A 0.125 g piece of vanadium reacts with nitric acid to produce 50.0 mL of a yellow solution of vanad — Stoichiometry / Titration / Electron Configuration Chemistry Question
Problem Context
A 0.125 g piece of vanadium reacts with nitric acid to produce 50.0 mL of a yellow solution of vanadium ions in their highest oxidation state.
Calculate the number of moles of vanadium dissolved and the molarity of vanadium ions in this solution.
Model Answer
0.049 M
Write the electron configuration of a neutral gaseous vanadium atom.
Model Answer
1s2 2s2 2p6 3s2 3p6 4s2 3d3
Give the oxidation state of vanadium in the yellow solution and outline your reasoning.
Model Answer
V is in +5 oxid st. due to loss of 4s and 3d electrons.
A 25.0 mL portion of this yellow solution is reduced with excess zinc amalgam under an inert atmosphere to give a violet solution. A 10.0 mL aliquot of this violet solution is titrated with a solution of 2.23 × 10–2 M KMnO4 in acid forming Mn2+. A volume of 13.20 mL of the MnO4– solution is required to convert the vanadium back to yellow. Determine the:
i. number of moles of MnO4– used in this titration,
ii. mole ratio of vanadium ions to MnO4– ions in this titration,
iii. oxidation number change for vanadium in this titration and the oxidation state of vanadium ions in the violet solution.
Model Answer
Mn goes from +7 → 2
When 2.00 mL of the violet solution are mixed with 1.00 mL of the original yellow solution, a green solution results. When this ratio is reversed a bright blue solution is formed. Determine the oxidation states of the green and blue vanadium ions. Support your answers with calculations.