Periodicity

Question 1

Why does oxygen have a lower first ionisation energy than nitrogen?

Answer 1

Oxygen is 1s2 2s2 2p4 & nitrogen is 1s2 2s2 2p 3.
With the box arrows thing oxygen fills up /\ /\ /\ / /. (Updown updown updown up up) and nitrogen fills up /\ /\ / / / (updown updown up up up)

As you move from nitrogen to oxygen you have 4 electrons in the p sub shell instead of 3. This means that there will be one pair because there are only three p orbitals in a p sub shell. These two electrons repel each other, increasing the atomic radius and decreasing the electrostatic force of attraction between the nucleus and the electron. Therefore this decreases the amount of energy need to overcome this attraction.

Question 2

Why is the first ionisation energy of boron less than the ionisation energy of beryllium?

Answer 2

Beryllium is 1s2 2s2 & boron is 1s2 2s2 2p1
With the box arrows thing beryllium fills up /\ /\ (updown updown) and boron fills up as /\ /\ / (updown updown up)

As you move from beryllium to boron there is a 5th electron in the 2p subshell. This means the electron configuration of boron is less stable than beryllium’s therefore less energy is required to remove one electron from boron. Beryllium has a stable configuration with the 2s sub shell completely full so more energy is required to remove an electron.

Question 3

Why does sulphur have a lower first ionisation energy than phosphorus?

Answer 3

As you move from phosphorus to sulphur you have 16 electrons in the 3p subshell instead of 15. The 16th electron fills up 3p subshell and it means the will be one pair of electrons because there are only 3 p orbitals in a p subshell. These two electrons repel each other, increasing the atomic radius and decreasing the electrostatic force of attraction between the nucleus and the electron. Therefore this decreases the amount of energy need to overcome this attraction.

Question 4

Why is the first ionisation energy of gallium significantly lower than that of zinc?

Answer 4

Because the last electron (the 31st) of gallium is in the 4p orbital instead of, like zinc where the last electron (the 30th) is in the 4s orbital. When you move from the 4s orbital out to the 4p orbital you move slightly further out from the nucleus AND the 4p orbital is shielded by the 4s orbital so this means the distance from the nucleus (atomic radius) increases, so that means the electrostatic attraction between the nucleus and electrons is weaker which means that less energy is needed to overcome that attraction and remove the electron.

Question 5

Why does lithium have a lower first ionisation energy than neon?

Answer 5

Lithium, although it has the same number of shells and similar shielding it has a lower atomic number than neon which means lower nuclear charge than neon. This means that there is a weaker attraction between the nucleus and outer shell electrons and therefore it requires less energy to overcome this.

Question 6

Explain the general trend in ionisation energy across period 2?

Answer 6

Across the period there is the same number of shells so SIMILAR SHIELDING(remember to say this!!!). As nuclear charges increases it causes a stronger force of attraction on the outer most electrons so more energy is required to overcome the electrostatic attraction.

Question 7

Explain the general trend in ionisation energy across period 3?

Answer 7

Same as period 2 - Across the period there is the same number of shells so similar shielding. As nuclear charges increases it causes a stronger force of attraction on the outer most electrons so more energy is required to overcome the electrostatic attraction.

Question 8

What areas of the periodic table have trends and acceptions in ionisation energies?

Answer 8

General trend across period 2 (acceptions Be - B)

General trend across period 3 (acceptions Mg - Al)

Question 9

Mg to Al

Answer 9

Aluminium’s outer electron is in a 3 p orbital rather than 3s like in Mg, so the electrons are found slightly further from the nucleus and the 3p orbital has additional shielding provided by the 3s2 electrons, these factors (shielding and distance) are strong enough to override effect of increased nuclear charge so we get a drop in ionisation energy

Question 10

Neon to helium

Answer 10

Down group new shell of electrons so further out, increases shielding, larger atomic radius, weaker attraction, lower IE

Question 11

Going across the periods 2 and 3 from left to right

Answer 11

Increase in melting and boiling points because the atomic number increases whereas the shielding remains the same so attraction between the two nuclei gets stronger.

