SODIUM IODIDE NaI

Properties of sodium iodide NaI:

White, when stored in the light turn yellow due to oxidation. It melts without decomposition, on further heating decomposes. Readily soluble in water (hydrolysis no). Typical reductant.

 

Obtaining sodium iodide NaI:

2 Na + E2 = 2 NaE (normal temp., Е = F, Cl; Вr, I).

2 NaOH (hot) + I2 + H2O2 =2 NaI + O2↑ + 2H2O

2 NаОН (cold) + I2 + H2S(gas) = 2 NaI + S↓ + 2H2O.

2 NaOH (diluted) + FeI2 = 2NaI + Fe(OH)2 ↓ (in the atmosphere of  N2)

2 NaOH (diluted) + 2AgNO3 = Ag2O ↓ + H2O + 2NaNO3.

NaIO3 + H2O + 2 Fe = NaI + 2 FeO(OH)↓ (boiling, in air).

 

Reactions with sodium iodide NaI:

2 NaI = 2Na + I2 (t>1400° С).

NaI·2H2O = NaI + 2Н2О (t> 68,9° С, vacuum).

NaI (diluted) + 4 Н2О = [Na(H2O)4]+ + I-(рН 7).

8 NaI (solid) + 9 H2SO4 (conc.) = 4 I2↓ + H2S↑ + 4Н2О + 8 NaHSO4 (30—50° С)

2 NaI (solid) + 4 HNO3 (conc.) = I2↓ + 2NO2↑ + 2H2O + 2NaNO3 (boiling, impurity of НIO3).

6 NaI + 2 H2O + O2 →τ 4NaOH + 2 Na[I(I)2] (normal temp.,on the light)

4 NaI + 4 НСl (diluted.) + O2 = 2I2↓ + 4 NaCl + 2 H2O (normal temp.,on the light).

2 NaI (cold) + E2 = 2NaE + I2↓ (E = Cl, Br)

Nal (hot) + 3  H2O + 3Cl2 = NaIO3 + 6HCl.

NaI + 4 NH3 (liquid) = [Na(NH3)4]I ↓ ↔ [Na(NH3)4]+ + I-

2 NaI + 3 H2SO4 (conc.) + MnO2 = 2NaHSO4 + I2↓ + 2H2O.

NaI + 4 NaOH + 4 NaNO3 = Na5IO6 + 4NaNO2 + 2H2O (300—330° C).

NaI + 3H2O → Electrolysis 3Н2↑(cathode) + NaIO3 (anode).