2A. 9

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Aashka Popat 1A
Posts: 50
Joined: Sat Sep 07, 2019 12:16 am

2A. 9

Postby Aashka Popat 1A » Sun Oct 27, 2019 10:24 pm

This is the question: Which M2+ ions (where M is a metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d7; (b) [Ar]3d6; (c) [Kr]4d4; (d) [Kr]4d3?

I don't really know how to approach this question, could anyone explain it to me?

Marty Hockey
Posts: 47
Joined: Wed Sep 18, 2019 12:19 am

Re: 2A. 9

Postby Marty Hockey » Sun Oct 27, 2019 10:30 pm

Essentially all you need to do is follow the electron configuration to the according element. So |Ar|3d7 would take you to the row following Argon (the 4th row) and since it is only 3d7 you just need to count 7 elements from the furthest d point which would be from Scandium to Cobalt, leaving you with the answer Co+2. The same process follows for the remaining questions.

Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Re: 2A. 9

Postby Jessica Booth 2F » Sun Oct 27, 2019 10:36 pm

This is the question: Which M2+ ions (where M is a metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d7; (b) [Ar]3d6; (c) [Kr]4d4; (d) [Kr]4d3?
a) Because there is a 2+ charge we need to add 2 electrons to the given number of valence electrons, which in this case is 7. Using the fact that is in the period below Ar and that it has 9 valence electrons we can say that the metal is Co, so the ion will be Co2+
b) again we add 2 to the number of given valence electrons, 2+6 =8. The metal will have 8 valence electrons and be in the period below Ar, so the metal is Fe, and the ion is Fe2+
c) add 2 to the 4 given valence of electrons for a total of 6. The element with 6 valence electrons in the period below Kr will be Mo and the ion will be Mo2+
d) add 2 to the given 3 electrons to get 5 valence electrons total. The element in the period below Kr with 5 valence electrons is Nb so the ion will be Nb2+


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