Ratio of magnetic moment of the ions `Co^(2+) and Cr^(3+)` is :

Options

A. `sqrt(15)`
B. 1
C. 3
D. 2

Explanation

To find the ratio of the magnetic moments of the ions \( \text{Co}^{2+} \) and \( \text{Cr}^{3+} \), we will follow these steps:

### Step 1: Determine the electronic configuration of \( \text{Co}^{2+} \)
Cobalt (Co) has an atomic number of 27. Its electronic configuration is:
\[ \text{Co}: [\text{Ar}] \, 3d^7 \, 4s^2 \]
When cobalt loses 2 electrons to form \( \text{Co}^{2+} \), the electrons are removed first from the 4s orbital, leading to:
\[ \text{Co}^{2+}: [\text{Ar}] \, 3d^7 \, 4s^0 \]

### Step 2: Determine the number of unpaired electrons in \( \text{Co}^{2+} \)
In the \( 3d^7 \) configuration, the distribution of electrons in the \( d \) orbitals can be represented as follows:
- \( 3d \) orbitals can hold a maximum of 10 electrons.
- The filling of the \( 3d \) orbitals follows Hund's rule, which states that every orbital in a subshell is singly occupied before any orbital is doubly occupied.

The configuration of \( 3d^7 \) will have:
- 3 orbitals with 1 electron each (3 unpaired)
- 2 orbitals with 2 electrons each (paired)

Thus, \( \text{Co}^{2+} \) has **3 unpaired electrons**.

### Step 3: Determine the electronic configuration of \( \text{Cr}^{3+} \)
Chromium (Cr) has an atomic number of 24. Its electronic configuration is:
\[ \text{Cr}: [\text{Ar}] \, 3d^5 \, 4s^1 \]
When chromium loses 3 electrons to form \( \text{Cr}^{3+} \), the electrons are removed first from the 4s and then from the 3d orbitals, leading to:
\[ \text{Cr}^{3+}: [\text{Ar}] \, 3d^3 \, 4s^0 \]

### Step 4: Determine the number of unpaired electrons in \( \text{Cr}^{3+} \)
In the \( 3d^3 \) configuration, the distribution of electrons in the \( d \) orbitals will be:
- 3 orbitals with 1 electron each (3 unpaired)

Thus, \( \text{Cr}^{3+} \) also has **3 unpaired electrons**.

### Step 5: Calculate the magnetic moments
The magnetic moment (\( \mu \)) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

For both \( \text{Co}^{2+} \) and \( \text{Cr}^{3+} \):
- \( n = 3 \)

Calculating the magnetic moment for both ions:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 6: Find the ratio of magnetic moments
Since both \( \text{Co}^{2+} \) and \( \text{Cr}^{3+} \) have the same magnetic moment:
\[ \text{Ratio of } \mu_{\text{Co}^{2+}} : \mu_{\text{Cr}^{3+}} = \sqrt{15} : \sqrt{15} = 1 : 1 \]

### Final Answer
The ratio of the magnetic moments of \( \text{Co}^{2+} \) and \( \text{Cr}^{3+} \) is **1:1**.

---