QUESTION 1
a) Mg = 1s2 2s2 2p6 3s2 and Ar = 1s2 2s2 2p6 3s2 3p6
b) The values of the 1st and 2nd IEs for Mg are much smaller than for Ar because Mg has 2 valence electrons that are not as tightly bound to the nucleus, while Ar has a full octet of electrons which is very stable. Mg also has lower affinity for it’s valence electrons due to its position on the far left of the periodic table. Ar is unlikely to lose or gain any electrons due to its stability. Mg however could lose these 2 electrons and become stable with the octet in the lower energy level. {Ar has a full set of s and p orbitals which leads to this stability}The 3rd IE is much larger due to the removal of a core electron in Mg that is very tightly bound to the positive charge of the nucleus due to its proximity to the nucleus.
c) In the reaction with Cl, the Mg is more willing to give up the valence electrons in the s orbitals to the electronegative Cl atoms. This then will form MgCl2 due to 2 valence electrons being donated to Cl in order to create two full octets in Cl as well as Mg. Ar has a full subshell, therefore, does not participate in bonding with Cl. Once Cl gains 1 electron each, then it has full s and p sublevels. (Requires 2 Cl because Cl only forms Cl-, whereas Mg froms Mg2+)
d) QCl, due to the extreme difference in IE between the first and second ionization. This means that there is 1 valence electron able to be removed, but the next one is a core electron, which is closer to the nucleus so more tightly held and thus requiring more energy to be stripped from the atom.
QUESTION 2
a) Across the period from Li to Ne the number of protons is increasing in the nucleus hence the nuclear charge is increasing with a consequently stronger attraction for electrons and an increase in I.E.
b) The electron ionized in the case of Be is a 2s electron whereas in the case of B it is a 2p electron. 2p electrons are higher in energy than 2s electrons because 2p electrons penetrate the core to a lesser degree.
c) The electron ionized in O is paired with another electron in the same orbital, whereas in N the electron comes from a singly-occupied orbital. The ionization energy of the O electron is less because of the repulsion between two electrons in the same orbital.
d) The ionization energy of Na will be less than those of both Li and Ne because the electron removed comes from an orbital which is farther from the nucleus, therefore less tightly held.
MULTIPLE CHOICE:
30. D
- Decreasing atomic number (less effective nuclear charge), greater distance between outermost electrons and nucleus
31. B
- Essentially means that electrons are farther away from the nucleus
33. D
- Farthest to the right/closest to top of the periodic table
34. E
- Farthest to the left/closest to bottom of the periodic table
35. A
- Only one valence electron; very weak pull from the nucleus, easier to lose it to to gain the full octet
36. D
- Full octet; very difficult to pull away electrons
37. E
- From right to left across periodic table
38. A
- Higher effective nuclear charge/closer to the full octet
39. D
- Only one valence electron (very easy to pull away)
40. D
- Once the first electron of an alkali is stripped away, then the remaining electrons are of the core group and very difficult to pull away from the nucleus (esp. in Na, in which they will not be very far from the nucleus)
41. A
- Represents the ion from the first ionization of calcium losing another electron (another ionization)
a) Mg = 1s2 2s2 2p6 3s2 and Ar = 1s2 2s2 2p6 3s2 3p6
b) The values of the 1st and 2nd IEs for Mg are much smaller than for Ar because Mg has 2 valence electrons that are not as tightly bound to the nucleus, while Ar has a full octet of electrons which is very stable. Mg also has lower affinity for it’s valence electrons due to its position on the far left of the periodic table. Ar is unlikely to lose or gain any electrons due to its stability. Mg however could lose these 2 electrons and become stable with the octet in the lower energy level. {Ar has a full set of s and p orbitals which leads to this stability}The 3rd IE is much larger due to the removal of a core electron in Mg that is very tightly bound to the positive charge of the nucleus due to its proximity to the nucleus.
c) In the reaction with Cl, the Mg is more willing to give up the valence electrons in the s orbitals to the electronegative Cl atoms. This then will form MgCl2 due to 2 valence electrons being donated to Cl in order to create two full octets in Cl as well as Mg. Ar has a full subshell, therefore, does not participate in bonding with Cl. Once Cl gains 1 electron each, then it has full s and p sublevels. (Requires 2 Cl because Cl only forms Cl-, whereas Mg froms Mg2+)
d) QCl, due to the extreme difference in IE between the first and second ionization. This means that there is 1 valence electron able to be removed, but the next one is a core electron, which is closer to the nucleus so more tightly held and thus requiring more energy to be stripped from the atom.
QUESTION 2
a) Across the period from Li to Ne the number of protons is increasing in the nucleus hence the nuclear charge is increasing with a consequently stronger attraction for electrons and an increase in I.E.
b) The electron ionized in the case of Be is a 2s electron whereas in the case of B it is a 2p electron. 2p electrons are higher in energy than 2s electrons because 2p electrons penetrate the core to a lesser degree.
c) The electron ionized in O is paired with another electron in the same orbital, whereas in N the electron comes from a singly-occupied orbital. The ionization energy of the O electron is less because of the repulsion between two electrons in the same orbital.
d) The ionization energy of Na will be less than those of both Li and Ne because the electron removed comes from an orbital which is farther from the nucleus, therefore less tightly held.
MULTIPLE CHOICE:
30. D
- Decreasing atomic number (less effective nuclear charge), greater distance between outermost electrons and nucleus
31. B
- Essentially means that electrons are farther away from the nucleus
33. D
- Farthest to the right/closest to top of the periodic table
34. E
- Farthest to the left/closest to bottom of the periodic table
35. A
- Only one valence electron; very weak pull from the nucleus, easier to lose it to to gain the full octet
36. D
- Full octet; very difficult to pull away electrons
37. E
- From right to left across periodic table
38. A
- Higher effective nuclear charge/closer to the full octet
39. D
- Only one valence electron (very easy to pull away)
40. D
- Once the first electron of an alkali is stripped away, then the remaining electrons are of the core group and very difficult to pull away from the nucleus (esp. in Na, in which they will not be very far from the nucleus)
41. A
- Represents the ion from the first ionization of calcium losing another electron (another ionization)