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The Periodic Table - Group 17 (Halogens)

Grade 12A LevelChemistry

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

🔑Concepts

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The Halogens are the elements in Group 17 of the Periodic Table, consisting of FF, ClCl, BrBr, II, and AtAt.

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They are non-metals that exist as diatomic molecules, represented by the general formula X2X_2.

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Down the group, the melting and boiling points increase because the molecules become larger, leading to stronger London dispersion forces (Van der Waals forces).

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Physical states at room temperature (25∘C25^\circ C): F2F_2 and Cl2Cl_2 are gases, Br2Br_2 is a liquid, and I2I_2 is a solid.

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The atomic radius increases down the group as more electron shells are added.

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Electronegativity decreases down the group because the increased atomic radius and shielding effect make it harder for the nucleus to attract a bonding pair of electrons.

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Reactivity decreases down the group because the outer shell is further from the nucleus, meaning the electrostatic attraction for an incoming electron is weaker.

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Halogens act as oxidizing agents; their oxidizing power decreases down the group (F2>Cl2>Br2>I2F_2 > Cl_2 > Br_2 > I_2).

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Displacement reactions occur when a more reactive halogen displaces a less reactive halide ion from its salt solution (e.g., Cl2Cl_2 will displace Br−Br^- and I−I^-).

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Silver nitrate (AgNO3AgNO_3) followed by dilute/concentrated ammonia (NH3NH_3) is used to identify halide ions: AgClAgCl (white, dissolves in dilute NH3NH_3), AgBrAgBr (cream, dissolves in conc. NH3NH_3), and AgIAgI (yellow, insoluble).

📐Formulae

2M+X2→2MX2M + X_2 \rightarrow 2MX

X2+2Y−→2X−+Y2 (where X is more reactive than Y)X_2 + 2Y^- \rightarrow 2X^- + Y_2 \text{ (where } X \text{ is more reactive than } Y)

H2(g)+X2(g)→2HX(g)H_2(g) + X_2(g) \rightarrow 2HX(g)

Ag+(aq)+X−(aq)→AgX(s)Ag^+(aq) + X^-(aq) \rightarrow AgX(s)

Cl2(g)+H2O(l)⇌HCl(aq)+HClO(aq)Cl_2(g) + H_2O(l) \rightleftharpoons HCl(aq) + HClO(aq)

💡Examples

Problem 1:

Predict the observation when chlorine water (Cl2(aq)Cl_2(aq)) is added to a solution of potassium iodide (KI(aq)KI(aq)). Write the ionic equation for the reaction.

Solution:

The solution will turn from colorless to brown/dark red (due to the formation of I2I_2). Ionic equation: Cl2(aq)+2I−(aq)→2Cl−(aq)+I2(aq)Cl_2(aq) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq).

Explanation:

Chlorine is more reactive than iodine because it has a smaller atomic radius and higher electronegativity. Therefore, Cl2Cl_2 displaces I−I^- ions from the solution, oxidizing them to molecular iodine (I2I_2).

Problem 2:

Explain why the boiling point of I2I_2 is significantly higher than that of Cl2Cl_2.

Solution:

The boiling point of I2I_2 is higher due to stronger intermolecular forces.

Explanation:

Both Cl2Cl_2 and I2I_2 are non-polar diatomic molecules held together by London dispersion forces. I2I_2 has a much larger electron cloud (more electrons) than Cl2Cl_2, which leads to stronger instantaneous dipole-induced dipole attractions, requiring more thermal energy to overcome.

Problem 3:

Calculate the mass of silver chloride formed when 50 cm350 \text{ cm}^3 of 0.1 mol/dm30.1 \text{ mol/dm}^3 AgNO3AgNO_3 reacts with excess NaClNaCl. (Atomic masses: Ag=108,Cl=35.5Ag=108, Cl=35.5)

Solution:

Mass = 0.718 g0.718 \text{ g}.

Explanation:

First, calculate moles of AgNO3AgNO_3: n=c×V=0.1×0.050=0.005 moln = c \times V = 0.1 \times 0.050 = 0.005 \text{ mol}. Since the ratio of Ag+Ag^+ to AgClAgCl is 1:11:1, 0.005 mol0.005 \text{ mol} of AgClAgCl is formed. Molar mass of AgCl=108+35.5=143.5 g/molAgCl = 108 + 35.5 = 143.5 \text{ g/mol}. Mass =0.005×143.5=0.7175 g= 0.005 \times 143.5 = 0.7175 \text{ g}.