"Chemical Bonding and Molecular Structure" — Practice

Question 1

$	ext{CH}_3	ext{Cl}$ has a dipole moment of $1.87$ D and $	ext{CH}_3	ext{F}$ of $1.85$ D, despite fluorine being far more electronegative than chlorine. Explain.

Accepted Answer

Dipole moment is the product of the partial charge and the bond length.

The $	ext{C}-	ext{F}$ bond has the larger charge separation but is much **shorter** ($138$ pm against $178$ pm).

The two effects nearly cancel, leaving the two moments almost identical.

This is why electronegativity difference alone never determines a dipole moment; the bond length matters equally.

Question 2

Arrange $	ext{AgCl}$, $	ext{NaCl}$, $	ext{MgCl}_2$ and $	ext{AlCl}_3$ in order of increasing covalent character, with reasons.

Accepted Answer

$	ext{NaCl}$: singly charged cation of noble-gas configuration, most ionic.

$	ext{MgCl}_2$: higher charge and smaller cation, more covalent.

$	ext{AlCl}_3$: higher charge still and smaller again.

$	ext{AgCl}$: although singly charged, $	ext{Ag}^+$ has a $4d^{10}$ pseudo noble gas configuration that screens poorly, making it a very strong polariser.

Order: $	ext{NaCl}<	ext{MgCl}_2<	ext{AlCl}_3<	ext{AgCl}$, which matches their solubilities in water.

Question 3

Predict the shape and approximate bond angles of $	ext{I}_3^-$, $	ext{BrF}_5$ and $	ext{XeO}_3$.

Accepted Answer

$	ext{I}_3^-$: central iodine has $2$ bonds and $3$ lone pairs, five domains. Lone pairs equatorial, giving a **linear** ion at $180^{\circ}$.

$	ext{BrF}_5$: $5$ bonds and $1$ lone pair, six domains. **Square pyramidal**, with angles slightly under $90^{\circ}$.

$	ext{XeO}_3$: $3$ bonds and $1$ lone pair, four domains. **Trigonal pyramidal**, about $103^{\circ}$.

In each case the lone pair compresses the remaining angles below the ideal value.

Question 4

Using Bent's rule, predict which positions the fluorine atoms occupy in $	ext{PF}_2	ext{Cl}_3$ and explain.

Accepted Answer

In a trigonal bipyramid the equatorial hybrids have more $s$ character than the axial ones, which are nearly pure $p$.

Bent's rule says the **more electronegative** substituent takes the orbital with **less** $s$ character.

Fluorine is more electronegative than chlorine, so the two fluorines occupy the **axial** positions and the three chlorines the equatorial ones.

The same rule places fluorines axially in $	ext{PF}_3	ext{Cl}_2$ and hydrogens equatorially in $	ext{PF}_4	ext{H}$.

Question 5

Compare the bond angles in $	ext{H}_2	ext{O}$ ($104.5^{\circ}$), $	ext{H}_2	ext{S}$ ($92.1^{\circ}$) and $	ext{H}_2	ext{Se}$ ($91^{\circ}$), and explain the trend.

Accepted Answer

Oxygen is small and electronegative, so its $2s$ and $2p$ orbitals are close in energy and hybridise effectively, giving an angle near tetrahedral.

Sulphur and selenium have a much larger $s$-$p$ energy gap, so hybridisation is poor and they bond using nearly pure $p$ orbitals, whose natural angle is $90^{\circ}$.

The lone pairs on the heavier atoms therefore reside largely in the $s$ orbital, contributing little angular strain.

The angle collapses from $104.5^{\circ}$ towards $90^{\circ}$ and changes very little thereafter.

Question 6

Write the molecular orbital configuration of $	ext{B}_2$ and $	ext{C}_2$, giving bond order and magnetism for each.

Accepted Answer

$	ext{B}_2$ has $6$ valence electrons: $\sigma_{2s}^{2}\sigma^{*}_{2s}{}^{2}\pi_{2p}^{2}$.

The last two electrons occupy the two degenerate $\pi$ orbitals singly, so $	ext{B}_2$ is **paramagnetic** with bond order $\dfrac{4-2}{2}=1$.

$	ext{C}_2$ has $8$: $\sigma_{2s}^{2}\sigma^{*}_{2s}{}^{2}\pi_{2p}^{4}$, bond order $2$, **diamagnetic**.

Remarkably, $	ext{C}_2$'s double bond consists of two $\pi$ bonds and no $\sigma$ bond at all — a result valence bond theory cannot produce.

