Atomic structure

Question 1

Anomalous electronic configuration of Cr

Answer 1

The electronic configuration of Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1
By having one electron each in the 3d and 4s orbitals, inter-electronic repulsion is minimised.

Question 2

Anomalous electronic configuration of Cu

Answer 2

The electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1
The fully filled 3d subshell is unusually stable due to the symmetrical charge distribution around the metal centre.

Question 3

Trend in atomic radii across a period

Answer 3

Across a period,
 the number of electronic shells remains the same.
 the number of protons increases and hence nuclear charge increases.
 number of electrons also increases but these electrons are added to the same outermost shell, and hence shielding effect remains approximately constant.
 effective nuclear charge increases.
 electrostatic attraction between the nucleus and the valence electrons increases, resulting in a decrease in the size of the electron cloud.
Hence atomic radii decrease across a period.

Question 4

Atomic radii down a group

Answer 4

Down a group,
 the number of electronic shells increases
 distance between the nucleus and the valence electrons
increases
 shielding experienced by valence electrons increases
Despite the increasing nuclear charge,
 electrostatic attraction between the nucleus and the
valence electrons decreases, resulting in an increase in
the size of the electron cloud.
Hence atomic radii increase down a group

Question 5

Cationic Radius

Answer 5

The radius of a cation is always smaller than that of the parent atom.
Both the cation and its parent atom have the same number of protons and hence have the same nuclear charge.
 However, the cation has one less electronic shell than its parent atom
Electrostatic attraction between the nucleus and the valence electrons increases, resulting in a decrease in size of the electron cloud.

Question 6

Anionic Radius

Answer 6

The radius of an anion is always greater than that of the parent atom
 Both the anion and its parent atom have the same number of protons and hence have the same nuclear charge.
 However, the anion has more electrons than its parent atom.
 With more electrons, electron–electron repulsion increases
 Electrostatic attraction between the nucleus and the valence electrons decreases, resulting in an increase in the size of the electron cloud.

Question 7

1st Ionisation energy definition

Answer 7

The first ionisation energy of an element M is the energy required to remove 1 mole of electrons from 1 mole of gaseous M atoms to form 1 mole of gaseous M+ ions

Question 8

IE across a period

Answer 8

Across a period,
 the number of electronic shells remain the same.
 number of protons increases and hence nuclear charge increases.
 number of electrons also increases but these electrons are added to the same outermost shell, and hence shielding effect remains approximately constant.
 effective nuclear charge increases.
 electrostatic attraction between the nucleus and the valence electrons increases, resulting in an increase in the energy required to remove the valence electron from an atom.
Hence the first ionisation energies of the elements generally increase across a period

Question 9

ionic radii trend from Na+, Mg2+ and Al3+

Answer 9

Na+, Mg2+ and Al3+ are isoelectronic species and hence their valence electrons
experience the same shielding effect.
 However, nuclear charge increases from Na+ to Al3+.
 Effective nuclear charge increases from Na+ to Al3+
 Electrostatic attraction between the nucleus and the valence electrons increases from Na+ to Al3+, resulting in the decrease in the size of the electron cloud.
Explanation for the trend observed from P3– to Cl– is the same as above

Question 10

ionic radii from Al3+ to P3-

Answer 10

From Al3+ to P3–,
 nuclear charge and shielding effect increase.
As the number of electronic shells increases,
 distance between the nucleus and the valence electrons increases
 electrostatic attraction between the nucleus and the valence electrons decreases, resulting in an increase in the size of the electron cloud.
Hence the ionic radius of P3- is bigger than that of Al3+.
Ionic radii of isoelectronic ions decrease across a period.

Question 11

irregularity: group 2 to group 13 (B vs Be and Al vs Mg)

Answer 11

The 3p electron to be removed from Al is at a higher energy level than the 3s electron to be removed from Mg. (Be: 2s, B: 2p)
 Hence less energy is required to remove the 3p electron in Al than the 3s electron in Mg.
 First ionisation energy of Al is lower than that of Mg.

Question 12

IE down a group

Answer 12

Down a group,
 the number of electronic shells increases
 distance between the nucleus and the valence electrons increases
 shielding experienced by valence electrons increases
Despite the increasing nuclear charge,
 electrostatic attraction between the nucleus and the valence electrons decreases, resulting in a decrease in the energy required to remove a valence electron.
Hence first ionisation energies decrease down a group.

Question 13

Question shows you the IE trend and asks you which group this element is from

Answer 13

A large jump in the __th and __th ionisation energies is observed.
Significantly more energy is required to remove the 6th electron as it is located in an inner shell that is nearer to the nucleus, and hence experiences a stronger electrostatic attraction to the nucleus.
 There are __ electrons in the valence shell.
 E is likely to be in Group ___

Question 14

Electronegativity definition

Answer 14

The electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electrons

Question 15

Electronegativity across a period

Answer 15

Across a period,
 the number of shells remain the same.
 number of protons increases and hence nuclear charge increases.
 number of electrons also increases but these electrons are added to the same outermost shell, and hence shielding effect remains approximately constant.
 effective nuclear charge increases.
 electrostatic attraction between the nucleus and the bonding electrons increases
Hence electronegativity increases across a period

Question 16

Electronegativity down a group

Answer 16

Down a group,
 the number of electronic shells increases
 distance between the nucleus and the bonding electrons increases
 shielding experienced by bonding electrons increases
Despite the increasing nuclear charge,
 electrostatic attraction between the nucleus and the
 bonding electrons decreases
Hence electronegativity decreases down a group.

Question 17

electronic configuration

Answer 17

1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2

Question 18

irregularity (group 15 vs 16)

Answer 18

 The 3p electron to be removed from S is a paired electron while that to be removed from P is an unpaired electron.
 Due to inter–electronic repulsion between paired electrons in the same orbital, less energy is required to remove the paired 3p electron from S.
 Hence the first ionisation energy of S is lower than that of P.