Chemistry Help...What is the groud state electron configuation of the following elements?

Question

A) Sn2+
B)Au+
C)Zn2+
D)N3-
E)Se2-
F)Br-
G)Ba2+
H)In3+
I) F-

Answer 1

A)1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 Sn+2
B) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d9 Au+2
C) 1s2 2s2 2p6 3s2 3p6 3d10 Zn+2
D) 1s2 2s2 N-3
E) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 Se-2
F) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 3d10 4p6 Br-
G) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Ba+2
H) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 In+3
I) 1s2 2s2 2p6 F-

Answer 2

You'll need to become very comfortable with Hund's rule for the order of filling shells and subshells of electrons. You should also be able to relate position of the element in the periodic table with its ground state electron configuration.

So Sn2+ will have 48 electrons, and will have the same electron configuration as Cd.

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10

Adding all the superscripts should give you 48 electrons.

All of the others are done in the same manner. Just be sure you add or remove electrons from the atomic number to give you the correct corresponding charge.

I think you can do the rest of them.

Answer 3

The noble gasoline configurations have 2, 10, 18, 36, fifty 4 & 86 electrons, happening the periodic table. One question for the different element is what number electrons could it earnings or lose to develop into and ion and get to a form of numbers. the respond contains the decrease form. subsequently nitrogen theoretically could lost 5 electrons to get to 2 (maximum unlikely!) or earnings 3 to get to eight (greater in all probability). yet another question is not any count if nitrogen could proportion a number of its electrons with different atom(s). If it shares 3 electrons, it might effectively have 8 returned. remember that non-metals can proportion or settle for; metals can in easy terms donate. Ba and Li are ordinary to respond to; N and S are somewhat greater complicated.

Answer 4

Zn+2 loses its (2) "s" electrons:
1s^2 2s^2 2p^6 3s^2 3p^6 3d^10

Br-1 gains one electron, as the Halogens prefer to do:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6
It will now have a filled valence shell. All the Halogen ions will end in p^6.
Hope this helps.