The element having the lowest atomic number and a ground state electronic configuration of `(n-1) d^(6)ns^(2)` is placed in:

Options

A. fifth period
B. sixth period
C. fourth period
D. third period

Answer

To determine the period in which the element with the ground state electronic configuration of `(n-1)d^6 ns^2` is placed, we can follow these steps:

### Step 1: Understand the Electronic Configuration
The electronic configuration given is `(n-1)d^6 ns^2`. Here, `n` represents the principal quantum number of the outermost shell (valence shell), while `(n-1)d` refers to the d-subshell that is one principal quantum number lower than the outermost shell.

### Step 2: Identify the Value of n
In the configuration `(n-1)d^6 ns^2`, we can see that:
- The `d` subshell has 6 electrons, which means it is filled from the `(n-1)` level.
- The `s` subshell has 2 electrons, which corresponds to the outermost shell.

### Step 3: Determine the Period
The value of `n` indicates the period of the element in the periodic table. Since the `s` subshell corresponds to the outermost shell:
- If `n = 4`, then the configuration would be `3d^6 4s^2`.

### Step 4: Conclusion
Since `n` is equal to 4, the element with the configuration `(n-1)d^6 ns^2` is placed in the **4th period** of the periodic table.

### Final Answer
The element is placed in the **4th period**.

---