Question 12

Going down the group melting and boiling point????????

Answer 12

Decreases because extra shell of electrons and larger atomic radius due to more shielding so greater distance between the nuclei of the atoms and so weaker force of electrostatic attraction so less energy required to overcome these forces.

Question 13

If you are comparing melting points what should you think about?

Answer 13

Shielding (shielding and electrons)
Atomic radius
Nuclear charge

Question 14

[Ionisation graph] Going down a long way

Eg. He down to Li

Answer 14

New group - additional shell - more shielding and electrons - larger atomic radii- weaker attraction between nucleus and outer shell electrons- less ionisation energy required to over come this attraction

Question 15

[Ionisation graph] Going up a long way

Eg. Na up to Ar

Answer 15

Going across groups - increasing atomic number - increasing nuclear charge - same number of shells - stronger attraction between nucleus and outer shell electrons - higher ionisation energy required to break this attraction.

Question 16

[Ionisation graph] Which jumps downwards are due to repelling?

Answer 16

P down to S
N down to O
As down to Se

Question 17

[Ionisation graph] Which jumps downwards are down to a further out p orbital and shielding of electrons?

Answer 17

Mg to Al

Zn to Ga

Question 18

Why does Selenium have a lower ionisation energy than Arsenic?

Answer 18

Selenium is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 and has 34 electrons /\ /\ /\/\/\ /\ /\/\/\ /\/\/\/\/\ /\ /\ / /

Arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 and has 33 electrons /\ /\ /\/\/\ /\ /\/\/\ /\/\/\/\/\ /\ / / /

The 34th electron of selenium has to go into the 4p orbital because it can hold six electrons. This means there is one pair of electrons in the 4p orbital compared to arsenic which doesn’t have any pairs of electrons as only 3 electrons are in its 4p orbital. These two electrons repel each other, increasing the atomic radius and decreasing the electrostatic force of attraction between the nucleus and the electron. Therefore this decreases the amount of energy need to overcome this attraction.

Question 19

[ionisation graph] Which jumps downwards are due to unstable electron configurations?

Answer 19

Be down to B

Question 20

Ionisation energy is always

Answer 20

EXOTHERMIC

Question 21

Why do Li and Na have similar ionisation energies?

Answer 21

There are 11 protons in a sodium atom but only 3 in a lithium atom, so the nuclear charge is much greater in Sodium. However, sodium doesn’t have a much larger IE because this increased nuclear charge is cancelled out by sodium’s outer shell’s electron greater distance from the nucleus + additional shielding of shells and electrons which means a weaker electrostatic attraction between the e- and nucleus.

Question 22

Why is there a large jump between the sixth and seventh ionisation energy of oxygen?

Answer 22

Oxygen has 8 electrons,( /\ ) - 1s part of first circle (/\ /\ / /) - 2s and 2p part of second circle. The first six removed starting from the outer shell are more easily removed because they are being removed from the same shell. However the 7th electron moves into a different shell, which is the closest one to the nucleus. This means there is a smaller ionic radius and so there is a very strong electrostatic attraction between this electron and the nucleus. There are also no more electrons shielding the 7th electron. Therefore lots more energy is required to overcome this attraction.

Question 23

First IE equation

Answer 23

X(g) —> X+(g) + e-

Question 24

2nd IE equation

Answer 24

X+(g) —> X2+(g) + e-

Question 25

On the periodic table, the element directly below will usually have a

Answer 25

Lower ionisation energy

Eg. Gallium is directly below Aluminium and it has a slightly lower IE
Eg.sodium is directly below lithium and it has a slightly lower IE

Question 26

How to work out an element using its successive ionisation energies if you know the period?

