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Group 16 Chemistry Notes
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1
Question
Which elements make up Group 16 (the chalcogens) listed in the notes, and what is special about oxygen in Earth's crust?
Answer
Group 16 consists of O (oxygen), S (sulfur), Se (selenium), Te (tellurium), and Po (polonium). Oxygen is the most abundant element in Earth's crust.
2
Question
Give formulas for common minerals/compounds: gypsum, epsom salt, baryte, zinc blende, and copper pyrite as written in the notes.
Answer
Gypsum: \(CaSO_4\cdot 2H_2O\) Epsom salt: \(MgSO_4\cdot 7H_2O\) Baryte (barite): \(BaSO_4\) Zinc blende (sphalerite): \(ZnS\) Copper pyrite (chalcopyrite): \(CuFeS_2\)
3
Question
Classify the elements O, S, Se, Te, and Po by broad chemical character as noted.
Answer
O and S are nonmetals; Se and Te are metalloids (or possess intermediate character); Po (polonium) is a radioactive metal.
4
Question
How does atomic radius change down Group 16 (O → Po) according to the notes?
Answer
Atomic radius increases down the group from O to Po.
5
Question
What is the trend in electron gain enthalpy (more negative = more exothermic) down Group 16 as listed in the notes?
Answer
Electron gain enthalpy becomes less exothermic (decreases in magnitude) down the group: S is most exothermic, then Se, then Te, then Po; oxygen is the least exothermic (less exothermic than sulfur) due to electron–electron repulsion in small 2p orbitals.
6
Question
State the ionization enthalpy order for Group 16 given in the notes and explain the main reason for this trend.
Answer
Ionization enthalpy decreases down the group: O > S > Se > Te > Po. The decrease is due to increasing atomic size which reduces the hold of the nucleus on valence electrons.
7
Question
What is the electron negativity trend for Group 16 described in the notes?
Answer
Electronegativity decreases down the group: O > S > Se > Te > Po. There are no exceptions noted in these notes.
8
Question
How does boiling point change down Group 16 according to the notes?
Answer
Boiling point generally increases down the group from O to Po.
9
Question
What is the melting point order for Group 16 hydrides (H2O, H2S, H2Se, H2Te) as given in the notes and which effect causes H2O to be an outlier?
Answer
Melting point order: Te (highest) > Po > Se > S > O. For hydrides specifically, H2S < H2Se < H2Te < H2O. H2O is an outlier with much higher melting/boiling points due to strong hydrogen bonding (intermolecular H-bonding).
10
Question
Explain why H–E (hydrogen–element) bond distance changes down Group 16 and how that affects bond energy.
Answer
H–E bond distance increases down the group (H2O < H2S < H2Se < H2Te). Longer bonds are weaker, so bond energy decreases down the group; H–O bond is the shortest and strongest, H–Te is the longest and weakest.
11
Question
Describe the anomalous behaviour of oxygen compared with other Group 16 elements as summarised in the notes.
Answer
Oxygen shows anomalous behaviour because of its very small size, high electronegativity, high bond energies for O–H and O–O, and lack of vacant d-orbitals. These properties give O different physical and chemical behaviours than heavier congeners (e.g., strong hydrogen bonding, unusual electron gain enthalpy trends).
12
Question
Write the general reactions of Group 16 elements with hydrogen and the hydrides formed.
Answer
General combination: E + H2 → H2E (where E = O, S, Se, Te, Po). Examples: O2 + H2 → H2O; S + H2 → H2S; also H2Se, H2Te, H2Po (hydrides of Se, Te, Po).
13
Question
Give the trend of melting points for the hydrides H2S, H2Se, H2Te, and H2O from lowest to highest as in the notes.
Answer
Lowest to highest: H2S < H2Se < H2Te < H2O. The exception is H2O, which is much higher due to hydrogen bonding.
14
Question
How does boiling point trend compare with melting point trend for Group 16 hydrides according to the notes?
Answer
The boiling point follows the same trend as melting point: H2S < H2Se < H2Te < H2O, with H2O significantly higher because of hydrogen bonding.
15
Question
What is stated about the formation enthalpies (ΔHf) of H2O, H2S, H2Se, and H2Te and their relative ordering including approximate values from the notes?
Page 4
Answer
The formation enthalpies indicate H2O is most exothermic: H2O (≈ −286 kJ/mol) < H2S (≈ −20 kJ/mol) < H2Se (≈ +73 kJ/mol) < H2Te (≈ +100 kJ/mol). The trend shows formation becomes less favorable (less exothermic/more endothermic) down the group.
16
Question
In the nitrogen group (Group 15) and similarly in Group 16, what trend is noted for hydride formation energies down the group?
