Question

question 1 (4 points) saved which one of the following sets of quantum numbers represents an electron with the highest energy? n = 3, l = 2, m_l = -2, m_s = +1/2 n = 4, l = 1, m_l = 0, m_s = -1/2 n = 4, l = 2, m_l = -1, m_s = -1/2 n = 5, l = 0, m_l = 0, m_s = +1/2 n = 4, l = 0, m_l = 0, m_s = +1/2

Explanation:

To determine the electron with the highest energy, we use the principle that the energy of an electron in an atom is mainly determined by the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)). The energy increases with increasing \(n\), and for the same \(n\), it increases with increasing \(l\) (since \(l = 0\) for \(s\)-orbitals, \(l = 1\) for \(p\)-orbitals, \(l = 2\) for \(d\)-orbitals, \(l = 3\) for \(f\)-orbitals, and the energy order is \(s < p < d < f\) for the same \(n\)).

Let's analyze each option:

1. \(n = 3, l = 2\): \(n = 3\), \(l = 2\) ( \(d\)-orbital).
2. \(n = 4, l = 1\): \(n = 4\), \(l = 1\) ( \(p\)-orbital).
3. \(n = 4, l = 2\): \(n = 4\), \(l = 2\) ( \(d\)-orbital).
4. \(n = 5, l = 0\): \(n = 5\), \(l = 0\) ( \(s\)-orbital).
5. \(n = 4, l = 0\): \(n = 4\), \(l = 0\) ( \(s\)-orbital).

Now, compare the energy levels:

• For \(n = 3, l = 2\): Energy is lower than \(n = 4\) cases (since \(n\) is smaller).
• For \(n = 4, l = 1\) ( \(p\)-orbital) vs. \(n = 4, l = 2\) ( \(d\)-orbital): \(d\)-orbitals have higher energy than \(p\)-orbitals for the same \(n\), so \(n = 4, l = 2\) is higher than \(n = 4, l = 1\).
• For \(n = 5, l = 0\) ( \(s\)-orbital) vs. \(n = 4, l = 2\) ( \(d\)-orbital): The energy of \(n = 5, l = 0\) ( \(5s\)) is lower than \(n = 4, l = 2\) ( \(4d\))? Wait, no—actually, the energy order can have exceptions (e.g., \(4d\) vs. \(5s\)), but in general, for multi - electron atoms, the energy of \(4d\) is higher than \(5s\)? Wait, no, the Aufbau principle tells us that \(5s\) fills before \(4d\), meaning \(5s\) has lower energy than \(4d\). Wait, let's recall the energy order: The energy of orbitals follows the order based on \(n + l\) value. For an orbital, the \(n + l\) value determines its energy level (lower \(n + l\) means lower energy; if \(n + l\) is the same, lower \(n\) means lower energy).

Let's calculate \(n + l\) for each:

• \(n = 3, l = 2\): \(n + l=3 + 2 = 5\)
• \(n = 4, l = 1\): \(n + l = 4+1 = 5\)
• \(n = 4, l = 2\): \(n + l=4 + 2 = 6\)
• \(n = 5, l = 0\): \(n + l=5 + 0 = 5\)
• \(n = 4, l = 0\): \(n + l=4 + 0 = 4\)

The \(n + l\) value for \(n = 4, l = 2\) is \(6\), which is higher than the others (others have \(n + l = 4\) or \(5\)). So the orbital with \(n = 4, l = 2\) has the highest energy among these options.

Answer:

\(n = 4, l = 2, m_l=-1, m_s = - 1/2\) (the third option: \(n = 4, l = 2, m_l=-1, m_s=-1/2\))