IMO Practice Test — The p-Block Elements (Group 13 & 14)
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
The number of 3-centre 2-electron bonds in diborane is:
1
2
3
4
Explanation: Two B-H-B bridge bonds are 3c-2e; the four terminal B-H bonds are 2c-2e.
Question 2 of 14
Which statement about the inert pair effect is INCORRECT?
It stabilises lower oxidation states down a group
It is caused partly by poor d/f shielding
It makes Tl3+ very stable
M-X bond energy decreasing down the group contributes to it
Explanation: The inert pair effect makes Tl+ (not Tl3+) more stable; Tl3+ is actually a strong oxidant.
Question 3 of 14
Total number of electrons involved in bonding within diborane is:
8
10
12
14
Explanation: B2H6 has 2(3) + 6(1) = 12 valence electrons forming six bonds (four 2c-2e + two 3c-2e).
Question 4 of 14
Which property does NOT increase down Group 14?
Atomic radius
Metallic character
Stability of +4 state
Stability of +2 state
Explanation: The +4 state becomes LESS stable down the group (inert pair effect); +2 becomes more stable.
Question 5 of 14
In C60 fullerene, the carbon atoms are arranged in:
only hexagons
12 pentagons and 20 hexagons
only pentagons
squares and triangles
Explanation: Buckminsterfullerene has 12 pentagonal and 20 hexagonal faces, like a football.
Question 6 of 14
Which pair best illustrates a diagonal relationship?
B and C
B and Si
Al and Si
C and Si
Explanation: Boron resembles silicon (the diagonal partner), not its own-group neighbour aluminium.
Question 7 of 14
The hybridisation of carbon in diamond and graphite respectively is:
sp2, sp3
sp3, sp2
sp, sp2
sp3, sp3
Explanation: Diamond is sp3 (3-D network); graphite is sp2 (layered).
Question 8 of 14
Why does boric acid behave as a monobasic acid?
It loses three protons
It accepts one OH- from water, releasing one H3O+
It donates one proton directly
It is a strong acid
Explanation: B(OH)3 + 2H2O → [B(OH)4]- + H3O+; one H3O+ released, so monobasic Lewis acid.
Question 9 of 14
The correct order of catenation tendency is:
C > Si > Ge > Sn
Si > C > Ge > Sn
Sn > Ge > Si > C
Ge > C > Si > Sn
Explanation: Bond energy falls down the group: C >> Si > Ge ≈ Sn > Pb.
Question 10 of 14
Which one acts as a reducing agent owing to oxidation-state stability?
Pb4+
Tl3+
Sn2+
Al3+
Explanation: Sn is most stable as Sn4+, so Sn2+ readily loses electrons — a reducing agent.
Question 11 of 14
Aluminium chloride exists as a dimer Al2Cl6 because:
Al is ionic
Al completes its octet via chlorine bridge (coordinate) bonds
Cl is electropositive
it is a network solid
Explanation: Electron-deficient Al accepts a lone pair from a bridging Cl, forming the dimer to complete its octet.
Question 12 of 14
Zeolites act as shape-selective catalysts because they:
are very dense
have pores/cavities of molecular dimensions
are metallic
have no Al
Explanation: Their molecular-sized cavities admit only certain molecules, giving shape selectivity in cracking.
Question 13 of 14
The acidic strength of Group 13 oxides as the metallic character increases down the group:
increases
decreases (becomes more basic/amphoteric)
stays constant
becomes strongly acidic
Explanation: B2O3 is acidic, Al2O3 amphoteric, Tl2O basic — oxides become less acidic / more basic down the group.
Question 14 of 14
Which best explains why carbon cannot form CCl62- but Si forms SiF62-?
C is more electronegative
C lacks valence d-orbitals so its covalency is limited to 4
Si is a non-metal
C is larger
Explanation: Silicon has vacant 3d orbitals to expand its octet to six; carbon does not, capping covalency at 4.