Corrected exercises: chemical bonding and molecular geometry (202-SN1-RE)

Nonpolar covalent if $\Delta EN<0.4$; polar covalent if $0.4\leq\Delta EN<1.7$; ionic if $\Delta EN\geq 1.7$. These thresholds are conventions: ionic character grows without a jump.

$Na-Cl$: $\Delta EN=|0.93-3.16|=2.23\geq 1.7$, ionic bond. $O-H$: $|3.44-2.20|=1.24$, polar covalent. $C-H$: $|2.55-2.20|=0.35<0.4$, nonpolar covalent. $N-H$: $|3.04-2.20|=0.84$, polar covalent.

Oxygen is more electronegative than hydrogen (3.44 versus 2.20), so it pulls the bonding pair more strongly and carries the partial negative charge $\delta^{-}$; hydrogen carries $\delta^{+}$.

Sodium $[Ne]3s^{1}$ loses its single valence electron and becomes $Na^{+}$ with configuration $[Ne]$; chlorine $[Ne]3s^{2}3p^{5}$ gains one and becomes $Cl^{-}$ with configuration $[Ar]$. Both ions reach a noble-gas configuration, and it is the complete transfer of an electron that distinguishes ionic bonding from covalent bonding. For $MgCl_{2}$: magnesium gives one electron to each of the two chlorines and becomes $Mg^{2+}$ with no pair at all; each $Cl^{-}$ ion carries four lone pairs and a charge of $-1$. The Lewis structure is therefore written as an $Mg^{2+}$ ion between two independent chloride ions, with no bond line: in an ionic compound there is no shared pair.

Since the charges are identical ($+1$ and $-1$), only the distance matters, and the lattice energy varies as $\dfrac{1}{r}$: the smaller and closer the ions, the stronger the crystal. Increasing distances: $NaF$ (231 pm) < $NaCl$ (276 pm) < $KCl$ (314 pm), hence decreasing energies $NaF>NaCl>KCl$. Checking the ratios: we predict $\dfrac{E(NaF)}{E(NaCl)}=\dfrac{276}{231}=1.19$ and measure $\dfrac{923}{788}=1.17$; we predict $\dfrac{E(KCl)}{E(NaCl)}=\dfrac{276}{314}=0.879$ and measure $\dfrac{699}{788}=0.887$. The agreement is excellent.

In $MgO$ the charges are $+2$ and $-2$, so the product $Q_{1}Q_{2}$ is four times larger than in $NaCl$. The predicted ratio is $\dfrac{4}{1}\times\dfrac{276}{212}=5.21$, and the real ratio is $\dfrac{3795}{788}=4.82$: the order of magnitude is right. The charge dominates by far. Going from single to double charges multiplies the energy by 4, whereas typical distances vary by only a few tens of percent. The practical rule is therefore to look at the charges first and to use the radii to break ties only when the charges are equal, as in b).

Melting an ionic crystal means overcoming the attraction between the ions: the melting point follows the lattice energy. Since that of $MgO$ is almost five times larger, it is no surprise that it takes more than 2000 °C more to melt it. As for the paradox, it is resolved by the complete energy bookkeeping. Ionizing magnesium, vaporizing the reactants and forcing two electrons onto oxygen do cost a great deal of energy, but the lattice energy released when the crystal forms, 3795 kJ/mol, exceeds the sum of all these costs. This is exactly the reasoning of the Born-Haber cycle: the stability of $MgO$ does not come from the ions taken in isolation, which are unfavourable, but from their packing into a lattice. And it is precisely because the charges are double that this gain is large enough to compensate for everything.

$CO_{2}$: $4+2(6)=16$ valence electrons. $H_{2}O$: $2(1)+6=8$. $NO_{3}^{-}$: $5+3(6)+1=24$ (one electron is added for the negative charge).

$CO_{2}$: the structure is $O=C=O$, two double bonds; the central carbon is surrounded by $2\times 4=8$ bonding electrons (octet satisfied, no lone pair); each oxygen carries two lone pairs and shares a double bond, that is $4+4=8$ electrons (octet satisfied). Total: $16$ electrons, consistent with a).

