Removing an electron from Au (problem2.47)

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Yinhan_Liu_1D
Posts: 51
Joined: Sat Sep 24, 2016 3:00 am

Removing an electron from Au (problem2.47)

Postby Yinhan_Liu_1D » Sat Oct 22, 2016 1:26 pm

When removing one electron from Au, why is it not in 5d orbitals but the only one in 6s orbitals?
Pt, the one that is directly before Au has 5d^9, right?

I would appreciate your reply.

Rochelle Ellison 2H
Posts: 29
Joined: Wed Sep 21, 2016 3:00 pm

Re: Removing an electron from Au (problem2.47)

Postby Rochelle Ellison 2H » Sat Oct 22, 2016 6:15 pm

When forming cations, the outermost electrons are removed. In Au the outermost electron is found in the 6s state since typically the 6s state is considered to be higher in energy and therefore farther away from the nucleus than the 5d state. This means that it is easier to remove electrons from the 6s state as the electrons are less tightly held by the nucleus. So when Au+ is formed the easiest electron to remove is the outermost one in the 6s state.

Yinhan_Liu_1D
Posts: 51
Joined: Sat Sep 24, 2016 3:00 am

Re: Removing an electron from Au (problem2.47)

Postby Yinhan_Liu_1D » Sat Oct 22, 2016 7:39 pm

Rochelle Ellison 1L wrote:When forming cations, the outermost electrons are removed. In Au the outermost electron is found in the 6s state since typically the 6s state is considered to be higher in energy and therefore farther away from the nucleus than the 5d state. This means that it is easier to remove electrons from the 6s state as the electrons are less tightly held by the nucleus. So when Au+ is formed the easiest electron to remove is the outermost one in the 6s state.


Thank you. Got it! ;)


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