Once the elements sit in their right places, their measurable properties — size, the energy to pull an electron off, the energy released when one is added — vary in smooth, predictable waves. Two ideas explain almost all of it: the effective nuclear charge ($Z_{eff}$) felt by an outer electron, and how strongly inner electrons screen (shield) that pull.
Effective nuclear charge and screening
An outer electron does not feel the full nuclear charge $Z$; inner electrons repel it and partially cancel the attraction. The net charge it experiences is $Z_{eff} = Z - S$, where $S$ is the screening constant. Across a period electrons are added to the same shell, which screens poorly, so $Z_{eff}$ rises sharply. Down a group a new, larger shell is added, so the outer electron is both farther out and better screened.
Atomic radius
Atomic radius (covalent or metallic) decreases across a period because rising $Z_{eff}$ pulls the same shell inward, and increases down a group because each element adds a shell. For example, in period 2 the covalent radius falls from Li ($152\,\text{pm}$) to F ($72\,\text{pm}$); in group 1 it grows from Li to Cs.
Ionic radius and isoelectronic species
A cation is smaller than its parent atom (it loses a shell and the remaining electrons feel a higher $Z_{eff}$); an anion is larger (added electrons increase repulsion). Isoelectronic species have the same number of electrons but different nuclear charges. Among them, the larger the nuclear charge $Z$, the smaller the ion. For the isoelectronic set $\text{O}^{2-}, \text{F}^-, \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}$ (all 10 electrons), size order is $\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$.
Ionisation enthalpy
The first ionisation enthalpy $\Delta_i H_1$ is the energy needed to remove the most loosely bound electron from one mole of gaseous atoms: $\text{X}(g) \rightarrow \text{X}^+(g) + e^-$. It increases across a period (higher $Z_{eff}$, smaller size, harder to remove) and decreases down a group (outer electron farther away and well screened).
Irregularities: in period 2, B ($ns^2np^1$) has a lower $\Delta_i H_1$ than Be ($ns^2$) because removing the single $p$ electron is easier than breaking a stable filled $s$ subshell. Similarly O ($2p^4$) has a lower value than N ($2p^3$): nitrogen's half-filled $2p^3$ is extra stable, and oxygen must lose an electron from a paired orbital, which has electron–electron repulsion. Successive ionisation enthalpies always increase ($\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3$) because removing an electron from an increasingly positive ion is harder.
Electron gain enthalpy
The electron gain enthalpy $\Delta_{eg} H$ is the energy change when an electron is added to a gaseous atom: $\text{X}(g) + e^- \rightarrow \text{X}^-(g)$. It usually becomes more negative across a period (atoms more eager for an electron) and less negative down a group. Halogens have the most negative values. An odd twist: chlorine's $\Delta_{eg} H$ ($-349\,\text{kJ mol}^{-1}$) is more negative than fluorine's ($-328\,\text{kJ mol}^{-1}$), because F is so small that adding an electron causes strong repulsion in its compact $2p$ shell.
Electronegativity
Electronegativity is the tendency of an atom in a bond to attract the shared electron pair. It is not directly measurable (it relates to but is not the same as electron gain enthalpy). On the Pauling scale it increases across a period and decreases down a group; fluorine ($4.0$) is the most electronegative element. Electronegativity is a useful guide to bond polarity and the metallic/non-metallic character of elements.
Arrange the isoelectronic species $\text{N}^{3-}, \text{O}^{2-}, \text{F}^-, \text{Na}^+, \text{Mg}^{2+}$ in increasing order of ionic radius.
Solution- All five species have 10 electrons, so they are isoelectronic.
- For isoelectronic species, a higher nuclear charge $Z$ pulls the electron cloud in more tightly, giving a smaller radius.
- List the nuclear charges: $\text{N}^{3-}$ ($Z=7$), $\text{O}^{2-}$ ($Z=8$), $\text{F}^-$ ($Z=9$), $\text{Na}^+$ ($Z=11$), $\text{Mg}^{2+}$ ($Z=12$).
- Smallest radius goes with largest $Z$.
Answer: $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-} < \text{N}^{3-}$.
Explain why the first ionisation enthalpy of boron is lower than that of beryllium.
Solution- Be has configuration $1s^2\,2s^2$ — a completely filled, stable $2s$ subshell.
- B has configuration $1s^2\,2s^2\,2p^1$; its outermost electron is a single $2p$ electron.
- The $2p$ electron is at slightly higher energy and is shielded by the $2s$ pair, so it is easier to remove.
- Removing it does not disturb a stable filled subshell, unlike in Be.
Answer: The loosely held, well-shielded $2p^1$ electron of boron is easier to remove than an electron from beryllium's stable filled $2s^2$, so $\Delta_i H_1(\text{B}) < \Delta_i H_1(\text{Be})$.
Why is the first ionisation enthalpy of nitrogen greater than that of oxygen?
Solution- N has configuration $1s^2\,2s^2\,2p^3$ — a half-filled $2p$ subshell, which is extra stable.
- O has configuration $1s^2\,2s^2\,2p^4$ — one $2p$ orbital now holds a pair.
- The paired electrons in O experience extra electron–electron repulsion, making one easier to remove.
- Removing an electron from N would break its stable half-filled arrangement, which costs more energy.
Answer: Nitrogen's stable half-filled $2p^3$ and oxygen's inter-electronic repulsion in a paired orbital make $\Delta_i H_1(\text{N}) > \Delta_i H_1(\text{O})$.
Account for the fact that the electron gain enthalpy of chlorine is more negative than that of fluorine.
Solution- Both gain an electron to complete their valence $p$ subshell.
- Fluorine's valence shell ($n=2$) is very small and compact.
- Adding an electron to this small shell causes strong electron–electron repulsion, releasing less energy than expected.
- Chlorine's larger $n=3$ shell accommodates the new electron with less repulsion, so more energy is released.
Answer: Because of high electron repulsion in F's small $2p$ shell, $\Delta_{eg} H$(Cl) $= -349\,\text{kJ mol}^{-1}$ is more negative than $\Delta_{eg} H$(F) $= -328\,\text{kJ mol}^{-1}$.
The first three ionisation enthalpies of Mg are $738$, $1451$ and $7733\,\text{kJ mol}^{-1}$. Explain the large jump after the second.
Solution- Mg has configuration $1s^2\,2s^2\,2p^6\,3s^2$; its two valence electrons are in $3s$.
- The first two electrons removed are the $3s$ electrons, giving the stable $\text{Mg}^{2+}$ noble-gas core.
- The third electron must come from the inner $2p^6$ shell, which is much closer to the nucleus and feels a far higher $Z_{eff}$.
- Removing a core electron requires far more energy, producing the sharp jump to $7733\,\text{kJ mol}^{-1}$.
Answer: The big jump after $\Delta_i H_2$ confirms Mg has only two valence electrons; the third comes from a tightly held inner shell.
Why does atomic radius decrease across period 2 from Li to F, while it increases down group 1 from Li to Cs?
Solution- Across a period, electrons enter the same shell while nuclear charge $Z$ increases.
- Screening by same-shell electrons is poor, so $Z_{eff}$ rises and the shell is pulled inward, shrinking the radius.
- Down a group, each element adds a new principal shell, so the outer electron is farther from the nucleus.
- The added inner shells also shield the outer electron, reducing the effective pull.
Answer: Rising $Z_{eff}$ contracts the atom across a period, whereas the addition of new shells (with greater shielding) expands it down a group.