Central oxygen (one double bond, one single bond, one lone pair): $FC=6-2-\tfrac{6}{2}=+1$. Doubly bonded oxygen (two lone pairs): $FC=6-4-\tfrac{4}{2}=0$. Singly bonded oxygen (three lone pairs): $FC=6-6-\tfrac{2}{2}=-1$. Sum: $(+1)+0+(-1)=0$, which is indeed consistent with a neutral molecule.

In $NO_{3}^{-}$, nitrogen forms a double bond with one of the three oxygens and single bonds with the other two, but nothing physically distinguishes the doubly bonded oxygen: it can be placed on any of the three. These three equivalent structures are resonance forms, and the real ion is their hybrid, with three identical N–O bonds (intermediate between single and double).

$BeCl_{2}$: $2+2(7)=16$ valence electrons; with two single bonds, beryllium is surrounded by only 4 electrons. $BF_{3}$: $3+3(7)=24$ electrons; boron is surrounded by only 6 electrons. These are incomplete octets, and they are real: these molecules exist and are very reactive, precisely because they lack electrons. One might think of a double bond $B=F$ to give boron an octet, but the formal charge calculation rules it out: boron would then be surrounded by 8 bonding electrons, $FC=3-0-\tfrac{8}{2}=-1$, and the doubly bonded fluorine would carry $FC=7-4-\tfrac{4}{2}=+1$. A positive charge would thus be placed on fluorine, the most electronegative element in the periodic table: this is chemically absurd. The incomplete-octet structure, in which all formal charges are zero, wins.

$PCl_{5}$: $5+5(7)=40$ valence electrons, and phosphorus is surrounded by 10 electrons. $SF_{6}$: $6+6(7)=48$ electrons, and sulfur is surrounded by 12. These are expanded octets, reserved for elements of the third period and beyond. Nitrogen cannot do this for two reasons that point the same way: it is small, and five fluorine atoms could not fit around it without crowding; above all, its valence shell $n=2$ has only the $2s$ and $2p$ subshells, that is four orbitals and at most eight electrons, with no $d$ subshell available. Phosphorus, in $n=3$, has $3d$ orbitals and has no such limit. This is why $NF_{3}$ exists but not $NF_{5}$, whereas $PF_{3}$ and $PF_{5}$ both exist.

$NO$: $5+6=11$ valence electrons; $NO_{2}$: $5+2(6)=17$. These numbers are odd, so it is impossible to pair all the electrons: each molecule is left with an unpaired electron. These are free radicals, paramagnetic and very reactive. Direct consequence: they tend to dimerize in order to pair this electron. Nitrogen dioxide dimerizes in this way to $N_{2}O_{4}$, which has $2(5)+4(6)=34$ valence electrons, an even number this time: all the electrons are paired, and $N_{2}O_{4}$ is diamagnetic and colourless where $NO_{2}$ is brown.

The boron of $BF_{3}$ has an empty orbital and an incomplete octet; the nitrogen of $NH_{3}$ has a lone pair. Nitrogen therefore supplies both electrons of the bond by itself: this is a coordinate covalent bond, also called a dative bond. Once formed, it is in every respect identical to an ordinary covalent bond; only its origin differs. The formal charges are calculated as usual: for boron, $FC=3-0-\tfrac{8}{2}=-1$; for nitrogen, $FC=5-0-\tfrac{8}{2}=+1$. The adduct is therefore written $F_{3}B^{-}-N^{+}H_{3}$, with an overall charge of zero. Boron now has its octet, and the behaviour of $BF_{3}$ in this reaction is exactly that of a Lewis acid, that is, a pair acceptor.

$CH_{4}$: 4 bonding pairs, $AX_{4}$, tetrahedral geometry, angle 109.5°. $NH_{3}$: 3 bonding pairs + 1 lone pair, $AX_{3}E$, trigonal pyramidal, angle ≈ 107°. $H_{2}O$: 2 bonding pairs + 2 lone pairs, $AX_{2}E_{2}$, bent, angle ≈ 104.5°. On the figure, the double strokes represent the lone pairs.

