In the general electronic configuration `(n-2)^f(1-14)` `(n-1)^d(0-1) ns^2`,if the value of n=7,the configuration will be of

Options

A. Lanthanoids
B. Actinides
C. Transition elements
D. p-block elements

Answer

To solve the question regarding the electronic configuration given by the formula `(n-2)^f(1-14) (n-1)^d(0-1) ns^2` with the value of n=7, we can follow these steps:

### Step 1: Substitute the value of n
Given the formula, we need to substitute n with 7:
- For the f subshell: `(n-2) = (7-2) = 5`, so we have `5f^(1-14)`.
- For the d subshell: `(n-1) = (7-1) = 6`, so we have `6d^(0-1)`.
- For the s subshell: `ns^2 = 7s^2`.

Putting it all together, we get the electronic configuration:
\[ 5f^{(1-14)} \, 6d^{(0-1)} \, 7s^2 \]

### Step 2: Identify the elements based on the configuration
Now, let's analyze the configuration:
- The `5f` subshell can hold up to 14 electrons, which corresponds to the actinides series.
- The `6d` subshell can hold up to 10 electrons, but in this case, it can have 0 or 1 electron.
- The `7s` subshell can hold 2 electrons.

### Step 3: Determine the period and block
Since the highest principal quantum number (n) is 7, this configuration belongs to the 7th period of the periodic table.

- The filling of the `5f` subshell indicates that we are dealing with the actinides, as they are characterized by the filling of the 5f orbitals.
- The `6d` subshell being partially filled (0 or 1 electron) also supports that we are in the actinide series, as the transition metals (d-block) would have a more filled d subshell.

### Conclusion
Based on the analysis, the electronic configuration corresponds to elements in the actinide series. Therefore, the configuration will belong to the actinides.

### Final Answer
The configuration will be of **actinides**.

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