Consider the following two reactions:
`Cd^(2+)(aq.)+4CH_(3)NH_(2) overset(K_(1)) to 'A',DeltaG_(1)^(@)`
`Cd^(2+)(aq.)+2H_(2)NCH_(2)CH_(2)NH_(2) overset(K_(2)) to 'B',DeltaG_(2)^(@)`. According to given information the correct statement(s) is/are:

Options

A. `DeltaG_(2)^(@)` is more negative than `DeltaG_(1)^(@)`
B. Compound 'A" is optically inactive
C. Compound 'B' is optically active
D. Formation constant `K_(2)` is greater than formation constant `K_(1)`

Answer

To solve the problem, we need to analyze the two reactions involving cadmium ions and the ligands provided. We will evaluate the stability of the complexes formed and the thermodynamic properties associated with these reactions.

### Step-by-Step Solution:

1. **Identify the Ligands and Their Properties:**
- The first ligand is CH₃NH₂ (methylamine), which is a monodentate ligand. This means it can donate one pair of electrons to the cadmium ion.
- The second ligand is H₂NCH₂CH₂NH₂ (ethylenediamine), which is a bidentate ligand. This means it can donate two pairs of electrons to the cadmium ion.

**Hint:** Understand the difference between monodentate and bidentate ligands, as it affects the stability of the complexes formed.

2. **Write the Reactions:**
- For the first reaction:
\[ \text{Cd}^{2+}(aq) + 4 \text{CH}_3\text{NH}_2 \overset{K_1}{\rightarrow} A \]
- For the second reaction:
\[ \text{Cd}^{2+}(aq) + 2 \text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 \overset{K_2}{\rightarrow} B \]

**Hint:** Pay attention to the stoichiometry of the ligands in each reaction.

3. **Analyze the Stability of the Complexes:**
- Bidentate ligands (like ethylenediamine) typically form more stable complexes than monodentate ligands (like methylamine) due to the chelation effect. This means that the formation constant \( K_2 \) for the bidentate ligand will be greater than \( K_1 \) for the monodentate ligand.

**Hint:** Recall that the chelation effect increases the stability of complexes formed by bidentate ligands.

4. **Evaluate the Gibbs Free Energy Changes:**
- The Gibbs free energy change (\( \Delta G \)) is related to the stability of the complex. A more negative \( \Delta G \) indicates a more stable complex. Since \( K_2 > K_1 \), it follows that:
\[ \Delta G_2 < \Delta G_1 \]
This means \( \Delta G_2 \) is more negative than \( \Delta G_1 \).

**Hint:** Remember that a more negative \( \Delta G \) corresponds to a more favorable reaction.

5. **Determine Optical Activity:**
- A complex is optically active if it has non-superimposable mirror images. Since both complexes A and B have symmetrical ligands (in the case of A, all ligands are the same, and in B, the bidentate ligand is symmetrical), both complexes are optically inactive.

**Hint:** Check the symmetry of the complexes to determine optical activity.

6. **Conclusion:**
- From the analysis, we can conclude the following statements:
1. \( \Delta G_2 \) is more negative than \( \Delta G_1 \) (True).
2. Compound A is optically inactive (True).
3. Compound B is optically inactive (True).
4. The formation constant \( K_2 \) is greater than \( K_1 \) (True).

Therefore, all statements are correct.

### Final Answer:
All statements regarding the reactions and complexes formed are correct.