Elements of Groups 16, 17 and 18
Exhaustive Inorganic Theory, Periodic Trends, Structures & Fully Solved Board PYQs
1. Introduction and Electronic Configuration
The elements of groups 16, 17, and 18 are p-block elements. Their valence shell electronic configuration is $ns^2 np^{4-6}$.
| Group | Common Name | Elements | General Outer Configuration |
|---|---|---|---|
| Group 16 | Chalcogens (Ore-forming) | Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po) | $ns^2 np^4$ |
| Group 17 | Halogens (Salt-producing) | Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At) | $ns^2 np^5$ |
| Group 18 | Noble/Inert Gases | Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) | $ns^2 np^6$ (except He: $1s^2$) |
2. General Trends in Physical Properties
Atomic and Ionic Radii
As we move down the group, atomic radii increase due to the addition of new electronic shells. Across the period, atomic radii decrease from Group 16 to 17 due to increasing effective nuclear charge. Noble gases (Group 18) have larger radii than Group 17 because their radii are measured as Van der Waals radii (since they don't form covalent bonds easily).
Ionization Enthalpy
Decreases down the group as atomic size increases. It is exceptionally high for Group 18 due to their stable octet configuration.
Electron Gain Enthalpy
Halogens (Group 17) have maximum negative electron gain enthalpy because they are just one electron short of a stable noble gas configuration. It becomes less negative down the group.
Electronegativity
Decreases down the group. Fluorine is the most electronegative element in the periodic table (4.0 on Pauling scale), followed by Oxygen (3.5).
3. Anomalous Behavior of First Elements (Oxygen and Fluorine)
The first element of each group differs significantly from the rest of the members. This is due to:
- Small atomic size.
- High electronegativity.
- High ionization enthalpy.
- Absence of d-orbitals in their valence shell. (Crucial reason)
Anomalous Behavior of Oxygen
- Oxygen is a diatomic gas ($O_2$), while others (like Sulfur $S_8$) are polyatomic solids.
- Oxygen exhibits hydrogen bonding in its compounds (like $H_2O$), others do not.
- Maximum covalency of Oxygen is 4 (usually 2), while others can expand their covalency to 6 (e.g., $SF_6$) because they have vacant d-orbitals.
- Oxygen is paramagnetic, others are diamagnetic.
Anomalous Behavior of Fluorine
- Fluorine is the most reactive halogen.
- It exhibits only -1 oxidation state, while other halogens show +1, +3, +5, +7 states.
- $HF$ is a liquid due to strong hydrogen bonding, while $HCl, HBr, HI$ are gases.
- Cannot form polyhalide ions (like $I_3^-$) due to the absence of d-orbitals.
4. Chemical Properties (Hydrides and Oxides)
Hydrides of Group 16 ($H_2E$)
Elements form hydrides of the type $H_2O, H_2S, H_2Se, H_2Te, H_2Po$.
- Acidic Character: Increases down the group ($H_2O < H_2S < H_2Se < H_2Te$). Because bond length increases down the group, the $H-E$ bond dissociation enthalpy decreases, making it easier to release $H^+$.
- Thermal Stability: Decreases down the group for the same reason (decreasing bond strength).
- Reducing Character: Increases down the group. $H_2O$ is not a reducing agent, while $H_2Te$ is a strong reducing agent.
Hydrides of Group 17 (Hydrogen Halides, $HX$)
Trend of Acidic Strength: $HF < HCl < HBr < HI$. (Due to decreasing $H-X$ bond dissociation enthalpy down the group).
Oxidation States
- Group 16: Common oxidation states are -2, +2, +4, +6. Oxygen mostly shows -2 (except in $OF_2$ where it is +2, and peroxides where it is -1). Stability of +6 decreases and +4 increases down the group due to the inert pair effect.
- Group 17: All show -1. Cl, Br, I also show +1, +3, +5, +7 due to availability of vacant d-orbitals.
5. Interhalogen Compounds
When two different halogens react with each other, they form a series of compounds known as interhalogen compounds. The general formula is $XX'_n$, where X is the larger/less electronegative halogen, X' is the smaller/more electronegative halogen, and n = 1, 3, 5, 7.
| Type | Examples | Structure / Shape | Hybridization of Central Atom |
|---|---|---|---|
| $XX'$ | $ClF, BrF, ICl$ | Linear | $sp^3$ (3 lone pairs) |
| $XX'_3$ | $ClF_3, BrF_3, ICl_3$ | T-shaped | $sp^3d$ (2 lone pairs) |
| $XX'_5$ | $ClF_5, BrF_5, IF_5$ | Square Pyramidal | $sp^3d^2$ (1 lone pair) |
| $XX'_7$ | $IF_7$ | Pentagonal Bipyramidal | $sp^3d^3$ (0 lone pairs) |
Reactivity: Interhalogen compounds are generally more reactive than pure halogens (except Fluorine) because the $X-X'$ bond is weaker than the $X-X$ bond in halogens due to dissimilar sizes.
6. Noble Gases and Xenon Compounds
Noble gases were considered inert because of their completely filled $ns^2 np^6$ configuration. However, in 1962, Neil Bartlett prepared the first noble gas compound.
He observed that $PtF_6$ oxidizes $O_2$ to form $O_2^+PtF_6^-$. Since the first ionization enthalpy of Xenon ($1170 \text{ kJ/mol}$) is very close to that of Oxygen ($1175 \text{ kJ/mol}$), he reacted Xenon with $PtF_6$ and successfully prepared a red solid: $Xe^+[PtF_6]^-$.
