atomic structure

Question 1

which 2 cases are the 4s orbitals not fully-filled first (3d filled first?

Answer 1

24Cr and 29Cu

Question 2

define ionisation energy

Answer 2

energy required to remove 1 mole of electrons from 1 mole of gasous atoms or ions

Question 3

what are the 2 factors affecting ionisation energy?

Answer 3

1. effective nuclear charge

2. atomic radius

Question 4

what are the 2 factors affecting effective nuclear charge?

Answer 4

1. nuclear charge
   increase NC
   increase attraction on the elctrons
   more energy required to remove the valence electron
2. shielding effect
   decrease number of PQS
   decrease shielding effect
   valence e shielded from the attraction of nucleus by less number of inner PQS
   more energy required to remove valence electron

larger effective nuclear charge (ENC)
valence e held more strongly to nucleus
more energy required to remove valence e

Question 5

define ENC

Answer 5

effective nuclear charge is the net attractive force of the nucleus on the valence e after taking into account sheilding effect

Question 6

what is atomic radius dependent on?

Answer 6

1. number of filled PQS
   increase number of filled PQS
   atomic radius increase
2. ENC
   increase ENC
   atomic radius decrese

larger atomic radius
valence e located further away
less energy required to remove the valence e
lower ionisation energy

Question 7

explain Be to B (or Mg to Al) anomaly

Answer 7

Be has electronic configuration 1s2 2s2
B has electronic configuration 1s2 2s2 2p1

The 2p electron to be removed from B has a higher energy that the 2s electron to be removed from Be

hence the 2p electron of B requires less energy to be removed

Therefore the 1st IE of B is lower than the 1st IE of Be

Question 8

explain N to O (or P to S) anomaly

Answer 8

N has electronic configuration of 1s2 2s2 2p3
O has electronic configuratio of 1s2 2s2 2p4

COULOMBIC REPULSION between the paired 2p electrons in O makes it easier to remove one of the paired 2p electrons than an unpaired 2p electron from N

therefore, less energy required

Hence, first 1E in O is lower than the 1st IE in N

Question 9

explain 1st IE down the group

Answer 9

1st IE decreases
down the group, atomic radius increases
shielding effect increases due to increase in number of PQS, which largely cancels out the increase in nuclear charge
hence valence e became increasingly less attracted to the positive nucleus and less energy is required to remove the valence e