Elements of Groups 16, 17 & 18
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The general outer electronic configuration of Group 16 elements (Chalcogens) is $ns^2 np^4$.
Answer: Fluorine (F). It is the most electronegative element and lacks d-orbitals, so it only exhibits a -1 oxidation state in its compounds.
Answer: The first noble gas compound prepared was Xenon hexafluoroplatinate, with the formula $Xe^+[PtF_6]^-$.
Answer: Helium (He) is used because it is a very light gas and is non-combustible.
2 Mark Questions (Short Answer-I)
Answer:
- Oxidation State: Fluorine shows only a -1 oxidation state, whereas other halogens show positive oxidation states (+1, +3, +5, +7) as well.
- Electron Gain Enthalpy: The negative electron gain enthalpy of fluorine is less than that of chlorine due to the very small size of the F atom and resulting strong interelectronic repulsions.
- Hydrogen Bonding: HF is a liquid at room temperature due to strong intermolecular hydrogen bonding, whereas HCl, HBr, and HI are gases.
Answer:
Oxygen has a small atomic size and high electronegativity. It forms stable $p\pi-p\pi$ multiple bonds, existing as diatomic $O_2$ molecules. The intermolecular forces between $O_2$ molecules are weak van der Waals forces, hence it is a gas.
Sulfur, due to its larger size, cannot form stable $p\pi-p\pi$ multiple bonds. Instead, it forms single bonds with other sulfur atoms, resulting in puckered, eight-membered ring molecules ($S_8$). Because $S_8$ molecules are much larger, the van der Waals forces of attraction between them are much stronger, making sulfur a solid at room temperature.
Answer:
Uses of Neon:
- Used in discharge tubes and fluorescent bulbs for advertisement display signs (Neon signs).
- Used in botanical gardens and green houses.
Uses of Argon:
- Used to provide an inert atmosphere in high-temperature metallurgical processes (e.g., arc welding of metals/alloys).
- Used in filling incandescent electric bulbs to prolong the life of the filament.
3 Mark Questions (Short Answer-II / Structures)
Answer:
Definition:
Compounds formed when two different halogens react with each other are called interhalogen compounds. They have the general formula $XX'_n$ (e.g., $ClF_3, BrF_5$).
Reason for Higher Reactivity:
Interhalogen compounds are generally more reactive than pure halogens (except Fluorine). This is because the $X-X'$ bond in interhalogens is weaker and more polar than the $X-X$ covalent bond in pure halogens, due to the difference in size and electronegativity of the two different halogen atoms.
Structure and Hybridization of $ClF_3$:
- The central Chlorine atom undergoes $sp^3d$ hybridization.
- It contains 3 bond pairs (with Fluorine) and 2 lone pairs of electrons.
- To minimize lone pair-lone pair repulsions (according to VSEPR theory), the two lone pairs occupy the equatorial positions.
- This results in a T-shaped geometry for the molecule.
Answer:
Oxygen, the first member of Group 16, differs from the rest of the members due to its small size, high electronegativity, and the absence of d-orbitals in its valence shell. Its anomalous properties include:
- Physical State and Atomicity: Oxygen exists as a diatomic gas ($O_2$) at room temperature, whereas other members of the group are polyatomic solid non-metals (e.g., Sulfur exists as $S_8$).
- Hydrogen Bonding: Due to its high electronegativity, oxygen forms strong intermolecular hydrogen bonds in its compounds (like $H_2O$), causing water to be a liquid. Other elements do not form hydrogen bonds (e.g., $H_2S$ is a gas).
- Covalency: The maximum covalency of oxygen is strictly limited to 4 (in practice, rarely exceeds 2) because it lacks vacant d-orbitals. Other members can expand their octet to show covalencies of 4 and 6 (e.g., $SF_6, TeF_6$).
- Magnetic Nature: Oxygen ($O_2$) is paramagnetic (contains two unpaired electrons in antibonding molecular orbitals), while other elements of the group are diamagnetic.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark manufacturing process combined with a 1-mark reasoning question).
(b) Why does $H_2S$ act only as a reducing agent while $SO_2$ acts as both an oxidizing and a reducing agent? [1 Mark] March 2015, March 2019, Oct 2022
Answer (a): Contact Process
The manufacture of $H_2SO_4$ by the Contact process involves three major steps:
Step 1: Production of Sulfur dioxide ($SO_2$)
Sulfur or iron pyrites ($FeS_2$) are burnt in an excess of air to produce sulfur dioxide.
$S_8(s) + 8O_2(g) \rightarrow 8SO_2(g)$
Step 2: Catalytic Oxidation of $SO_2$ to $SO_3$ (Key Step)
This is the most crucial step. Sulfur dioxide is catalytically oxidized by atmospheric oxygen to sulfur trioxide ($SO_3$) in the presence of Vanadium pentoxide ($V_2O_5$) catalyst at a temperature of 720 K and a pressure of 2 bar. The reaction is reversible and exothermic.
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -196.6 \text{ kJ/mol}$
Step 3: Absorption of $SO_3$ to form Oleum
$SO_3$ gas is not dissolved directly in water (as it forms a highly exothermic, uncontrollable acid fog). Instead, it is absorbed in concentrated $H_2SO_4$ (98%) to form Oleum (fuming sulfuric acid, $H_2S_2O_7$).
$SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)$
Oleum is then diluted with a calculated amount of water to obtain $H_2SO_4$ of the desired concentration.
$H_2S_2O_7(l) + H_2O(l) \rightarrow 2H_2SO_4(aq)$
Answer (b): Reasoning
In $H_2S$, sulfur is in its lowest possible oxidation state (-2). Therefore, it can only lose electrons (get oxidized) to increase its oxidation state, acting only as a reducing agent.
In $SO_2$, sulfur is in an intermediate oxidation state (+4). It can decrease its oxidation state to 0 or -2 (acting as an oxidizing agent) or increase its oxidation state to +6 (acting as a reducing agent).
(b) Draw the structures of $XeF_4$ and $XeF_6$. [2 Marks] Oct 2013, March 2020
Answer (a): Deacon's Process
Deacon's process is a commercial method for the preparation of chlorine gas. It involves the oxidation of hydrogen chloride ($HCl$) gas by atmospheric oxygen in the presence of Cuprous chloride ($CuCl_2$) as a catalyst at a temperature of 723 K.
$4HCl(g) + O_2(g) \xrightarrow{CuCl_2, \text{ } 723 \text{ K}} 2Cl_2(g) + 2H_2O(g)$
Answer (b): Structures of Xenon Fluorides
1. Structure of $XeF_4$
Hybridization: $sp^3d^2$
Geometry: Square Planar.
Xe has 4 bond pairs with F and 2 lone pairs. The lone pairs occupy the axial positions (above and below the plane) to minimize repulsion, leaving the 4 F atoms at the corners of a square.
2. Structure of $XeF_6$
Hybridization: $sp^3d^3$
Geometry: Distorted Octahedral.
Xe has 6 bond pairs with F and 1 lone pair. The presence of the single lone pair distorts the regular octahedral geometry, causing it to become a distorted octahedral structure.
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