All three have four electron pairs around the central atom (tetrahedral basic arrangement). But a lone pair, closer to the nucleus, takes up more room than a bonding pair and pushes the neighbouring pairs away more strongly. $CH_{4}$ has no lone pair (109.5°); $NH_{3}$ has one, which squeezes the angle to ≈ 107°; $H_{2}O$ has two, hence ≈ 104.5°. The more lone pairs, the smaller the angle.

$CO_{2}$: $AX_{2}$, linear, 180°. $BF_{3}$: $AX_{3}$, trigonal planar, 120°. $SF_{6}$: $AX_{6}$, octahedral, 90°.

In $CO_{2}$ (linear, $O=C=O$), the two $C=O$ bonds are polar, but their dipole moments are opposite and of equal magnitude: they cancel, and the molecule is nonpolar. In $H_{2}O$ (bent), the two polar $O-H$ bonds are not aligned: their dipole moments partly add and give a polar molecule. It is the geometry, not the polarity of the bonds, that decides.

Nonpolar: $CO_{2}$ (symmetric linear) and $CCl_{4}$ (symmetric tetrahedral, the four $C-Cl$ dipoles cancel). Polar: $H_{2}O$ (bent) and $NH_{3}$ (pyramidal, the lone pair breaks the symmetry).

$CH_{4}$: nonpolar, so London (dispersion) forces only. $HCl$: polar molecule, dipole-dipole forces in addition to London forces. $H_{2}O$: hydrogen bonding (H bonded to O), the strongest, in addition to London forces.

$H_{2}O$ and $H_{2}S$ have the same bent shape, but oxygen is electronegative and small enough to form hydrogen bonds between water molecules, which sulfur does not allow. These hydrogen bonds, much stronger than the dipole-dipole forces of $H_{2}S$, make water boil at a much higher temperature despite its lower molar mass. Molar mass (hence London forces) cannot by itself explain boiling points once hydrogen bonding comes into play.

$SF_{4}$: $6+4(7)=34$ valence electrons. Four bonds use 8 electrons and the four fluorines carry $4\times 6=24$ nonbonding electrons, which leaves $34-8-24=2$ electrons, that is one lone pair on sulfur: five domains, notation $AX_{4}E$. $ClF_{3}$: $7+3(7)=28$ electrons; $28-6-18=4$, that is two lone pairs: $AX_{3}E_{2}$. $XeF_{2}$: $8+2(7)=22$ electrons; $22-4-12=6$, that is three lone pairs: $AX_{2}E_{3}$. All three molecules have five domains and therefore derive from the same bipyramid.

A lone pair takes up more room than a bonding pair, and the most costly repulsions are those at 90°, far greater than those at 120°. It is therefore enough to count the neighbours at 90°. In an axial position, the lone pair has the three equatorial domains at 90° from it: three strong repulsions. In an equatorial position, it has only the two axial domains at 90°; the other two equatorial domains are at 120°, hence much less bothersome: only two strong repulsions. Since $2<3$, the equatorial position is systematically preferred. The same count explains why, in an octahedron, where all positions are equivalent (four neighbours at 90° everywhere), the question does not arise for a first lone pair.

$SF_{4}$, $AX_{4}E$: the lone pair takes an equatorial position, leaving two axial fluorines and two equatorial fluorines, which gives the seesaw shape (or distorted tetrahedron). The axial-axial angle drops below 180°, to about 173°, and the equatorial-equatorial angle below 120°, to about 102°, both pushed by the lone pair. $ClF_{3}$, $AX_{3}E_{2}$: the two lone pairs occupy two equatorial positions, leaving one equatorial fluorine and two axial fluorines: the molecule is T-shaped, with angles close to 87° instead of 90°. $XeF_{2}$, $AX_{2}E_{3}$: the three lone pairs take the whole equator and the two fluorines remain axial, at 180° from each other: the molecule is perfectly linear.

