Final answer:
A ground state atom of Arsenic (As) could not have an electron with the configuration B) n = 4, l = 2, ml = 0, ms = +1/2 because arsenic's outermost electrons in its ground state are in the 4p orbital, not the 4d orbital. All other listed configurations are possible for arsenic in its ground state.
Step-by-step explanation:
To determine which electron configuration a ground-state atom of Arsenic (As) could not have, we must understand quantum numbers and how they apply to electron configurations within an atom. These configurations are described using four quantum numbers: n (the principal quantum number), l (the azimuthal or angular momentum quantum number), ml (the magnetic quantum number), and ms (the spin quantum number). Each of these numbers must adhere to specific rules, and no two electrons can have the same set of four quantum numbers within an atom, as dictated by the Pauli exclusion principle. Arsenic is in the fourth period of the periodic table with a ground-state electron configuration ending in 4s23d104p3. For the electron to be in its ground state in arsenic, 'n' must be either 1, 2, 3, or 4 since arsenic's electrons fill these shells. For each value of 'n', 'l' can be any integer ranging from 0 to n-1. For example, with n = 3, l can be 0 (3s), 1 (3p), or 2 (3d). The magnetic quantum number 'ml' can range from -l to +l, while the spin quantum number 'ms' can be either +1/2 or -1/2. Considering these rules, option B) n = 4, l = 2, ml = 0, ms = +1/2 is an electron configuration that a ground state atom of arsenic could not have, because, with n = 4 and l = 2, the electron would reside in a 4d orbital. However, arsenic does not have any electrons in the 4d orbitals in its ground state; its outermost electrons are in the 4p orbital. All other options A), C), D), and E) are possible configurations within arsenic's ground state.