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3.1: Electron Configurations - Chemistry LibreTexts

3.1: Electron Configurations – Chemistry LibreTexts – Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3 s 2 3 p 1 , is analogous to its family member boron, [He]2 s 2 2 p 1 .atoms have filled s or p subshell so adding electron would push to next higher energy subshell and instead of releasing energy, atom would absorb energy trend of electron affinity tend to more neg. values as progress to right across periodic table – more energy is released and stabler neg. ions are formedWhich Of The Following Atoms Or Ions Does Not Have A Filled Outer Subshell? Ca S Zn2+ S2 Ca2+ Question: Which Of The Following Atoms Or Ions Does Not Have A Filled Outer Subshell?

Chemistry Chapter 8 You'll Remember | Quizlet – Evidence from the formation of ions. This last bit about the formation of the ions is clearly unsatisfactory. We say that the 4s orbitals have a lower energy than the 3d, and so the 4s orbitals are filled first. We know that the 4s electrons are lost first during ionisation.-all group 8A elements with n ≥ 2 have filled outer ns and np subshells while the transition metal chromium has an incompletely filled ____ subshell. p; d. alkali and alkali earth metals. the energy required for the complete removal of 1 mol of electrons from 1 mol of gaseous atoms or ions is called _____ energy. for a neutral elementQuestion: Sort The Following Atoms/ions According To Whether They Have Half-filled Or Filled Subshells. Zn2+ Half-filled Subshell In An Ion Drag Answer Here HE Fe Mo Filled Subshell In An Ion Drag Answer Here Fe3+ Ar Filled Subshell In A Neutral Atom Drag Answer Here Half-filled Subshell In A Neutral Atom Drag Answer Here

Chemistry Chapter 8 You'll Remember | Quizlet

Solved: Which Of The Following Atoms Or Ions Does Not Have – The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s 2 configuration, is analogous to its family member beryllium, [He]2s 2.Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s 2 3p 1, is analogous to its family member boron, [He]2s 2 2p 1.Lanthanide (or rare earth) elements have atoms or ions with partially filled. f subshells. The colors of the visible spectrum are blue, green, orange, red, violet, and yellow. An electron gains energy in the transition from a 6s subshell to a 5d subshell. true. Arrange the following ions in order of decreasing ionic radius: Al3+, Mg2(All electrons in the outer shell are valence electrons!) Atoms tend to form ions or chemical bonds in order to end up with filled outer "s" and "p" subshells. This is called the "octet" rule. (Not all chemical bonds follow this – it's a RULE OF THUMB, not a scientific law!)

