(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.

(i) The order : NH₃ < H₂O < HF
(ii) Strength of H bond depends upon the atomic size and electronegativity of the other atom to which H atom is covalently bonded. 
Since electronegativity of F > O > N, the strength of H-bond is in the order H-F......H > H - O......H > H-N......H.
H-bonds are much weaker than covalent bonds.

When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium
\( 2 \mathrm{NaOH}+\mathrm{D}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOD}+\mathrm{HOD} \)
\(\mathrm{HCl}+\mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{DCl}+\mathrm{HOD} \)
\(\mathrm{NH}_{4} \mathrm{Cl}+4 \mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{ND}_{4} \mathrm{Cl}+4 \mathrm{HOD}\)
These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
For example, when D₂O is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
\(\mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{D}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{DPO}_{2}+\mathrm{HDO}\)

a) Oxidation is performed in acidic medium:
Eg : H₂O₂ Oxidises FeSO₄ to Fe (SO₄)₃
H₂O₂ + 2H⁺+ 2e^- ➝ 2H₂O (E⁰ = +1.77 V)
2FeSO₄ + H₂SO₄ + H₂O₂ ➝ Fe₂(SO₄)₃ + 2H₂O
b) Reduction is performed in basic medium:
Eg : H₂O₂ reduces KMnO₄ to MnO₂
HO₂^- + OH^- ➝ H₂O + 2e^- (E⁰ = +0.08 V)
\(2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{O}_{2}\)

Interstitial hydrides have metallic bonding. These are non - stoichiometric; their composition varies with temperature and pressure. The crystal lattice expands due to the inclusion of dihydrogen and there is distortion of crystal lattice. So they have lower density than the parent metal. Recent studies have shown that except the hydrides of Ni, Pd, Ce & Ac, other hydride are different from the parent metals.

a) HCl is a gas. It is an electron rich covalent hydride. It is a covalent molecule with strong covalent bond. However the forces between the molecules are not as strong.
b) NaH is a solid. It is an ionic hydride of group I sodium metal. It is an insoluble hydride

Hydrogen has electronic configuration 1s¹. So it resembles ns¹ general valence shell configuration of alkali metals.
Its similarity with alkali metals are:
1. Form uni positive (H⁺) ions like Na⁺, K⁺, Cs⁺
2. Form halides, (HX), oxides (H₂O) peroxides (H₂O₂) and sulphides (H₂S) like (Nax, Na2O, Na₂O₂, Na₂S)
3. Acts as reducing agent. 
4. Has +1 oxidation state.
5. If it is placed in the halogen family it should be the most electro negative element which is not true.

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.