krit.club logo

Periodicity - The periodic table

Grade 11IBChemistry

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

🔑Concepts

•

The Periodic Table is arranged by increasing atomic number (ZZ). Horizontal rows are called periods and vertical columns are called groups.

•

Elements in the same group have the same number of valence electrons and share similar chemical properties. For example, Group 1 elements have a valence configuration of ns1ns^1.

•

Effective Nuclear Charge (ZeffZ_{eff}): This is the net positive charge experienced by valence electrons. It increases across a period because the number of protons increases while the inner-shell shielding remains relatively constant.

•

Atomic Radius: Decreases across a period (due to increasing ZeffZ_{eff} pulling electrons closer) and increases down a group (due to the addition of new principal energy levels, nn).

•

Ionic Radius: Cations (Mn+M^{n+}) are always smaller than their parent atoms because they lose electrons and sometimes an entire outer shell. Anions (Xn−X^{n-}) are always larger than their parent atoms due to increased electron-electron repulsion.

•

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

•

Electronegativity: A measure of the ability of an atom at attract a bonding pair of electrons in a covalent bond. It increases across a period and decreases down a group. Fluorine (FF) is the most electronegative element.

•

Electron Affinity: The energy change when one mole of electrons is added to one mole of gaseous atoms. It generally becomes more exothermic (more negative) across a period.

•

Metallic Character: Decreases across a period and increases down a group. Metals generally have low IEIE and low electronegativity, tending to form cations.

•

Oxide Periodicity: Across Period 3, oxides transition from basic (Na2ONa_2O, MgOMgO), to amphoteric (Al2O3Al_2O_3), to acidic (SiO2SiO_2, P4O10P_4O_{10}, SO3SO_3, Cl2O7Cl_2O_7).

📐Formulae

X(g)→X+(g)+e−(First Ionization Energy)X(g) \rightarrow X^+(g) + e^- \quad (\text{First Ionization Energy})

X(g)+e−→X−(g)(First Electron Affinity)X(g) + e^- \rightarrow X^-(g) \quad (\text{First Electron Affinity})

Zeff=Z−S(where S is the shielding constant)Z_{eff} = Z - S \quad (\text{where } S \text{ is the shielding constant})

2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)(Alkali Metal + Water)2Li(s) + 2H_2O(l) \rightarrow 2LiOH(aq) + H_2(g) \quad (\text{Alkali Metal + Water})

Cl2(aq)+2KI(aq)→2KCl(aq)+I2(aq)(Halogen Displacement)Cl_2(aq) + 2KI(aq) \rightarrow 2KCl(aq) + I_2(aq) \quad (\text{Halogen Displacement})

💡Examples

Problem 1:

Explain why the first ionization energy of Sulfur (SS) is lower than that of Phosphorus (PP), despite Sulfur being further to the right in Period 3.

Solution:

Phosphorus has an electron configuration of [Ne]3s23p3[Ne] 3s^2 3p^3, where the 3p3p subshell is half-filled (one electron in each orbital). Sulfur has a configuration of [Ne]3s23p4[Ne] 3s^2 3p^4.

Explanation:

In Sulfur, the fourth 3p3p electron must pair up with another electron in a 3p3p orbital. This creates inter-electron repulsion, which makes it easier to remove the electron compared to the stable, half-filled subshell of Phosphorus. Therefore, the IE1IE_1 of SS is lower.

Problem 2:

Arrange the following species in order of increasing radius: Mg2+Mg^{2+}, F−F^-, Na+Na^+.

Solution:

Mg2+<Na+<F−Mg^{2+} < Na^+ < F^-

Explanation:

These are isoelectronic species, meaning they all have the same number of electrons (1010 electrons, configuration 1s22s22p61s^2 2s^2 2p^6). The radius depends on the number of protons (ZZ). Mg2+Mg^{2+} has 1212 protons, Na+Na^+ has 1111, and F−F^- has 99. The higher the atomic number, the stronger the nuclear pull on the same number of electrons, resulting in a smaller radius.

Problem 3:

Predict the acid-base nature of the product formed when P4O10P_4O_{10} reacts with water and write the equation.

Solution:

The product is H3PO4H_3PO_4 (phosphoric acid), which is acidic. Equation: P4O10(s)+6H2O(l)→4H3PO4(aq)P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq).

Explanation:

Non-metal oxides are generally acidic. When they react with water, they form oxyacids. P4O10P_4O_{10} is a typical acidic oxide from the right side of Period 3.