Compounds of Xenon with Fluorine
$XeF_2$
Hybridization: $sp^3d$
Lone Pairs: 3 (equatorial)
Shape: Linear
$XeF_4$
Hybridization: $sp^3d^2$
Lone Pairs: 2 (axial)
Shape: Square Planar
$XeF_6$
Hybridization: $sp^3d^3$
Lone Pairs: 1
Shape: Distorted Octahedral
7. Manufacture of Important Compounds
Manufacture of Sulfuric Acid ($H_2SO_4$) - Contact Process
It involves three main steps:
- Production of $SO_2$: By burning sulfur or roasting iron pyrites in air.
$S_8 + 8O_2 \rightarrow 8SO_2(g)$ - Catalytic Oxidation of $SO_2$ to $SO_3$: This is the key step. It is an exothermic and reversible reaction.
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -196.6 \text{ kJ}$
Catalyst used: Vanadium pentoxide ($V_2O_5$) at 720 K and 2 bar pressure. - Absorption of $SO_3$: $SO_3$ is absorbed in concentrated $H_2SO_4$ to produce Oleum ($H_2S_2O_7$), which is then diluted with water to get $H_2SO_4$ of desired concentration.
$SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$ (Oleum)
$H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$
Manufacture of Chlorine ($Cl_2$) - Deacon's Process
Oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of $CuCl_2$ (catalyst) at 723 K.
$4HCl + O_2 \xrightarrow{CuCl_2, 723K} 2Cl_2 + 2H_2O$
8. Solved Textbook Reasoning Questions
Q1: Why is $H_2O$ a liquid and $H_2S$ a gas at room temperature?
Answer:
Oxygen is highly electronegative and small in size. Therefore, water ($H_2O$) molecules undergo extensive intermolecular hydrogen bonding. This strong force of attraction keeps the molecules together, making it a liquid. Sulfur is larger and less electronegative, so $H_2S$ cannot form hydrogen bonds; it only has weak Van der Waals forces, making it a gas.
Q2: Why does Ozone ($O_3$) act as a powerful oxidizing agent?
Answer:
Ozone is thermodynamically unstable with respect to oxygen. It readily decomposes to liberate nascent oxygen $[O]$.
$O_3 \rightarrow O_2 + [O]$
The liberation of highly reactive nascent oxygen makes ozone a powerful oxidizing agent (stronger than $O_2$). Example: It oxidizes black lead sulfide ($PbS$) to white lead sulfate ($PbSO_4$).
9. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the general electronic configuration of Group 16 elements. (Oct 2013, March 2018)
Answer: The general outer electronic configuration is $ns^2 np^4$.
Q2. Name the element of Group 17 which shows only -1 oxidation state. (March 2016, Oct 2021)
Answer: Fluorine (F).
Q3. Give the formula of the first noble gas compound prepared by Neil Bartlett. (March 2015, March 2022)
Answer: $Xe^+[PtF_6]^-$ (Xenon hexafluoroplatinate).
2 Mark Questions (SA-I)
Q4. State any two anomalous properties of Fluorine. (March 2014, Oct 2019)
- Fluorine shows only a -1 oxidation state, while other halogens show positive oxidation states (+1, +3, +5, +7).
- Fluorine has a lower electron gain enthalpy than chlorine, unlike the normal trend, due to its very small size and interelectronic repulsions.
Q5. Write two uses of (a) Helium and (b) Argon. (March 2017, March 2020)
Uses of Helium: 1. Used in filling weather balloons because it is light and non-combustible. 2. Used in gas mixtures by deep-sea divers to dilute oxygen (prevents "bends").
Uses of Argon: 1. Used to provide an inert atmosphere in high-temperature metallurgical processes (arc welding). 2. Used in filling incandescent electric bulbs.
3 Mark Questions (SA-II)
Q6. What are interhalogen compounds? Why are they more reactive than halogens? Give the structure of $ClF_3$. (Oct 2015, March 2019, March 2023)
Definition: Compounds formed when two different halogens react with each other are called interhalogen compounds (e.g., $ClF_3, BrF_5$).
Reactivity: They are more reactive than pure halogens (except Fluorine) because the $X-X'$ bond in interhalogens is weaker than the $X-X$ bond in halogens. This is due to the difference in size and electronegativity, making the bond polarized and easier to break.
Structure of $ClF_3$: The central atom Cl undergoes $sp^3d$ hybridization. It has 3 bond pairs and 2 lone pairs. According to VSEPR theory, to minimize repulsion, the lone pairs occupy equatorial positions, resulting in a T-shaped molecule.
Q7. Describe the contact process for the manufacture of Sulfuric acid with necessary chemical equations. (Oct 2014, March 2021)
The contact process involves three main steps:
Step 1: Production of Sulfur dioxide ($SO_2$)
Sulfur or iron pyrites are burnt in the air to produce $SO_2$.
$S_8 + 8O_2 \rightarrow 8SO_2(g)$
Step 2: Catalytic Oxidation of $SO_2$ to $SO_3$
This is the key step. $SO_2$ is oxidized by atmospheric oxygen in the presence of Vanadium pentoxide ($V_2O_5$) catalyst at 720 K and 2 bar pressure.
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad (\Delta H = -196.6 \text{ kJ})$
Step 3: Absorption of $SO_3$
$SO_3$ gas is absorbed in concentrated $H_2SO_4$ to produce Oleum (fuming sulfuric acid). Oleum is then diluted with a calculated amount of water to get $H_2SO_4$ of desired concentration.
$SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$ (Oleum)
$H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$
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