Ionization energy
It is the energy required to remove the least tightly bound electron from a neutral atom in the gas phase.
What is the periodic trend in ionization energy?
Ionization energy is lowest at the bottom left corner of the table, and increases going up and to the right, with
the highest ionization energy being for Helium.
The trend is the same in regards to effective nuclear charge and atomic radii. In ther words, the smaller atomic
radius, the greater ionization energy and effective nuclear charge.
What is the real reason that a half-filled subshell is so stable?
A half-filled subshell serves to maximize the stabilizing interactions, while minimizing the destabilizing
interactions among electrons.
,What is the exchange interaction?
Is it stabilizing or destabilizing?
The exchange interaction, πe, is the stabilizing interaction, and it is a result of electrons pairing in degenerate
orbitals with parallel spin.
What is harder to ionize, a high energy electron or a low energy electron?
A high energy electron is easier to ionize. It already contains more energy, so the input to remove it from the
atom is less.
If a 3d series metal is ionized, what orbital loses the electron? What happens to the valence configuration?
If a d series metal has the configuration 4s^2 3d^n, the first electron to be ionized will come from the 4s
orbital.
The remaining s orbital will then enter a d orbital, giving the configuration of the +1 cation 4s^0, 3d^(n+1).
,Describe Slater's Rules
Slater’s Rules tell us what the effective nuclear charge will be.
Formally, it is written as Zeff= Z – σ.
Z is the atomic number, and σ=Σ(si)
(Si) is the number of electrons in a given subshell multiplied by a weighting coefficient.
The weighting coefficients are:
1) A. For s and p valence electrons,
(n-2) groups or lower contribute 1.00 to si (weighting coefficient of 100%). 1.00xne
B. (n-1) groups contribute 0.85 to si. 0.85xne
C. N(s,p) groups (the valence groups) contribute 0.35 to si. 0.35xne
2) For d and f valence electrons:
A. electrons in the valence shell contribute 0.35 to si. 0.35xne
B. electrons in any lower group contribute 1.00 to si.1.00xne
, What is shielding?
When electrons in inner shells weaken the attraction between electrons in the outer shell
What is the Lanthanide Contraction?
It is the reduction in atomic radius following the lanthanide series, contrary to the overall trend observed for
the periodic table. The lanthanides are elements 57 through 71, and are the first appearance of filled f orbitals.
F orbitals are poor at shielding, so any electrons added into the subshell following the f orbital experience a
higher effective nuclear charge, shrinking their radius over what is to be expected from overall trends.
What is penetration?
Describes when an electron of a higher atomic orbital is found within the shell of electrons of lower atomic
number. That is to say that an electron of higher energy is found within an orbital of lower energy.
It is the energy required to remove the least tightly bound electron from a neutral atom in the gas phase.
What is the periodic trend in ionization energy?
Ionization energy is lowest at the bottom left corner of the table, and increases going up and to the right, with
the highest ionization energy being for Helium.
The trend is the same in regards to effective nuclear charge and atomic radii. In ther words, the smaller atomic
radius, the greater ionization energy and effective nuclear charge.
What is the real reason that a half-filled subshell is so stable?
A half-filled subshell serves to maximize the stabilizing interactions, while minimizing the destabilizing
interactions among electrons.
,What is the exchange interaction?
Is it stabilizing or destabilizing?
The exchange interaction, πe, is the stabilizing interaction, and it is a result of electrons pairing in degenerate
orbitals with parallel spin.
What is harder to ionize, a high energy electron or a low energy electron?
A high energy electron is easier to ionize. It already contains more energy, so the input to remove it from the
atom is less.
If a 3d series metal is ionized, what orbital loses the electron? What happens to the valence configuration?
If a d series metal has the configuration 4s^2 3d^n, the first electron to be ionized will come from the 4s
orbital.
The remaining s orbital will then enter a d orbital, giving the configuration of the +1 cation 4s^0, 3d^(n+1).
,Describe Slater's Rules
Slater’s Rules tell us what the effective nuclear charge will be.
Formally, it is written as Zeff= Z – σ.
Z is the atomic number, and σ=Σ(si)
(Si) is the number of electrons in a given subshell multiplied by a weighting coefficient.
The weighting coefficients are:
1) A. For s and p valence electrons,
(n-2) groups or lower contribute 1.00 to si (weighting coefficient of 100%). 1.00xne
B. (n-1) groups contribute 0.85 to si. 0.85xne
C. N(s,p) groups (the valence groups) contribute 0.35 to si. 0.35xne
2) For d and f valence electrons:
A. electrons in the valence shell contribute 0.35 to si. 0.35xne
B. electrons in any lower group contribute 1.00 to si.1.00xne
, What is shielding?
When electrons in inner shells weaken the attraction between electrons in the outer shell
What is the Lanthanide Contraction?
It is the reduction in atomic radius following the lanthanide series, contrary to the overall trend observed for
the periodic table. The lanthanides are elements 57 through 71, and are the first appearance of filled f orbitals.
F orbitals are poor at shielding, so any electrons added into the subshell following the f orbital experience a
higher effective nuclear charge, shrinking their radius over what is to be expected from overall trends.
What is penetration?
Describes when an electron of a higher atomic orbital is found within the shell of electrons of lower atomic
number. That is to say that an electron of higher energy is found within an orbital of lower energy.
