})_2(P)$. So, in the first reaction, $X$ is $\mathrm{CaC}_2$.

$\mathop {Ca{C_2}}\limits_{(X)} + 2{D_2}O\buildrel {} \over \longrightarrow {C_2}{D_2} + \mathop {Ca{{(OD)}_2}}\limits_{(P)} $

Aluminium carbide $(\mathrm{Al}_4 \mathrm{C}_3)$ reacts with heavy water ($\mathrm{D}_2 \mathrm{O})$ to give $\mathrm{CD}_4$ and $\mathrm{Al}(\mathrm{OD})_3$.

So, $Y$ is $\mathrm{Al}_4 \mathrm{C}_3$.

$\mathop {A{l_4}{C_3}}\limits_{(Y)} + 12{D_2}O\buildrel {} \over \longrightarrow 3C{D_4} + \mathop {4Al{{(OD)}_3}}\limits_{(Q)} $

∵ (I) $\mathrm{Zn}+2 \mathrm{HCl}$ (dil.) $\longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2$ and ∵ (II)

$ \mathrm{Zn}+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 $

while calcium hydrogen carbonate, i.e. calcium bicarbonate $\left[\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2\right]$. On heating give

$ \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \xrightarrow{\Delta} \mathrm{CaCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} $

As it only exsist in solution and will not give $\mathrm{H}_2$.

Hence, (I) and (II) only give $\mathrm{H}_2$, i.e. dihydrogen and option (c) is the correct answer.

$\because \mathrm{H}_2 \mathrm{O}_2$ show oxidising as well as reducing action.

Oxidising nature (i.e. act as oxidising agent)

(i) In acidic solution,

$ \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} $

(ii) In alkaline solution,

$ \mathrm{H}_2 \mathrm{O}_2+2 e^{-} \longrightarrow 2 \mathrm{OH}^{-} $

Reducing nature (i.e. act as reducing agent)

$ \begin{aligned} &\mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}^{+}+\mathrm{O}_2+2 e^{-}\\ &\text { (in acidic medium) } \end{aligned} $

$ \begin{aligned} &\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+2 e^{-}\\ &\text { (in alkaline medium) } \end{aligned} $

Hence, (i), (ii), (iii) and (iv) all are possible and (c) is the correct answer.

Primary components by synthesis gas are hydrogen and carbon monoxide.

It is also known as syn gas. It is used as a fuel gas. Hence, option (a) is the correct option.

$22.4 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2$ means I mL of $\mathrm{H}_2 \mathrm{O}_2$

$ \begin{array}{ll} 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & +\mathrm{O}_2 \\ 2 \mathrm{~mol} & 1 \mathrm{~mol} \\ 68 \mathrm{~g} & 22400 \mathrm{~mL} \text { at NTP } \end{array} $

$\therefore 22400 \mathrm{~mL} \mathrm{O}_2$ is obtained by 68 g $\mathrm{H}_2 \mathrm{O}_2$

$\therefore 22.4 \mathrm{~mL} \mathrm{O}_2$ is obtained by

$ \frac{68 \times 22.4}{22400}=0.068 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2 $

or $\mathrm{l} \mathrm{mL} \mathrm{H}_2 \mathrm{O}_2$ solution $=0.068 \mathrm{~g}$

$\therefore 100 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2$ solution

$ =0.0680 \times 100=6.8 \% $

Hard water does not produce lather with soap is called hard water. Hardness is due to the presence of bicarbonate, chlorides and sulphates of Ca and Mg .

$ \begin{aligned} & M^{2+}+2 \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa} \longrightarrow \\ & \left(M=\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\right) \\ & \left(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}\right)_2 M+2 \mathrm{Na}^{+} \\ & \text {Sodium stearate (soap) } \end{aligned} $

$\mathrm{Ca}^{2+}$ or $\mathrm{Mg}^{2+}$ ion present in hard water react with soap to form precipitate of Ca and Mg salt of fatty acids and hence, no lather is produced.