Summary of Electron Configurations for Ions
This video tutorial explains how to write electron configurations for ions, focusing on where electrons are removed from (for cations) or added to (for anions). The instructor uses an orbital diagram and explains the key principles: Aufbau principle, Pauli exclusion principle, and Hund's rule. For a foundational review of these concepts, see our Orbital Diagrams & Electron Configuration: Step-by-Step Guide.
Key Principles
- Aufbau Principle: Electrons fill orbitals from lowest to highest energy.
- Pauli Exclusion Principle: A maximum of two electrons per orbital, with opposite spins.
- Hund's Rule: For orbitals of equal energy, place one electron in each before pairing, with the same spin.
- Octet Rule: Atoms gain, lose, or share electrons to achieve a full outer shell (8 valence electrons).
Main Group Elements (s- and p-block)
For main group elements, electrons are removed from or added to the highest energy level (n).
- Cations (Positive Ions): Remove electrons from the highest n value.
- Example: Magnesium (Mg) → Mg2+
- Neutral: 1s2 2s2 2p6 3s2
- Ion: Lose two 3s electrons → 1s2 2s2 2p6 (like Ne)
- Example: Magnesium (Mg) → Mg2+
- Anions (Negative Ions): Add electrons to the highest n value.
- Example: Phosphorus (P) → P3−
- Neutral: 1s2 2s2 2p6 3s2 3p3
- Ion: Gain three electrons → 3p6 → 1s2 2s2 2p6 3s2 3p6 (like Ar)
- Example: Phosphorus (P) → P3−
Transition Elements (d-block) – Important Exception
For transition metals, the 4s orbital has a higher principal quantum number (n=4) than the 3d (n=3), so electrons are removed from 4s first when forming cations. This is a critical detail explored further in our Easy Method to Write Electron Configurations Using Orbital Diagrams.
- Example 1: Chromium (Cr) → Cr3+
- Neutral: 1s2 2s2 2p6 3s2 3p6 4s2 3d4
- Ion: Remove two 4s electrons + one 3d electron → 1s2 2s2 2p6 3s2 3p6 3d3
- Example 2: Iron (Fe) → Fe2+ and Fe3+
- Neutral: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
- Fe2+: Remove two 4s electrons → 1s2 2s2 2p6 3s2 3p6 3d6
- Fe3+: Remove two 4s + one 3d electron → 1s2 2s2 2p6 3s2 3p6 3d5
Anion Example for Main Group
- Example: Bromine (Br) → Br−
- Neutral: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
- Ion: Gain one electron → 4p6 → 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (like Kr)
How to Write Electron Configurations for Ions (Step-by-Step)
- Write the neutral atom's configuration using the Aufbau order (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...). For a quick reference on using the periodic table for this, check out How to use the periodic table to write electron configuration easily.
- Identify the charge (ion type): Cation (+) or Anion (-).
- For Cations (+):
- Remove electrons from the orbital with the highest principal quantum number (n) first.
- For transition metals, remove 4s electrons before 3d electrons.
- For Anions (-):
- Add electrons to the highest energy (n) orbital (usually the p orbital).
- Check for noble gas configuration.
Key Takeaway
- Main groups (s/p-block): Remove/add from highest n level.
- Transition metals (d-block): Always remove 4s electrons (highest n) before 3d electrons when forming cations.
