Question

write the ground-state electron configuration for each element.
part 1 of 2
au:
part 2 of 2
as:

Explanation:

Part 1: Ground - state electron configuration of Au (Gold)

Step 1: Determine the atomic number of Au

Gold (Au) has an atomic number of 79. This means a neutral Au atom has 79 electrons.

Step 2: Recall the noble gas shorthand method

We can use the noble gas that comes before Au in the periodic table to simplify the electron configuration. The noble gas before Au is Xenon (Xe), which has an electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}$.

Step 3: Fill the remaining electrons

After Xe, we need to fill the remaining $79 - 54=25$ electrons. The electron configuration for the electrons beyond Xe follows the order of orbitals. The electron configuration of Au is $[Xe]4f^{14}5d^{10}6s^{1}$. The reason for the $6s^{1}$ and $5d^{10}$ (instead of the expected $5d^{9}6s^{2}$) is due to the stability associated with a completely filled d - orbital and a half - filled s - orbital (or in this case, a single electron in the s - orbital and a filled d - orbital, which is a more stable configuration for Au).

Part 2: Ground - state electron configuration of As (Arsenic)

Step 1: Determine the atomic number of As

Arsenic (As) has an atomic number of 33, so a neutral As atom has 33 electrons.

Step 2: Use the noble gas shorthand method

The noble gas before As is Argon (Ar), with an electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$.

Step 3: Fill the remaining electrons

We need to fill $33 - 18 = 15$ electrons beyond Ar. The order of filling orbitals is $4s$, then $3d$, then $4p$. The $4s$ orbital can hold 2 electrons, the $3d$ orbital can hold 10 electrons, and the remaining $15-(2 + 10)=3$ electrons go into the $4p$ orbital. So the electron configuration of As is $[Ar]4s^{2}3d^{10}4p^{3}$.

Answer:

s

• Au: $\boldsymbol{[Xe]4f^{14}5d^{10}6s^{1}}$
• As: $\boldsymbol{[Ar]4s^{2}3d^{10}4p^{3}}$