Understanding Equilibrium Direction

The question asks us to identify the chemical equilibria from the given options that are NOT favoured to proceed in the forward direction (to the right). Chemical equilibrium represents a state where the rates of the forward and reverse reactions are equal. However, the position of equilibrium can lie more towards the reactants (not favoured to the right) or more towards the products (favoured to the right). The favourability of a reaction can often be determined by considering factors like the relative stability of reactants versus products, especially in terms of bonding preferences (like Hard-Soft Acid-Base principle) or the formation of insoluble precipitates.

Let's examine each given equilibrium reaction:

Analyzing Equilibrium (A)

The reaction is: $\text{As}_2\text{S}_5 + 5\text{HgO} \rightleftharpoons \text{As}_2\text{O}_5 + 5\text{HgS}$

This reaction involves the exchange of anions ($\text{S}^{2-}$ and $\text{O}^{2-}$) between cations ($\text{As}^{5+}$ and $\text{Hg}^{2+}$). We can analyze this using the Hard-Soft Acid-Base (HSAB) principle.

• $\text{As}^{5+}$ is a small ion with a high positive charge, making it a hard acid.
• $\text{Hg}^{2+}$ is a larger metal ion, making it a soft acid.
• $\text{O}^{2-}$ is a small anion with high electronegativity, making it a hard base.
• $\text{S}^{2-}$ is a larger, more polarizable anion, making it a soft base.

According to the HSAB principle, hard acids prefer to bond with hard bases, and soft acids prefer to bond with soft bases, forming more stable compounds. Let's look at the combinations in reactants and products:

• Reactants: $\text{As}_2\text{S}_5$ (Hard acid $\text{As}^{5+}$ with Soft base $\text{S}^{2-}$), $\text{HgO}$ (Soft acid $\text{Hg}^{2+}$ with Hard base $\text{O}^{2-}$)
• Products: $\text{As}_2\text{O}_5$ (Hard acid $\text{As}^{5+}$ with Hard base $\text{O}^{2-}$), $\text{HgS}$ (Soft acid $\text{Hg}^{2+}$ with Soft base $\text{S}^{2-}$)

The products $\text{As}_2\text{O}_5$ and $\text{HgS}$ represent stable hard-hard and soft-soft combinations, respectively. The reactants $\text{As}_2\text{S}_5$ and $\text{HgO}$ represent less stable hard-soft and soft-hard combinations. Therefore, the formation of products is favoured.

This equilibrium is favoured to go to the right.

Analyzing Equilibrium (B)

The reaction is: $\text{La}_2(\text{CO}_3)_3 + \text{Bi}_2\text{S}_3 \rightleftharpoons \text{La}_2\text{S}_3 + \text{Bi}_2(\text{CO}_3)_3$

This is another reaction involving anion exchange ($\text{CO}_3^{2-}$ and $\text{S}^{2-}$ anions between $\text{La}^{3+}$ and $\text{Bi}^{3+}$ cations). Let's use the HSAB principle:

• $\text{La}^{3+}$ is a relatively large ion with moderate charge density (Group 3 metal), often considered a hard acid.
• $\text{Bi}^{3+}$ is a heavier metal ion, often considered a soft or borderline acid.
• $\text{CO}_3^{2-}$ is a small, non-polarizable anion system, a hard base.
• $\text{S}^{2-}$ is a larger, polarizable anion, a soft base.

Let's look at the combinations:

• Reactants: $\text{La}_2(\text{CO}_3)_3$ (Hard acid $\text{La}^{3+}$ with Hard base $\text{CO}_3^{2-}$), $\text{Bi}_2\text{S}_3$ (Soft/Borderline acid $\text{Bi}^{3+}$ with Soft base $\text{S}^{2-}$)
• Products: $\text{La}_2\text{S}_3$ (Hard acid $\text{La}^{3+}$ with Soft base $\text{S}^{2-}$), $\text{Bi}_2(\text{CO}_3)_3$ (Soft/Borderline acid $\text{Bi}^{3+}$ with Hard base $\text{CO}_3^{2-}$)

The reactants involve stable hard-hard and soft-soft (or borderline-soft) combinations. The products involve less stable hard-soft and soft-hard (or borderline-hard) combinations. Therefore, the formation of reactants is favoured.

This equilibrium is NOT favoured to go to the right.

Analyzing Equilibrium (C)

The reaction is: $\text{CdSO}_4 + \text{CaS} \rightleftharpoons \text{CdS} + \text{CaSO}_4$

This reaction involves the exchange of anions ($\text{SO}_4^{2-}$ and $\text{S}^{2-}$ between $\text{Cd}^{2+}$ and $\text{Ca}^{2+}$). We can consider the solubility of the resulting salts.

Generally, sulfide salts of transition metals like cadmium are very insoluble. Calcium sulfate has limited solubility in water.

The formation of a highly insoluble precipitate ($\text{CdS}$) from soluble reactants is a strong driving force for a reaction to proceed in the forward direction.

This equilibrium is favoured to go to the right due to the precipitation of $\text{CdS}$.

Analyzing Equilibrium (D)

The reaction is: $\text{BeF}_2 + \text{HgI}_2 \rightleftharpoons \text{BeI}_2 + \text{HgF}_2$

This reaction involves the exchange of halogens ($\text{F}^-$ and $\text{I}^-$) between cations ($\text{Be}^{2+}$ and $\text{Hg}^{2+}$). Let's use the HSAB principle:

• $\text{Be}^{2+}$ is a very small ion with a high charge density, making it a hard acid.
• $\text{Hg}^{2+}$ is a larger metal ion, making it a soft acid.
• $\text{F}^-$ is a small anion with high electronegativity, making it a hard base.
• $\text{I}^-$ is a larger, more polarizable anion, making it a soft base.

Let's look at the combinations:

• Reactants: $\text{BeF}_2$ (Hard acid $\text{Be}^{2+}$ with Hard base $\text{F}^-$), $\text{HgI}_2$ (Soft acid $\text{Hg}^{2+}$ with Soft base $\text{I}^-$)
• Products: $\text{BeI}_2$ (Hard acid $\text{Be}^{2+}$ with Soft base $\text{I}^-$), $\text{HgF}_2$ (Soft acid $\text{Hg}^{2+}$ with Hard base $\text{F}^-$)

The reactants involve stable hard-hard and soft-soft combinations. The products involve less stable hard-soft and soft-hard combinations. Therefore, the formation of reactants is favoured.

This equilibrium is NOT favoured to go to the right.

Identifying Equilibria NOT Favoured Right

Based on the analysis:

• Equilibrium (A): Favoured to the right.
• Equilibrium (B): NOT favoured to the right.
• Equilibrium (C): Favoured to the right.
• Equilibrium (D): NOT favoured to the right.

The equilibria that are NOT favoured to go to the right are (B) and (D).