For more practice with ion formation and compound formulas, see How to Write Ionic Compound Formulas: Step-by-Step Guide and Understanding Polyatomic Ions and Ionic Compound Formulas.
hello everyone and welcome back my name is mr cobalt and in this video i'm going to go over how to write electron
configurations for ions and show where those electrons are coming from so that way we don't make a mistake with regard
to like making sure right and writing the right electronic configuration to show where those electrons are coming
from all right so here i have my orbital diagram i'm going to be filling in this orbital diagram uh using the
different principles and rules so for example i'm going to be using aufbau principle right so to build up from low
energy to high energy i'm going to be using the polyexclusion principle so that way um to show that
two electrons maximum can go into each orbital each line is an orbital and those uh those electrons should be
uh opposite spin so spin up and spin down in each orbital um also i'll be using hund's rule right remember hund's
rule says that if you have any orbitals of equal energy when you're adding electrons to each of those orbitals you
add them one at a time and they have to have the same spin so maybe spin up spin up spin up before you start pairing them
up okay so let's get into this so um here we have uh some atoms i'm going to
write out the electron configurations for the um the neutral atoms right and then we're
going to see where those electrons come from and what that electron configuration for the ion would look
like okay so let's start with magnesium magnesium is number 12 on the periodic
table so we need to put 12 electrons in there because we're assuming neutrality right now
so let's get into this so 12 electrons so i'm going to put starting from low energy
spin up spin down to in in in the 1s and then
the 2s gets spin up spin down so that's 4 and i'm going to add one at a time spin up
spin up spin up and so there's three and then i can
start pairing them up spin down spin down spin down okay so that so far that's
10 and i need two more so i'm going to put spin up and spin down so that's 12. so
now i have my 12 electrons and so here um
we got to figure out where the electrons are going to be coming from so if you know the periodic table if you
know where this element lies you can figure out the charge from there remember the octet rule um
all atoms want to be like noble gases so they're going to lose or gain electrons based upon uh which is easier to do so
they want to get that full outer shell that valence shell they want that to be full and they're going to lose or gain
electrons to get that full shell and so here you can see that the last electrons the highest energy
level is three and so here to get that full shell with your s and
your p you need either six more electrons so magnesium can either add six electrons
to that full shell or it could lose these two electrons and then next shell would then be the full shell so 2s 2p
together would be your full 8 and so it's a lot easier to remove two electrons than it is to gain
gain 6. so magnesium will lose those two electrons so lose these two electrons here
from from the 3s which is the at this point the highest energy level is three and
you can tell it's the highest because you got the three there so that's where the electrons are going
to be pulled from so they're always going to be pulled from the highest energy according to
the shell number here the energy level number here all right so you lose those two
electrons there right so then
if we write out the uh the electron configuration for magnesium before we move those electrons would be
1 s 2 2 s 2 2 p 6
3 s 2. so we lost the 3 s electron so this would be the electron configuration for
the neutral atom but we lost those two electrons so we would remove the 3s2 and so that
would be the electron configuration for the ion magnesium with a two plus charge
okay let's try the next one so we have phosphorus phosphorus is number 18 i'm sorry number 15 on the periodic table so
we're going from let us add our electrons back oops
so we have spin up spin down so here we have 12 electrons we need three more electrons to get 15.
so we're going to add one two
three notice i'm i'm following hund's rule so those three go are going to go one at a time singly into
the three equal orbitals and that's it so we have 12 13 14 15. so now
we can figure out what phosphorus is going to do so we have it could either gain three
more electrons or it can get to get that full outer cell right that third full outer cell with the s and the p
or it could uh lose these three electrons lose those two
electrons so lose five electrons to then have a full outer shell it's going to be a lot harder to remove for five
electrons than it is to gain three so it will gain three electrons from some other atom or atoms
and so spin down spin down spin down so if we write the
electron configuration for the atom before it gained electrons it would be