Answer 26

Look at the ionisation energy where there is a big jump. 
Eg. 1st= 10kJ mol-1, 
2nd= 20kJ mol-1, 
3rd= 30kJ mol-1, 
4th= 300kJ mol-1, 
5th= 310kJ mol-1, 
6th= 320kJ mol-1

Here there is a big jump at the 4th ionisation energy therefore this element is in group 4. If we know it is in period 2 then it must be boron. Boron has a jump in ionisation energy at the 4th IE because starting from the outer shell electron, by the 4th electron it is being removed from the most inner orbital which is closest to the nucleus so there is a stronger attraction.

Question 27

Why does ionisation energy decrease down a group?

Answer 27

As you go down the group there are more shells which means more electron shielding. This means a larger atomic radius so the force of attraction between the nucleus and outer shell electron is much weaker. Therefore less energy is required to remove the outer shell electron.

Question 28

Explain the trend in reactivity down group 2 elements?

Answer 28

Reactivity increases as you go down the group because as you go down the ionisation energy decreases (because larger atoms, atomic radius and more electron shielding) which means the two outer shell electrons [2 because elements are are in group 2] that are less attracted the nucleus so are more easily lost.

Question 29

Explain why there is a significant increase in first IE from Hydrogen to Helium

Answer 29

In both hydrogen and helium the electron is being removed from the same S orbital (which is within the S subshell) so this means same shielding, screening and inner shell electrons. However helium has two protons whereas hydrogen only has 1 proton which means that there is a stronger force of attraction between the two protons in helium and its outer electron that needs to be removed compared to the 1 proton in hydrogen and its outer electron which must be removed. So more energy is required to remove the electron in helium so first IE is higher.

Question 30

Why are 2nd IEs much larger than 1st IEs?

Answer 30

Because it takes more energy to remove an electron from an already positive ion rather than a neutral atom

Question 31

State the Aufbau principle

Answer 31

Before occupying higher energy levels, electrons must fill atomic orbitals with the lowest available energy first

Question 32

State Hund’s rule

Answer 32

every orbital in a sub level is singly occupied before any of the orbitals are doubly occupied

Question 33

How many F orbitals in an F subshell? and how many electrons?

Answer 33

7 F orbitals with 2 electrons each so in total 14 electrons

Question 34

What is in quantum shell n = 1 ?

grand total of electrons?

Answer 34

1 s subshell (so 1 s orbital containing 2e-)

= 2 ELECTRONS

Question 35

What is in quantum shell n = 2 ?

grand total of electrons?

Answer 35

1 s subshell (so 1 s orbital containing 2e-)

1 p subshell (so 3 p orbitals containing 2e- each so in total 6e-)

= 2 + 6 = 8 ELECTRONS

Question 36

What is in quantum shell n = 3 ?

grand total of electrons?

Answer 36

1 s subshell (so 1 s orbital containing 2e-)

1 p subshell (so 3 p orbitals containing 2e- each so in total 6e-)

1 d subshell (so 5 p orbitals containing 2e- each so in total 10e-)

= 2 + 6 + 10 = 18 ELECTRONS

Question 37

What is in quantum shell n = 4 ?

grand total of electrons?

Answer 37

1 s subshell (so 1 s orbital containing 2e-)

1 p subshell (so 3 p orbitals containing 2e- each so in total 6e-)

1 d subshell (so 5 p orbitals containing 2e- each so in total 10e-)

1 f subshell (so 7 f orbitals containing 2e- each so in total 14e-)

= 2 + 6 + 10 + 14 = 32

Question 38

How to draw an s, p, d and f orbital?

Answer 38

s = circle 
p = eight figure (dumbbell shape)
d = dash
f = dash

Question 39

first ionisation energy definition

Answer 39

the energy required for one mole of gaseous

atoms to form one mole of gaseous ions with a single positive charge

Question 40

Why are successive energies always larger?

Answer 40

because when an electron is removed an ion forms, this ion increases the attraction on the remaining electrons and so larger IE’s are required to remove these electrons

Question 41

Can you remove the very last electron, closest to the nucleus from an atom?

Answer 41

X

find out later

Question 42

periodicity definition

Answer 42

repeating patterns in properties across different periods