Hydrogen Questions in English

(A) Producer gas is a mixture of $CO$ and $N_2$.
Its calorific value is low due to the high percentage of non-combustible nitrogen gas $(N_2)$ present in the mixture.

(A) The oxidation state of hydrogen is $-1$ in metal hydrides (where hydrogen is bonded to a less electronegative element) and $+1$ in compounds where hydrogen is bonded to a more electronegative non-metal.
In $LiAlH_4$,hydrogen is present as a hydride ion $(H^-)$,so its oxidation state is $-1$.
In $H_2O$,hydrogen is bonded to oxygen (which is more electronegative),so its oxidation state is $+1$.
Therefore,$X = LiAlH_4$ and $Y = H_2O$.

(B) Protium,deuterium,and tritium are isotopes of hydrogen with the following notations:
Protium: $^1_1H$ (Number of neutrons $= 1 - 1 = 0$)
Deuterium: $^2_1H$ (Number of neutrons $= 2 - 1 = 1$)
Tritium: $^3_1H$ (Number of neutrons $= 3 - 1 = 2$)
The sum of the total number of neutrons $= 0 + 1 + 2 = 3$.

(C) Among the isotopes of hydrogen,$^1H$ (Protium),$^2H$ (Deuterium),and $^3H$ (Tritium),only $^3H$ (Tritium) is radioactive.
It emits low-energy $\beta$-particles and has a half-life of $12.33 \ \text{years}$.

(B) $(i)$ Ortho and para hydrogens have different nuclear spin. In $H_2$ molecule,two protons in two $H$ atoms with parallel spin are called ortho-hydrogen.
$(ii)$ Para-hydrogen is more stable than ortho-form at low temperatures,whereas ortho-hydrogen is more stable at and above room temperature. Therefore,the statement that ortho-form is more stable than para-form is not universally correct.
$(iii)$ The percent composition of ortho and para changes with the temperature. Thus,their ratio also changes.
$(iv)$ They have slightly different boiling points.
Hence,$(B)$ is the answer.

(A) Ortho-hydrogen and para-hydrogen are nuclear spin isomers of the hydrogen molecule $(H_2)$.
In ortho-hydrogen,the nuclear spins of the two protons are parallel,whereas in para-hydrogen,the nuclear spins are antiparallel.
Because they have the same electronic structure,their chemical properties are identical.
However,due to the difference in their nuclear spin states,they exhibit different physical properties,such as different thermal conductivities and specific heats.

(D) In $N_2H_4$ (hydrazine),each nitrogen atom is bonded to two hydrogen atoms and one nitrogen atom,with one lone pair. The steric number for each $N$ atom is $3 + 1 = 4$,which corresponds to $sp^3$ hybridisation.
In $H_2O_2$ (hydrogen peroxide),each oxygen atom is bonded to one hydrogen atom and one oxygen atom,with two lone pairs. The steric number for each $O$ atom is $2 + 2 = 4$,which corresponds to $sp^3$ hybridisation.
Therefore,both compounds show similarity in the hybridisation of their central atoms $(sp^3)$.