EAN of Cr in `[Cr (NH_3)_6] CI_3` is

Options

A. 32
B. 33
C. 34
D. 35

Answer

To find the Effective Atomic Number (EAN) of Chromium (Cr) in the complex \([Cr(NH_3)_6]Cl_3\), we can follow these steps:

### Step 1: Determine the oxidation state of Chromium (Cr)
In the complex \([Cr(NH_3)_6]Cl_3\), the chloride ions (\(Cl^-\)) have a charge of -1 each. Since there are three chloride ions, the total charge contributed by the chloride ions is:
\[ 3 \times (-1) = -3 \]
To neutralize this charge, the overall charge of the complex must be +3. Therefore, the oxidation state of Chromium in this complex is +3.

### Step 2: Calculate the number of electrons lost by Chromium
The oxidation state of +3 indicates that Chromium has lost 3 electrons.

### Step 3: Determine the contribution of the ligands
The ligands in this complex are six ammonia (\(NH_3\)) molecules. Each ammonia molecule donates a lone pair of electrons (2 electrons). Thus, the total number of electrons donated by the six ammonia ligands is:
\[ 6 \times 2 = 12 \text{ electrons} \]

### Step 4: Calculate the total number of electrons around Chromium
The total number of electrons around Chromium can be calculated using the formula:
\[ \text{Total Electrons} = \text{Atomic Number of Cr} - \text{Electrons Lost} + \text{Electrons Gained from Ligands} \]
The atomic number of Chromium (Cr) is 24. Substituting the values we have:
\[ \text{Total Electrons} = 24 - 3 + 12 \]
\[ \text{Total Electrons} = 24 - 3 + 12 = 33 \]

### Step 5: Conclusion
Thus, the Effective Atomic Number (EAN) of Chromium in the complex \([Cr(NH_3)_6]Cl_3\) is 33.