The energy of 6s subshell is ............than that of 4f subshell.

Answer

To determine whether the energy of the 6s subshell is lower or higher than that of the 4f subshell, we can use the n + l rule. Here’s a step-by-step solution:

### Step 1: Understand the n + l Rule
The n + l rule states that the energy of an electron subshell can be determined by adding the principal quantum number (n) and the azimuthal quantum number (l). The subshell with the lower n + l value has lower energy.

### Step 2: Determine n and l for the 6s subshell
For the 6s subshell:
- The principal quantum number (n) is 6.
- The azimuthal quantum number (l) for an s subshell is 0.

Now, we calculate n + l:
\[ n + l = 6 + 0 = 6 \]

### Step 3: Determine n and l for the 4f subshell
For the 4f subshell:
- The principal quantum number (n) is 4.
- The azimuthal quantum number (l) for an f subshell is 3.

Now, we calculate n + l:
\[ n + l = 4 + 3 = 7 \]

### Step 4: Compare the n + l values
Now we compare the n + l values:
- For 6s: n + l = 6
- For 4f: n + l = 7

Since 6 < 7, this indicates that the 6s subshell has lower energy than the 4f subshell.

### Step 5: Conclusion
Thus, we conclude that the energy of the 6s subshell is lower than that of the 4f subshell.

### Final Answer
The energy of the 6s subshell is lower than that of the 4f subshell.

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