Question

Help with this. When you get a chance

Answer 1

upper right corner: should be s and p block only, notice how Argon stops before the dblock
if you are having trouble you can think about the two elements that actually have that configuration,
Neon and Argon
[He] 2s2 2p6
[Ne] 3s2 3p6

see if you can figure out which choice accurately describes both configurations

Answer 2

for the next one below it, you almost have it, but the d needs to have 10 electrons since both those elements have a full d block (notice how you have to go all the way across the d block to get those two elements)

Answer 3

for the one below that you have a fully filled f-block so the one with f14 is your only possibility

Answer 4

finally for the lower left corner you only have 7 electrons in the f block so you only have two chocies

ns2(n-1)f7 or ns2(n-2)f7
looking at the actual electron configurations for Europium and Americium

[Xe] 4f7 6s2
[Rn] 5f7 7s2

should make it clear whether it's n-1 or n -2 for the f block

Answer 5

upper right corner:

[He] 2s2 2p6
[Ne] 3s2 3p6

since n is the same for the p block and s block it should just be

ns2 np6

Answer 6

for the one below it you originally had

ns2(n-1)d5np4

you just needed to fill up the d block to get

ns2(n-1)d10np4

the f-block is not involved

Answer 7

for the third one on the right (where thallium would be)
you have a full f-block
so ns2(n-2)f14(n-1)d10np1 is your only possibility

Answer 8

also I didn't really catch this before but the second one (where scandium and yttrium would be) it needs to be ns2(n-1)d1

because the outer configuration is 4s2 3d1
and 5s2 4d1

Answer 9

almost

notice how the bottom configurations for americium and europium are

[Xe] 4f7 6s2
[Rn] 5f7 7s2

so it has to be n - 2 since 4 is 2 less than 6; 5 is 2 less than 7

Answer 10

otherwise it's good

Answer 11

okay so n-2 instead of n-1

Answer 12

yes