Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour ?

Options

A. `[Ni(NH_(3))_(6)]^(2+)`
B. `[Zn(NH_(3))_(6)]^(2+)`
C. `[Cr(NH_(3))_(6)]^(2+)`
D. `[Co(NH_(3))_(6)]^(2+)`

Explanation

To determine which of the given complexes is an outer orbital complex and exhibits paramagnetic behavior, we will analyze the oxidation state, electron configuration, hybridization, and magnetic properties of each complex. Let's go through the steps using the example of the complex [Ni(NH3)6]²⁺.

### Step 1: Determine the Oxidation State
Assume the oxidation state of nickel (Ni) is \( X \). The ammonia (NH3) is a neutral ligand, so it contributes 0 to the charge. The overall charge of the complex is +2.

\[ X + 6(0) = +2 \implies X = +2 \]

**Hint:** To find the oxidation state, set up an equation based on the charge of the complex and the contributions of the ligands.

### Step 2: Write the Electron Configuration
Nickel has an atomic number of 28. The ground state electron configuration of Ni is:

\[ \text{Ni: } [Ar] 4s^2 3d^8 \]

In the +2 oxidation state, we remove two electrons (the 4s electrons are removed first):

\[ \text{Ni}^{2+}: [Ar] 3d^8 \]

**Hint:** Remember that when determining the electron configuration in an ion, remove electrons from the outermost shell first.

### Step 3: Analyze the d-Orbital Filling
In the 3d subshell, there are 8 electrons. The d-orbitals can be filled as follows:

\[ \text{3d: } \uparrow \downarrow \, \uparrow \downarrow \, \uparrow \downarrow \, \uparrow \, \uparrow \]

This shows that there are 2 unpaired electrons in the 3d orbitals.

**Hint:** Use Hund's rule to fill the d-orbitals before pairing the electrons.

### Step 4: Determine Hybridization
Ammonia (NH3) is a strong field ligand and can cause pairing of electrons. However, since we are considering the outer orbital complex, we will use the 4s and 4p orbitals along with the 3d orbitals for hybridization.

The hybridization for an octahedral complex with 6 ligands is:

\[ \text{Hybridization: } sp^3d^2 \]

This indicates that the complex is formed from the outer orbitals.

**Hint:** For octahedral complexes, the hybridization is typically \( sp^3d^2 \) if outer orbitals are involved.

### Step 5: Determine Geometry and Magnetic Properties
The geometry of the complex is octahedral due to the \( sp^3d^2 \) hybridization. Since there are 2 unpaired electrons, the complex exhibits paramagnetic behavior.

**Hint:** Count the number of unpaired electrons to determine if the complex is paramagnetic (unpaired) or diamagnetic (all paired).

### Conclusion
The complex [Ni(NH3)6]²⁺ is an outer orbital complex and exhibits paramagnetic behavior due to the presence of unpaired electrons.

**Final Answer:** The correct option is [Ni(NH3)6]²⁺.