Number of unpaired electrons present in`[Ni(H_(2)O)_(6)]^(3+)`

Options

A. Two
B. One
C. Four
D. Three

Answer

To find the number of unpaired electrons in the complex \([Ni(H_2O)_6]^{3+}\), we can follow these steps:

### Step 1: Determine the oxidation state of Nickel (Ni)
The complex \([Ni(H_2O)_6]^{3+}\) has a total charge of \(+3\). The water molecules (H2O) are neutral ligands, so they do not contribute to the charge. Let the oxidation state of Ni be \(x\).

Setting up the equation:
\[ x + 6(0) = +3 \implies x = +3 \]

### Step 2: Write the electron configuration of Ni in the +3 oxidation state
Nickel (Ni) has an atomic number of 28. The ground state electron configuration of Ni is:
\[ [Ar] 4s^2 3d^8 \]
When Ni is in the +3 oxidation state, it loses 3 electrons. The 4s electrons are lost first, followed by one 3d electron. Thus, the electron configuration becomes:
\[ [Ar] 3d^7 \]

### Step 3: Determine the arrangement of the 3d electrons
The 3d subshell can hold a maximum of 10 electrons. For 7 electrons, the arrangement in the 3d subshell is:
\[ \text{3d: } \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \downarrow \]
This shows that there are 3 unpaired electrons (the first five electrons occupy separate orbitals before pairing occurs due to Hund's rule).

### Step 4: Consider the ligand field
Water (H2O) is a weak field ligand, which means it does not cause significant pairing of electrons in the d-orbitals. Therefore, the arrangement of the 3d electrons remains as calculated, with no additional pairing.

### Step 5: Conclusion
The number of unpaired electrons in \([Ni(H_2O)_6]^{3+}\) is 3.

### Final Answer:
The number of unpaired electrons present in \([Ni(H_2O)_6]^{3+}\) is **3**.

---