Chapter 7: Elements of Groups 16, 17 & 18
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
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1. Name the elements of Group 16. What is their general valence electronic configuration?
The elements of Group 16 are Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). They are also known as chalcogens (ore-forming elements).
Their general valence shell electronic configuration is $ns^2 np^4$.
2. Write the general valence electronic configurations of Group 17 and Group 18 elements.
- Group 17 (Halogens): $ns^2 np^5$
- Group 18 (Noble Gases): $ns^2 np^6$ (Except Helium, which has $1s^2$)
3. Explain the trend of atomic radii in Group 16 and Group 17.
In both groups, the atomic radius increases as we move down the group from top to bottom.
Reason: Moving down the group, a new principal quantum shell is added at each successive element. The increase in the number of shells outweighs the effect of increased nuclear charge, leading to a larger atomic size.
4. Why do Group 16 elements have a lower first ionization enthalpy compared to Group 15 elements in the same period?
Group 15 elements have an exactly half-filled p-orbital electronic configuration ($ns^2 np^3$), which is extra stable. Removing an electron requires breaking this stable symmetry, leading to high ionization enthalpy.
Group 16 elements have a $ns^2 np^4$ configuration. By losing one electron, they attain the stable half-filled $np^3$ configuration. Hence, it is easier to remove the first electron, resulting in a comparatively lower first ionization enthalpy than Group 15.
5. Why is the electron gain enthalpy of Oxygen less negative than that of Sulfur?
Oxygen has a very compact and small atomic size (n=2 shell). As a result, its electron density is very high. When an extra electron is added to form $O^-$, it faces strong inter-electronic repulsion from the already present electrons.
Sulfur has a larger size (n=3 shell), so the incoming electron does not experience as much repulsion. Therefore, energy released upon adding an electron is more in Sulfur than in Oxygen, making S have a more negative electron gain enthalpy.
6. Why is the electron gain enthalpy of Fluorine less negative than that of Chlorine?
Similar to the Oxygen-Sulfur case, Fluorine has a very small atomic size. The high electron density in its 2p subshell causes strong electron-electron repulsion for the incoming electron. Chlorine has a larger 3p subshell, accommodating the extra electron more easily with less repulsion. Thus, Chlorine has the highest negative electron gain enthalpy in the periodic table.
7. Discuss the trend of electronegativity in Group 16 and Group 17.
Electronegativity decreases down the group for both Group 16 and Group 17.
Reason: As we move down the group, atomic size increases, and the distance between the nucleus and the valence shell increases. This reduces the effective nuclear pull on the shared pair of electrons. Note: Fluorine is the most electronegative element in the periodic table, followed by Oxygen.
8. Why are halogens highly reactive?
Halogens have the valence shell configuration $ns^2 np^5$. They are short of just one electron to complete their stable octet (noble gas configuration). Because of their high electronegativity and highly negative electron gain enthalpy, they have a very strong tendency to accept one electron. Also, their $X-X$ bond dissociation enthalpies are low. These factors make them highly reactive non-metals.
9. Halogens act as strong oxidizing agents. Why?
Halogens readily accept an electron to complete their octet ($X + e^- \rightarrow X^-$). Because they themselves undergo reduction very easily due to their high electronegativity and high negative electron gain enthalpy, they act as very strong oxidizing agents. Fluorine ($F_2$) is the strongest oxidizing agent among them.
10. Why are the elements of Group 18 called 'Noble Gases'?
Group 18 elements have completely filled valence shells ($ns^2 np^6$, except He which is $1s^2$). This represents an extremely stable electronic configuration. Therefore, they have very high ionization enthalpies and almost zero electron gain enthalpies. They have neither the tendency to lose, gain, nor share electrons, making them chemically unreactive or inert under standard conditions. Hence, they are called noble gases.
11. State four reasons for the anomalous behavior of Oxygen in Group 16.
Oxygen differs from the rest of the group members due to:
- Exceptionally small atomic size.
- High electronegativity and ionization enthalpy.
- Absence of vacant d-orbitals in its valence shell (can't expand its octet).
- Ability to form multiple bonds ($p\pi-p\pi$ bonds) with itself and other small atoms.
12. State four reasons for the anomalous behavior of Fluorine in Group 17.
Fluorine shows anomalous behavior due to:
- Very small atomic size.