Page 4
Answer
In Group 15: NH3 formation is exothermic, while heavier hydrides (PH3, AsH3, SbH3, BiH3) become progressively less exothermic and eventually endothermic. Similarly in Group 16: H2O and H2S formation are exothermic, whereas H2Se, H2Te, H2Po formation becomes endothermic down the group.
17
Question
What are the approximate bond angles for H2O, H2S, H2Se, and H2Te listed in the notes, and what rule explains the trend?
Page 4
Answer
Bond angles listed: H2O ≈ 104°, H2S ≈ 92°, H2Se ≈ 96°, H2Te ≈ 90°. The trend (smaller bond angles down the group except water) is explained by Bent’s/Valence Shell Electron Pair Repulsion (VSEPR) and the effect known as Dragna? (notes state ‘Dragas Rule’ — likely referring to decreased lone pair–bond pair repulsion down the group).
18
Question
What general types of oxides do Group 16 elements form with oxygen, and what physical states are noted for some (O3, SO2, SeO2)?
Page 5
Answer
Group 16 elements form oxides of types EO2 or E2O3 (e.g., SO2, SO3, SeO2, TeO2, etc.). Ozone (O3) and sulfur dioxide (SO2) are gases; SeO2 is a solid. All Group 16 oxides are acidic in character.
19
Question
How does the reducing property of oxides change from SO2 to TeO2 according to the notes, and what special behavior is noted for TeO2?
Page 5
Answer
Reducing property decreases from SO2 to TeO2 (i.e., SO2 is a better reducing agent than TeO2). TeO2, instead of acting as a reducing agent to reach +6, tends to act as an oxidizing agent to reach +2 oxidation state; TeO2 is more stable in +4 state due to inert pair effect.
20
Question
Which oxides do S, Se, and Te form in higher oxidation states as listed in the notes?
Page 5
Answer
They form the corresponding trioxides: SO3, SeO3, and TeO3 (i.e., oxides in higher oxidation states such as +6).
21
Question
Summarize the statement at the end of the oxides page regarding acidity.
Page 5
Answer
All Group 16 oxides are acidic in nature.
22
Question
What types of halides do Group 16 elements form, and what general trend in stability is given?
Page 6
Answer
Group 16 elements form halides of type EX2, EX4, and EX6. Stability of halides decreases in the order F > Cl > Br > I (i.e., fluorides are most stable). Among EX6 species, only hexafluorides are generally stable (e.g., SF6 is stable; other hexahalides are not).
23
Question
What physical states of the tetrafluorides SF4, SeF4, and TeF4 are recorded in the notes?
Page 6
Answer
SF4 is a gas, SeF4 is a liquid, and TeF4 is a solid.
24
Question
What hybridisation is suggested for these Group 16 fluorides (EX4) in the notes?
Page 6
Answer
The notes state that these fluorides have sp3d hybridisation (to account for expanded octet geometry in central atoms forming EX4 or EX6 structures).
25
Question
Which dihalides are said to form for Group 16 elements (except oxygen), and in what molecular form do they exist according to the notes?
Page 6
Answer
All elements except oxygen form dichlorides and dibromides — these exist as dimers such as S2F2, S2Cl2, S2Br2, Se2Cl2, Se2Br2. They can undergo disproportionation (formation of different oxidation state products).
26
Question
Write the disproportionation example given in the notes for Se2Cl2.
Page 7
Answer
Reaction: 2 Se2Cl2 → SeCl4 + 3 Se (i.e., Se in different oxidation states, product SeCl4 (Se +4) and elemental Se (0)).
27
Question
List key physical properties of alpha-sulfur (α-sulfur, S8) described in the notes.
Page 7
Answer
α-Sulfur (yellow rhombic S8): yellow in color, insoluble in water, readily soluble in carbon disulfide (CS2) because it is nonpolar, formed by evaporating 'roll sulfur' in CS2, consists of crown-shaped S8 rings, and exists as a yellow crystalline form.
28
Question
Describe how β-sulfur (monoclinic S8) is formed and one of its distinguishing features according to the notes.
Page 7
Answer
β-Sulfur is formed by melting α-sulfur in a dish and cooling it until crystals form; specifically, melt α-sulfur, cool a bit (two holes in the crust are made and liquid poured out) to form colorless crystals of β-sulfur. Its S8 molecules are packed differently to give a different crystal structure compared to α-sulfur.
29
Question
What shape do S8 molecules adopt and how does the note characterise their shape?
Page 7
Answer
S8 molecules adopt a crown-shaped (crown-shaped ring) conformation; both α and β forms have this S8 crown shape.
30
Question
According to the notes, what happens to sulfur at elevated temperatures (~1000 K)?
Page 7
Answer
At elevated temperatures around 1000 K, S2 is formed and it is paramagnetic (similar to O2). The note indicates S2 is what forms at high temperature.