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Chemistry 23 Online
zarkam21:

Draw the lewis dot diagram for the following atoms F He S

Vocaloid:

any ideas on F? based on its location on the ptable how many valence electrons do you think it has? as a reminder these are neutral atoms, not ions

zarkam21:

7

Vocaloid:

awesome, so you'd draw 7 electrons (3 pairs of 2, plus 1 leftover)

Vocaloid:

thoughts on helium?

zarkam21:

2

Vocaloid:

good helium is a special case, it only has a 1s orbital so the electrons have to be paired up

Vocaloid:

thoughts on sulfur?

zarkam21:

6

Vocaloid:

good sulfur follows the normal rule for pairing, fill up 4 electrons separately then pair up the last 2

Vocaloid:

|dw:1540096403486:dw|

zarkam21:

CH4

Vocaloid:

any ideas? have you tried adding up all the valance electrons for 1 carbon 4 hydrogens?

zarkam21:

|dw:1540096455806:dw|

Vocaloid:

perfect

zarkam21:

But isn't there dots

Vocaloid:

there are no free electrons so there shouldn't be any

Vocaloid:

this is a covalent compound so all the bonds involved shared electrons, represented by the lines

zarkam21:

Sel2

Vocaloid:

thoughts? any attempts to add up the valence electrons?

Vocaloid:

well, anyway, selenium has 4 valence e's and each iodine has 7, so 6 + 7 + 7 = 20, divide by 2 to get 10 electron pairs so set up your compound where Se is in the middle, and I is bonded to the left and right, and fill up enough lone pairs so that the whole compound has 10 electron pairs total

zarkam21:

|dw:1540102685656:dw|

Vocaloid:

that's only five pairs, so you gotta fill up the iodine atoms completely to get 10 total pairs

Vocaloid:

10 total pairs = 20 total electrons

zarkam21:

|dw:1540102838122:dw|

Vocaloid:

almost just one more bond on the first SeI

Vocaloid:

also when you make a double bond you have to erase an electron pair

Vocaloid:

... wait

zarkam21:

but thats still 16 :(

Vocaloid:

oh, of course, selenium has an expanded octet so it can actually take more than 8, so put the last electrons on the selenium atom itself

Vocaloid:

|dw:1540103171322:dw|

zarkam21:

I3^-

Vocaloid:

well, each iodine gives 7, so 3*7, then you gotta add 1 because we have a -1 charge, so 22 electrons or 11 pairs

Vocaloid:

as a hint this one only has single bonds

zarkam21:

|dw:1540103405053:dw|

Vocaloid:

yeah that should be it, just two things 1. the pair of electrons on the right side shouldn't be in between the bond, they need to be off to the right side 2. since this is an ion you also need to draw a set of brackets around the whole thing, and write the charge at the top right

zarkam21:

|dw:1540103565219:dw|

Vocaloid:

|dw:1540103559925:dw|

Vocaloid:

yup good

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