Question

How do I determine this?

Answer 1

"stable nonmetal ions" as a hint, a stable metal ion will have a full outer valence shell, the same as a noble gas. any ideas what number this would be?

Answer 2

8

Answer 3

good

Answer 4

for the second one, the metal will lose all of its valence electrons, going back to 0

Answer 5

Add charges, single electron dots, and/or pairs of dots as appropriate to show the Lewis symbols for the most stable ion of each element. Treat n=4 as the valence shell in all cases.

Answer 6

hm, good on the metals, but for nonmetals you should be looking to gain electrons and thus have a negative charge

for example, Se has 6 valence electrons and wants to gain 2 to have a full octet, giving Se2- not Se6+, similar logic with Br

Answer 7

-3?

Answer 8

Br has 7 electrons, it will gain 1 more to become Br-

Answer 9

so just Br-

Answer 10

I thought it was based off of n=4 so 7-4 = -3?

Answer 11

because of the gaining of electrons

Answer 12

n = 4 is the period number not the group #, idk why they even give you that

Answer 13

ions are based on valence electrons which are based on group #'s

Answer 14

OKay so could you just explain that one more time Br has 7 valance electrons. Where do I go from here

Answer 15

Br is a nonmetal ---> it gains electrons to get a full set of 8 --> it already has 7 valence electrons --> so it gains 1 more --> so it goes from a neutral charge to a -1 charge

Answer 16

thus, Br-

Answer 17

OH okay a full set is always 8 and the elements work towards getting a full set which determiend how many electorns they lose and gain ?

Answer 18

* full set is 8 but nonmetals will gain to get a full set of 8; metals will lose electrons to get 0*

Answer 19

Got it

Answer 20

Thanks so much

Answer 21

I did this and it says its incorrect

Answer 22

huh. weird.

Answer 23

Add charges, single electron dots, and/or pairs of dots as appropriate to show the Lewis symbols for the most stable ion of each element. Treat n=4 as the valence shell in all cases.

Answer 24

These are the directions for the problem

Answer 25

oh, of course, it's also asking for the electron dots

Answer 26

OKay so just the original electron dots?

Answer 27

like for k 1 for ca 2

Answer 28

nonmetals should have a full set of 8, metals should have 0

Answer 29

so Se and Br should have 8 dots, the rest shouldn't have any I believe

Answer 30

Incorrect :S

Answer 31

Maybe it should just be the original

Answer 32

?

Answer 33

I'll try to ask someone else cause i'm stumped

Answer 34

Okay

Answer 35

ok apparently ars-enic and gallium don't always follow the octet rule

Answer 36

OH really

Answer 37

So we are back to square one ? NOne of them have dots

Answer 38

BUt thats what I did before

Answer 39

no, se and br should definitely have 8 dots each

I just looked it up and ars-enic apparently prefers +3 not +5

Answer 40

OKay is that only with Cookienic or

Answer 41

is ga and ca okay

Answer 42

yeah those should be ok

Answer 43

Ugh still incorrect wth

Answer 44

the only other thing I can think of is ars-enic is sometimes -3 but I wouldn't input that yet.

Answer 45

If thats the only thing I mean I guess I could try it

Answer 46

idk i don't want you to lose points yet

Answer 47

what do you have inputed right now

Answer 48

oh

well if Ars-enic has 5 valence electrons and loses 2 to become As3+ then I believe it should still have 2 unpaired electrons

Answer 49

so 2 single dots

Answer 50

yeah, 2 unpaired electrons on opposite ends of the atom. that's all i can think of.

Answer 51

still incorrect

Answer 52

its okay ill just move on thanks though

Answer 53

This was the answer if you were curious

Answer 54

oh, I see, they wanted the 3- on ars-enic.