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How to Write the Electron Configuration for an Element in Each Block – We'll go over the two different ways to
write the electron configuration but don't worry I'll go over everything step
by step.
Hello everyone I'm Melissa Maribel your personal tutor and here's
what you really need to know for electron configuration. Electron
configuration helps us see how electrons are arranged in atomic orbitals for a
specific element. There are four types of sub shells s, p, d and f the s subshell
has one orbital that can hold up to two electrons. The p subshell has three
orbitals that can hold up to six electrons. The d subshell has five
orbitals that can hold up to 10 electrons and the f subshell has seven
orbitals that can hold up to fourteen electrons. The periodic table has all
four types of subshells on it we refer to it as a block. These first two groups
are our s block and helium is also part of the s block, on the opposite side is our
p block, the inner transition metals are our d block and finally at the bottom we
have our f block. Make sure you know these, this is the
specific order that we follow for electron configuration and yes the order
does matter. Let's do an example of an element in each block. Example 1: s block
to find the electron configuration of any element we always start from
hydrogen and make our way going from left to right to the element we are
trying to find which in this case is beryllium. So writing our electron
configuration since we start with the first row hydrogen is 1s1, we move on
to helium which is 1s2 so we have 1s2. Next we are on the second row but
still in the s block so 2s1,2, 2 s2 and we stop since we reached the element
we wanted. This is the electron configuration of beryllium. Example 2: p
block we'll find the electron configuration of sulfur by starting
from hydrogen and making our way to sulfur. So 1s1, 2, 1 s2 is the
first part, come back around on to the second row 2s1, 2, 2 s2 is the second
part, keep going straight across to the p block, 2p, 1, 2, 3, 4, 5, 6, 2p6 is next, come
back around onto the 3rd row to 3s1, 2 so 3s2, keep going straight across to
the p block in row 3 and we will count up until we get to sulfur so 3p, 1, 2, 3, 4,
3p4, this is our full electron configuration for sulfur. There is an
abbreviated, shorthand or condensed electron configuration that you will
need to know how to find. Let's find it for sulfur, we always use the noble gas
that is in the row before the element we are looking for in this case it's
neon and we place the noble gas in brackets. Think of the noble gas as a
placeholder or the new place to start so we will continue after neon and write
the remaining parts of the electron configuration for sulfur. So 3s2 & 3p4, you know by placing our neon first we are actually accounting for this entire
portion of the full electron configuration since that portion is the
electron configuration of neon. Let's do another example. Example 3: d block we'll
find the electron configuration of iron by starting from hydrogen and making our
way to iron. So 1s2, 2s2, keep going straight across to 2p6 come
back around to 3s2, straight across to 3p6, come back around to the
4th row to 4s2. Now we are in the d block and the coefficient or number in
front is always 1 less than the row it is in. So instead of 4d this is actually
3d and we will count up until we get to iron so 3d, 1, 2, 3, 4, 5, 6, 3d6 this is
our full electron configuration for iron. Now let's find the condensed electron
configuration using the noble gas in the previous row which is argon we'll
continue from argon to iron. Do 4s2 & 3d6 and this is our condensed electron
configuration. Example 4: f block we'll find the electron configuration of
plutonium by starting from hydrogen and making our way to plutonium so 1s2, 2s2
keep going straight across to 2p6, come back around to the 3s2, straight across
the 3p6, come around to the 4th row to 4s2, remember the coefficient or
number in front is always 1 number less then the row it is in for the d block
so we have 3d10. We're still in the fourth row so 4p6 come around to the
fifth row to 5s2 we're in the d block so 4d10, still in the fifth row to 5p6,
come around to the sixth row to 6s2 and we have a break in our periodic table, your
clue is the change in atomic numbers we are now in the f block which is at the
bottom. Now the coefficient or number in front is always two numbers less than
the row it is in for the f block. So instead of 6f we will write 4f and
this is 4f14, now we go back up to the d block
we know this because the atomic numbers have to go in order here we had an
atomic number of 70 so next is 71, so 5d10 continue to 6p6, come around to the
seventh row to 7s2. There is a break in the table again to the f block so 5f,1,
2, 3 ,4, 5, 6 , 5f6. Here's the full electron configuration. Now let's find the
condensed electron configuration using the noble gas that is before plutonium,
so plutonium is at the bottom but would have been in the seventh row so we will
use the noble gas in the six row which is radon and continue from there so 7s2 and 5f6 and here's the condensed electron configuration. Now if you would
like help with your homework online tutoring or other helpful resources I
have all of that and more in the description box and if you want to make
sure you're ready for your next exam I created a practice quiz video just for
you with of course step-by-step answers you can find that right here and
remember stay determined you can do this! .

