Classification of Elements and Periodicity

Fluorine is the most electronegative element there is. So it must release the most energy on gaining an electron. Which halogen actually has the most negative electron gain enthalpy?

Chlorine, at $-349$ kJ mol${}^{-1}$, against fluorine's $-328$.

Fluorine's $2p$ subshell is tiny. Packing an extra electron into that compact shell forces it into a region already crowded with seven others, and the electron-electron repulsion partly cancels the nuclear attraction. Chlorine's $3p$ shell is roomier, so the same electron arrives with less of a repulsion penalty.

That is the shape of every question in this chapter. The smooth trends are easy and rarely asked. What Advanced examines is the exceptions — and each one is a place where electron-electron repulsion, or exchange energy, or a poorly shielding inner shell, beats the smooth pull of the nucleus.

1. Effective nuclear charge does almost all the work

Across a period, each added proton increases $Z$ by one while each added electron enters the same shell and screens only about $0.35$ of it. The net pull therefore rises steadily:

$Z_{eff}\text{rises across a period},Z_{eff}\text{is nearly constant down a group}$

Down a group the outer electron enters a new shell, and each complete inner shell screens almost perfectly, so $Z_{eff}$ barely moves while $n$ jumps. Size is then decided by $n$.

Everything else follows: atoms shrink across a period and grow down a group; ionisation enthalpy rises across and falls down; electronegativity does the same. Learn the cause and the trends need no memorising.

Illustration 1

Explain why the radius falls from sodium ($186$ pm) to chlorine ($99$ pm) but rises from fluorine ($64$ pm) to iodine ($133$ pm).

Across period 3, eight protons are added while the added electrons all enter $n=3$ and screen poorly, so $Z_{eff}$ climbs from about $2.2$ to $6.1$ and the shell is pulled in.

Down group 17, each step adds a full new shell. The inner shells screen almost completely, so $Z_{eff}$ stays near $6$ while $n$ goes from $2$ to $5$.

One variable rises, the other jumps. Which of $Z_{eff}$ and $n$ is changing tells you the direction of every size trend without recalling a single number.

2. Radii: four kinds, and isoelectronic series

The word "radius" means four different measured quantities, and questions exploit the difference.

Van der Waals radii exceed covalent radii substantially, which is why noble gases appear anomalously large in tables that quote them.

For an isoelectronic series — species with the same electron count — size is decided entirely by nuclear charge:

${\text{N}}^{3-}>{\text{O}}^{2-}>{\text{F}}^{-}>\text{Ne}>{\text{Na}}^{+}>{\text{Mg}}^{2+}>{\text{Al}}^{3+}$

All have ten electrons; the one with the most protons holds them tightest.

Illustration 2

Arrange ${\text{O}}^{2-}$, ${\text{F}}^{-}$, ${\text{Na}}^{+}$ and ${\text{Mg}}^{2+}$ in order of increasing radius, and explain.

All four have ten electrons, so compare nuclear charges: $8<9<11<12$.

${\text{Mg}}^{2+}<{\text{Na}}^{+}<{\text{F}}^{-}<{\text{O}}^{2-}$

The anion is always bigger, the cation always smaller. Removing an electron reduces repulsion and often empties a whole shell, while adding one increases repulsion at unchanged nuclear charge.

Illustration 3

A table lists the atomic radius of neon as $160$ pm and that of fluorine as $64$ pm. Explain the apparent contradiction.

The two numbers are not the same quantity. Fluorine's is a covalent radius, half the bond length in ${\text{F}}_2$.

Neon forms no bonds, so only a van der Waals radius can be quoted, measured from the closest non-bonded approach of two atoms — always much larger.

Comparing them directly is meaningless. Fluorine's own van der Waals radius is about $147$ pm, which is properly smaller than neon's.

Always check which radius a table is quoting. Noble gases look anomalously large in every periodic trend for exactly this reason, and the anomaly is an artefact of the definition rather than of the chemistry.

3. Ionisation enthalpy and its four anomalies

Ionisation enthalpy rises across a period and falls down a group, but four well-known reversals interrupt it — and each has a specific cause.

Beryllium exceeds boron. Beryllium loses a $2s$ electron; boron loses a $2p$, which is higher in energy and less penetrating.

Nitrogen exceeds oxygen. Nitrogen's $2p^3$ is half-filled and carries maximum exchange energy; oxygen's $2p^4$ must place two electrons in one orbital, and the pairing repulsion makes that fourth electron easier to remove.

The same two patterns repeat as magnesium exceeding aluminium and phosphorus exceeding sulphur in period 3.

Successive ionisation enthalpies always increase, and the position of the large jump identifies the group. Sodium's second ionisation enthalpy is nine times its first, because the second electron must come from the neon core.

Illustration 4

The successive ionisation enthalpies of an element are $738$, $1451$, $7733$ and $10540$ kJ mol${}^{-1}$. Identify the group.

The jump between the second and third values is a factor of $5.3$, far larger than any other step.

So two electrons are removed easily and the third comes from a noble-gas core: the element belongs to group 2.