Question 7

Arrange $	ext{N}_2$, $	ext{N}_2^+$, $	ext{N}_2^-$ and $	ext{N}_2^{2-}$ in order of increasing bond length.

Accepted Answer

$	ext{N}_2$ has $10$ valence electrons and bond order $3$.

$	ext{N}_2^+$ loses a **bonding** $\sigma_{2p}$ electron: bond order $2.5$.

$	ext{N}_2^-$ gains an **antibonding** $\pi^{*}$ electron: bond order $2.5$.

$	ext{N}_2^{2-}$ gains two: bond order $2$.

Bond length: $	ext{N}_2<	ext{N}_2^+\approx	ext{N}_2^-<	ext{N}_2^{2-}$. Note that the cation and anion have equal bond orders for opposite reasons.

Question 8

$	ext{CN}^-$, $	ext{CO}$ and $	ext{NO}^+$ all have bond order $3$. Explain, and predict the bond order of $	ext{CN}$ and $	ext{NO}$.

Accepted Answer

All three have $14$ valence electrons and are isoelectronic with $	ext{N}_2$, giving bond order $3$ and diamagnetism.

$	ext{CN}$ has $13$: one fewer bonding electron, bond order $2.5$, paramagnetic.

$	ext{NO}$ has $15$: one electron in $\pi^{*}$, bond order $2.5$, paramagnetic.

Both radicals are unstable for that reason, and both are stabilised by gaining or losing that single electron to reach fourteen.

Question 9

Explain why trisilylamine is planar while trimethylamine is pyramidal, and state a consequence for basicity.

Accepted Answer

In trimethylamine the nitrogen keeps its lone pair in an $sp^{3}$ orbital and the molecule is pyramidal.

In trisilylamine, silicon has empty $d$ orbitals that can accept the nitrogen lone pair by back bonding. Nitrogen rehybridises to $sp^{2}$ to make its lone pair available in a $p$ orbital, and the molecule flattens.

Because the lone pair is delocalised into silicon, trisilylamine is a **very weak base**, whereas trimethylamine is a good one.

The same effect widens the $	ext{Si}-	ext{O}-	ext{Si}$ angle in siloxanes to about $144^{\circ}$.

Question 10

Explain why $	ext{HF}$ boils at $19.5^{\circ}$C while $	ext{HCl}$ boils at $-85^{\circ}$C, even though $	ext{HCl}$ has the greater molar mass.

Accepted Answer

$	ext{HF}$ forms strong hydrogen bonds, because fluorine is small and highly electronegative and carries lone pairs.

$	ext{HCl}$ can only manage dipole-dipole attraction, since chlorine is too large and insufficiently electronegative to support hydrogen bonding.

The hydrogen-bond network in liquid $	ext{HF}$ must be broken to boil it, which costs far more energy than the extra dispersion forces available to the heavier $	ext{HCl}$.

The same anomaly appears for water against hydrogen sulphide and for ammonia against phosphine.

Question 11

Explain why $	ext{LiF}$ has a higher lattice enthalpy than $	ext{CsI}$, and why $	ext{CaO}$ exceeds both by a wide margin.

Accepted Answer

Lattice enthalpy is proportional to the charge product divided by the sum of the ionic radii.

$	ext{LiF}$ and $	ext{CsI}$ have the same charge product of $1$, but lithium and fluoride are both small while caesium and iodide are both large, so the denominator is much smaller for $	ext{LiF}$.

$	ext{CaO}$ pairs $2+$ with $2-$, giving a charge product of $4$ — a fourfold increase in the numerator.

Since charge enters as a product and radius only as a sum, the charge effect dominates, and $	ext{CaO}$ has by far the highest lattice enthalpy of the three.

Question 12

Draw the two principal resonance contributors of the nitrate ion and explain why all three $	ext{N}-	ext{O}$ bonds are identical in length.

Accepted Answer

Each contributor has one $	ext{N}=	ext{O}$ double bond and two $	ext{N}-	ext{O}^-$ single bonds, with nitrogen carrying a formal charge of $+1$ and two oxygens $-1$ each.

Three such structures exist, related by which oxygen carries the double bond, and all three are equivalent in energy.

The real ion is a single structure that is the average of all three, so each bond has a bond order of $\dfrac43$ and all three lengths are identical at $124$ pm.

Resonance forms are contributors to one structure, not species interconverting in equilibrium.

Practice — "Chemical Bonding and Molecular Structure"

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