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CH 221 Chapter Six Part II Concept Guide
1. Writing Electron Configurations
Question What is the complete electron configuration of the zirconium atom?
Approach With increasing atomic number, electrons occupy the subshells available in each main energy level in order, with few exceptions: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
Solution: Forty electrons must be accommodated. The total electron configuration for Zr is: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^2 5s^2.
This configuration may also be written as: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^2.
2. Writing Electron Configurations
Question What is the complete electron configuration for the arsenic atom?
Approach With increasing atomic number, electrons occupy the subshells available in each main energy level in order, with few exceptions: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
Solution: Thirty-three electrons must be accommodated. The total electron configuration for As is: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^3.
This configuration may also be written as: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3.
3. Writing Electron Configurations
Question What is the noble gas notation for the electron configuration for the rubidium atom?
Approach With increasing atomic number, electrons occupy the subshells available in each main energy level in specific order, with few exceptions: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. The noble gas notation substitutes the symbol of the noble gas for the corresponding noble gas core in the electron configuration.
Solution: The noble gas just prior to Rb in the periodic table is Kr. The noble gas notation for the electron configuration for Rb is: [Kr]5s^1.
4. Writing Electron Configurations
Question What is the complete electron configuration for the Br¯? With which element in the periodic table is it isoelectric?
Approach With increasing atomic number, electrons occupy the subshells available in each main energy level in specific order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
Solution: The anion Br¯ is formed by the addition of 1 electron to the lowest energy orbital that has a vacancy. The electron configuration for Br¯ is: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6.
Cl¯ is isoelectric with krypton.
5. Electron Configurations of Transition Metal Cations
We will look at iron, a transition metal, and two of its cations, and compare their electron configurations. We will compare what we would expect from main group behavior against the real behavior of the cations. We'll examine the real behavior closely to understand why it occurs, and then come to some conclusions about why the real behavior differs from our expectations.
We will be examining Fe, Fe2+, and Fe3+. The electron configuration of neutral Fe is [Ar]4s^2 3d^6.
Question: What will be the electron configurations of Fe2+^ and of Fe3+?
Expectation (based on main group behavior): If Fe loses electrons to form cations, we might expect the electrons removed to be those that were added last. We would expect:
Fe2+: [Ar]4s^2 3d^4 Fe3+: [Ar]4s^2 3d^3
Real Behavior:
Fe2+: [Ar]3d^6 Fe3+: [Ar]3d^5
Elaboration: Most transition metal atoms have electron configurations with two electrons in an s orbital, and some electrons in the d subshell of the shell below (with a lower value of n) that of the s orbital.
When transition metal atoms lose electrons to form ions, the first electrons lost are those in the s orbital. The electron configuration of Fe2+^ is [Ar]3d^6. The 4s electrons have been lost, the 3d electrons remain. If a third electron is lost, forming Fe3+, it will be removed from the 3d subshell, forming an ion with an electron configuration [Ar]3d^5.
Approach Atomic size increases moving down a group, and decreases moving left to right along a period.
Solution: (a) Silicon is the second element in the carbon family, and lead is the fifth element. Silicon atoms should be smaller as size increases down a representative periodic group. (b) Cesium and lead are in the same period, with lead further to the right in the period. Size decreases across the periods, and therefore cesium atoms should be larger than lead atoms. (c) Rhodium and iridium are the second and third members of a d-transition metal group. Elements in the second and third transition series are very similar in size due to the lanthanide contraction, therefore, these atoms should be similar in size. (d) Titanium and vanadium are adjacent elements in the same d-transition period. The decrease in size across the periods for transition elements is gradual. Atoms of these two elements should be similar in atomic radius.
7. Ionic Radii
Problem Place the following species in order of increasing radius: Cl Cl-^ Cl+
Approach For species having the same number of protons, the more electrons, the larger the species.
Solution: The positively charged chlorine atom is smaller than neutral chlorine; the former has one fewer electron than neutral chlorine. The negatively charge chlorine atom is the largest of all three; it has one additional electron than neutral chlorine.
Cl+^ < Cl < Cl-
8. Ionic Radii
Problem Arrange the following ions in order of increasing radius: F-^ Mg2+^ Cl-^ Be2+^ S^2 - Na+
Approach Ionic size increases moving down a group, and decreases moving left to right along a period. For isoelectric species, the greater the number of protons, the smaller the species.
Solution: Mg2+, Na+, F-^ all have the same number of electrons, thus they are similar in size. Mg2+, however, has the largest number of protons, therefore it is the smallest of these three ions. Likewise, S2-^ is larger than Cl-^ because it has fewer protons. S2-^ and Cl-^ have an additional shell relative to Mg2+, Na+, and F-, therefore S2-^ and Cl-^ are both larger than these three ions. Be2+^ has one fewer shell, and is therefore smaller than Mg2+, Na+, and F-.
Be2+^ < Mg2+^ < Na+^ < F-^ < Cl-^ < S^2 -