1 s2 2s2 2p6
3s2 and it would be three p three
right so before we added those three electrons it was three p three but when it gains those three electrons it will
be three p six so you add a 6 there and so that would be the electron configuration for
phosphorus with a 3 minus charge okay
let's move to the next one so calcium calcium is number 20 on the periodic table so here we have
uh right now what is this we have 20 plus
6 i think that's 26 let's count so we have 2 4
and this is 6 here that's 10 sorry that's 18. so we have 18 here so 18
uh calcium is number 20 so we need two more electrons
so we add two more electrons here and so now we write the lewis dot structure i'm not
i'm sorry i keep mixing up lewis dot structure we're gonna uh write the electron configuration for uh calcium
before we take away any electrons or gain any electrons so the electron configuration is 1 s
2 2 s 2 2 p 6
s two three p six and four s two okay so
that is the electron configuration for calcium as a neutral atom again we come here
notice that the highest electron our electrons are in the 4s 4 is the highest number we have
those electrons are going to removed first so you so far the pattern is that the electrons are being removed from
the highest energy level so 4s here they were removed they were added to the highest to make that full shell
here they were subtracted from the 3s right to re uh so that 3s was the
highest one so here uh the 4s is the highest so those electrons are going to be the ones
to remove be removed again you could add six electrons but that
would be hard to do it'd be easier to lose those two so calcium tends to lose two electrons as you know it's in group
two so if you know the relationship of charge and the periodic table group two elements tend to have a two plus charge
and that's why so these electrons would be removed and so since these electrons would be
removed then we would erase the 4s2 here and that would be the electron
configuration for calcium two plus so that would be the two plus charge there
so so for the positive ions we we are removing the electrons
first from the outer uh the most um how should i say the uh highest energy shell
okay let's try chromium let's do chromium this is where it kind of gets interesting
so chromium is number 24. so we got to put those electrons back so that's 20 and
now the next highest energy level is going to be the d and so we have four electrons put in so we're going to put
them in following hundrel so spin up spin up spin up i skipped one
spin up so our four electrons are all same spin they're one in each orbital
so this would be the orbital diagram for chromium so let's write out the
the electron configuration so here would be 1 s
2 2 s 2 2 p 6 3s2 3p6 4s2
and 3d4 okay so this is the electron configuration for the neutral atom of
chromium okay so what would be the ion um where are these electrons going to come from
remember what i said the electrons are going to come from the uh the highest energy shell
and that is the number that is determined by the number in front of the letter right so you'll notice here
that the 4s these electrons here are in energy shell number four even though here according to the orbital diagram
these are higher energy level so here even though this is a 3d it looks higher
the electrons are being taken from the 4s this is the higher number so the electrons are taken
from this here okay so we remove those electrons not these so
we these are always removed first so let's assume let's uh chromium is a transition element it can have
various charges let's assume that it has a plus one uh let's say it has a plus three charge okay so if we give this a
three plus charge we're going to need to remove three electrons the first two electrons are
removed from here not here so we don't remove three electrons from
here first right that would be the wrong way to go about it the electrons would be first removed
from the 4s because that's a higher number and then you would remove from the 3d
so these would be removed so remove those that's two and then for the three plus we need one
more electron so we would remove one of the electrons here and so now the electron configuration
would be 1s 1s2 2s2 2p6 3s2 3p6 3d3 so then we would get rid of
the 4s2 and replace that with 3d3
so that is the electron configuration for the chromium 3 plus ion this is typically what happens when we're
dealing with the transition elements so you got to be careful when you're figuring out what the electron
configuration for ions that are are the transition elements all right so what about uh
iron iron again is a transition element it is number 26 so let's put in
26 so we have 20 1 2 3 4
5 6 and again the uh
for the neutral atom the electron configuration would be 1 s 2 2 s 2
2 p 6 3 s 2 3 p 6
3 oops 4 s 2
3 d and that would be uh six three d
six but once again just like before um
we have our two electrons here in the 4s but we have electrons in the 3d when we're trying to figure out the charge or
