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Chapter 18 Solutions
Physical Chemistry
- What is the energy of the He+ electron with n=4? (k=2.18×10−18 J), please explain distribution steps into parenthesis.arrow_forwardGive the ground-state electron configu rations of (a) H2 - . (b) Li2, (c) Be2, (d) C2, (e) N2, and (f) O2 .arrow_forwardCalculate the energy (in eV) of the process: Br(g) + Na(g) → Br- + Na . Data: Ionization energy of Na = 5.00 eV Electronic affinity of Br = -3.50 eV Interaction energy between Na+ and Br- ions = 4.40 earrow_forward
- Using photoelectron spectroscopy, the ionization energy of the least tightly bound valence electron on P was determined to be 10.5 eV. Calculate the Zeffective for this electron.arrow_forwardUsing photoelectron spectroscopy, the ionization energy of the least tightly bound valence electron on Al was determined to be 6.00 eV. Calculate the Zeffective for this electron.arrow_forwardUse the data given below to construct a Born-Haber cycle to determine the electron affinity of Br. ΔH°(kJ mol-1) K(s) → K(g) 89 K(g) → K⁺(g) + e⁻ 419 Br2(l) → 2Br(g) 193 K(s) + ½Br2(g) → KBr (s) -394 KBr(s) → K⁺(g) + Br⁻(g) 674 A) -885 kJ B) -464 kJ C) +367 kJ D) -325 kJ E) +246 kJarrow_forward
- Identify the list in which the atoms are ordered by increasing Zeff for a valence electron calculated by the simple or approximate method? A) C<B<N<O B) B<N<C<O C) B<C<N<O D) O<N<C<B E) N<B<O<Carrow_forwardUse the following data and the Born-Haber cycle to calculate the first ionization energy (ΔH IE1) of K(g) K(s) → K(g) 89 kJ mol-1 Cl(g) + e– → Cl–(g) -349 kJ mol-1 K(s) + ½ Cl2(g) → KCl(s) -437 kJ mol-1 K(g) → K+(g) + e– ΔH IE1 Cl2(g) → 2Cl(g) 244 kJ mol-1 K+(g) + Cl–(g) → KCl(s) -717 kJ mol-1 Use the bond energies listed in the table below to estimate the standard enthalpy of the reaction: CCl3CHCl2(g) + 2HF(g) → CCl3CHF2(g) + 2HCl(g) C – H 412 kJ mol-1 C – C 348 kJ mol-1 C – Cl 338 kJ mol-1 H – Cl 431 kJ mol-1 C – F 484 kJ mol-1 H – F 565 kJ mol-1arrow_forwardUse the following data and the Born-Haber cycle to calculate the first ionization energy (ΔH IE1) of K(g) K(s) → K(g) 89 kJ mol-1 Cl(g) + e– → Cl–(g) -349 kJ mol-1 K(s) + ½ Cl2(g) → KCl(s) -437 kJ mol-1 K(g) → K+(g) + e– ΔH IE1 Cl2(g) → 2Cl(g) 244 kJ mol-1 K+(g) + Cl–(g) → KCl(s) -717 kJ mol-1arrow_forward
- Identify the states (s, aq, l, g) of each term in this equation. Cr2O72- + U4+ → Cr3+ + UO22+arrow_forwardWhat is the ground state electron configuration for phosphide ion (p3-)? A. Ne 3s2 B. Ne 3s2 3p3 C. Ne 3p3 D. Ne 3s2 3p6arrow_forwardButadiene can be roughly modeled as a linear “box” with L = 0.424 nm. There are four electrons, two of which can go in each particle in a box state (two electrons in nX= 1; two electrons in nX= 2). The lowest excitation involves promoting an electron from nX= 2 to nX= 3. What is the energy of that excitation?arrow_forward
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