MC test: Further aspects of covalent bonding

Multiple choice test on 14.1 Further aspects of covalent bonding & structure

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

Which atomic orbitals can combine to form a sigma (σ) bond?

I. s + s

II. s + p_x

III. p_x + p_x

I, II and III

II and II only

I and III only

I and II only

When the two p_x orbitals meet 'head on' they form a σ bond. When two p_y or two p_z orbitals meet 'sideways on' they form a pi (π) bond.

What does a triple bond between two carbon atoms consist of?

three σ bonds

three π bonds

one π and two σ bonds

one σ and two π bonds

The σ bond lies along the axis directly between the two carbon nuclei, one of the two π bonds lies to the front and behind and the other lies above and below this axis.

Which species follows the octet rule?

PF₅

BCl₃

NO has an odd number of electrons, the octet is expanded in PF₅ and is incomplete in BCl₃. CO follows the octet rule by using a coordinate (dative) bond.

Which false assumption is made when determining the formal charge?

All the atoms in a molecule are covalently bonded.

All the atoms in a molecule have the same ionization energy.

All the atoms in a molecule have the same electron affinity.

All the atoms in a molecule have the same electronegativity.

To determine the formal charge it is assumed that all the atoms in a molecule or ion have the same electronegativity and yet clearly that is not the case.

The Lewis structure of sulfur trioxide can be drawn following the octet rule using resonance hybrids or in a single format by expanding the octet. One of the resonance hybrid forms and the expanded octet form are shown.

Which statements are correct?

I. The total formal charge in both formats is zero.

II. The formal charge on each of the singly bonded O atoms in the resonance format is − 1.

III. The formal charge on the S atom in both formats is zero.

II and III

I and II

I and III

I, II and III

Both formats give a total formal charge of zero but the formal charge on the S atom in the resonance format is +2 which cancels out the two −1 charges on the singly bonded oxygen atoms. The formal charge of all the doubly bonded O atoms is zero. The expanded octet form is preferred as the individual atoms have the lowest formal charges.

Why does benzene not readily undergo an addition reaction with chlorine?

The electrons in the σ bonds repel the chlorine molecules.

The electrons in the pi bonds repel the chlorine molecules.

Considerable energy is required to overcome the delocalization energy of the benzene ring.

The three C=C double bonds in benzene are too strong.

The delocalization energy of the six p electrons in the π bonding over all six carbon atoms in the benzene ring is approximately 150 kJ mol^-1 and this must be overcome for an addition reaction to occur.

Which statements are correct about the ethanoate ion, H₃CCOO⁻?

I. The bond lengths and strengths of the two carbon to oxygen bonds are equal.

II. Two p electrons form a delocalized π bond over the carbon and both oxygen atoms.

III. The bonding in the ion can be described using two Lewis resonance structures.

I, II and III

I and III only

I and II only

II and III only

Resonance hybrids and the delocalized form are two equivalent ways of showing the bonding

The action of sunlight on freons such as CCl₂F₂ in the stratosphere can cause destruction of the ozone layer. Which equation describes the initial reaction that occurs?

CCl₂F₂ → CClF₂· + Cl·

CCl₂F₂ → CF₂· + Cl₂

CCl₂F₂ → CClF₂⁺ + Cl⁻

CCl₂F₂ → CCl₂F· + F·

The C−F is too strong to be broken by sunlight. It is the C−Cl bond that is broken homolytically to form free radicals.

Why does the steady state equilibrium between oxygen and ozone in the ozone layer prevent much of the sun's uv radiation reaching the Earth's surface?

Short wavelength UV light is required to break the bonds homolytically in ozone and longer wavelength UV light is required to break the bonds heterolytically in oxygen.

Short wavelength UV light is required to break the bonds homolytically in oxygen and longer wavelength UV light is required to break the bonds homolytically in ozone.

Short wavelength UV light is required to break the bonds heterolytically in oxygen and longer wavelength UV light is required to break the bonds homolytically in ozone.

Short wavelength UV light is required to break the bonds homolytically in ozone and longer wavelength UV light is required to break the bonds homolytically in oxygen.

Both reactions are homolytic. O=O is the stronger bond so requires shorter wavelength (higher energy) UV light to break compared to the O to O bond in ozone which has a bond order of 1.5.

Which compound contains 8σ bonds and 2π bonds?

H₃CCHCH₂

H₃CCCH

H₃CCH₂CN

H₃CCONH₂

There are seven single bonds in H₃CCH₂CN giving seven σ bonds and the triple bond between C and N is made up of one σ bond and two π bonds. H₃CCCH also has two π bonds but only six σ bonds. H₃CCHCH₂ and H₃CCONH₂ both only have one π bond.

Total Score:

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MC test: Further aspects of covalent bonding

Multiple choice test on 14.1 Further aspects of covalent bonding & structure

MC test: Further aspects of covalent bonding