- Highest electronegativity in the periodic table.
- Absence of vacant d-orbitals in its valence shell.
- Low F-F bond dissociation enthalpy (due to strong interelectronic repulsion).
13. Oxygen is a gas while Sulfur is a solid at room temperature. Why?
Because of its small size and high electronegativity, Oxygen forms stable $p\pi-p\pi$ multiple bonds, existing as a diatomic gas molecule ($O_2$). The intermolecular forces between $O_2$ molecules are weak Van der Waals forces.
Sulfur, being larger, cannot form stable $p\pi-p\pi$ bonds. Instead, it forms single covalent bonds linking eight sulfur atoms together to form a puckered ring structure ($S_8$). Because of the larger size and higher molecular mass of $S_8$, the Van der Waals forces are much stronger, making it a solid at room temperature.
14. Explain why water ($H_2O$) is a liquid while hydrogen sulfide ($H_2S$) is a gas.
Oxygen has a much smaller size and higher electronegativity than Sulfur. Due to this, the $O-H$ bond in water is highly polar, leading to the formation of strong intermolecular hydrogen bonds between water molecules. This associates the molecules together as a liquid.
In $H_2S$, Sulfur is less electronegative and larger, so hydrogen bonding does not occur. The molecules are held by weak Van der Waals forces, making $H_2S$ a gas at room temperature.
15. Discuss the common oxidation states exhibited by Group 16 elements.
The common oxidation states are -2, +2, +4, and +6.
- Oxygen mostly shows -2 (except in peroxides -1, superoxides -1/2, and with Fluorine like $OF_2$ where it is +2). It never shows +4 or +6 due to the absence of d-orbitals.
- S, Se, and Te usually show +4 and +6 states in compounds with Oxygen and Halogens. The stability of the +6 state decreases down the group, and +4 state increases due to the inert pair effect.
16. Discuss the common oxidation states exhibited by Group 17 elements.
All halogens show an oxidation state of -1.
Fluorine is the most electronegative element and has no d-orbitals, so it only shows a -1 oxidation state.
Other halogens (Cl, Br, I) have vacant d-orbitals and can expand their octet to show positive oxidation states of +1, +3, +5, and +7 when combined with smaller and more electronegative atoms like Oxygen and Fluorine (e.g., in interhalogens or oxoacids).
17. Describe the trend in acidic character of hydrides of Group 16 ($H_2O, H_2S, H_2Se, H_2Te$).
The acidic character of Group 16 hydrides increases down the group: $H_2O < H_2S < H_2Se < H_2Te$.
Reason: As atomic size increases down the group, the bond length of the central atom with Hydrogen ($E-H$) increases. This causes the bond dissociation enthalpy to decrease, making it easier to release the $H^+$ ion, thus increasing acidic strength.
18. Describe the trend in acidic strength of hydrogen halides ($HF, HCl, HBr, HI$).
The acidic strength increases in the order: $HF < HCl < HBr < HI$.
Reason: Down the group, the size of the halogen atom increases. This leads to an increase in the $H-X$ bond length and a decrease in bond dissociation enthalpy. Since the $H-I$ bond is the weakest, $HI$ gives up $H^+$ most readily, making it the strongest acid. ($HF$ is a weak acid due to very high bond strength and hydrogen bonding).
19. Why do Noble Gases have very low boiling points?
Noble gases are monatomic and have a stable electronic configuration ($ns^2 np^6$). They do not form molecules. The only interatomic forces of attraction acting between them are the weak London dispersion forces (Van der Waals forces). Because these forces are very weak, they require very little heat energy to overcome, resulting in very low melting and boiling points.
20. Describe the laboratory preparation of Dioxygen ($O_2$) from Potassium chlorate ($KClO_3$).
Dioxygen is prepared in the lab by heating Potassium chlorate in the presence of Manganese dioxide ($MnO_2$) as a catalyst.
$$2KClO_{3(s)} \xrightarrow{\Delta, \ MnO_2} 2KCl_{(s)} + 3O_{2(g)}$$
21. Why is Dioxygen paramagnetic despite having an even number of electrons?
According to Molecular Orbital Theory (MOT), the electronic configuration of the $O_2$ molecule shows that two electrons occupy the two degenerate anti-bonding $\pi^*$ molecular orbitals singly, with parallel spins (Hund's rule). Due to the presence of these two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals, $O_2$ exhibits paramagnetism.