Electronegativity and bonding | Periodic table | Chemistry | Khan Academy – In the previous video, we
defined the term electron affinity.
And we looked at
electron affinities for lithium,
beryllium, and boron. So we talked about
their respective values and what they meant in terms of
structure of the neutral atom and of the ion that formed. In this video,
we're going to talk about trends for
electron affinity. And in terms of a group
trend, some textbooks will tell you there's
a modest group trend. Some textbooks will tell
you there is no group trend. At any rate, there is
more of a clear trend going across a period. And so that's the one that
we're going to look at instead. And so if we analyze
some of these numbers– we'll start with boron here. So boron gives off 27
kilojoules per mole of energy when you add an electron
to the neutral atom. When you move onto
carbon here, more energy is released– 122
kilojoules per mole. And if I look at oxygen,
141 is given off. So an increase. And finally, fluorine,
328 is given off. So even more energy
is given off. And remember, the more
energy it's given off, the more of an attraction the
nucleus has for the electron. So therefore, the more affinity
the atom has for the electron. So electron affinity increases. And that bothers
students sometimes because they think about
this negative sign. But remember, that negative sign
just means energy is given off. So just think
about the magnitude if the negative sign
is confusing to you. So fluorine would give off
328 kilojoules per mole. So that would be the most. So once again, as you go
across a period, in general, there's an increase
in electron affinity. And so there are
several exceptions to that general trend
that we talked about in the previous video. We talked about why
beryllium doesn't have an affinity
for an electron. And we also talked
about the reason that this number and
this number don't quite fit up with our trend. And so first, let's talk about
why that general trend is true. So as you go across
a period, you're increasing in terms
of nuclear charge. So boron has five
protons in the nucleus. Carbon has six protons
in the nucleus. So that's an increase in
the number of protons. And there's not a
whole lot of difference in terms of electron shielding. And so your effective
nuclear charge increases as you
go across a period. And so the effective
nuclear charge is the charge that the electron
that you're adding would feel. And so therefore, since your
effective nuclear charge is increasing, the
electron that you're adding to the neutral
atom would feel more of an attractive force. And so therefore, the atom
has increased affinity for that electron. And so that's the reason for the
general trend that we see here. So the reason for these
numbers for boron, for carbon, for oxygen,
and for fluorine– increased number of
protons in the nucleus. Let's think about
one of the exceptions to that general trend. We can see nitrogen has an
electron affinity of 0, which means it's going to
take energy to add an electron to the
neutral nitrogen atom. And let's think
about that in terms of electron configurations. So for nitrogen, nitrogen's
electron configuration would be 1s2, 2s2, and 2p3. So if you want to put that
into orbital notation, we'd have the 1s orbital
here, 2s orbital, and then we have 3p orbitals in
the second energy level. So let's fill in
those electrons. So we have– let's go ahead
and change colors here. We have two electrons
in the 1s orbital, so those would be
our two electrons. Two electrons in the 2s orbital. And then when we get
to the p orbitals, we have three p electrons. So we're going to put one
electron in this p orbital, one in this one, and then
finally, one in this one. So that's the neutral atom. If we wanted to add an
electron to nitrogen– so we can see that we have a
half-filled p subshell here. So if we're going
to add an electron, we're going to add an electron
to one of these p orbitals. And let's say we add the
electron to this one over here. So the electron that
we added in magenta has to occupy the same space
as the electron in green. So when we spin pair
those electrons there, that increases the amount of
electron-electron repulsion, because those two electrons
are occupying the same orbital there. And so that, of course–
that extra repulsion means that this is not
energetically favorable. So if we were to add an
electron to nitrogen– we'd have a negative charge now. This would no longer be
2p3, this would be 2p4. And again, that extra
electron-electron repulsion is the reason that
you usually see given in textbooks for the
fact that nitrogen does not have an affinity
for an electron. So let's talk about neon next. So neon's over here. And neon also does not have
an affinity for an electron. So an electron
affinity of zero here. So let's think about
why that is the case. So the number of
protons in neon is 10. Its atomic number is 10. So there are 10
protons in the nucleus. So 10 positive charges in
the nucleus, like that. And there are 10 electrons. So if we think about
where those electrons are, there are two electrons
in the inner shell– so here's my very
crude representation of the inner shell there,
with two electrons. And then eight
electrons in this shell here– so in this
second energy level, eight electrons in
this shell like that. And so if you were to think
about adding an electron to neon– let's go ahead
and write neon's electron configuration here. So neon would be 1s2, 2s2, 2p6. And so let's just
talk about where I'm getting those numbers
for my diagram here. So the two electrons
in the inner shell would be these two electrons. And the eight in
this outer shell here would be the
two and the six. So 2 and 6 give us 8. And so this is our electron
configuration for neon. And just a little picture
about what it would look like. If you wanted to add
an electron to neon, your second energy
level is full. So you have to move to
the third energy level. So you have to open
up a new shell. So if I wanted to add
an electron to neon, it would have to go
into a new energy level. So the electron would have to
go into the 3s orbital, so 3s1. This would make this
negative charge now. And we know that it's
a new energy level. So we can talk
about the electron we're adding being, on average,
further away from the nucleus. So that would be the picture. And we know that
this positive 10 is going to try to attract
this electron that we added. However, you have all of these
inner shell electrons here shielding this other electron. And so because of
these electrons in here repelling it and
screening the outer electron from that positive
10 charge, the atom has no affinity
for that electron in magenta that we are adding. And so that's the
reason for the value that you see here for the
electron affinity of neon. So it doesn't have an
affinity for an electron. And so this anion
would be very unstable. And so the idea is,
of course, that neon lacks the sufficient
nuclear charge– doesn't have enough positive
charge in the nucleus to attract the electron
and magenta that we'll be adding to form
the anion here. And so that's the idea. And so hopefully this just gives
you a little bit more insight into the general trend
for electron affinity, and also some of the exceptions. .

Chem 200 Ionic Compounds – .