the electron configuration for our iron we remove electrons from the 4s before the 3d because 4 is higher than 3. so
this is the higher energy shell so remove electrons here first before we start removing electrons there
so again iron can be a plus it could be a plus charge be a plus two it could be a plus three
so if uh if iron is a plus three charge right once again
right if it's a plus two charge then we would just remove those two electrons and then the electron configuration
would be 1s2 2s2 2p6 3s2 3p6 and then 3d6 if it's a 3 then we remove these plus 1
over there so if we move those electrons and remove one here right
so now the electron configuration is going to be if we remove the 4s to here and we replace the 3d
3d6 with 3d5 and oops color
so that's going to be 3 d 5
then that would be the electron configuration for for iron
and then finally for bromine like so bromine is
number 35 so here we add
our two electrons 20 25 26 27 28 29 30
and then we start adding here 31 32 so the 3d then the 4p and so now
we have bromine bromine as one two three oh wait hold on a second
uh 35 so we have one two three four uh 10
18 20 that's going to be uh
20 30 and then 32
okay so here um bromine i looked uh 35
oh 35 i forgot electrons 30 i said 32 but this is 35 so we have 33 34
and 35 that's much better that's less confusing okay so now here
you can see that the energy level the highest energy level here is four and it's almost full it's got seven
valence electrons in there and so now bromine can either add one or it can
lose seven it's a lot easier to lose to gain one electron so it's going to gain that one electron
like that and so the electron configuration would be
1s2 2s2 2p6 3s2 3p6 4s2 3d10 and then 4p6 so let's write that so it's going to be
1 s2 2s2 2p6
3s2 p six uh
four s two three d ten and then four s uh no i'm sorry four p
six and so that would be the electron configuration of bromine with a minus
one charge and you can see that they have a full outer shell there so that's it for this
video i hope that is helpful for uh writing the electron configuration for ions you got to be really careful
about the transition elements remember that you're going to remove electrons first from the highest energy level so
especially for the transition elements that's going to be probably the s orbital so you
so you're going to remove electrons first from those and then if you need to still remove electrons then you start
removing from the d uh d orbital or d sub sub level uh i hope this was enjoyable i hope you
got a lot from this video please make sure you hit that like button
[Music] share this video with your friends make comments in the comment section let me
know what you think subscribe to my channel hit that notification bell
and uh let me know what you think ask me some questions down below
thanks for joining me have a great day
For main group element cations, remove electrons from the orbital with the highest principal quantum number (n). For example, to form Mg²⁺ from neutral Mg (1s²2s²2p⁶3s²), remove the two 3s electrons, resulting in 1s²2s²2p⁶, which is isoelectronic with neon.
When forming cations from transition metals, electrons are always removed from the 4s orbital before the 3d orbital, even though the 4s fills before 3d in neutral atoms. For instance, to form Fe²⁺ from neutral Fe (1s²2s²2p⁶3s²3p⁶4s²3d⁶), remove the two 4s electrons first, resulting in 1s²2s²2p⁶3s²3p⁶3d⁶.
To form a +3 transition metal cation, first remove both 4s electrons, then remove one 3d electron. For example, Cr³⁺ from neutral Cr ([Ar]4s²3d⁴): remove two 4s electrons and one 3d electron, giving [Ar]3d³.
Anions add electrons to the highest energy-level orbital, typically the p orbital. For example, bromine (Br) gains one electron to form Br⁻. The neutral configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵ becomes 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶, matching krypton's configuration.
The Aufbau principle (electrons fill lowest-energy orbitals first), Pauli exclusion principle (max two electrons per orbital with opposite spins), and Hund's rule (fill degenerate orbitals singly before pairing) apply to both neutral atoms and ions. For ions, the key is to remove or add electrons from the highest principal quantum number orbital.
Even though 4s fills before 3d in neutral atoms, the 4s orbital has a higher principal quantum number (n=4) than 3d (n=3). When forming cations, electrons are removed from the highest n orbital first. Since 4s electrons are farther from the nucleus, they are easier to remove when forming positive ions.
After writing the ion's electron configuration, compare it to the nearest noble gas. For main group cations (like Ca²⁺), the configuration should match the previous noble gas (argon). Anions (like S²⁻) should match the next noble gas (argon). Transition metal cations typically do not achieve noble gas configurations.
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