How many of the following statements is `//` are correct ?
(1)The order of splitting energy is `PtCl_(4)^(2-) gt PdCl_(4)^(2-) gt NiCl_(4)^(2-) ` ( consider only magnitude)
(2)`[Ni(CO)_(4)]` is diamagnatic whereas `[Ni(H_(2)O)_(6)]^(2+)` is paramagnetic
(3) `[Ni(CN)_(4)]^(2-) rarr dsp^(2)` hybridized and paramagnetic
(4) The magnetic moment of `K_(3) [Fe(CN)_(6)]` is `sqrt(3)` B.M

Options

A. The order of splitting energy is `PtCl_(4)^(2-) gt PdCl_(4)^(2-) gt NiCl_(4)^(2-) ` ( consider only magnitude)
B. `[Ni(CO)_(4)]` is diamagnatic whereas `[Ni(H_(2)O)_(6)]^(2+)` is paramagnetic
C. `[Ni(CN)_(4)]^(2-) rarr dsp^(2)` hybridized and paramagnetic
D. The magnetic moment of `K_(3) [Fe(CN)_(6)]` is `sqrt(3)` B.M

Answer

To determine how many of the given statements are correct, we will evaluate each statement one by one.

### Statement 1:
**The order of splitting energy is \( \text{PtCl}_4^{2-} > \text{PdCl}_4^{2-} > \text{NiCl}_4^{2-} \) (consider only magnitude).**

- **Analysis:**
- \( \text{Pt}^{2+} \) (5d) has the highest splitting energy due to its higher principal quantum number and the presence of more d-orbitals.
- \( \text{Pd}^{2+} \) (4d) has a moderate splitting energy.
- \( \text{Ni}^{2+} \) (3d) has the lowest splitting energy.

Thus, the order of splitting energy is indeed \( \text{PtCl}_4^{2-} > \text{PdCl}_4^{2-} > \text{NiCl}_4^{2-} \).

**Conclusion:** This statement is correct.

### Statement 2:
**\[ \text{[Ni(CO)}_4] \] is diamagnetic whereas \[ \text{[Ni(H}_2\text{O)}_6]^{2+} \] is paramagnetic.**

- **Analysis:**
- In \[ \text{[Ni(CO)}_4] \], CO is a strong field ligand, leading to pairing of all electrons. Thus, it is diamagnetic.
- In \[ \text{[Ni(H}_2\text{O)}_6]^{2+} \], water is a weak field ligand, which does not cause pairing of all d-electrons. Hence, it has unpaired electrons and is paramagnetic.

**Conclusion:** This statement is also correct.

### Statement 3:
**\[ \text{[Ni(CN)}_4]^{2-} \] is dsp\(^2\) hybridized and paramagnetic.**

- **Analysis:**
- CN\(^-\) is a strong field ligand, which causes pairing of electrons in nickel.
- The oxidation state of Ni in \[ \text{[Ni(CN)}_4]^{2-} \] is +2, and its electronic configuration is \[ \text{[Ar]} 3d^8 \].
- With pairing, it will not have unpaired electrons, making it diamagnetic, not paramagnetic.
- The hybridization is actually dsp\(^2\), but the statement about being paramagnetic is incorrect.

**Conclusion:** This statement is incorrect.

### Statement 4:
**The magnetic moment of \[ K_3[\text{Fe(CN)}_6] \] is \(\sqrt{3}\) B.M.**

- **Analysis:**
- In \[ K_3[\text{Fe(CN)}_6] \], Fe is in the +3 oxidation state and has the electronic configuration of \[ \text{[Ar]} 3d^5 \].
- CN\(^-\) is a strong field ligand, leading to pairing of electrons. Therefore, all electrons are paired, resulting in no unpaired electrons.
- The magnetic moment can be calculated using the formula \(\sqrt{n(n+2)}\). Here, \(n = 0\) (no unpaired electrons), so the magnetic moment is 0, not \(\sqrt{3}\).

**Conclusion:** This statement is incorrect.

### Summary of Results:
- Statement 1: Correct
- Statement 2: Correct
- Statement 3: Incorrect
- Statement 4: Incorrect

### Total Correct Statements:
There are **2 correct statements** out of the 4 provided.

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