In the followin electronic configuration,
`ns^(2)(n-1)d^(0-1)(n-2)f^(1-14)`
If value of `(n-1)=5` the configuration will be of `:-`

Options

A. Lanthanides
B. `d-`block
C. Actinides
D. `s-`block

Explanation

To solve the question regarding the electronic configuration given as `ns^(2)(n-1)d^(0-1)(n-2)f^(1-14)` with the condition that `(n-1) = 5`, we can follow these steps:

### Step 1: Determine the value of n
From the equation `(n-1) = 5`, we can find the value of n:
\[ n = 5 + 1 = 6 \]

### Step 2: Substitute n into the electronic configuration
Now that we have n, we can substitute it into the electronic configuration:
- \( n = 6 \)
- Therefore, the configuration becomes:
\[ 6s^{2} (5d^{0-1}) (4f^{1-14}) \]

### Step 3: Analyze the configuration
The configuration now reads:
- \( 6s^{2} \): This indicates that there are 2 electrons in the 6s subshell.
- \( 5d^{0-1} \): This indicates that there can be either 0 or 1 electron in the 5d subshell.
- \( 4f^{1-14} \): This indicates that there can be between 1 to 14 electrons in the 4f subshell.

### Step 4: Identify the element type
Now we need to identify which group of elements this configuration corresponds to:
- The presence of \( 4f \) electrons indicates that we are dealing with either lanthanides or actinides.
- For lanthanides, the configuration typically ends with \( 4f^{1-14} \).
- For actinides, the configuration typically ends with \( 5f^{1-14} \).

### Step 5: Conclusion
Since we have \( 4f^{1-14} \) in our configuration, it indicates that the configuration corresponds to the lanthanides. Therefore, the answer to the question is:

**The configuration will be of: Lanthanides.**

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