It is magnesium. The size of the jump, not the absolute values, is what carries the information, which is why this question type can be answered without knowing a single tabulated number.

Illustration 5

Explain why the first ionisation enthalpy of aluminium ($577$) is lower than that of magnesium ($738$), and why gallium's ($579$) is almost identical to aluminium's despite being lower in the group.

Magnesium loses a $3s$ electron from a filled subshell; aluminium loses a $3p$, which is higher and shielded by the $3s$ pair.

Gallium follows the ten $3d$ elements. The $3d$ electrons shield poorly, so $Z_{eff}$ on gallium's $4p$ electron is unusually high and cancels the expected drop with increasing $n$.

Poor $d$ shielding is a recurring theme. It also explains the small size of gallium relative to aluminium and reappears in the lanthanoid contraction below.

4. Electron gain enthalpy

The first electron gain enthalpy is usually negative, but the second is always positive, because an electron must now be forced onto a species that is already negative.

$\text{O}\left(g\right)+e^{-}\to {\text{O}}^{-}\left(g\right):-141{\text{kJ mol}}^{-1}$ ${\text{O}}^{-}\left(g\right)+e^{-}\to {\text{O}}^{2-}\left(g\right):+780{\text{kJ mol}}^{-1}$

which is why oxide ions exist only in lattices, where the lattice energy pays that cost.

Three patterns are examined. Chlorine beats fluorine, and sulphur beats oxygen, for the compactness reason in the hook. Noble gases have positive values, since the electron must enter a new shell. And group 2 elements have near-zero or positive values, because their $n{s}^2$ subshell is already full.

Illustration 6

Explain why the electron gain enthalpy of sulphur is more negative than that of oxygen, but that of selenium is less negative than sulphur's.

Oxygen's $2p$ shell is compact, so the incoming electron suffers heavy repulsion. Sulphur's $3p$ is roomier and takes the electron more comfortably.

Beyond sulphur, size continues to increase but the compactness penalty is already gone, so the ordinary trend takes over: the incoming electron is further from the nucleus and less strongly bound.

The anomaly is confined to the second period. Once past it, the smooth trend resumes, which is exactly why the second-period elements need separate treatment.

5. Electronegativity: three scales, three meanings

Electronegativity is not measurable directly, so it is defined operationally, and the three standard scales measure different things.

Mulliken's is the most transparent: an atom that both holds its own electrons tightly and attracts others strongly is electronegative. All three agree on the ordering even though the numbers differ.

Electronegativity also depends on hybridisation and oxidation state: an $sp$ carbon is more electronegative than an $s{p}^3$ carbon, because the $s$ orbital penetrates closer to the nucleus.

Illustration 7

An element has an ionisation energy of $17.4$ eV and an electron affinity of $3.6$ eV. Find its Mulliken electronegativity and comment.

${\chi }_M=\frac{17.4+3.6}{2}=10.5$ eV

Dividing by the standard factor of about $2.8$ to compare with Pauling values gives roughly $3.7$.

That is near the top of the scale, consistent with a highly electronegative element such as oxygen or fluorine. The two atomic properties reinforce rather than oppose each other, which is the signature of electronegativity.

Illustration 8

Explain why ethyne is far more acidic than ethene or ethane, in terms of electronegativity.

The carbon in ethyne is $sp$ hybridised, with $50\%$ $s$ character; in ethene $s{p}^2$ with $33\%$; in ethane $s{p}^3$ with $25\%$.

More $s$ character means the electron pair sits closer to the nucleus, so the carbon is more electronegative and holds the resulting carbanion's lone pair more comfortably.

$p{K}_a$ values run about $25$, $44$ and $50$ respectively.

Electronegativity is not a fixed property of an element. It depends on the hybridisation and the oxidation state, which is why the same carbon atom behaves quite differently in three hydrocarbons.

6. Lanthanoid contraction and the second-row twins

Filling the $4f$ subshell across the lanthanoids adds fourteen protons while the $4f$ electrons screen very poorly. The result is a steady contraction of about $10$ pm from lanthanum to lutetium, and it has a striking consequence for the elements that follow.

Zirconium and hafnium have essentially identical radii and therefore nearly identical chemistry, which is why they are among the hardest pairs of elements to separate. The same holds for niobium and tantalum, and for molybdenum and tungsten.

The contraction also explains why the third transition series has higher ionisation enthalpies than the second, reversing the usual downward trend.

Illustration 9

Give three consequences of the lanthanoid contraction.

Zirconium and hafnium are nearly identical in size, so they occur together and are separated only with difficulty by solvent extraction or ion exchange.

Third-row transition metals have higher ionisation enthalpies and densities than their second-row partners, reversing the trend seen elsewhere in the table.

The lanthanoids themselves become progressively harder to separate, since their ionic radii differ by only about one picometre per element.

All three trace to the same cause: fourteen protons added while the $4f$ electrons doing the screening are diffuse and ineffective at it.

