TPRH Gen Chem Passage 5 Question 5

[MCATMountain]

There's a table with the following elements:

Lithium, Strontium, Sodium, Barium, Potassium, and Copper(I).

The question is

"The reduction of any of the ions in the table with a single eletron would result in a species that is:

A) Negatively charged
B) Paramagnetic
C) Dimagnetic
D) Radioactive "


I put D because the others didn't make sense and I couldn't eliminate D. Negatively charged didn't make sense because Lithium reduced is a neutral atom, therefore no negative charge. Paramagnetic wouldn't work because lithium plus an electron would make a 2s2 configuration and therefore the electrons would be paired (Paramagnetic = at least one unpaired electron). Dimagnetic also wouldn't work because Barium plus an electron would create a 4f1 configuration and would be an unpaired electron (Dimagnetic = no unpaired electrons).


The book says that it's B, paramagnetic because addition of an electron to any of these species would either create a neutral atom, or an odd number of electrons.


If anyone can shed light on this, it'd be greatly appreciated. Thanks in advance.

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[orderofthescar]

They're asking about the reduction of the ion forms, not the neutral elements.

So Lithium ion has an electron configuration of [He]. If you reduce the ion, you'll get the uncharged helium element, [He] 2s^1. Because this electron is unpaired, a reduced Li ion is paramagnetic.

Barium ion has a 2+ oxidation state and its configuration is [Xe]. If you reduce the ion with a single electron, it will become [Xe] 6s^1, which is unpaired. Therefore, a reduced Ba ion is also paramagnetic.

If you go through the rest of the ions you'll find that after adding one electron, that electron is unpaired and therefore the element is paramagnetic hope that helps! I could see where it gets confusing -- the description in the passage mentions the listed elements are in ion form, but they neglected to denote that in the table.

 

[MCATMountain]

They're asking about the reduction of the ion forms, not the neutral elements.

So Lithium ion has an electron configuration of [He]. If you reduce the ion, you'll get the uncharged helium element, [He] 2s^1. Because this electron is unpaired, a reduced Li ion is paramagnetic.

Barium ion has a 2+ oxidation state and its configuration is [Xe]. If you reduce the ion with a single electron, it will become [Xe] 6s^1, which is unpaired. Therefore, a reduced Ba ion is also paramagnetic.

If you go through the rest of the ions you'll find that after adding one electron, that electron is unpaired and therefore the element is paramagnetic hope that helps! I could see where it gets confusing -- the description in the passage mentions the listed elements are in ion form, but they neglected to denote that in the table.

thanks a lot. I'll have a look at the passage to see if it mentioned ion form.