Question

Help with electron configuration

Answer 1

?

Answer 2

electron configuration of what chimic element ?

Answer 3

there are so much

Answer 4

Gallium

Answer 5

do you think it hence ,in this way :

Gallium. Electronic configuration. 1s22s22p63s23p63d104s24p1.

Answer 6

I think it's also important to understand where that comes from

Answer 7

first we locate gallium on the periodic table

Answer 8

now, starting from hydrogen, the first row is the 1s block

2 electrons gives us

1s2
for Helium, then we keep going

Answer 9

next row is the 2s block (Li and Be) giving us

1s2 2s2

from Boron to Neon is the 2p block, thats 6 electrons going across the row, so

1s2 2s2 2p6

Answer 10

same logic for Na to Mg (the 3s block) and Al to Ar (the 3p block)

so 1s2 2s2 2p6 3s2 3p6

now it gets a bit tricky once you get into the d block. the 3d (Sc to Zn) block is lower in energy than the 4s block (K and Ca) so we have

1s2 2s2 2p6 3s2 3p6 3d10 4s2

finally, gallium is the first element in the 4p block. we stop at gallium giving us 4p1 as the final term in the configuration

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

this can also be expressed in terms of noble gas configuration as

[Ar] 3d10 4s2 4p1
basically, we write the first noble gas that comes before Gallium (which is argon), and the electron configuration for the elements after argon

Answer 11

there is another method to do this, which you may or may not have learned



basically start from the first diagonal and work your way along the diagonals until you have the right # of electrons (in this case, 31)

Answer 12

Yes I did learn about feeling certain orbitals first

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
next row is the 2s block (Li and Be) giving us

1s2 2s2

from Boron to Neon is the 2p block, thats 6 electrons going across the row, so

1s2 2s2 2p6
\(\color{#0cbb34}{\text{End of Quote}}\)
I get where you are getting the 1s from because that is where the element is located on the periodic table but why is it 1s2

Answer 14

hydrogen = 1s1
helium = 1s2

Answer 15

same logic with Li and Be

Li = 1s2 2s1
Be = 1s2 2s2

Answer 16

Wait I though Helium was 1p

Answer 17

an s orbital has 2 electrons so naturally the 1s orbital must include both hydrogen and helium

Answer 18

okay so an l,l,ml,mz values for Gallium would be


4,3,-1,+1/2

Answer 19

n,l,ml,mz **

Answer 20

actually it would be

1,0,0,-1/2

Answer 21

hm, you were a little closer the first time

n = principal quantum number = energy level = 4 since we're in a 4p orbital
l = 1 since it's in a p-orbital
ml = -1 since it's the first electron in the p orbital
spin = actually it can be either -1/2 or +1/2

Answer 22

its l , n-1?

Answer 23

isnt l, n-1?

Answer 24

or is it just the letter

Answer 25

no, l can take all possible values from 0 to n-1

Answer 26

s orbital = l = 0
p orbital = l = 1
d orbital = l = 2
f orbital = l = 3, etc.

Answer 27

oh okay so it can actually be 4-1=3,2,1,0?

Answer 28

yes, those are the possible values for n = 4

since we have a p orbital, l = 1

Answer 29

4p
n=4
l=1
ml=-1
mz=+1/2

Answer 30

okay so n is just the number before the letter and the rest like l and ml are determined by the the subshell not the number before the letter?

Answer 31

sort of yeah

Answer 32

n will give you the possible values of (l) but yes, the spdf will determine which l-value it is

for m-values, you need to list all the possible l-values in order, and determine which one it should be based on which element you're dealing with

for the gallium example, the possible m-values are

-1,0,1

since gallium is the first electron in the 4p orbital, it gets -1
the next element will get 0 and the element after that will get m = 1

Answer 33

okay so my values are correct?

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) @zarkam21
4p
n=4
l=1
ml=-1
mz=+1/2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 35

yeah that should be good

Answer 36

isn't this right for boron?

Answer 37

the 5 electrons

Answer 38

and the first two are filled first , that would equal

Answer 39

yeah that should be right

Answer 40

its not

Answer 41

do I fill all of 2p

Answer 42

hm. weird. no, you wouldn't fill 2p.

Answer 43

nvm I got it. The ones that were empty I wasnt supposed to fill

Answer 44

after 4s is it 4p?

Answer 45

or 3 d?

Answer 46

after 4s is 3d

Answer 47

hm. now that I think of it I did swap 4s and 3d on the gallium configuration >>

Answer 48

Use this tool to generate the electron configuration of Firenic (As).

Answer 49

almost

in the 4p orbital, remember that electrons don't really like being paired up unless they have to

so each subshell only gets 1 electron

Answer 50

oh right i have to fill them as single first and then doulbe if i have to

Answer 51

good start just missed one

#3: 3, 2,-2, 1/2 is fine

if l = 2, then m can be any integer from -2 to 2 so #3 is fine

Answer 52

others should be fine

Answer 53

ground state of a neutral beryllium atom, Be

Answer 54

hm. i'd have to think about this one for a bit but Be only has 4 electrons.

Answer 55

but wouldnt it fall in the d category so 2?

Answer 56

Be is in the 2s orbital

Answer 57

Thats what I meant. SO n value woud be 2 and l would be -1,0,1

Answer 58

anyway the quantum numbers for Be are:

n = 2
l = 0 (this is an s-orbital electron)
m, therefore, must also be 0
spin can be +1/2 or -1/2 as usual

Answer 59

remember, Be still has the same electrons that Helium and hydrogen would have
so the n = 1 electrons are also in Be

Answer 60

so putting this all together

the 2s electrons:

(2, 0, 0, 1/2) and (2, 0, 0, -1/2)

the 1s electrons:

(1, 0 , 0, 1/2) and (1, 0, 0, -1/2)

anything with (2, 1....) is a p-orbital electron. that would be Boron, carbon. etc. and would not be included in Be.

Answer 61

(2, 0, 0, 1/2) and (2, 0, 0, -1/2)

(1, 0 , 0, 1/2) and (1, 0, 0, -1/2)

Answer 62

yes, those would be the 4 electrons included in Be