electron configuration and ionisation energy

Question 1

general overview (what each bit is split up into)

Answer 1

electrons move around the NUCLEUS in shells
these SHELLS are divided into SUB-SHELLS
these SUB-SHELLS have different numbers of ORBITALS

Question 2

SHELLS

Answer 2

electrons move around the nucleus in SHELLS
each SHELL is given a principal quantum number
(is intuitive as 1st electron shell the principal quantum number is 1 and 2nd is 2 and so on)

Question 3

SUB-SHELLS

Answer 3

these SHELLS are divided into SUB-SHELLS
different electron shells have different numbers of SUB-SHELLS
sub cells can either be

s sub-shells
p sub-shells
d sub-shells
f sub-shells

Question 4

what SUBSHELLS do each SHELL contain

Answer 4

the 1st shell contains 1 SUB-SHELL- 1s
the 2nd shell contains 2 SUB-SHELLS- 2s and 2p
the 3rd shell contains 3 SUB-SHELLS- 3s,3p and 3d
the 4th shell contains 4 SUB-SHELLS 4s,4p,4d and 4f

Question 5

how to remember the order the sub shells come in

Answer 5

S-stop
P-putting
D-dick
F-first

Question 6

ORBITALS

Answer 6

these SUB-SHELLS have different numbers of ORBITALS which can each hold up to

2 ELECTRONS

Question 7

how many electrons can each orbital hold

Answer 7

2

Question 8

how to remember how many ORBITALS are in a SUB-SHELL

Answer 8

s- 1 orbital
p - 3 orbitals
d - 5 orbitals
f - 7 orbitals

just the order of sub-shells in the order of ODD NUMBERS

Question 9

how to count how many electrons in each shell

Answer 9

use the SHELLS to find which SUB-SHELLS make it up
then look at these SUB-SHELLS and find the ORBITALS that make up each SUBSHELL and use the ODD NUMBER fact to work out the number of ORBITALS per SHELL 1,3,5,7
then because each orbital can hold 2 electrons you have 2 electrons per orbital so jump add it up

SHELL 1 = 1s so it has ONE orbital so (2x1)= 2 electrons
SHELL 2 = 2s+2p so it has (2x1) + (3x2)= 8 electrons
as a p sub-shell always has 3 orbitals
SHELL 3 = 3s+3p+3d = (2x1)+(3x2)+(5x2)=18
as a d subshell always has 5 orbitals
SHELL 4= 4s+4p+4d+4f= (2x1)+(3x2)+(5x2)+(7x2)=32
as an f sub-shell always has 7 orbitals

Question 10

what must electrons have opposite off

Answer 10

they must have opposite spin

Question 11

how to write the shorthand electron configuration

Answer 11

write the nearest noble gas before the element instead of all the subshells
then just add the last subshell on
eg berryllium (4electrons)= 1s2 2s2 and nearest noble gas is helium
so the shorthand electron configuration is [HE] 2s2
eg sulfur (16 electrons)= 1s2,2s2,2p6,3s2,3p6,4s2 and nearest noble gas before it is NE
so shorthand it is [NE]4s2

Question 12

what is the key info about the 3d subshell and 4s subshell

Answer 12

initially the 4s subshell has lower energy than the 3d subshell so electrons fill the 4s first
however
when the electrons have filled the 4s subshell, it now has a higher energy than the 3d subshell

so therefore when an element looses electrons it looses electrons from the highest energies first so it looses electrons from the 4s before the 3d

Question 13

what state do all things need to be in the equation for ionisation energy

Answer 13

all need to be gaseous

Question 14

what is the equation for the first 4 ionisation energies for Mg

Answer 14

1st ionisation energy Mg(g) –> Mg+(g) + e-
2nd ionisation energy Mg+(g) –> Mg2+(g) +e-
3rd ionisation energy Mg2+(g) –> Mg3+(g) + e-
4th ionisation energy Mg3+(g) –> mg4+(g) +e-

all element’s follow that pattern
start with the charge of the ionisation before as the starting charge on LHS and then you create the element with the charge the same as the number on the ionisation energy on the RHS and they need to both be in gaseous state
also an electron is removed so is added to equation

Question 15

what is the detailed definition of second ionisation energy

Answer 15

the amount of energy required to remove 1 mole of electrons from one mole of 1+ ions in their gaseous state to from one mole of 2+ ions in their gaseous state

Question 16

what are the three factors which affect ionisation energy

Answer 16

it all depends on the strength of attraction between the nucleus and the electrons

• the distance between the nucleus and the outermost electron (ATOMIC RADIUS) as the atomic radius increases the force of attraction decreases as they are further apart
• the charge on the nucleus (NO. OF PROTONS) the greater the no. of protons means the greater the charge means the greater force of attraction between the -electrons and the +nucleus
• shielding, outer electrons are repelled by electrons in inner shells so therefore reduce the attraction between the outer electrons and the nucleus due to the shielding by other electrons

Question 17

trends in ionisation energy

Answer 17

slight increase in ionisation energy after each electron is removed if they are in the same shell
as when an electron is removed the electrons in the same shell move slightly closer towards nucleus so have a slightly stronger attraction.

however we get a huge increase in ionisation energy when we a remove an electron from a different shell, this is because the shell will be closer to the nucleus (atomic radius) and experience less shielding, so will have a greater attraction compared to the electrons in the previous shell we were removing electrons from o

Question 18

how does the ionisation energy change going down a group and why

Answer 18

it decreases as:

the atomic radius increases as you go down group (as No. of shells for each element increases as you go down a group due to increase in electrons) which means the outer electrons are further away from the nucleus so there is less attraction between so it takes less energy to remove

also as the number of electrons shells increases as you go down the group so more shielding will take place

