The set of quantum numbers n=4 , l=0 m=0 and `s=+1/2` correspond to the most loosely bound , ground state electron of which one of the following atoms

Options

A. Na
B. Cl
C. Cr
D. Rb

Explanation

To determine which atom corresponds to the quantum numbers \( n=4 \), \( l=0 \), \( m=0 \), and \( s=+\frac{1}{2} \), we will follow these steps:

### Step 1: Identify the Quantum Numbers
The quantum numbers provided are:
- Principal quantum number (\( n \)) = 4
- Azimuthal quantum number (\( l \)) = 0
- Magnetic quantum number (\( m \)) = 0
- Spin quantum number (\( s \)) = +\(\frac{1}{2}\)

### Step 2: Determine the Orbital Type
The azimuthal quantum number \( l \) indicates the type of orbital:
- \( l = 0 \) corresponds to an s orbital.

### Step 3: Identify the Electron Configuration
Since \( n = 4 \) and \( l = 0 \), we are dealing with the 4s orbital. The quantum number \( m = 0 \) confirms that we are in the only available orbital for \( l = 0 \), which is the 4s orbital.

### Step 4: Determine the Number of Electrons in the Atom
The spin quantum number \( s = +\frac{1}{2} \) indicates that we are considering one electron in the 4s orbital.

### Step 5: Identify the Atom with the Configuration Ending in 4s¹
The 4s orbital is filled before the 3d orbitals in the periodic table. The electron configuration for various elements can be analyzed:
- For Sodium (Na, atomic number 11): \( 1s^2 2s^2 2p^6 3s^1 \) (does not have 4s)
- For Chlorine (Cl, atomic number 17): \( 1s^2 2s^2 2p^6 3s^2 3p^5 \) (does not have 4s)
- For Chromium (Cr, atomic number 24): \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1 \) (has 4s¹)
- For Rubidium (Rb, atomic number 37): \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10} 4p^6 \) (has 5s¹)

### Step 6: Conclusion
Among the options analyzed, Chromium (Cr) has the electron configuration ending in \( 4s^1 \), which corresponds to the most loosely bound ground state electron with the given quantum numbers.

Thus, the answer is **Chromium (Cr)**.

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