The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are
- \( n = 4, \, l = 2, \, m = -1, \, s = +\frac{1}{2} \)
- \( n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2} \)
- \( n = 3, \, l = 0, \, m = 1, \, s = +\frac{1}{2} \)
- \( n = 2, \, l = 0, \, m = 0, \, s = +\frac{1}{2} \)
The Correct Option is B
Approach Solution - 1
The question asks us to identify the set of quantum numbers for the electron in the outermost orbital of potassium, which has an atomic number of 19. Let's solve this step-by-step:
- Electronic Configuration: The atomic number 19 corresponds to the element potassium (K). Its electronic configuration can be written as: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1\) This indicates that the outermost electron is in the 4s orbital.
- Examine Quantum Numbers:
- Principal Quantum Number (n): This describes the shell and energy level of the electron. Here, \(n = 4\) since the outermost electron is in the 4s orbital.
- Azimuthal Quantum Number (l): Represents the subshell/angular momentum. For s-orbitals, \(l = 0\).
- Magnetic Quantum Number (m): Represents the orientation of the orbital. For \(l = 0\), the only possible value of \(m\) is \(0\).
- Spin Quantum Number (s): Represents the electron's spin, which can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). Typically, atoms are in their lowest energy state, so we often assume \(s = +\frac{1}{2}\).
- Conclusion: Therefore, the correct set of quantum numbers for the outermost electron in potassium is \(n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}\).
After evaluating the options:
- \(n = 4, \, l = 2, \, m = -1, \, s = +\frac{1}{2}\) - Incorrect; \(l = 2\) is for d-orbitals, not applicable to s-orbitals.
- \(n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}\) - Correct; corresponds to the outer electron in the 4s orbital.
- \(n = 3, \, l = 0, \, m = 1, \, s = +\frac{1}{2}\) - Incorrect; 4s is the outermost orbital, not 3s.
- \(n = 2, \, l = 0, \, m = 0, \, s = +\frac{1}{2}\) - Incorrect; This refers to the 2s subshell, not the outermost shell for potassium.
Thus, the correct answer is \( n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2} \).
Approach Solution -2
To determine the quantum numbers for the electron in the outermost orbital of potassium (atomic number 19), we first need to consider its electronic configuration. Potassium is the first element in the fourth period of the periodic table, so its electron configuration is:
\(\text{K:} \, 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1\)
Here, the electron in the outermost shell is in the \(4s\) orbital. We now assign the quantum numbers to this electron:
- Principal Quantum Number (n): This represents the shell in which the electron resides. For the \(4s\) orbital, \(n = 4\).
- Azimuthal Quantum Number (l): This represents the subshell of the electron and depends on the type of orbital:
- \(s\) orbital has \(l = 0\)
- \(p\) orbital has \(l = 1\)
- \(d\) orbital has \(l = 2\)
- \(f\) orbital has \(l = 3\)
- Magnetic Quantum Number (m): This describes the orientation of the orbital, ranging from \(-l\) to \(+l\). For \(l = 0\), \() must be \\).
- Spin Quantum Number (s): This describes the electron's spin, which can either be \(+\frac{1}{2}\) or \(-\frac{1}{2}\). Typically, the first electron in an orbital is assigned \(+\frac{1}{2}\) spin.
Based on the above reasoning, the four quantum numbers for the outermost electron in potassium are:
\(n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}\)
This matches the given option: \( n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2} \), making it the correct answer.
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