1E.21 (omit c)

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Inderpal Singh 2L
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1E.21 (omit c)

Postby Inderpal Singh 2L » Sun Nov 15, 2020 11:34 am

How many unpaired electrons are predicted for the ground-state configuration of each of the following atoms: (a) Bi; (b) Si; (c) Ta; (d) Ni?

Please help :)

Sandy Lin 1L
Posts: 87
Joined: Wed Sep 30, 2020 9:48 pm

Re: 1E.21 (omit c)

Postby Sandy Lin 1L » Sun Nov 15, 2020 11:48 am

In Bi, there 3 unpaired electrons because on the periodic table it is in the third column of the p block which has 3 orbitals (according to Hund's rule each orbital must have one electron before you can add a second). In Si, there are 2 unpaired electrons because it is in the second column of the p block. Ni would have 2 unpaired electrons because it has 8 electrons and it is in the d block. This would result in 6 electrons being paired and the last two remaining unpaired.

Justin Lin 1B
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Re: 1E.21 (omit c)

Postby Justin Lin 1B » Sun Nov 15, 2020 11:57 am

The ground state configuration for Ta would be [Xe] 4f14 5d3, 6s2. If we look at the 5d orbital, there are 3 unpaired electrons present.

Raashi Chaudhari 3B
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Joined: Wed Sep 30, 2020 9:31 pm

Re: 1E.21 (omit c)

Postby Raashi Chaudhari 3B » Sun Nov 15, 2020 9:27 pm

It helps to draw this out visually for me. So, for the Bi it is in the p block and has 5 valence electrons. You know that the p block has 3 subshells, so draw out the subshells and begin to evenly distribute the electrons to each subshell (Hund's Rule). After doing that you will be able to see that there are two full subshells and one subshell is not full. That one subshell that is not full is the only one that is an unpaired electron which is why the answer is 1. You can follow these steps for the other elements. Hope this helped!

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