Question 19

what is the general trend for ionisation energy across a period and why

Answer 19

it generally increases (2exceptions)

this is because as we move across a period the NUCLEAR CHARGE increases because the NO.of protons increases, this increases the attraction between the nucleus and the electrons
this also causes the electrons to be pulled in more as their attraction increases so causes the ATOMIC RADIUS to decrease also

both the INCREASED ATOMIC CHARGE and the DECREASED ATOMIC RADIUS mean the outer electrons are more attracted to the nucleus so causing the FIRST IONISATION ENERGY to increases as we go across a period

Question 20

the ionisation energy going across a period has two anomalies why ( only talking about the first in this)

Answer 20

the first anomaly occurs due to the fact that an element (eg boron in period 2, or AL in period 3)
has its 1 outer electron in the p subshell

this means less energy is required to remove the outer electron from the p sub-shell than is required to remove the outer electrons from the s- subshell of the element before it in the period so therefore has a lower ionisation energy requirement

eg.
electron config of
Mg- 1s2 2s2 2p6 3s2
Al- 1s2 2s2 2p6 3s2 3p1

because Al is after Mg across the period you would expect it to have a higher ionisation energy.
but because Al has its single outer electron in its 3p1 subshell that requires less energy to remove than Mg outer electron in 3s2 (due to it being a higher energy) so therefore causing Al to have a lower ionisation energy than Mg which is an anomaly to the trend

this is the same explanation about the first anomaly for each period it would just be about different elements.

Question 21

what is the second anomaly in regards to first ionisation energy when going across a period.

Answer 21

in general across a period the first ionisation energy increases until the first anomaly where it decreases slightly , then after it increases until it hits the second anomaly where it decreases again then increases again after

to explain the second anomaly it is easier to write the electron configuration as electrons in boxes.

lets use the example of N and O in period 2
O is after N in the period so if it followed the trend it would have a higher first ionisation energy

if you drew the electrons in boxes for N and O you would notice that they are the same apart from in subshell 2p
N= 2p3 (so one electron per box as 2p has 3 orbitals)
O=2p4 ( so has an electron in each box like N but has one more so that pairs up with the first electron in the first orbital)

because these electrons are paired up with each other in the same orbital they experience repulsion from each other so therefore this repulsion sort of helps push this outer electron out so therefore decreases the ionisation energy

basically because O has a pair of electrons in the same orbital which repel each other it requires less energy to remove the outer electron than it does to remove the outer electron of N.
hence why lower ionisation energy even though further across the period

this is the same explanation just for different elements for the second anomaly but always in the p shell across all the period

Question 22

generally why first ionisation energy decreases down a group even though the nuclear charge increases

Answer 22

although the nuclear charge increases, its affect is outweighed by the increased radius ( each element down a group is in a different shell further away from the nucleus) and as more inner shells shielding also (inner electrons repel outer electrons off)

so overall, down a group nuclear attraction decreases so first ionisation energy decreases

Question 23

generally trend of first ionisation energy across a period why does it increases

Answer 23

because all are in the same shell across a period the first ionisation energy decreases
-same shielding as all in same shell
-increased nuclear charge
-atomic radius decreases
outweigh this fact and overall GOING ACROSS A PERIOD FIRST IONISATION ENERGY DECREASES

Question 24

when are the first and second falls in first ionisation energy across a period even though there should be a general increases

Answer 24

first fall is marked by the beginning of the filling of the p subshell
there is a single electron in p and because p has a higher energy than the previous elements outer electron in a full 2s subshell it is lost easier so therefore the element further down the period has a lower first ionisation energy.

second fall is marked by the the electrons beginning to pair in the p orbital
when there is 4 electrons in the p subshell it means that last electrons pairs with the other electron already in the first orbital
these electrons repel each other so therefore less energy is require to remove this outer electron
so the period further down the period has a lower ionisation energy that the element previous which is an anomaly to general pattern

Question 25

explain why successive ionisation energies always increase

Answer 25

because as each electron is removed the outer shell is drawn close to the nucleus, nuclear attraction is greater so more energy is required to remove the outer electron

Question 26

explain the general trend in first ionisation energy from Na-Ar (across a period)

Answer 26

nuclear charge increases from Na to Ar and outer electrons are in the same shell so experience similar shielding,
the atomic radius decreases and this results in an increases in nuclear attraction for the outer electrons and therefore an increases in first ionisation energy.

Question 27

explain the sharp drop in first ionisation energy between Ne and Na (Ne is at the end of period 3 and Na starts period 4)

Answer 27

sharp drops reflects addition of a new shell which results in an increase of atomic radius and electron shielding.
this causes a decreases in attraction between outer electrons and nucleus, decreasing the first ionisation energy

Question 28

explain the trend in first ionisation energy shown by He,Ne,Ar (down a group)

Answer 28

first ionisation energy decreases as we go from He to Ne to Ar due to increase of shells which leads to an increases in atomic radius and an increase in electron shielding.
this causes a decrease in attraction between outer electrons and nucleus and therefore causes a decreases to first ionisation energy

Question 29

explain why Al has a lower ionisation energy than mg

Answer 29

the 3p sub-shell in Al has a higher energy level than the 3s subshell in Mg so the 3p electron is easier to remove

Question 30

explain why S has a lower first ionisation energy than P

Answer 30

P has 3 electrons in the 3p subshell, one electron in each 3p orbital
S has 4 electrons in the 3p subshell, one extra electron so it pairs with another electron in an orbital.
the paired electrons in S repel each other, making it easier to remove the outer electron from S than P as P has no paired electrons in the 3p subshell