IMO Practice Test — The p-Block Elements
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
The total number of P–H bonds in H3PO2, H3PO3 and H3PO4 taken together is:
1
2
3
4
Explanation: H3PO2 has 2 P–H, H3PO3 has 1, H3PO4 has 0; total = 3.
Question 2 of 14
Which one of these is NOT a correct reason for choosing ~700 K (not a lower temperature) in the Haber process?
a higher temperature increases the reaction rate
the iron catalyst is most active around this range
a low temperature would shift equilibrium away from NH3
a low temperature would make the rate impractically slow
Explanation: The synthesis is exothermic, so a LOW temperature actually FAVOURS NH3; the moderate temperature is a kinetic compromise, making this statement the false reason.
Question 3 of 14
Among NH3, PH3, AsH3 and SbH3, the hydride with the highest bond angle is:
NH3
PH3
AsH3
SbH3
Explanation: Bond angle decreases down the group with electronegativity; NH3 has the largest (~107°).
Question 4 of 14
The number of lone pairs of electrons on the central atom in BrF5 is:
0
1
2
3
Explanation: BrF5 has 5 bond pairs and 1 lone pair (sp3d2), giving a square pyramidal shape.
Question 5 of 14
Hot concentrated H2SO4 reacts with copper to liberate:
H2
SO2
H2S
SO3
Explanation: Cu + 2H2SO4(conc.) → CuSO4 + SO2 + 2H2O; S is reduced from +6 to +4.
Question 6 of 14
Among the halogens, the one with the most negative (most exothermic) electron-gain enthalpy is:
F
Cl
Br
I
Explanation: Chlorine, not fluorine, has the most negative electron-gain enthalpy; the small F atom has high electron–electron repulsion.
Question 7 of 14
The shape of XeO3 is:
trigonal planar
pyramidal
T-shaped
square planar
Explanation: XeO3 has 3 bond pairs and 1 lone pair (sp3), giving a pyramidal shape.
Question 8 of 14
Which reaction correctly represents partial hydrolysis of XeF6?
XeF6 + 3H2O → XeO3 + 6HF
XeF6 + H2O → XeOF4 + 2HF
XeF6 + 6H2O → Xe + 6HF + 3H2O2
XeF6 → XeF4 + F2
Explanation: Partial hydrolysis replaces some F by O: XeF6 + H2O → XeOF4 + 2HF.
Question 9 of 14
Assertion: the +5 oxidation state becomes less stable down Group 15. Reason: the inert-pair effect makes the ns2 electrons increasingly reluctant to bond. Choose the correct option.
Both A and R true, R explains A
Both true, R does not explain A
A true, R false
A false, R true
Explanation: Down the group the inert-pair effect stabilises +3 over +5, exactly the reasoning given, so R correctly explains A.
Question 10 of 14
The interhalogen with a pentagonal bipyramidal shape is:
ClF3
BrF5
IF7
ICl
Explanation: IF7 has 7 bond pairs (sp3d3) and adopts a pentagonal bipyramidal shape.
Question 11 of 14
Which property explains why F2 is a stronger oxidising agent than Cl2 despite a less negative electron-gain enthalpy?
higher F–F bond enthalpy
low F–F bond enthalpy and high hydration enthalpy of F-
larger atomic size of F
lower electronegativity of F
Explanation: The weak F–F bond and the very high hydration enthalpy of the small F- ion make F2 the strongest oxidiser.
Question 12 of 14
On dissolving SO3 in concentrated H2SO4 during the Contact process, the product formed is:
dilute H2SO4
oleum (H2S2O7)
H2SO3
H2S2O8
Explanation: SO3 + H2SO4 → H2S2O7 (oleum), which is then diluted to give H2SO4.
Question 13 of 14
The brown gas evolved when copper reacts with concentrated nitric acid is:
NO
N2O
NO2
N2
Explanation: Concentrated HNO3 with Cu gives brown NO2: Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O.
Question 14 of 14
The number of Xe–F bonds and lone pairs respectively in XeF4 is:
4 and 0
4 and 1
4 and 2
6 and 1
Explanation: XeF4 has 4 bonding pairs and 2 lone pairs, hence square planar (sp3d2).