Periodic Table

Question 1

Ionisation energy

Answer 1

Measures how easily an atom LOSES ELECTRONS to form POSITIVE IONS.

Question 2

First ionisation energy

Answer 2

This is the energy required to remove ONE ELECTRON from each atom in ONE MOLE of gaseous ATOMS to form one mole of gaseous 1+ IONS.

Question 3

Factor affecting the IN ~ ATOMIC RADIUS

Answer 3

• The greater the distance between the nucleus and the outer electrons, the lower the nuclear attraction.
• The force of attraction ,F, falls off sharply with increasing distance between the nucleus and electrons ,d.
• F is DIRECTLY PROPORTIONAL to1/d^2

Question 4

Factor affecting IN ~ NUCLEAR CHARGE

Answer 4

• The greater the number of PROTONS in the nucleus of the atom , the greater the ATTRACTION between the nucleus and the outer electrons.

Question 5

Factor affecting IN ~ ELECTRON SHIELDING

Answer 5

• Inner shell electrons REPEL outer shell electrons and REDUCE the attraction between the nucleus and the outer electrons.
• The repulsion is called the SHIELDING EFFECT.

Question 6

SUCCESSIVE ionisation energy and why it gets HIGHER

Answer 6

Measures the amount of energy required to remove each electron in turn.

Each successive ionisation energy is higher because:

• The REPULSION between the remaining electrons DECREASES.
• The POSITIVE NUCLEAR CHARGE outweighs the negatively charged electrons more and more each time an electron is removed.
• The remaining electrons are DRAWN CLOSER to the nucleus because they are attracted more STRONGLY.

Question 7

SECOND ionisation energy

Answer 7

How much energy is needed for ONE MOLE of 1+ ions to form one mole of 2+ ions.

( The NUMBER of the ionisation energy is the same as the CHARGE on the ion produced.)

Question 8

Nuclear Shielding

Answer 8

Shells of electrons found between the positive nucleus and the outermost electrons REDUCE the ability of the nucleus to attract the electrons by SHIELDING or SCREENING the nucleus’ charge.

Question 9

TREND in first ionisation energy across a PERIOD

Answer 9

• It INCREASES due to the increase in NUCLEAR CHARGE.
• Each element has an EXTRA PROTON so attraction for all electrons increases across the period.
• Shielding of the nuclear charge DOES NOT increase because electrons are added to the SAME SHELL.

Question 10

Why is there a SHARP DECREASE in first ionisation energy going from He to Li, Ne to Na and Ar to K ?

Answer 10

The outermost electron:

• is located in the next shell and further away from the nucleus.
• is SHIELDED from the nuclear charge due to the complete inner shells of electrons.
• feels LESS nuclear attraction and hence requires LESS ENERGY to be removed.

Question 11

TRENDS in ATOMIC RADII across a period

Answer 11

• atomic radii DECREASES across the period.
• Each element has an EXTRA PROTON.
• the increase in nuclear charge across the period ATTRACTS all electrons more strongly and pulls them slightly CLOSER to the nucleus.
• The number of SHELLS remains the SAME across the period so there is NO EXTRA SHIELDING.

Question 12

Trend in FIRST IONISATION ENERGY down a group

Answer 12

• The number of SHELLS INCREASES.
• The DISTANCE between the outer electrons and nucleus INCREASES.
• The number of INNER SHELLS INCREASES so SHIELDING of the nuclear charge also increases.
• The increase in electron shielding OUTWEIGHS the increase in nuclear charge down the group.
• The ATTRACTION between the nucleus and outer electrons DECREASES so LESS ENERGY is required to remove the outer electrons.
• The first ionisation energies DECREASES down a group.

Question 13

Trend in ATOMIC RADII down a group

Answer 13

• The number of SHELLS INCREASES.
• The DISTANCE of the outer electrons from the nucleus INCREASES.
• The number of INNER SHELLS INCREASES and so SHIELDING of the nuclear charge for the outer electrons increases.
• The overall ATTRACTION between the nucleus and outer electrons DECREASES and so electrons are NOT pulled closer to the nucleus.
• The atomic radii INCREASES down a group.

Question 14

Microtrends in first ionisation energies across periods 2 & 3

Answer 14

• The GENERAL TREND in first ionisation energy shows a general INCREASE across both periods.
• However, the first ionisation energy DROPS in TWO PLACES when going across both periods and the places are IDENTICAL for both periods.
• The periodic cause for the drops is due to the presence of SUB-SHELLS, their ENERGIES and how ORBITALS fill with electrons.

