When the value of the azimuthal quantum number is 3 the maximum and the minimum values of the spin multiplicites are

Options

A. 4,3
B. 8,1
C. 1,3
D. 8,2

Answer

To solve the problem regarding the maximum and minimum values of spin multiplicities when the azimuthal quantum number (l) is 3, we will follow these steps:

### Step 1: Identify the orbital type
The azimuthal quantum number \( l = 3 \) corresponds to the f-orbital.

### Step 2: Determine the maximum number of electrons
Each f-orbital can hold a maximum of 14 electrons. This is because:
- Each orbital can hold 2 electrons (one with spin up and one with spin down).
- There are 7 orbitals in the f-subshell (since \( l = 3 \)), thus \( 7 \times 2 = 14 \).

### Step 3: Calculate the maximum spin multiplicity
The spin multiplicity is given by the formula:
\[ \text{Spin multiplicity} = 2s + 1 \]
where \( s \) is the total spin angular momentum.

To find \( s \):
- The maximum number of unpaired electrons occurs when all electrons have the same spin direction (all spin up).
- For 14 electrons, the maximum number of unpaired electrons is 7 (as they can be arranged with all spins up).
- Therefore, \( s = \frac{7}{2} \).

Now substituting \( s \) into the spin multiplicity formula:
\[ \text{Maximum spin multiplicity} = 2 \left(\frac{7}{2}\right) + 1 = 7 + 1 = 8 \]

### Step 4: Calculate the minimum spin multiplicity
For the minimum spin multiplicity, we consider the scenario where all the electrons are paired:
- In this case, all 14 electrons can be paired, leading to no unpaired electrons. Thus, \( s = 0 \).

Substituting \( s \) into the spin multiplicity formula:
\[ \text{Minimum spin multiplicity} = 2(0) + 1 = 1 \]

### Final Answer
Thus, the maximum spin multiplicity is 8 and the minimum spin multiplicity is 1.

### Summary
- Maximum spin multiplicity: 8
- Minimum spin multiplicity: 1

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