Q.Will CCl4 give a white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

The key concept here is the thermal decomposition of hydrogen carbonates (bicarbonates) and the differential solubility of the resulting hydroxides and carbonates. Both Mg(HCO₃)₂ and Ca(HCO₃)₂ decompose on heating, but the final precipitates differ because MgCO₃ is not stable in hot water — it hydrolyzes further to Mg(OH)₂, whereas CaCO₃ is stable and precipitates directly.

Let’s walk through the chemistry step by step.

1. General decomposition of bicarbonates When any metal bicarbonate is heated, it decomposes into the metal carbonate, water, and carbon dioxide:

This is the first reaction for both calcium and magnesium bicarbonates.

2. Calcium case — stable carbonate For calcium:

Calcium carbonate is sparingly soluble and does not decompose further under normal boiling conditions. So the precipitate from calcium is $\text{CaCO}_3$.

3. Magnesium case — further hydrolysis For magnesium, the initially formed MgCO₃ is not stable in hot water. It undergoes hydrolysis:

Magnesium hydroxide is even less soluble than magnesium carbonate, so the final precipitate is $\text{Mg(OH)}_2$, not MgCO₃.

4. Why the difference? …

The question asks which metal nitrate does not liberate NO₂ gas on heating. This is a classic problem about the thermal decomposition of nitrates, and the pattern depends on the metal’s electropositive character.

Concept & Intuition:

When you heat a metal nitrate, the nitrate ion (NO₃⁻) can break down in different ways. For most metals (those less reactive than magnesium), the nitrate decomposes to the metal oxide, nitrogen dioxide (NO₂), and oxygen. But for the most reactive metals — the alkali metals (Group 1) — the nitrate is more stable and decomposes differently: it forms the metal nitrite and oxygen, without producing NO₂. So the metal whose nitrate does not give NO₂ must be an alkali metal.

Let’s check each option:

1. Option (A): Pb (Lead) Lead is a heavy metal, not an alkali or alkaline earth. Its nitrate decomposes as:

This clearly liberates brown NO₂ gas. So (A) is not the answer.

2. Option (B): Ba (Barium) Barium is an alkaline earth metal (Group 2). Except for beryllium and magnesium, most alkaline earth nitrates decompose to the oxide, NO₂, and O₂:

So Ba(NO₃)₂ also gives NO₂. Not the answer.

3. Option (C): Li (Lithium) Lithium is an alkali metal (Group 1). However, lithium is the smallest alkali metal and behaves a bit differently from its heavier cousins. Lithium nitrate decomposes to lithium oxide, NO₂, and O₂ — similar to alkaline earth metals:

So LiNO₃ does liberate NO₂. Thus (C) is not the answer. …

The concept here is vector cancellation of bond dipoles. A molecule’s overall dipole moment is the vector sum of the individual bond polarities. Even if a bond is polar, the molecule can be nonpolar if its shape arranges those bond dipoles symmetrically so they cancel out. This depends entirely on molecular geometry — not just on the presence of polar bonds.

Let’s go through each molecule step by step.

1. CCl$_4$ — Tetrahedral geometry. The four C–Cl bonds are polar (chlorine is more electronegative), but they point to the corners of a regular tetrahedron. Their vector sum is exactly zero. So dipole moment = 0.

2. BF$_3$ — Trigonal planar geometry. The three B–F bonds are polar, but they lie at 120° angles in a plane. The vectors cancel perfectly. Dipole moment = 0.

3. CHCl$_3$ — This is chloroform. It is also tetrahedral, but here the four substituents are not identical: three Cl atoms and one H atom. The symmetry is broken — the bond dipoles do not cancel. The resultant points roughly along the C–H bond direction (toward the chlorines). Dipole moment ≠ 0.

4. CS$_2$ — Linear molecule, S=C=S. The two C=S bonds are polar, but they point in exactly opposite directions. Their vectors cancel. Dipole moment = 0.

5. NH$_3$ — Trigonal pyramidal geometry (lone pair on nitrogen). The three N–H bonds are polar, and the lone pair also contributes. The symmetry is not perfect — the resultant points upward through the nitrogen. Dipole moment ≠ 0. …

This question tests your knowledge of the characteristic reactions of sulphur dioxide. Sulphur dioxide ($\mathrm{SO_2}$) is an acidic gas that can undergo various reactions, including oxidation and addition reactions. In this specific case, it reacts with chlorine ($\mathrm{Cl_2}$) in a direct combination reaction.

Charcoal, which is a form of carbon, acts as a catalyst in this reaction. Catalysts speed up the rate of a chemical reaction without being consumed themselves. Here, charcoal provides a surface for the reactant molecules ($\mathrm{SO_2}$ and $\mathrm{Cl_2}$) to adsorb onto, bringing them into close proximity and facilitating their reaction to form the product.

1. Identify the reactants and catalyst: We have sulphur dioxide ($\mathrm{SO_2}$) and chlorine ($\mathrm{Cl_2}$) as reactants. Charcoal (carbon) is specified as the catalyst.
2. Determine the reaction type: This is a direct combination reaction where two simpler substances combine to form a more complex one. Sulphur dioxide is oxidized by chlorine, and chlorine is reduced.
3. Predict the product: When sulphur dioxide reacts with chlorine, they combine to form sulphuryl chloride. In sulphuryl chloride, the sulphur atom is bonded to two oxygen atoms and two chlorine atoms. The oxidation state of sulphur in $\mathrm{SO_2}$ is $+4$, and in $\mathrm{SO_2Cl_2}$ it is $+6$.
4. Write the balanced chemical equation: The reaction proceeds as follows:

The product, sulphuryl chloride, is a colourless fuming liquid. …

Concept and Intuition

Noble gases, elements in Group 18 of the periodic table, are famously known for their chemical inertness. This inertness stems from their stable electronic configuration, typically an $ns^2np^6$ octet (or $1s^2$ for Helium), which means their valence shells are completely filled. This configuration results in very high ionization energies (making it difficult to remove electrons) and nearly zero electron affinities (making it difficult to gain electrons), leading to a reluctance to participate in chemical bonding.

However, the term "inert" is not absolute. As we move down the noble gas group, the atomic size increases, and the outermost electrons are further from the nucleus. This leads to a decrease in ionization energy. For heavier noble gases like Krypton (Kr) and especially Xenon (Xe), the ionization energy becomes low enough that they can be coaxed into forming compounds, particularly with highly electronegative elements such as Fluorine (F) and Oxygen (O). These highly electronegative elements can effectively draw electron density from the noble gas atom, facilitating bond formation.

The first noble gas compound, XePtF$_6$, was synthesized in 1962, shattering the long-held belief of absolute inertness. Since then, numerous compounds of Xenon and Krypton, and even a few of Radon, have been discovered.

Step-by-Step Analysis

Let's examine each option based on these principles:

1. Analyze Option (A) XeOF$_4$:

   Xenon (Xe) is a relatively heavy noble gas with a lower ionization energy compared to lighter noble gases. It can expand its octet and form bonds with highly electronegative elements like Fluorine and Oxygen. XeOF$_4$ (Xenon oxytetrafluoride) is a well-known and stable compound where Xenon is in a $+6$ oxidation state. It has a square pyramidal molecular geometry. Therefore, XeOF$_4$ exists.

2. Analyze Option (B) NeF$_2$: …