Question

18 electron count

Answer 1

@abb0t

hey dude, can you confirm some answers for me?

A) Pt2+ d8, 16 e
B) W1+ d5, 16 e
C) Ir3+ d6, 18 e

Answer 2

What's your question? :P

Answer 3

Oh man, I rmr doing this!! :) hapticity

Answer 4

lol are those counts right? also is the binding on the drawing above correct? no charge, and electron contribution of 2?

Answer 5

yaa hapticity, i'm just reviewing making notes so i don't forget this pellet lol

Answer 6

I used to make a chart like this

Answer 7

let's see, for A, B, and C right?

Answer 8

yeah, and maybe F, which i got Fe d8, 18 e. Assuming the bridging carbonyls provide one electron to each iron, which i'm not sure about.

Answer 9

yeah, there's no bridging on A, B, or C. Only one I see is on F

Answer 10

2 electrons are being shared from the H-C=C-H right? or was it 4?

Answer 11

i think 4.. but i'm not sure lol.. if it's 4 that would make it 18 e

Answer 12

i mean i think 2

Answer 13

I think A is 11 +2 +3 = 16

Answer 14

yeah, I think it's 2 b/c i rmr acetylene was 4.

Answer 15

cuz it's the same thing as:

Answer 16

2 electons being shared on the metal

Answer 17

okay yeah, that makes sense, because it can be draw either way, good stuff

Answer 18

now, onto B :)

Answer 19

its easier to rmr my prof telling me "all neutral molecules give 2 elctrons" :3

Answer 20

okay cool. yeah, i think most molecules except like bridging ligands do

Answer 21

B. i got 6 + 10 + 2 = 18 e

Answer 22

2(5) = 10 carbonyl
6 = tungsten
2 = carbon double bond

and it's a neutral molecule.

Answer 23

so the carbon double bond would be neutral, not a -1 charge?

Answer 24

Well, it's kind of line an alkene, right? So I would think it's neutral. But i'm not 100% sure :)

Answer 25

i think you're right though, the double bond is 4 e donor, so it's 18 e

Answer 26

Hmm...that polyhydride for C is throwing me off, haha. but I am going to guess that it's 4 since it CAN also go:

So 9 + 4 + 5 - 1 = 17

Answer 27

Cp carries a -1 charge.

Answer 28

I've seen it written as \(\eta^4 - C_5H_6 \)

Answer 29

idk. tell me what you think :)

Answer 30

hmm for the hydride, i think the charge is 0, just like in A, so..



yea?