IMO Practice Test — General Principles and Processes of Isolation of Elements
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
Two sulphide ores A and B are in one lump. Adding NaCN floats only A. Then A is most likely:
$PbS$
$ZnS$
$CuFeS_2$
$FeS_2$
Explanation: NaCN depresses $ZnS$ (as $[Zn(CN)_4]^{2-}$), so the floated ore is $PbS$.
Question 2 of 14
A metal oxide line on the Ellingham diagram suddenly changes slope to a steeper rise. This usually marks:
melting of the oxide
boiling/melting of the metal (a phase change)
a redox change
no physical meaning
Explanation: A phase change of the metal alters $\Delta S$, kinking the line; e.g. the Zn line steepens after Zn boils.
Question 3 of 14
Above the C/CO and metal-line crossing temperature, the better reducing agent is:
the metal
oxygen
carbon
neither
Explanation: Beyond the crossing point carbon’s $\Delta G^0$ is more negative, so carbon reduces the metal oxide.
Question 4 of 14
For $\Delta G^0 = -RT\ln K$, doubling $T$ at constant $K$ would make $|\Delta G^0|$:
halve
unchanged
become zero
double
Explanation: At fixed $K$, $\Delta G^0$ is proportional to $T$, so its magnitude doubles.
Question 5 of 14
A semiconductor-grade rod must be 99.9999% pure. The correct method is:
zone refining
liquation
froth flotation
roasting
Explanation: Zone refining sweeps impurities into a moving melt to give ultra-pure Si/Ge.
Question 6 of 14
In $4Au + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Au(CN)_2] + 4NaOH$, gold is:
reduced
oxidised
unchanged
a catalyst
Explanation: Gold goes from 0 to +1; it is oxidised while $O_2$ is reduced.
Question 7 of 14
Why is $CO$ rather than $C$ the main reducer in the cooler upper blast-furnace zone?
CO is cheaper
C is solid
at lower T the C/CO line lies above the FeO line so C is ineffective; CO works
CO is magnetic
Explanation: Below the crossing temperature carbon cannot reduce the oxide, but $CO \rightarrow CO_2$ still can.
Question 8 of 14
The position of the $Mg/MgO$ line below the C/CO line at low T implies that:
C reduces MgO at low T
Mg is unreactive
MgO is an ore
Mg can reduce $CO$ to C at low T
Explanation: Where Mg’s line is lower, Mg is the stronger reducer and can even reduce $CO_2$/$CO$.
Question 9 of 14
A volatile carbonyl used in refining decomposes on heating. Increasing temperature shifts $M + 4CO \rightleftharpoons M(CO)_4$:
toward the free metal
toward the carbonyl
no shift
toward CO only
Explanation: Carbonyl formation is exothermic, so heating decomposes it to the pure metal (Mond process).
Question 10 of 14
Cryolite is added in aluminium extraction chiefly to:
oxidise alumina
lower the fusion temperature and raise conductivity
act as cathode
remove silica
Explanation: It dissolves $Al_2O_3$, dropping the working temperature to ~950 $^\circ C$ and conducting current.
Question 11 of 14
An oxide ore lighter than its gangue cannot be concentrated by hydraulic washing because:
it is magnetic
it dissolves in water
gravity separation needs the ore to be denser than gangue
it floats
Explanation: Hydraulic washing relies on the ore being heavier so the lighter gangue is washed away.
Question 12 of 14
In the thermite reaction iron is obtained molten because the reaction is:
endothermic
electrolytic
photochemical
highly exothermic
Explanation: The large negative $\Delta H$ releases enough heat to melt the iron produced.
Question 13 of 14
Roasting $PbS$ partly to $PbO$ then heating without air gives Pb by:
self-reduction ($2PbO + PbS \rightarrow 3Pb + SO_2$)
carbon reduction
electrolysis
liquation
Explanation: The remaining sulphide reduces the oxide it just formed — self-reduction.
Question 14 of 14
For two metals whose Ellingham lines cross at 1500 K, at 1600 K the metal that can reduce the other’s oxide is the one whose line at 1600 K is:
higher
lower
flat
irrelevant
Explanation: The lower line (more negative $\Delta G^0$) belongs to the better reducing metal at that temperature.