Online Test — Chemical Bonding and Molecular Structure
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
The octet rule was put forward in the work of:
Bohr and Sommerfeld
Kossel and Lewis
Heisenberg
Avogadro
Explanation: Kossel and Lewis (1916) linked bonding to attaining a stable octet like the noble gases.
Question 2 of 20
The formal charge on each oxygen in O=C=O is:
0
+1
-1
+2
Explanation: FC = 6 - 4 - (1/2)(4) = 0 for each oxygen.
Question 3 of 20
Which species has an expanded octet?
BF3
CH4
H2O
SF6
Explanation: SF6 has 12 electrons around sulphur, exceeding the octet using d-orbitals.
Question 4 of 20
Lattice enthalpy is greatest for:
NaCl
KCl
CsCl
MgO
Explanation: MgO has doubly charged, small ions, giving the highest lattice enthalpy.
Question 5 of 20
According to Fajans' rules, covalent character increases with:
a large cation
a small anion
a large, easily polarised anion
low ionic charge
Explanation: A large polarisable anion (and small, highly charged cation) increases covalent character.
Question 6 of 20
The molecule with zero net dipole moment is:
H2O
NH3
CO2
HF
Explanation: CO2 is linear, so its two bond dipoles cancel.
Question 7 of 20
The shape of CH4 predicted by VSEPR is:
square planar
pyramidal
linear
tetrahedral
Explanation: Four bond pairs and no lone pairs give a tetrahedral shape at 109.5 degrees.
Question 8 of 20
The bond angle in water is about:
120 degrees
109.5 degrees
104.5 degrees
90 degrees
Explanation: Two lone pairs compress the angle to about 104.5 degrees.
Question 9 of 20
The hybridisation of carbon in ethyne (C2H2) is:
sp
sp3
sp2
sp3d
Explanation: Each carbon forms 2 sigma bonds and 2 pi bonds, so it is sp hybridised (linear).
Question 10 of 20
PCl5 has the hybridisation and shape:
sp3, tetrahedral
sp3d, trigonal bipyramidal
sp3d2, octahedral
sp2, trigonal planar
Explanation: Five bond pairs need sp3d hybridisation, giving a trigonal bipyramidal shape.
Question 11 of 20
A pi bond is formed by:
axial overlap of orbitals
transfer of electrons
overlap of two s-orbitals
lateral (sideways) overlap of p-orbitals
Explanation: Pi bonds arise from sideways overlap of unhybridised p-orbitals.
Question 12 of 20
All C-O bonds in CO32- are equal in length because of:
hydrogen bonding
resonance
ionic bonding
expanded octet
Explanation: Resonance delocalises the double bond equally over all three C-O bonds.
Question 13 of 20
The bond order of H2 is:
1
0
0.5
2
Explanation: Bond order = (2 - 0)/2 = 1.
Question 14 of 20
The bond order of N2 is:
1
2
3
2.5
Explanation: Bond order = (10 - 4)/2 = 3, a triple bond.
Question 15 of 20
Molecular orbital theory predicts that O2 is:
diamagnetic with bond order 3
diamagnetic with bond order 1
paramagnetic with bond order 3
paramagnetic with bond order 2
Explanation: O2 has bond order 2 and two unpaired pi* electrons, so it is paramagnetic.
Question 16 of 20
An antibonding molecular orbital has:
lower energy than the atomic orbitals
a node between the nuclei
high electron density between the nuclei
no node
Explanation: Antibonding MOs have a node between the nuclei and lie higher in energy.
Question 17 of 20
Hydrogen bonding is strongest when H is bonded to:
F, O or N
C
P
S
Explanation: Small, highly electronegative F, O and N give the strongest hydrogen bonds.
Question 18 of 20
Ice floats on water because hydrogen bonding makes ice:
denser than water
less dense than water
ionic
non-polar
Explanation: An open cage-like H-bonded structure leaves large gaps, so ice is less dense than liquid water.
Question 19 of 20
The hybridisation of nitrogen in NH3 is:
sp
sp2
sp3d
sp3
Explanation: Three bond pairs and one lone pair around N give sp3 hybridisation (pyramidal).
Question 20 of 20
Which intermolecular force is the strongest?
London dispersion
dipole-dipole
hydrogen bonding
ion-induced dipole
Explanation: Hydrogen bonding is the strongest of the intermolecular attractions.