Hydrogen and It's Compounds

Permanent hardness of water is caused by the presence of soluble salts of calcium and magnesium, such as calcium chloride ($\mathrm{CaCl}_2$) and magnesium sulfate ($\mathrm{MgSO}_4$). These salts cannot be removed by simple boiling. One effective method to remove permanent hardness is to convert these soluble salts into insoluble compounds. This can be achieved by adding certain chemicals to the water that react with the calcium and magnesium ions to form precipitates.

Sodium carbonate ($\mathrm{Na}_2 \mathrm{CO}_3$), also known as washing soda, is commonly added for this purpose. When sodium carbonate is added to hard water, it reacts with the calcium and magnesium ions to form insoluble carbonates, which can then be removed by filtration. The reactions are as follows:

$\mathrm{Ca}^{2+} + \mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3 \downarrow + 2\mathrm{Na}^+$

$\mathrm{Mg}^{2+} + \mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{MgCO}_3 \downarrow + 2\mathrm{Na}^+$

In these reactions, calcium and magnesium ions combine with carbonate ions from sodium carbonate to form insoluble calcium carbonate ($\mathrm{CaCO}_3$) and magnesium carbonate ($\mathrm{MgCO}_3$) precipitates, respectively.

Therefore, the correct answer is:

Option B

$\mathrm{Na}_2 \mathrm{CO}_3$

The correct answer is Option C: ZnH₂. Here's why:

Interstitial hydrides are metallic hydrides formed when hydrogen atoms occupy the interstitial spaces (voids) within the crystal lattice of a transition metal. These hydrides are typically non-stoichiometric, meaning the ratio of metal to hydrogen atoms is not fixed and can vary depending on the conditions.

Let's analyze the options:

A) NH₃ (Ammonia): Ammonia is a covalent compound with a strong N-H bond. It does not form an interstitial hydride.

B) CH₄ (Methane): Methane is a covalent compound with strong C-H bonds. It does not form an interstitial hydride.

C) ZnH₂ (Zinc Hydride): Zinc is a transition metal, and its hydride, ZnH₂, is a classic example of an interstitial hydride. Hydrogen atoms occupy the interstitial sites within the zinc crystal lattice.

D) H₂O (Water): Water is a covalent compound with strong O-H bonds. It does not form an interstitial hydride.

Therefore, only ZnH₂ fits the definition of an interstitial hydride.

The correct answer is Option D: All of these. Let's break down why each statement is true:

I. T₂ > D₂ > P₂ (order of boiling point)

This is correct. Boiling point is directly related to the strength of intermolecular forces. The heavier the isotope, the stronger the London dispersion forces (temporary dipoles) between molecules. Thus, T₂ (tritium) has the highest boiling point followed by D₂ (deuterium) and then P₂ (protium).

II. T₂ > D₂ > P₂ (order of bond energy)

This is also correct. Bond energy is related to the strength of the covalent bond. Heavier isotopes have slightly stronger bonds due to the increased mass of the nucleus, leading to a stronger electrostatic attraction between the nuclei and the shared electrons. Therefore, T₂ has the highest bond energy, followed by D₂ and then P₂.

III. T₂ = D₂ = P₂ (order of bond length)

This is correct. Bond length is primarily determined by the size of the atoms involved. Since all three isotopes of hydrogen have the same number of protons and electrons, their atomic radii are essentially identical. Consequently, the bond lengths in T₂, D₂, and P₂ are very similar, and for practical purposes, we can consider them equal.

IV. T₂ < D₂ < P₂ (order of reactivity with Cl₂)

This is also correct. The reactivity of hydrogen isotopes with chlorine is inversely proportional to the strength of the H-H bond. As we discussed earlier, T₂ has the strongest bond, followed by D₂ and P₂. Therefore, P₂ is the most reactive with Cl₂, followed by D₂ and then T₂.

In summary, all four statements accurately reflect the trends in properties for the isotopes of hydrogen. Heavier isotopes generally exhibit stronger intermolecular forces, stronger bonds, and slightly lower reactivity.

ortho and para hydrogen similar to each other in their chemical properties but they have different physical properties like boiling point, thermal conductivity, due to their difference in overall spins.

Dielectric constant of H₂O > D₂O. Therefore, H₂O is more effective solvent. B. P. of D₂O > B. P. of H₂O.