16 This question is about the Group 4 element germanium and some of its compounds.
(a) Naturally‑occurring germanium consists of five stable isotopes.
Explain what is meant by the term isotopes.
(2)
ANSWER
Atoms of same element that have the same atomic number i.e. proton number but different mass number i.e. neutron number.
(b) The mass spectrum of a sample of germanium is shown.
Calculate the relative atomic mass of this sample of germanium.
(2)
ANSWER
(c) Give the electronic configuration of a germanium atom.
(1)
ANSWER
Since the electron number for germanium is 32, therefore
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p2
(d) Germane is a compound with the formula GeH4.
It can be formed by the reaction shown.
Na2GeO3 + NaBH4 + H2O → GeH4 + 2NaOH + NaBO2
(i) Calculate the atom economy, by mass, for the formation of germane.
Use Ar of Ge = 72.6
(2)
ANSWER
Calculation of atom economy by using ∑Mr (all reactants) / ∑Mr (all products)
(ii) Germane can donate a proton to ammonia, forming the ion GeH3–
NH3 + GeH4 → NH4+ + GeH3–
Draw a dot‑and‑cross diagram of the ion GeH3–.
Show the outer electrons only.
(2)
ANSWER
(iii) Use your answer to (d)(ii) and electron‑pair repulsion theory to predict the name of the shape and bond angle of the ion GeH3 – .
(2)
ANSWER
Shape = Trigonal pyramidal
Bond Angle = 107°
(iv) Germane is toxic and the maximum permitted concentration is 0.640mgm–3 in air.
Calculate the maximum number of germane molecules permitted in a laboratory with a volume of 231m3 .
[Avogadro constant (L) = 6.02×1023mol–1]
(4)
ANSWER
Total mass = 0.640 x 10-3gm–3 x 231m3
= 0.1478 gram
Therefore,
Moles = mass/Mr
=0.1478/76.6
= 0.00193 mol
Hence,
No. of molecules = 0.00193 mol x 6.02×1023mol–1
= 1.16 x 1021 molecules
(e) A halide of germanium, GeX4, reacts with water as shown.
GeX4 + 2H2O → GeO2 + 4HX
1.50g of GeX4 was added to excess water.
The insoluble GeO2 was removed by filtration.
The solution of hydrogen halide formed was then added to excess magnesium carbonate, forming 335.5cm3 of carbon dioxide at room temperature and pressure (r.t.p.).
MgCO3 + 2HX → MgX2 + CO2 + H2O
[Molar volume of a gas at r.t.p. = 24000cm3mol–1]
(i) Calculate the number of moles of carbon dioxide formed.
(1)
ANSWER
(ii) Deduce the number of moles of HX formed when GeX4 reacted with the excess water.
(1)
ANSWER
By seeing the molar ratio between HX and CO2, following is observed
For HX: CO2 is 2 moles: 1 mole
Therefore
(iii) Deduce the number of moles of GeX4 that reacted with the excess water.
(1)
ANSWER
By seeing the molar ratio between GeX4 and HX, following is observed
For GeX4: HX is 1 mole: 4 moles
Therefore,
(iv) Calculate the molar mass of GeX4 and hence identify X.
(2)
ANSWER
By using ratio method,
1 mole of GeX4 = X gram of GeX4
0.0069896mol of GeX4 = 1.50g of Ge4 used
Therefore,
Now from above we can say that the molecular mass of GeX4 = 214.6
X
Hence, X is Chlorine (Cl).