$BrF_{5}$: $7+5(7)=42$ electrons; $42-10-30=2$, one lone pair, notation $AX_{5}E$, square pyramidal geometry, the four fluorines of the base being pushed slightly toward the apical fluorine (angles close to 85°). $XeF_{4}$: $8+4(7)=36$ electrons; $36-8-24=4$, two lone pairs, notation $AX_{4}E_{2}$, square planar geometry. The two lone pairs sit 180° apart because this is the only way, in an octahedron, to keep them as far apart as possible: side by side they would be at 90°, which is the most repulsive arrangement possible between two lone pairs. Polarity: $PCl_{5}$ (complete, symmetric bipyramid), $XeF_{2}$ (symmetric linear) and $XeF_{4}$ (symmetric square planar) are nonpolar, the bond dipole moments cancelling in pairs. $SF_{4}$ (seesaw) and $BrF_{5}$ (square pyramid) are polar, the lone pair breaking the symmetry. The general rule applies: it is lone pairs that nothing compensates by symmetry, not their number, that make a molecule polar ($XeF_{2}$, with three lone pairs, stays nonpolar).

A purely ionic bond means a complete transfer, so $Q=e=1.602\times 10^{-19}$ C, separated by $d=127.5$ pm $=1.275\times 10^{-10}$ m. Then $\mu_{ionic}=(1.602\times 10^{-19})(1.275\times 10^{-10})=2.043\times 10^{-29}$ C·m. In debyes: $\dfrac{2.043\times 10^{-29}}{3.336\times 10^{-30}}=6.12$ D.

The measured moment is only 1.08 D. The percent ionic character is $\dfrac{\mu_{measured}}{\mu_{ionic}}\times 100=\dfrac{1.08}{6.12}\times 100=17.6\%$. Since the moment is proportional to the charge, this proportion is directly that of the charge transferred: $\delta=0.176\,e$, or about $2.8\times 10^{-20}$ C. In other words, chlorine has not taken a whole electron but roughly a sixth of one. This is the physical meaning of $\delta^{+}$ and $\delta^{-}$, and the calculation gives it a numerical value.

$HF$: $\mu_{ionic}=(1.602\times 10^{-19})(9.17\times 10^{-11})=1.469\times 10^{-29}$ C·m $=4.40$ D, hence $\dfrac{1.82}{4.40}\times 100=41.3\%$. $HBr$: $\mu_{ionic}=2.265\times 10^{-29}$ C·m $=6.79$ D, hence $\dfrac{0.82}{6.79}\times 100=12.1\%$. $HI$: $\mu_{ionic}=2.578\times 10^{-29}$ C·m $=7.73$ D, hence $\dfrac{0.44}{7.73}\times 100=5.7\%$. Decreasing order: $HF$ 41.3% > $HCl$ 17.6% > $HBr$ 12.1% > $HI$ 5.7%. Note that the denominator grows with the size of the halogen while the numerator shrinks: the two effects add up, hence the very marked decrease.

The electronegativity differences are $HF$ 1.78; $HCl$ 0.96; $HBr$ 0.76; $HI$ 0.46. The ranking is exactly the same as that of the percentages, and the correlation is clear: the larger the electronegativity difference, the larger the charge actually displaced. This is what justifies the use of $\Delta EN$ as an indicator. On the other hand, the cutoff of 1.7 corresponds to no measurable break. The convention simply comes from the empirical curve below: $\Delta EN=1.7$ is the difference at which the ionic character reaches 50%, and the custom has been to call ionic anything beyond half. $HF$, with $\Delta EN=1.78$, exceeds this cutoff and should be called ionic, yet its bond is only 41% ionic: it is a gaseous, polar covalent molecule, not an ionic solid. The important conclusion is that there is no sharp boundary between covalent and ionic, but a continuum. The tabulated cutoffs are practical landmarks, not laws, and the percent ionic character calculated from the dipole moment is the only experimental measure of this gradation.