The crystalline structure of β-AgI (written as -AgI in text) is similar to that of ice, allowing it — Inorganic Chemistry — Solid State Chemistry Question
Silver iodide
The crystalline structure of β-AgI (written as -AgI in text) is similar to that of ice, allowing it to induce freezing by the process known as heterogeneous nucleation (cloud seeding). β-AgI is a bright yellow solid and has the wurtzite structure. [1]
When the solid silver iodide is exposed to sunlight, it will darken rapidly. What is the oxidation state of silver in the darkened solid? [1]
Model Answer
0
The reduction of Ag+ to Ag0 causes the change of color. [2]
What is the solubility trend in AgF, AgCl, AgBr, AgI? [1]
Model Answer
AgF > AgCl > AgBr > AgI
The stronger interaction between Ag+ and I– as well as the low hydration energy of I– result in the poor solubility of AgI. The smaller size of other halide ions led to poorer interaction with Ag+; hence, higher solubility. [3]
Consider:
(a) Ag+(aq) + e– ⟶ Ag(s) Eo = +0.80 V [1, 4]
(b) AgI(s) ⇌ Ag+(aq) + I–(aq) Ksp = 8.51×10-17 [4]
(c) Ag+(aq) + 3 I–(aq) ⇌ [AgI3]2–(aq) K = 1014 [4]
From the given information, find the standard reduction potential of [AgI3]2–. [4]
Model Answer
–0.028 V
ΔGo = –nFEo
For (a), ΔGo_a = –(–1)(96,500)(0.80) = –77,200 J (mol Ag)–1 [3]
ΔGo = –RT lnK
For (c), ΔGo_c = –(8.314)(298)(ln 1014) = –79,867 J mol–1 [3]
The reduction half reaction of [AgI3]2–:
(d) [AgI3]2–(aq) + e– ⇌ Ag(s) + 3 I–(aq) [3, 5]
Eqn (d) = (a) – (c) [5]
Therefore, ΔGo_d = ΔGo_a − ΔGo_c = –77,200 – (–79,867) = 2,667 J mol–1 [5]
Eo = −ΔGo_d / (nF) = –(2,667)/(96,500) = –0.028 V [5]
The salt [PPh3Me]2[AgI3] contains a triiodoargentate(I) ion [AgI3]2– with approximate trigonal planar geometry. (See reference: Bowmaker, G. A.; Camus, A.; Skelton, B. W.; White, A. H. J. Chem. Soc., Dalton Trans., 1990, 727-731.) The structure of [AgI3]2– is shown below.
[VISUAL]
Draw the crystal field splitting diagram for the d orbitals of silver and fill in all appropriate electrons. [2, 4]
Model Answer
For trigonal planar, the ligands lie in the xy plane, then the dx2-y2 and dxy orbitals that have their electron density concentrated in this plane will have the highest energy. The dxz and dyz orbitals have their electron density out of this xy plane, so their energies are the lowest. The dz2 orbital has its electron density mostly out of the xy plane, but there is a ring of electron density in the xy plane, so the dz2 orbital will have energy higher than the dxz and dyz orbitals but still lower than the dx2-y2 and dxy orbitals. In addition, the number of d-electrons for silver in [AgI3]2– is 10. Therefore, all d-orbitals should be filled. [5, 6]
[VISUAL] showing the coordinate axes (x, y, z) and the d-orbital splitting diagram with five orbitals split as two high-energy orbitals (dx2-y2, dxy), one middle-energy orbital (dz2), and two low-energy orbitals (dxz, dyz), all fully occupied with 10 d-electrons (5 pairs of electrons with opposite spins).