Step 1: Understand the magnetic quantum number (\(m_l\)).
The magnetic quantum number (\(m_l\)) describes the orientation of an orbital in space. For a given azimuthal (or angular momentum) quantum number \(l\), the possible values of \(m_l\) range from \(-l\) to \(+l\), including 0. Each unique \(m_l\) value corresponds to one specific orbital. Each orbital can hold a maximum of two electrons (according to the Pauli Exclusion Principle), one with spin \(+\frac{1}{2}\) and one with spin \(-\frac{1}{2}\).
Step 2: Analyze each given orbital type for the possibility of \(m_l = -2\).
3d orbital:
For a d-orbital, the azimuthal quantum number \(l = 2\). The possible values for \(m_l\) are \(-2, -1, 0, +1, +2\). Since \(m_l = -2\) is a possible value for a d-orbital, the 3d subshell contains an orbital with \(m_l = -2\).
This orbital can accommodate a maximum of 2 electrons.
6d orbital:
Similar to 3d, for a d-orbital, \(l = 2\).
The possible values for \(m_l\) are \(-2, -1, 0, +1, +2\).
Since \(m_l = -2\) is a possible value for a d-orbital, the 6d subshell contains an orbital with \(m_l = -2\).
This orbital can accommodate a maximum of 2 electrons.
5s orbital:
For an s-orbital, the azimuthal quantum number \(l = 0\).
The only possible value for \(m_l\) is \(0\).
Therefore, \(m_l = -2\) is NOT possible for a 5s orbital. It contributes 0 electrons to the count.
4f orbital:
For an f-orbital, the azimuthal quantum number \(l = 3\).
The possible values for \(m_l\) are \(-3, -2, -1, 0, +1, +2, +3\).
Since \(m_l = -2\) is a possible value for an f-orbital, the 4f subshell contains an orbital with \(m_l = -2\).
This orbital can accommodate a maximum of 2 electrons.
Step 3: Calculate the total maximum number of electrons.
Sum the maximum number of electrons from each orbital type that satisfies the condition \(m_l = -2\):
Total electrons = (electrons in 3d with \(m_l=-2\)) + (electrons in 6d with \(m_l=-2\)) + (electrons in 5s with \(m_l=-2\)) + (electrons in 4f with \(m_l=-2\))
Total electrons = 2 + 2 + 0 + 2 = 6 electrons. The final answer is $\boxed{6}$.
An element has two isotopes having atomic masses 10 and 15 u, respectively. If the percent abundance of lighter isotopes is 80%, then the average atomic mass of the element is:
The following data represents the frequency distribution of 20 observations:
Then its mean deviation about the mean is:
What is the molarity of a solution prepared by dissolving 5.85 g of NaCl in 500 mL of water?
(Molar mass of NaCl = 58.5 g/mol)