Question

stumped

Answer 1

@Vocaloid

Answer 2

1. there's only two choices (C and O) the less electronegative atom goes in the center and the more EN atoms go on the outer edges

so would C be the first blank or O?

Answer 3

c would be the first

Answer 4

for the second blank:

Answer 5

good, so carbon first blank and oxygen second blank

Answer 6

2. ideas about the bond type? it would either have to be covalent or ionic

Answer 7

covalent

Answer 8

*looks at answer choices* oh wait that's not one of the choices whoops

I guess they're looking for either single or double. any attempts to draw the lewis structure?

Answer 9

it would be double bonded

Answer 10

good, so double

Answer 11

c in the middle with 2 o's with double bonds

Answer 12

linear for the next geometry

Answer 13

3. ideas for the geometry? 2 bonds, 0 lone pairs

Answer 14

good

Answer 15

4. hybridization, any ideas? if you are stumped try calculating the steric #

Answer 16

sp3

Answer 17

hm, not quite, steric number is 2 (C is bonded to 2 other atoms, plus 0 lone pairs) so sp

Answer 18

how do you determin the steric NUm again?

Answer 19

i'd have to think a bit about 5 but the pi orbitals should result from the overlap of p orbitals

Answer 20

yeah well pi bonds are always p

Answer 21

yeah that should be it

Answer 22

What is the hybridization of the central iodine atom in I3−?

Answer 23

sp3d

Answer 24

good that's what i got too

Answer 25

it says : Express your answer in the orbital notations s, p, d, and f.

Answer 26

so that would still be right

Answer 27

yes

Answer 28

this can be tedious but you'd start with compound 1 (1 valence electron), put 1 electron in to the MO diagram. --> that one ends up unpaired so this is paramagnetic

then you'd continue this process w/ the other compounds

Answer 29

need to re-do some of them

for example, 6 is paramagnetic. you'd fill up the first 4 electrons but then the last 2 are unpaired b/c they occupy separate orbitals

Answer 30

as a hint: any odd number will be paramagnetic (if there are an odd number it's impossible for them all to be paired)

for even numbers you have to consider on a case-by-case basis

Answer 31

8 should be diamagnetic (they all get paired up w/in the first four orbitals) but everything else is good

Answer 32

6 should be paramagnetic per our earlier reasoning (unpaired electrons in 2pi bonding)

diamagnetic should contain: 2, 4, 8, 10, 14, 16; everything else is para

Answer 33

hm. :/

Answer 34

oh I see they wanted a different orbital diagram than the one they used before

Answer 35

yeah and thats what I based my previous answer off of

Answer 36

This

Answer 37

then yeah, your orig. should have been right

Answer 38

that's a good attempt but F2- fills up all of the anti-bonding orbitals and anti-bonding orbitals are not desired for stability. I'd put F2- at the bottom

F2+ keeps the antibonding orbitals empty so it's the most stable

F2 goes in between

so basically just reverse the order you have already

Answer 39

anything with a * symbol (sigma*, pi*) etc. are antibonding orbitals

Answer 40

not 100% sure on this but thinking the lithium might be diamagnetic

Answer 41

okay so lithium and carbon would be dia

Answer 42

I don't really feel super confident on these tbh

Answer 43

if I"m not mistaken B2(2+) has 8 total electrons, occupying the first four orbitals, therefore being dia

Answer 44

oh

since each C has 6 electrons -- 12 electrons --> +2 because it's an anion --> 14 total electrons --> fills up the first 7 orbitals to be diamagnetic

so if moving C2(2-) doesn't work idk where to go from there

Answer 45

so all three dia

Answer 46

no, B2(2+ is still para

Answer 47

ugh now its 3 out of 3 incorrect

Answer 48

oh in that case that makes it super easy, just switch all the para's to dia's and all the dia's to paras

Answer 49

ohhh I see where I went wrong, there's a -1 charge on the carbon not a -2 charge >_>

Answer 50

got it ughh thankssss