Atomic structure

Question 1

Trend in atomic radius down the group

Answer 1

Atomic radius increases down the group.

As number of quantum shells increases, the outermost electrons are further away from the nucleus, hence the atomic radius increases. Both the nuclear charge and the shielding effect increase down the group, hence the effective nuclear charge differs little down the group.

Question 2

Atomic radius across Periods 2 and 3

Answer 2

Atomic radius decreases across the period. Nuclear charge increases due to increase in the number of protons in the nucleus. Shielding effect remains relatively constant. Effective nuclear charge increases, resulting in stronger electrostatic forces of attraction between the nucleus and the outermost electron. Outermost electrons are pulled closer to the nucleus and hence a decrease in the atomic radius

Question 3

Across first row transition elements

Answer 3

Atomic radius is relatively invariant. Nuclear charge increases due to increasing number of protons. Electrons are added to the inner 3d sub shell, which contributes to the shielding effect. Shielding effect increases thereby nullifying, to a considerable extent, the influence of each additional proton in the nucleus. Effective nuclear charge remains almost constant.

Question 4

Trend in Ionic radius across period 3

Answer 4

The ions, Na+, Mg2+, Al3+ and Si4+ are said to be isoelectronic as they have the same number of electrons, which is 10. Similarly, P3-, S2- and Cl- are isoelectronic as they have 18 electrons each. Across the two isoelectronic series, nuclear charge increases and shielding effect is the same.

This results in an increase in the efffective nuclear charge and a stronger attraction between the outermost electrons and the nucleus, hence a decrease in ionic size across each series.

There is a sharp increase in ionic radius from the ccationic series of Na+ to Si4+ to the anionic series of P3- to Cl-, because the anions have one more quantum shell of electrons than the cations.

Question 5

1st Ionisation energy

Answer 5

First ionisation energy is the energy needed to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of unipositively charged gaseous ions

Question 6

2nd Ionisation energy

Answer 6

Second ionisation energy is the energy required to remove 1 mole of electrons from 1 mole of unipositively charged gaseous ions to from 1 mole of gaseous ions with double positive charge.

Question 7

Trend in the successive ionisation energies of the same element IE1< IE2 < IE3 < IE4

Answer 7

When an electron is removed from a neutral atom, the number of protons that exert an attraction for the remaining electrons remains the same. However, the shielding effect among the remaining electrons in the outermost shell is reduced since there is now one less electron. Hence effective nuclear charge increases. More energy is needed to remove another electron from the more positively charged ion, and hence a higher ionisation energy .

Question 8

Trend in the first IE of elements down a group.

Answer 8

First ionisation energy of elements decreases down a group.

Down the group, the number of quantum shells of electrons increases, hence the outermost electrons are further from the nucleus. Therefore, the electrostatic forces of attraction between the nucleus and the outermost electron is weaker and less energy is required to remove this electron.

Both the nuclear charge and the shielding effect increase down the group, so the effective nuclear charge differs little down the group.

Question 9

Trend in 1st IE across Periods 2 and 3

Answer 9

First ionisation energy of elements increases across a period.

Across the period, nuclear charge increases and shielding effect remains relatively the same. Hence, the effective nuclear charge increases and the electrostatic for between the outermost electrons and the nucleus become stronger, so more energy is required to remove the outermost electron.

Question 10

Anomalies of trend of 1st IE across periods 2 and 3 and explanation

Answer 10

1. Small dip between group 2 and group 13 elements

Al has lower 1st ionisation energy than Mg.

The 3p subshell of Al is further away from the nucleus than the 3s subshell. There is
weaker attraction between the nucleus and the outermost electron. Hence, less energy is required to remove the 3p subshell.

Small dip between Group 15 and Group 16 elements.

S has a lower 1st IE than P

All the 3p electrons in P are unpaired. In S, two of the 3p electrons are paired. There is some inter electronic repulsion between the paired electrons in the 3p subshell in S. Thus, less energy than expected is required to remove one of these paired electrons from S.

Question 11

Trend in 1st IE across transition elements

Answer 11

First ionisation energy remains relatively invariant.

First ionisation energy involves removal of a 4s electron.

The nuclear charge increases due to increasing number of protons while additional electrons are added to the inner 3d subshell, which contributes to the shielding effect. The shielding effect increases thereby nullifying the influence of each additional proton in the nucleus. Effective nuclear charge remains almost constant. Thus, energy required to remove outermost electron of each succeeding element remains relatively invariant.