Which of the following statements is false?

Options

A. Permanent magnetic moment of `[Cu(NH_(3))_(4)]^(2+)` is 1.732 B.M.
B. Equilibrium constant is the ratio of rate constants of forward and backward reactions
C. `[Ni(CO_(4)]` is tetrahedral
D. For forming `NCl_(5)'N'` adopts `sp^(3)d` hybrid state

Answer

To determine which statement is false, let's analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement 1:** "Permanent magnetic moment of Cu(NH3)4^2+ is 1.732 µB."

- **Analysis:**
- Copper (Cu) in Cu(NH3)4^2+ has an oxidation state of +2. The electron configuration of Cu is [Ar] 4s¹ 3d¹⁰. When it loses two electrons to become Cu²⁺, the configuration becomes [Ar] 3d⁹.
- NH3 is a strong field ligand and causes pairing of electrons in the d-orbitals. In this case, the 3d electrons will pair up, leaving one unpaired electron.
- The formula for the magnetic moment (µ) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where n is the number of unpaired electrons. For Cu²⁺, n = 1.
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.732 \, \mu_B \]
- **Conclusion:** This statement is true.

### Step 2: Analyze the second statement
**Statement 2:** "Equilibrium constant is the ratio of rate constant of forward and backward reaction."

- **Analysis:**
- For a reaction A ⇌ B, the equilibrium constant (K) is defined as:
\[ K = \frac{K_f}{K_b} \]
where K_f is the rate constant of the forward reaction and K_b is the rate constant of the backward reaction.
- **Conclusion:** This statement is true.

### Step 3: Analyze the third statement
**Statement 3:** "Ni(CO)4 is tetrahedral."

- **Analysis:**
- Nickel (Ni) in Ni(CO)4 has an oxidation state of 0, and the electron configuration is [Ar] 3d⁸ 4s².
- CO is a strong field ligand and causes pairing of the 3d electrons. The hybridization for Ni(CO)4 is sp³, leading to a tetrahedral geometry.
- **Conclusion:** This statement is true.

### Step 4: Analyze the fourth statement
**Statement 4:** "For forming NCl5, N atom is sp³ hybridized, which is not possible."

- **Analysis:**
- The electron configuration of nitrogen is 1s² 2s² 2p³. Nitrogen does not have d-orbitals available for hybridization since it is in the second period of the periodic table.
- To form NCl5, nitrogen would need to utilize d-orbitals, which it cannot do. Therefore, the statement that N is sp³ hybridized to form NCl5 is incorrect.
- **Conclusion:** This statement is false.

### Final Conclusion
The false statement is: **"For forming NCl5, N atom is sp³ hybridized, which is not possible."**

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