General Chemistry for Engineers Notes Class 26

General Chemistry for Engineers - CHEN 1201

Midterm 2: Tuesday, October 28th from 6:00-7:30pm
- Format: Closed book, closed notes
- Information Sheet: Provided with all necessary information
- Exam Locations: CHEM140 (this room) and CHEM142 (next door)
- Required Items:
- Scientific (non-programmable) calculator with batteries
- A couple of #2 pencils
- Student ID and student number
Prep Material:
- Practice Exam: Posted for review
- In-class Review: Scheduled for next Monday
- Review Session: With Engineering Fellows on Sunday afternoon in CHEM140
- Study Session: Wednesday, October 22nd 6:30-8:30pm (Tonight!)

Cationic Radii:
- Cations are smaller than their corresponding neutral atoms because they lose an outermost electron that is shielded from the nucleus.
Anionic Radii:
- Anions are larger than their corresponding neutral atoms because they gain an outermost electron which increases repulsive interactions in the electron shell.

Lithium Ion Configuration:
- Neutral Lithium: $Li: 1s^2 2s^1$
- Lithium Ion: $Li^+: 1s^2 2s^0$
Oxygen Configuration:
- Neutral Oxygen: $O: 1s^2 2s^2 2p^4$
- Oxide Ion: $O^{2-}: 1s^2 2s^2 2p^6$
Selected Ionic Radii Values:
- $S^{2-}$ (184 pm)
- $Cl^-$ (181 pm)
- $K^+$ (133 pm)
- $Ca^{2+}$ (99 pm)

Order of Ionic Radii (in increasing order):
- A. $Mg^{2+} < K^{+} < I < I^{-}$
- B. $K^{+} < Mg^{2+} < I < I^{-}$
- C. $K^{+} < Mg^{2+} < I^{-} < I$
- D. $I < I^{-} < Mg^{2+} < K^{+}$
- Correct Answer: Arrange in increasing atomic or ionic radius: $I, I^{-}, Mg^{2+}, K^{+}$
Order of Increasing Radius for Specific Ions: Ca$^{2+}$, S$^{2-}$, Cl$^{-}$
- A) $Ca^{2+} < Cl^{-} < S^{2-}$
- B) $Cl^{-} < Ca^{2+} < S^{2-}$
- C) $S^{2-} < Cl^{-} < Ca^{2+}$
- D) $Ca^{2+} < S^{2-} < Cl^{-}$
- E) $Cl^{-} < S^{2-} < Ca^{2+}$

Ionization Energy (IE):
- The energy required to remove an electron from an atom or ion in its gaseous state.
- Example Values:
- $IE_1 = 496 ext{ kJ/mol}$
- $IE_2 = 4560 ext{ kJ/mol}$
Reactions:
- $Na(g)
  ightarrow Na^{+}(g) + e^{-}$
- $Na^{+}(g)
  ightarrow Na^{2+}(g) + e^{-}$

Trends Moving Down a Group:
- As you move down the periodic table from top to bottom, the highest-n orbital (outermost electron shell) increases in size, making electrons easier to extract.
Trends Moving Across a Period:
- From left to right across a period, there is an increase in effective nuclear charge ($Z_{eff}$) leading to stronger attraction to electrons, making them more difficult to extract.

Group 3A:
- Ionization energy $IE_1$ for Beryllium (Be) is greater than that of Boron (B) due to the more shielding experienced by the first electron in the 2p subshell compared to the last electron in the 2s subshell.
Group 6A:
- Ionization energy $IE_1$ for Oxygen (O) is lower than that of Nitrogen (N) due to electron-electron repulsion in the same orbital.

Electron Affinity (EA):
- The energy change that occurs when an electron is added to a gaseous atom. Atoms typically release energy upon gaining electrons.
- Example:
- $Cl(g) + e^{-}
  ightarrow Cl^{-}(g)$
- EA = $-349 ext{ kJ/mol}$

Trends Across a Period:
- From left to right, as the effective nuclear charge ($Z_{eff}$) increases, the energy release of EA generally becomes larger as atoms become better at attracting additional electrons.
Trends Down a Group:
- Moving down a group, EA values decrease because as orbitals increase in size, the energy released upon gaining an electron diminishes.

Question: Which element has the most negative electron affinity?
- A. K
- B. Mg
- C. Se
- D. Xe
- E. B
Comparison of Electron Configurations:
- a) $1s^2 2s^2 2p^6 3s^2 3p^{1}$
- b) $1s^2 2s^2 2p^6 3s^2 3p^{5}$
- c) $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$
- Goal: Identify which configuration corresponds to the atom that has the largest negative number for its electron affinity.