6th edition 19b  [ENDORSED]

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6th edition 19b

Postby hannabarlow1A » Wed Jul 10, 2019 8:30 pm

In the 6th edition, problem 19 part b asks for the electron configuration and number of unpaired electrons for Sn (4+). Can someone explain why the answer [Kr]4d^85s^2? Thank you

David Zhang 1B
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Re: 6th edition 19b  [ENDORSED]

Postby David Zhang 1B » Wed Jul 10, 2019 9:40 pm

The 4+ means that 4 electrons have been removed. A natural Sn atom would have [Kr]4d^10 5s^2 5p^2. Taking away 4 electrons from the valence would make it [Kr] 4d^10.

Yasmin Olvera 1D
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Re: 6th edition 19b

Postby Yasmin Olvera 1D » Thu Jul 11, 2019 11:44 am

The configuration for Sn would be [Kr]4d^(10)5s^(2)5p^(2). HOWEVER, since it loses four electrons, it becomes Sn+4. The four lost would be those on the outer shells being 5s^(2)5p^(2). Hance the anser is [Kr]4d^(10).

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