Answer
See the explanation
Work Step by Step
a. In the ground state of mercury (\( \mathrm{Hg} \)), the atomic orbitals with \( n=3 \) are the 3s, 3p, and 3d orbitals. The maximum number of electrons that can occupy each of these orbitals is given by the formula \( 2(2l+1) \), where \( l \) is the azimuthal quantum number.
For the 3s orbital, \( l=0 \), so the maximum number of electrons is \( 2(2(0)+1) = 2 \) electrons.
For the 3p orbital, \( l=1 \), so the maximum number of electrons is \( 2(2(1)+1) = 6 \) electrons.
For the 3d orbital, \( l=2 \), so the maximum number of electrons is \( 2(2(2)+1) = 10 \) electrons.
b. The \( d \) atomic orbitals refer to the 3d, 4d and 5d orbitals. As mentioned above, the maximum number of electrons that can occupy the d orbitals is 30.
c. There are 4 sets of np orbitals. The number of electrons occupying the \( p \) atomic orbitals is $4(2)=8$.
d. The spin of an electron is described by the quantum number \( m_{s} \). Each orbital can accommodate two electrons with opposite spins, one with \( m_{s}=+\frac{1}{2} \) (spin "up") and the other with \( m_{s}=-\frac{1}{2} \) (spin "down"). Since the question specifically asks for the number of electrons with spin "up" (\( m_{s}=+\frac{1}{2} \)), we can say that the number of electrons with this spin is equal to the total number of electrons in the system. However, without additional information about the electron configuration of mercury, we cannot determine the exact number of electrons with spin "up" in the ground state of mercury.