Question 15

Trend in 1st I.E going from Li to Be

Answer 15

• The first ionisation energy INCREASES going from Li to Be.
• Be has one more PROTON than Li so its NUCLEAR CHARGE is greater.
• The extra electron in Be is going into the same sub-shell so there is NO EXTRA SHIELDING of the nuclear charge.
• The outermost electrons are closer to and attracted more strongly to the nucleus so more energy is required to remove and electron in Be than Li

Question 16

Trend in 1st I.E going from Be to B

Answer 16

• B has a LOWER first ionisation energy than Be.

-It is easier to remove an electron from the 2p sub-shell than the 2s sub-shell because the 2p sub-shell is HIGHER in energy and FURTHER AWAY from the nucleus.

Question 17

Trend in 1st I.E going from N to O

Answer 17

NITROGEN:
- has a HALF-FULL 2p sub-shell.
- Each p-orbital has one electron and the electrons spin in PARALLEL at 90 degrees to each other to MINIMISE repulsion.

OXYGEN:
- Hs a pair of electrons in one of the 2p orbitals.

• The repulsion between the paired electrons makes it EASIER to remove one of them in comparison with nitrogen, even though the NUCLEAR CHARGE of oxygen is LARGER.

Question 18

Trend in MELTING POINT across period 3:

   From sodium to aluminium

Answer 18

Melting point : INCREASES

• Na , Mg and Al exist as GIANT METALLIC LATTICE structures
• Each element has an EXTRA PROTON and so the nuclear charge increases and an EXTRA ELECTRON.
• The CHARGE on the metal ion INCREASES and the electrostatic attraction for the greater number of delocalised electrons increases.
• The metallic bonds become STRONGER going from Na to Al and MORE ENERGY is required to overcome these forces.

Question 19

Trend in MELTING POINT across period 3:

      Aluminium to silicon

Answer 19

Melting point: INCREASES

• Silicon exists as a GIANT COVALENT LATTICE structure , analogous to diamond.
• A LARGE AMOUNT of energy is required to break covalent bonds between Si atoms.

Question 20

Trend in MELTING POINT across period 3:

From P4 to S8 , S8 to Cl2, and Cl2 to Ar

Answer 20

• P4, S8 , Cl2 & Ar all have very LOW MP in comparison with Na , Mg , Al and Si.
• They exist as SIMPLE MOLECULAR structures with weak LONDON forces between the molecules as so LITTLE ENERGY is required to overcome these forces.

FROM P4 to S8:
- the melting point INCREASES
- sulfur has more electrons than phosphorus.
- It has larger TEMPORARY DIPOLES and STORNGER London forces which require more energy to overcome.

FROM S8 to Cl2 to Ar:
- the melting point DECREASES
- the number of electrons decreases
- the temporary dipoles become SMALLER
and the London forces become WEAKER.
- LESS energy is required to overcome the London forces

Question 21

Trend in ELECTRICAL CONDUCTIVITY across period 3

Answer 21

SODIUM , MAGNESIUM & ALUMINIUM:
- all metals
- have metallic bonding , in which positive metal ions are attracted to delocalised electrons.
- these delocalised electrons are free to move and can carry charge.
- the number of delocalised electrons from sodium to aluminium increases so the electrical conductivity INCREASES.

SILICON:
- a metalloid with a giant covalent structure
- the outer electrons in each atom are involved in strong covalent bonds and very few electrons have enough energy at room temp to enter high energy levels and bring about electrical conduction.
- Therefore , silicon is a POOR conductor as it has very few delocalised electrons.

PHOSPHORUS TO ARGON:
- their outer electrons are not free to move as they are involved in covalent bonding.
- Therefore they CANNOT conduct electricity.

Question 22

Trend in FIRST IONISATION ENERGY across
period 3

Answer 22

The general trend is that it will INCREASE due to the increase in nuclear charge.

However, there are other factors that outweigh this that will change the pattern.

MAGNESIUM TO ALUMINIUM:
- the highest energy electron in aluminium
is in a 3p sub-shell.
- this is higher in energy than the highest energy electron in magnesium which is in the 3s sub-shell.
- Therefore, less energy is needed to
remove it.

PHOSPHORUS TO SULFUR:
- phosphorus has a half full 3p sub-shell and each p orbital has on electron which spin in parallel to minimise repulsion.

• Sulfur has a PAIR of electrons in one of the 3p orbitals which REPEL.
• This makes it easier to remove an electron from sulfur , even though the nuclear charge of sulfur is LARGER.