How many electrons in an atom can have the designation \(1 p\), \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)

For each designation, we need to verify if the combination of quantum numbers is valid. We have three rules for the quantum numbers: 1. The principal quantum number (n) should be a positive integer (n=1,2,3,..). 2. The secondary quantum number (orbital shape) is given by letters s, p, d, and f, corresponding to quantum numbers l=0,1,2,3, respectively. The value of l can range from 0 to (n-1). 3. The magnetic quantum number (m) is associated with the orientation of the orbitals and can take integer values ranging from -l to +l. Now let's check the validity: 1. \(1p\) ➝ \(n=1\). For \(n=1\), the maximum value of l can be 0 (s-orbital only). Hence, the 1p designation is invalid. 2. \(6d_{x^{2}-y^{2}}\) ➝ \(n=6\). For \(n=6\), the maximum value of l can be 3 (f-orbitals). Since l value for d-orbitals is 2, this designation is valid. 3. \(4f\) ➝ \(n=4\). For \(n=4\), the maximum value of l can be 3 (f-orbitals). This designation is valid. 4. \(7p_{y}\) ➝ \(n=7\). For \(n=7\), the maximum value of s can be 6 (g-orbitals). The l value for a p-orbital is 1, which satisfies this condition, so the designation is valid. 5. \(2s\) ➝ \(n=2\). For \(n=2\), the maximum value of l can be 1. Since the s-orbital has l=0, this designation is valid. 6. \(n=3\) ➝ All orbitals with \(n = 3\) are valid.

For each valid designation, we have to find how many possible orientations exist for those orbitals, which will allow us to determine the number of electrons that can be accommodated in each of these orbitals, considering two electrons per orbital with opposite spins. 1. \(6d_{x^{2}-y^{2}}\) ➝ This orbital has a specific orientation, so it can accommodate 2 electrons. 2. \( 4f\) ➝ The f-orbitals correspond to l=3, which means \(m_l\) can take values from -3 to +3. There are a total of 7 different f-orbitals, and each orbital can accommodate 2 electrons, resulting in a total of 14 electrons. 3. \(7p_{y}\) ➝ This orbital has a specific orientation, so it can accommodate 2 electrons. 4. \(2s\) ➝ For the s-orbital, corresponding to an l=0, there is only one possible orientation. The 2s-orbital can accommodate 2 electrons. 5. \(n=3\) ➝ For the principal quantum number \(n = 3\), orbitals with l=0 (s-orbitals) and l=1 (p-orbitals) are possible. For the 3s orbitals, there is one possible orientation, and they can accommodate 2 electrons. For the 3p orbitals, there are three possible orientations, each of them can accommodate 2 electrons, resulting in a total of 6 electrons. Now, by adding the number of electrons for each valid designation, we find the total number of electrons that can have the given designations: 2 + 14 + 2 + 2 + (2 + 6) = 28 electrons.