`[Fe(e n)_(2)(H_(2)O)_(2)]^(2+) +e n to` complex (x). The correct statement about the complex (x) it is

Options

A. a low spin complex
B. diamagnetic
C. shows geometrical isomerism
D. (1) and (2) both

Explanation

To solve the question regarding the complex \([Fe(en)_2(H_2O)_2]^{2+} + en\) leading to complex \(x\), we will follow these steps:

### Step 1: Identify the initial complex
The initial complex is \([Fe(en)_2(H_2O)_2]^{2+}\). Here, \(Fe\) is in the +2 oxidation state, and the ligands are ethylenediamine (en) and water (H2O).

### Step 2: Determine the coordination number and geometry
In this complex, \(Fe\) is coordinated to 4 ligands (2 en and 2 H2O). Ethylenediamine is a bidentate ligand, which means it forms two bonds with the metal ion. The coordination number is 6 (2 from each en and 2 from H2O). The geometry of a coordination number of 6 is typically octahedral.

### Step 3: Analyze the ligand field
Ethylenediamine (en) is a strong field ligand, which means it has a strong tendency to cause pairing of electrons in the d-orbitals. This results in a low-spin configuration.

### Step 4: Determine the electronic configuration
For \(Fe^{2+}\), the electronic configuration is \([Ar] 3d^6\). In the presence of strong field ligands like en, the electrons will pair up in the lower energy \(t_{2g}\) orbitals, leading to a configuration of \(t_{2g}^6\) and no unpaired electrons.

### Step 5: Conclude the nature of the complex
Since all the electrons are paired, the complex is diamagnetic. Additionally, due to the strong field ligands and the pairing of electrons, it is classified as a low-spin complex.

### Final Answer
The correct statement about complex \(x\) is that it is a low-spin, diamagnetic complex.