The general electronic configuration of outer most and penultimate shell is given as `(n-1)s^2(n-1)p^6(n-1)d^x ns^2`. Then for an element with `n = 4` and `x = 7`
The number of unpaired electrons in the divalent cation of the element in isolated gaseous state is 

Options

A. 0
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to determine the number of unpaired electrons in the divalent cation of the element with the given electronic configuration. Let's break it down step by step.

### Step 1: Determine the electronic configuration
Given the general electronic configuration:
\[ (n-1)s^2 (n-1)p^6 (n-1)d^x ns^2 \]
Substituting \(n = 4\) and \(x = 7\):
\[ (4-1)s^2 (4-1)p^6 (4-1)d^7 4s^2 \]
This simplifies to:
\[ 3s^2 3p^6 3d^7 4s^2 \]

### Step 2: Write the complete electronic configuration
The complete electronic configuration for the element is:
\[ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2 \]

### Step 3: Identify the element
The element with this configuration corresponds to the atomic number 27, which is cobalt (Co).

### Step 4: Determine the cation configuration
For the divalent cation (Co²⁺), we need to remove 2 electrons. Electrons are removed first from the outermost shell (4s) before the d-orbitals:
\[ \text{Co} \rightarrow \text{Co}^{2+} : 3s^2 3p^6 3d^7 \]

### Step 5: Count the unpaired electrons
Now we look at the \(3d^7\) configuration. The filling of the d-orbitals follows Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up. The \(3d\) subshell can hold a maximum of 10 electrons. The \(3d^7\) configuration can be represented as follows:

- Each of the five \(d\) orbitals will have one electron first, and then two of them will pair up:
\[ \text{d-orbital filling: } \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \downarrow \]
This results in:
- 3 unpaired electrons (the three electrons that are in separate orbitals).

### Final Answer
The number of unpaired electrons in the divalent cation of the element in isolated gaseous state is **3**.

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