## ch 2, 47, part d

Remi Lathrop 1G
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### ch 2, 47, part d

Question 47 from chapter 2 asks us to predict which orbital an electron will be removed from to form a 1+ ion given certain elements. I understand parts a-c, but d asks about Au and the solutions manual states that the electron will be removed from the 6s orbital which confuses me since Au is not filling the d-orbital.

Why don't you remove an electron from the 5d orbital as opposed to the 6s orbital?

Kamryn Chang 1I
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am

### Re: ch 2, 47, part d

In the lecture, Dr. Lavelle talked about how there are two main exceptions to electron configurations. One of the examples that he gave was Cu. In this case, Au is similar to the Cu exception. The full $d^{10}$ subshell turns out to have a lower energy. Therefore, Au has the electron configuration of $[Xe]4f^{14}5d^{10}6s^{1}$, which is why an electron is removed from the 6s-orbital.

Ramya Lakkaraju 1B
Posts: 67
Joined: Fri Sep 29, 2017 7:03 am

### Re: ch 2, 47, part d

All the transition metals that are in the same group as chromium and copper will want to make both their s and p subshells half full so instead of s2 p4, it will be s1 p5