7. Diagonal relationships, the inert pair, and second-period anomalies

Diagonal relationships arise because moving right increases $Z_{eff}$ while moving down increases size, and the two effects can cancel. Three pairs behave remarkably alike:

• Li and Mg: both form nitrides directly, both give hydroxides and carbonates that decompose on heating, both form covalent organometallics
• Be and Al: both amphoteric oxides, both covalent chlorides that are Lewis acids, both passivated by nitric acid
• B and Si: both form volatile hydrides that hydrolyse, both give acidic oxides

The inert pair effect appears down groups 13 to 15: the $n{s}^2$ pair becomes increasingly reluctant to participate in bonding, so the lower oxidation state grows more stable. Thallium(I) is more stable than thallium(III), lead(II) than lead(IV), bismuth(III) than bismuth(V).

Second-period elements differ from their groups for three reasons: small size, high electronegativity, and no available $d$ orbitals. The last explains why nitrogen cannot form ${\text{NF}}_5$ while phosphorus forms ${\text{PF}}_5$ readily, and why oxygen cannot expand its octet.

Illustration 10

Explain why lithium resembles magnesium more than it resembles sodium.

Lithium's small size gives it an unusually high charge density, comparable with that of the doubly charged but larger magnesium ion.

Both therefore polarise anions strongly and form more covalent compounds than their group neighbours.

Both form nitrides directly with atmospheric nitrogen, both give carbonates that decompose on heating, and both form insoluble fluorides and carbonates — none of which sodium does.

Charge density, not charge, is the controlling quantity. That is what makes the diagonal cancellation work.

Illustration 11

Predict which is the more stable oxidation state and explain: thallium(I) or thallium(III); aluminium(I) or aluminium(III).

Thallium(I) is more stable. The $6s^2$ pair is strongly stabilised by relativistic contraction and by poor $4f$ and $5d$ shielding, so it resists participating in bonding.

Aluminium(III) is more stable. Aluminium is high in the group, where the inert pair effect has not yet set in.

The effect strengthens down a group, which is why group 13 runs from an exclusively trivalent aluminium to a predominantly monovalent thallium.

Illustration 12

Explain why nitrogen forms ${\text{NF}}_3$ but not ${\text{NF}}_5$, while phosphorus forms both ${\text{PF}}_3$ and ${\text{PF}}_5$.

Nitrogen is a second-period element with only $2s$ and $2p$ orbitals available, giving a maximum covalence of four and no way to accommodate ten bonding electrons.

Phosphorus has energetically accessible $3d$ orbitals, allowing expansion beyond the octet.

The same restriction explains why oxygen cannot form ${\text{OF}}_6$, and why fluorine has no positive oxidation state at all — there is nowhere for the extra pairs to go.

Illustration 13

Arrange ${\text{Li}}^{+}$, ${\text{Be}}^{2+}$, ${\text{Na}}^{+}$ and ${\text{Mg}}^{2+}$ by polarising power and predict which forms the most covalent chloride.

Polarising power scales as charge divided by radius squared, so higher charge and smaller size both help.

${\text{Be}}^{2+}$ is both doubly charged and the smallest: it is by far the strongest polariser.

Order: ${\text{Na}}^{+}<{\text{Li}}^{+}<{\text{Mg}}^{2+}<{\text{Be}}^{2+}$

${\text{BeCl}}_2$ is the most covalent, which is why it is a polymeric chain solid that sublimes rather than an ionic lattice.

Summary

• Almost every trend reduces to $Z_{eff}$: it rises across a period and is nearly constant down a group, where $n$ takes over.
• Four radius types differ; van der Waals radii are much the largest, which makes noble gases look anomalous in tables.
• Isoelectronic series: same electrons, so more protons means smaller. ${\text{N}}^{3-}>{\text{O}}^{2-}>{\text{F}}^{-}>{\text{Na}}^{+}>{\text{Mg}}^{2+}>{\text{Al}}^{3+}$.
• Ionisation enthalpy anomalies: Be > B and Mg > Al (removing $s$ versus $p$); N > O and P > S (half-filled stability versus pairing repulsion).
• The large jump in successive ionisation enthalpies identifies the group; sodium's second value is nine times its first.
• Poor $3d$ shielding makes gallium's ionisation enthalpy match aluminium's despite the extra shell.
• Chlorine beats fluorine and sulphur beats oxygen on electron gain enthalpy: the second-period shells are too compact.
• Second electron gain enthalpy is always positive; oxide ions survive only because lattice energy pays for them.
• Three electronegativity scales: Pauling from bond energies, Mulliken as $\frac{IE+EA}{2}$, Allred-Rochow as $\frac{Z_{eff}}{r^2}$. Electronegativity also rises with $s$ character.
• Lanthanoid contraction makes Zr and Hf, Nb and Ta, Mo and W nearly identical in size and hard to separate.
• Covalent, metallic, van der Waals and ionic radii are different quantities — never compare across types.
• Diagonal relationships (Li-Mg, Be-Al, B-Si) come from $Z_{eff}$ and size effects cancelling.
• Inert pair effect strengthens down groups 13 to 15: Tl(I) over Tl(III), Pb(II) over Pb(IV), Bi(III) over Bi(V).
• Second-period elements have no $d$ orbitals, so nitrogen forms ${\text{NF}}_3$ but never ${\text{NF}}_5$.