PASSAGE - I : The electron with the highest energy in an atom has the quantum numbers `n=4,l=1,m=0,s=1//2`
The element could be

Options

A. copper
B. silver
C. gallium
D. indium

Explanation

To determine the element with the highest energy electron characterized by the quantum numbers \( n=4, l=1, m=0, s=\frac{1}{2} \), we can follow these steps:

### Step 1: Identify the Orbital
The given quantum numbers are:
- \( n = 4 \) (principal quantum number)
- \( l = 1 \) (azimuthal quantum number, which corresponds to the p orbital)
- \( m = 0 \) (magnetic quantum number, which indicates the specific p orbital, in this case, the \( p_z \) orbital)
- \( s = \frac{1}{2} \) (spin quantum number)

Since \( l = 1 \), the electron is in a p orbital. Therefore, the highest energy electron is in the \( 4p \) orbital.

### Step 2: Determine the Electronic Configuration
The electron configuration leading up to the \( 4p \) orbital can be constructed as follows:
- Fill the lower energy orbitals first:
- \( 1s^2 \)
- \( 2s^2 \)
- \( 2p^6 \)
- \( 3s^2 \)
- \( 3p^6 \)
- \( 4s^2 \)
- \( 3d^{10} \)
- \( 4p^x \) (where \( x \) can be from 1 to 6, as a p orbital can hold a maximum of 6 electrons)

Thus, the complete electronic configuration can be written as:
\[ 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^x \]

### Step 3: Determine the Possible Atomic Numbers
The maximum number of electrons in the \( 4p \) orbital can be 6. Therefore, the atomic number of the element must be between:
- \( 31 \) (which corresponds to \( 4p^1 \))
- \( 36 \) (which corresponds to \( 4p^6 \))

### Step 4: Identify the Elements
Now, we can identify the elements with atomic numbers between 31 and 36:
- \( 31 \) - Gallium (Ga)
- \( 32 \) - Germanium (Ge)
- \( 33 \) - Arsenic (As)
- \( 34 \) - Selenium (Se)
- \( 35 \) - Bromine (Br)
- \( 36 \) - Krypton (Kr)

### Step 5: Conclusion
Since the quantum numbers indicate that the highest energy electron is in the \( 4p \) orbital, the element that fits this configuration is Gallium (Ga), which has an atomic number of 31.

### Final Answer
The element could be Gallium (Ga).

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