Hydrogen and It's Compounds

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.

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 $

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

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)} $