An atom has electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3P^(6) 3d^(3) 4s^(2)` . In which group wouold it be placed ?

Options

A. fifth group
B. fifteenth group
C. second group
D. third group

Explanation

To determine the group of the atom with the electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3) 4s^(2)`, we can follow these steps:

### Step 1: Identify the Atomic Number
The given electronic configuration can be broken down as follows:
- `1s^2` contributes 2 electrons
- `2s^2` contributes 2 electrons
- `2p^6` contributes 6 electrons
- `3s^2` contributes 2 electrons
- `3p^6` contributes 6 electrons
- `3d^3` contributes 3 electrons
- `4s^2` contributes 2 electrons

Now, we can calculate the total number of electrons:
\[ 2 + 2 + 6 + 2 + 6 + 3 + 2 = 23 \]
Thus, the atomic number (Z) of the element is 23.

### Step 2: Identify the Element
The element with atomic number 23 is Vanadium (V).

### Step 3: Determine the Block of the Element
Vanadium is located in the d-block of the periodic table because its last electron enters the d-subshell (3d).

### Step 4: Determine the Group Number
For d-block elements, the group number can be determined using the formula:
\[ \text{Group Number} = \text{Number of electrons in } (n-1)d + \text{Number of electrons in } ns \]
Where:
- \( n \) is the principal quantum number of the outermost shell.
- For Vanadium, \( n = 4 \) (since the outermost shell is 4s).

From the electronic configuration:
- Electrons in \( 3d \) (n-1) = 3
- Electrons in \( 4s \) = 2

Now, substituting the values:
\[ \text{Group Number} = 3 + 2 = 5 \]

### Conclusion
Therefore, Vanadium (atomic number 23) belongs to group 5 of the periodic table.

### Final Answer
The atom with the electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3) 4s^(2)` is placed in **Group 5**.

---