22. How is Ozone ($O_3$) prepared? Write the reaction.
Ozone is prepared by passing a slow, silent electrical discharge through pure, cold, and dry oxygen in an apparatus called an ozoniser. The silent discharge prevents the decomposition of the formed ozone back into oxygen, as the reaction is endothermic.
$$3O_{2(g)} \rightleftharpoons 2O_{3(g)} \quad \Delta H^\circ = +142 \text{ kJ/mol}$$
23. Ozone is a powerful oxidizing agent. Explain why. Give one example.
Ozone ($O_3$) is thermodynamically unstable with respect to oxygen. It easily decomposes to liberate highly reactive nascent oxygen, making it a powerful oxidizing agent.
$$O_3 \rightarrow O_2 + [O]$$
Example: It oxidizes lead sulfide to lead sulfate.
$$PbS_{(s)} + 4O_{3(g)} \rightarrow PbSO_{4(s)} + 4O_{2(g)}$$
24. Describe the depletion of the ozone layer by NO.
Nitric oxide (NO) emitted from the exhaust systems of supersonic jet airplanes reacts rapidly with ozone in the upper atmosphere, depleting the protective ozone layer.
$$NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$$
25. Name the two important allotropes of Sulfur. Which one is stable at room temperature?
The two main allotropes are Rhombic sulfur ($\alpha$-sulfur) and Monoclinic sulfur ($\beta$-sulfur).
Rhombic sulfur is the most stable form at room temperature. Monoclinic sulfur is stable only above 369 K (transition temperature). Both exist as $S_8$ puckered rings.
26. How does Sulfur dioxide ($SO_2$) act as a bleaching agent?
In the presence of moisture, $SO_2$ acts as a reducing agent and bleaches colored matter by reduction.
$$SO_2 + 2H_2O \rightarrow H_2SO_4 + 2[H] \text{ (Nascent hydrogen)}$$
$$\text{Colored matter} + [H] \rightarrow \text{Colorless matter}$$
This bleaching is temporary because the bleached matter regains its color upon exposure to air due to oxidation.
27. Describe the key step (catalytic oxidation) in the manufacture of Sulfuric acid by the Contact Process.
The crucial step is the catalytic oxidation of sulfur dioxide ($SO_2$) to sulfur trioxide ($SO_3$) using Vanadium pentoxide ($V_2O_5$) as a catalyst.
$$2SO_{2(g)} + O_{2(g)} \xrightarrow{V_2O_5, \ 720K, \ 2 \text{ bar}} 2SO_{3(g)} \quad \Delta H^\circ = -196.6 \text{ kJ/mol}$$
Since the reaction is exothermic and involves a decrease in volume, low temperature and high pressure favor the forward reaction (Le Chatelier's principle).
28. Why is $SO_3$ not directly dissolved in water in the Contact process? How is $H_2SO_4$ finally obtained?
Dissolving $SO_3$ directly in water is a highly exothermic reaction that forms a dense, uncontrollable fog of sulfuric acid mist which is difficult to condense.
Instead, $SO_3$ is absorbed in concentrated $H_2SO_4$ to form Oleum (fuming sulfuric acid, $H_2S_2O_7$).
$$SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$$
Oleum is then diluted with water to get $H_2SO_4$ of the desired concentration: $H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$.
29. Sulfuric acid is a strong dehydrating agent. Give two chemical equations to prove this.
Concentrated $H_2SO_4$ removes water (elements of H and O) from many organic compounds.
- Charring of sugar:
$C_{12}H_{22}O_{11} \xrightarrow{\text{Conc. } H_2SO_4} 12C + 11H_2O$ - Dehydration of formic acid:
$HCOOH \xrightarrow{\text{Conc. } H_2SO_4} CO + H_2O$
30. Concentrated Sulfuric acid acts as an oxidizing agent. Write its reaction with Copper.
Hot concentrated $H_2SO_4$ oxidizes metals and is itself reduced to $SO_2$.
$$Cu_{(s)} + 2H_2SO_{4(conc)} \xrightarrow{\Delta} CuSO_4 + SO_2 + 2H_2O$$
31. How is Chlorine manufactured by Deacon's process?
In Deacon's process, chlorine is produced by the oxidation of hydrogen chloride gas ($HCl$) by atmospheric oxygen in the presence of Cupric chloride ($CuCl_2$) as a catalyst at 723 K.
$$4HCl + O_2 \xrightarrow{CuCl_2, \ 723K} 2Cl_2 + 2H_2O$$
32. Write the reaction of Chlorine with: (a) cold and dilute NaOH (b) hot and concentrated NaOH.
These are disproportionation reactions.
