How many orbitals in an atom can have the designation \(5 p, 3 d_{z^{2}}\), \(4 d, n=5, n=4 ?\)

For each designation, we need to identify the four quantum numbers (n, l, m_l, and m_s) involved: - n (Principal Quantum Number): Represents the energy level of the electron and the relative distance of the electron from the nucleus. Allowed integer values are: n = 1, 2, 3, ... - l (Azimuthal Quantum Number): Represents the shape of the orbital and has integer values from 0 to (n-1). 0 = s, 1 = p, 2 = d, 3 = f, ... - m_l (Magnetic Quantum Number): Represents the orientation of the orbital and has integer values from -l to +l. For example, if l = 1, then m_l can be -1, 0, +1. - m_s (Spin Quantum Number): Represents the spin of an electron and has two values: +1/2 (up) and -1/2 (down).

Now we will work with each specified designation: 1. \(5p\): For this designation, n = 5 and l = 1. As a result, m_l can have values -1, 0, and +1. The spin quantum number (m_s) will double the number of orbitals since each orbital can hold 2 electrons (up and down). So, there are 2 x 3 = 6 orbitals in this case. 2. \(3d_{z^2}\): For this designation, n = 3 and l = 2. Since the question specifies the \(d_{z^2}\) orbital, we only need to consider one particular m_l value, which in this case is 0. There are 2 possible spins (up and down) for the electrons in this orbital. Therefore, there are 2 x 1 = 2 orbitals. 3. \(4d\): For this designation, n = 4 and l = 2. For l = 2, there are 5 possible m_l values: -2, -1, 0, +1, and +2. As before, there are 2 possible spins (up and down) for the electrons in each of these orbitals. Therefore, there are 2 x 5 = 10 orbitals in this case. 4. \(n=5\): For this designation, since only n is specified, we need to consider all possible combinations of l and m_l for n = 5. For n = 5, there can be l = 0, 1, 2, 3, 4. For each value of l, we can calculate the number of orbitals as: \[ \begin{aligned} \text{orbitals} & = 2 \cdot (1 + 3 + 5 + 7 + 9) \\ & = 2 \cdot (25) \\ & = 50 \end{aligned} \] 5. \(n=4\): Similarly, for n = 4, we will consider all possible combinations of l and m_l. For n = 4, there can be l = 0, 1, 2, 3. The calculation for the number of orbitals is: \[ \begin{aligned} \text{orbitals} & = 2 \cdot (1 + 3 + 5 + 7) \\ & = 2 \cdot (16) \\ & = 32 \end{aligned} \]

Now, we can add up the number of orbitals for each designation: Total orbitals = Orbitals for 5p + Orbitals for 3d_{z^2} + Orbitals for 4d + Orbitals for n=5 + Orbitals for n=4 = 6 + 2 + 10 + 50 + 32 = 100 So, there are 100 orbitals in total that can have the designation \(5p\), \(3d_{z^2}\), \(4d\), \(n=5\), and \(n=4\) in an atom.