High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
 At anode: 2OH^- ➝ H₂O + 1/2O₂ + 2e^-
At cathode: 2H₂O + 2e^- ➝ 2OH^- + H₂
Overall reaction: H₂O ➝ H₂ + 1/2O₂ 

(i) Hydrogen has the electronic configuration of 1s1 which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals (Na⁺,K⁺,Cs⁺)
2. It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
3. It also acts as a reducing agent.
4. lt is an electro positive element
However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1.314 kJ mol^-1 which is much higher than alkali metals.
Like the formation of halides (X ^-) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion(H⁺) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\( 1 / 2 \mathrm{H}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{H}^{-} \Delta \mathrm{H}=+36 \mathrm{kcal} \mathrm{mol}^{-1} \)
\(1 / 2 \mathrm{Br}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-} \Delta \mathrm{H}=-55 \mathrm{kcal} \mathrm{mol}^{-1}\)
Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

The element which occupies group number (16) and period number (2) is oxygen. (B) is D₂O which is used as a moderator in nuclear reactions.
So (A) must be deuterium, which is an isotope of hydrogen
    \(2\underset { (A) }{ { D }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (B) }{ { D }_{ 2 }O } \)
So (B) is D₂O
(A) adds to (C) as follows :
\(3 \mathrm{D}_{2}+\mathrm{C}_{3} \mathrm{H}_{6} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2}\)
So (D) is 1,2 - dideutero propane. 

The group metal - 1 which is present in common salt is sodium.
So (A) is sodium
Sodium reacts with hydrogen (B) to give, sodium hydride (C). In sodium hydride the hydrogen is present in -1 oxidation state.
\(2\underset { (A) }{ Na } +\underset { (B) }{ { H }_{ 2 } } \rightarrow 2\underset { (C) }{ NaH } \)
So (B) hydrogen and (C) is sodium hydride
H₂ reacts with oxygen gas (D) to give an universal solvent, water (E) follows:
\(2\underset { (B) }{ { H }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (D) }{ { H }_{ 2 }O } \)
So (E) is water. Water is the universal solvent
Water (E)  reacts with sodium  (A) follow to give (F), which is a strong base.
\(2\underset { (E) }{ { H }_{ 2 }O } +\underset { (A) }{ 2Na } \rightarrow 2\underset { (F) }{ NaOH } +{ { H }_{ 2 } } \)
So (F) is sodium hydroxide.