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The Periodic Table - Periodic trends

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Effective Nuclear Charge (ZeffZ_{eff}): The net positive charge experienced by an electron in a multi-electron atom. It is calculated as Zeff=Z−SZ_{eff} = Z - S, where ZZ is the atomic number and SS is the shielding constant.

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Atomic Radius: Decreases across a period due to increasing ZeffZ_{eff} pulling electrons closer to the nucleus. It increases down a group because new principal energy levels (shells) are added, increasing the distance between the nucleus and valence electrons.

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First Ionization Energy (IE1IE_1): The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+1+ ions. Generally increases across a period and decreases down a group.

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Electronegativity: A measure of the tendency of an atom to attract a bonding pair of electrons. Values increase across a period and decrease down a group, with Fluorine being the most electronegative element (4.04.0 on the Pauling scale).

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Ionic Radius: Cations (X+X^+) are always smaller than their parent atoms because the loss of electrons reduces inter-electron repulsion and sometimes removes an entire outer shell. Anions (X−X^-) are larger than their parent atoms due to increased electron-electron repulsion.

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Electron Affinity: The energy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous 1−1- ions. It becomes more exothermic (more negative) across a period as the nucleus attracts incoming electrons more strongly.

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Period 3 Trends: Elements transition from metallic (NaNa, MgMg, AlAl) to metalloid (SiSi) to non-metallic (PP, SS, ClCl, ArAr). This is reflected in their melting points and electrical conductivity.

📐Formulae

X(g)→X(g)++e−(ΔH=IE1)X_{(g)} \rightarrow X^{+}_{(g)} + e^{-} \quad (\Delta H = IE_1),

X(g)+→X(g)2++e−(ΔH=IE2)X^{+}_{(g)} \rightarrow X^{2+}_{(g)} + e^{-} \quad (\Delta H = IE_2),

X(g)+e−→X(g)−(ΔH=EA1)X_{(g)} + e^{-} \rightarrow X^{-}_{(g)} \quad (\Delta H = EA_1),

Zeff≈Z−SZ_{eff} \approx Z - S

💡Examples

Problem 1:

Explain why the first ionization energy of Phosphorus (PP, Z=15Z=15) is higher than that of Sulfur (SS, Z=16Z=16), despite the general trend of increasing ionization energy across Period 3.

Solution:

The electronic configuration of PP is [Ne]3s23p3[Ne] 3s^2 3p^3 and SS is [Ne]3s23p4[Ne] 3s^2 3p^4.

Explanation:

In Phosphorus, the 3p3p subshell is exactly half-filled with three unpaired electrons, which provides extra stability. In Sulfur, the fourth 3p3p electron must pair up in an orbital already containing one electron. The repulsion between these two electrons in the same pp-orbital makes it easier to remove the electron from SS than from PP, resulting in a lower first ionization energy for Sulfur.

Problem 2:

Compare the ionic radii of Na+Na^{+}, Mg2+Mg^{2+}, and Al3+Al^{3+}.

Solution:

r(Na+)>r(Mg2+)>r(Al3+)r(Na^{+}) > r(Mg^{2+}) > r(Al^{3+})

Explanation:

These ions are isoelectronic, meaning they all have the same electron configuration (1s22s22p61s^2 2s^2 2p^6). However, the number of protons increases from NaNa (1111) to MgMg (1212) to AlAl (1313). The higher nuclear charge in Al3+Al^{3+} exerts a stronger electrostatic pull on the same number of electrons, resulting in a significantly smaller ionic radius.

Problem 3:

Predict the trend in melting points for the elements NaNa, MgMg, and AlAl.

Solution:

Al>Mg>NaAl > Mg > Na

Explanation:

These are all metallic elements. Moving from NaNa to AlAl, the number of delocalized valence electrons increases (Na=1,Mg=2,Al=3Na=1, Mg=2, Al=3) and the ionic charge increases while the ionic radius decreases. This results in a much stronger metallic bond (stronger electrostatic attraction between cations and the 'sea' of delocalized electrons), leading to higher melting points.

Periodic trends - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry