12.1 Electrons in atoms

Question 1

an electron’s energy can be calculated using which 2 formulas?

Answer 1

1. E = hv
2. E = hc/λ

Question 2

what does h stand for?

Answer 2

planck’s constant = 6.63 x 10^34 J

Question 3

what are the periodic trends for 1st ionisation energy?

Answer 3

decreases down the group
increase across the period

Question 4

what are the 2 exceptions to the first ionisation energy periodic trend?

Answer 4

ionisation energies of the first p-orbital and the fourth p-orbital electrons are lower than the ionisation energy before them

Question 5

what is the trend of successive ionisation energies of an atom?

Answer 5

it increases as more electrons are removed due to less electrons being attracted more by the nucleus
significant increase occurs when removing electrons from a new shell or orbital

Question 6

define nuclear charge

Answer 6

cumulative total positive charge in the nucleus, directly linked to the number of protons present

Question 7

describe the trends of nuclear charges, down the group and across periods

Answer 7

• increases across a period
• increase down a group

Question 8

define electrostatic attraction

Answer 8

attractive force of the negative electrons to the positive nucleus, since opposite charges attract

Question 9

define shielding effect

Answer 9

reduction of the electrostatic attraction between valence electrons and the nucleus due to the blocking by inner electrons

Question 10

describe the trends of the shielding effect, down the groups and across periods

Answer 10

• increases down the groups
• remains the same across periods

Question 11

define first ionisation energy

Answer 11

the energy required to remove on electron from a mole of gaseous atoms, measured in kJ/mol
X(g) —> X^+(g) + e^-

Question 12

what are the 4 factors that affects the size of the first ionisation energy?

Answer 12

1. size of the nuclear charge
2. distance of outer electrons from the nucleus
3. shielding effect of inner electrons
4. spin pair repulsion

Question 13

explain the irregularities seen for Al and Mg where there is a huge dip in the ionisation energy graph in terms of electronic structure

Answer 13

• Al’s outer electron is in the 3p orbital whereas Mg’s is in a 3s orbital
• the 3p orbital is higher in energy than the 3s orbital
• therefore, less energy is required to remove the electron from the 3p orbital