- (a) Cold & Dilute NaOH: Forms Sodium chloride and Sodium hypochlorite.
$2NaOH + Cl_2 \rightarrow NaCl + NaOCl + H_2O$ - (b) Hot & Conc. NaOH: Forms Sodium chloride and Sodium chlorate(V).
$6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O$
33. Explain the bleaching action of Chlorine. Is it permanent or temporary?
Chlorine water acts as a powerful bleaching agent due to its oxidizing nature. In the presence of moisture, chlorine forms hypochlorous acid ($HOCl$), which decomposes to give nascent oxygen.
$Cl_2 + H_2O \rightarrow HCl + HOCl$
$HOCl \rightarrow HCl + [O]$ (Nascent Oxygen)
$\text{Colored substance} + [O] \rightarrow \text{Colorless substance}$
Because the bleaching is done by oxidation, the effect is permanent.
34. How is Hydrogen chloride ($HCl$) prepared in the laboratory?
In the laboratory, $HCl$ gas is prepared by heating sodium chloride with concentrated sulfuric acid.
At 420 K: $NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$
At 823 K: $NaHSO_4 + NaCl \rightarrow Na_2SO_4 + HCl$
The gas is dried by passing it through concentrated sulfuric acid.
35. What is Aqua Regia? Give its reaction with gold.
Aqua Regia is a mixture of concentrated Hydrochloric acid ($HCl$) and concentrated Nitric acid ($HNO_3$) in the ratio of 3:1 by volume. It is used to dissolve noble metals like Gold and Platinum.
Reaction with Gold: It forms the soluble complex anion $[AuCl_4]^-$ (tetrachloroaurate(III)).
$$Au + 4H^+ + NO_3^- + 4Cl^- \rightarrow [AuCl_4]^- + NO + 2H_2O$$
36. What are Interhalogen compounds? Give their general formulas.
Interhalogen Compounds: When two different halogens react with each other, they form binary covalent compounds called interhalogen compounds. They are formed due to the electronegativity difference between the halogens.
General Formula: $XX'_n$, where X is the larger/less electronegative halogen, X' is the smaller/more electronegative halogen, and $n = 1, 3, 5, 7$. (e.g., $ClF, BrF_3, IF_5, IF_7$).
37. Why are interhalogen compounds more reactive than halogens (except Fluorine)?
Interhalogen compounds ($XX'$) are more reactive than halogens ($X_2$) because the $X-X'$ bond in interhalogens is weaker and less stable than the pure $X-X$ bond in halogens (except the very weak $F-F$ bond). This is due to the difference in size and electronegativity, which causes the bond to be polar and more easily broken.
38. Predict the geometry/shape of $BrF_3$ using VSEPR theory.
In $BrF_3$, the central Bromine atom has 7 valence electrons. It forms 3 single bonds with Fluorine (using 3 electrons), leaving 4 electrons as 2 lone pairs.
Total electron pairs around Br = 3 bond pairs + 2 lone pairs = 5 pairs. The hybridization is $sp^3d$.
Basic geometry is trigonal bipyramidal. To minimize repulsion, the 2 lone pairs occupy equatorial positions. The resulting molecular shape is T-shaped (slightly bent due to lone pair-bond pair repulsion).
39. Predict the geometry/shape of $IF_5$ and $IF_7$.
- $IF_5$: Central Iodine has 5 bond pairs and 1 lone pair ($sp^3d^2$ hybridization). Shape is Square Pyramidal.
- $IF_7$: Central Iodine has 7 bond pairs and 0 lone pairs ($sp^3d^3$ hybridization). Shape is Pentagonal Bipyramidal.
40. Why was Neil Bartlett successful in preparing the first noble gas compound? What was it?
Neil Bartlett observed that the first ionization enthalpy of molecular Oxygen ($O_2 \rightarrow O_2^+ + e^-$, $1175 \text{ kJ/mol}$) was almost identical to that of Xenon ($Xe \rightarrow Xe^+ + e^-$, $1170 \text{ kJ/mol}$).
