Question

question 1
1 pts
which of the following is the electron configuration of silver?
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁹ 5p⁶ 6s²
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁹
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰ 4p⁶ 5s² 5d⁹

Explanation:

Step1: Recall silver's atomic number

Silver (Ag) has an atomic number of 47, which means it has 47 electrons.

Step2: Apply electron - filling rules

Electrons fill orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, etc. The s - subshell can hold 2 electrons, p - subshell can hold 6 electrons, and d - subshell can hold 10 electrons.
The electron - configuration of silver is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{10}$. But in the given options, the correct one based on the normal filling - order representation (ignoring the anomalous behavior for simplicity in this context) for a neutral atom's electron - configuration up to the correct number of electrons is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{9}$ which is due to the sequential filling of orbitals before considering the more stable configuration related to the d - subshell's full or half - full states in some cases.

Answer:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{9}$ (assuming the normal orbital - filling order as presented in the options and not the more stable actual configuration of silver which has a $5s^{1}4d^{10}$ due to stability reasons)