considering the electron of outermost orbital of `Cu` match the items given the column I with their values given in column II. `Column-I { (A) `orbital angular momentum`, (B)` angular momentum in an orbit`, (C)` spin angular momentum`}` `Column - II ( I. 4h, II. 0, III. 0.86h, IV. 1.73)`

Options

A. A(II),B(I), C(III)
B. A(III),B(IV),C(II)
C. A(I),B(IV),C(II)
D. A(I),B(II),C(III)

Answer

To solve the problem, we first need to understand the electronic configuration of copper (Cu) and then calculate the required angular momentum values for the outermost electron.

### Step 1: Determine the electronic configuration of Copper (Cu)
Copper has an atomic number of 29. The electronic configuration of copper is:
\[ \text{Cu: } [\text{Ar}] 3d^{10} 4s^1 \]
This means that the outermost electron is in the 4s orbital.

### Step 2: Identify the outermost electron
The outermost electron of copper is in the 4s orbital.

### Step 3: Calculate the orbital angular momentum (L)
The orbital angular momentum (L) for an electron in an s orbital (l = 0) is given by:
\[ L = \sqrt{l(l+1)}h \]
For the 4s orbital, l = 0:
\[ L = \sqrt{0(0+1)}h = 0 \]
Thus, the value for orbital angular momentum is:
**(B) 0**

### Step 4: Calculate the angular momentum in an orbit (L)
The angular momentum in an orbit for an electron can be calculated using the formula:
\[ L = n \cdot \frac{h}{2\pi} \]
For the outermost electron in the 4s orbital, n = 4:
\[ L = 4 \cdot \frac{h}{2\pi} = 2h \]
However, this is not one of the options provided. We will consider the effective angular momentum which is typically represented as:
\[ L = \sqrt{l(l+1)}h \]
For the 4s orbital, l = 0, so:
**(B) 0**

### Step 5: Calculate the spin angular momentum (S)
The spin angular momentum (S) is given by:
\[ S = \sqrt{s(s+1)}h \]
For an electron, s = 1/2:
\[ S = \sqrt{\frac{1}{2}(\frac{1}{2}+1)}h = \sqrt{\frac{1}{2} \cdot \frac{3}{2}}h = \sqrt{\frac{3}{4}}h = \frac{\sqrt{3}}{2}h \approx 0.86h \]
Thus, the value for spin angular momentum is:
**(C) 0.86h**

### Step 6: Match the values from Column I to Column II
Now we can match the values:
- (A) Orbital angular momentum → II. 0
- (B) Angular momentum in an orbit → I. 4h (but we found it to be 0, so this may not match)
- (C) Spin angular momentum → III. 0.86h

### Final Matching
- (A) → II. 0
- (B) → IV. 1.73 (this is not directly calculated, but could be a misinterpretation)
- (C) → III. 0.86h

### Summary of Matches
- (A) → II. 0
- (B) → IV. 1.73
- (C) → III. 0.86h