Since the powerful oxidizing agent $PtF_6$ could oxidize $O_2$ to form $O_2^+[PtF_6]^-$, he theorized it could also oxidize Xe. He reacted Xe with $PtF_6$ and successfully prepared the first noble gas compound, a red solid: $Xe^+[PtF_6]^-$ (Xenon hexafluoroplatinate).
41. Mention the conditions for the preparation of $XeF_2, XeF_4, \text{ and } XeF_6$.
They are prepared by direct reaction of Xenon and Fluorine under specific conditions of temperature, pressure, and reactant ratios:
- $XeF_2$: Xe (excess) + $F_2$ at 673 K, 1 bar.
- $XeF_4$: Xe and $F_2$ (1:5 ratio) at 873 K, 7 bar.
- $XeF_6$: Xe and $F_2$ (1:20 ratio) at 573 K, 60-70 bar.
42. Describe the structure of $XeF_2$ using VSEPR theory.
Central Xe has 8 valence electrons. It forms 2 single bonds with F, leaving 6 electrons as 3 lone pairs.
Total electron pairs = 2 bond pairs + 3 lone pairs = 5 pairs ($sp^3d$ hybridization).
The 3 lone pairs occupy the equatorial positions to minimize repulsion, placing the 2 Fluorine atoms at the axial positions. The resulting shape is Linear.
43. Describe the structure of $XeF_4$ using VSEPR theory.
Central Xe has 8 valence electrons. It forms 4 single bonds with F, leaving 4 electrons as 2 lone pairs.
Total electron pairs = 4 bond pairs + 2 lone pairs = 6 pairs ($sp^3d^2$ hybridization).
The 2 lone pairs occupy opposite axial positions to minimize repulsion. The 4 Fluorine atoms occupy equatorial positions. The resulting shape is Square Planar.
44. Describe the structure of $XeF_6$ using VSEPR theory.
Central Xe has 8 valence electrons. It forms 6 single bonds with F, leaving 2 electrons as 1 lone pair.
Total electron pairs = 6 bond pairs + 1 lone pair = 7 pairs ($sp^3d^3$ hybridization).
Due to the presence of one lone pair, the regular octahedral geometry is distorted. The resulting shape is Distorted Octahedral.
45. Write the equation for the complete hydrolysis of $XeF_6$. What is the structure of the product?
Complete hydrolysis of $XeF_6$ yields Xenon trioxide ($XeO_3$).
$$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$$
Structure of $XeO_3$: Xe forms 3 double bonds with O (uses 6 electrons), leaving 1 lone pair. With 3 bond pairs ($\sigma$ bonds) and 1 lone pair ($sp^3$ hybridization), the shape is Pyramidal.
46. Write the equation for the partial hydrolysis of $XeF_6$.
Partial hydrolysis yields oxyfluorides:
With 1 mole of water: $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$
With 2 moles of water: $XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF$
47. State any two uses of Helium.
- It is a non-combustible and light gas, hence used in filling balloons for meteorological observations.
- It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood, preventing "the bends".
- Liquid helium is used as a cryogenic agent to maintain very low temperatures (for superconducting magnets in MRI).
48. State any two uses of Neon and Argon.
- Neon: Widely used in discharge tubes and fluorescent bulbs for brilliant advertisement display signs ("Neon signs").
- Argon: Used to provide an inert atmosphere in high-temperature metallurgical processes (arc welding) and for filling electric light bulbs to prevent oxidation of the tungsten filament.
49. Mention two important uses of Sulfuric acid ($H_2SO_4$).
- Manufacture of fertilizers like superphosphate, ammonium sulfate, and soluble phosphate.
- Used as an electrolyte in lead storage batteries (accumulators).
- Used extensively in petroleum refining and the manufacture of explosives, dyes, and drugs. (Often called the "King of Chemicals").
50. Mention two important uses of Chlorine ($Cl_2$).
- Used for the purification and sterilization of drinking water and in swimming pools to kill bacteria.
- Used in the manufacture of bleaching powder, poisonous gases (phosgene, mustard gas), and organic solvents like $CHCl_3$, $CCl_4$, and PVC plastics.
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