Maharashtra Board Class 12 Chemistry Chapter 7 Elements of Groups 16 17 and 18 Solutions

Elements Of Groups 16, 17 And 18 Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 7 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 7 Exercise Solutions

1. Select Appropriate Answers For The Following.

Question i. Which of the following has the highest electron gain enthalpy?
(A) Fluorine
(B) Chlorine
(C) Bromine
(D) lodine
Answer: (B) Chlorine
In simple words: Chlorine has the highest electron gain enthalpy among halogens due to its larger atomic size compared to fluorine, which reduces electron-electron repulsion, allowing it to more easily accept an extra electron.

🎯 Exam Tip: Remember the exception in electron gain enthalpy for fluorine and chlorine; chlorine typically has a higher value due to reduced electron repulsion in its larger atomic orbitals.

Question ii. Hydrides of group 16 are weakly acidic. The correct order of acidity is
(A) H2O > H2S > H2Se > H2Te
(B) H2Te > H2O > H2S > H2Se
(C) H2Te > H2Se > H2S > H2O
(D) H2Te > H2Se > H2O > H2S
Answer: (C) H2Te > H2Se > H2S > H2O
In simple words: The acidic strength of hydrides in group 16 increases down the group because the bond strength decreases with increasing atomic size of the central element, making it easier to release a proton.

🎯 Exam Tip: Acidity of hydrides in a group is primarily determined by bond strength, which decreases as the central atom's size increases, making proton release easier.

Question iii. Which of the following element does not show oxidation state of +4 ?
(A) Ο
(B) S
(C) Se
(D) Te
Answer: (A) Ο
In simple words: Oxygen does not exhibit a +4 oxidation state because it lacks vacant d-orbitals to expand its octet.

🎯 Exam Tip: Elements in the second period (like Oxygen) cannot expand their octet due to the absence of d-orbitals, limiting their maximum oxidation states.

Question iv. Hl acid when heated with conc. H2SO4 forms
(A) HIO3
(B) KIO3
(C) I2
(D) KI
Answer: (C) I2
In simple words: Concentrated sulfuric acid acts as an oxidizing agent, oxidizing HI to elemental iodine when heated.

🎯 Exam Tip: Remember that concentrated H2SO4 is a strong oxidizing agent, especially when heated, and can oxidize hydrohalic acids (except HCl) to their respective halogens.

Question v. Ozone layer is depleted by
(A) NO
(B) NO2
(C) NO3
(D) N2O5
Answer: (A) NO
In simple words: Nitric oxide (NO) acts as a catalyst in the depletion of the ozone layer by reacting with ozone to form oxygen and nitrogen dioxide.

🎯 Exam Tip: Recall that catalytic cycles involving species like NO, CFCs, and hydroxyl radicals are key mechanisms for ozone depletion in the stratosphere.

Question vi. Which of the following occurs in liquid state at room temperature?
(A) HIO3
(B) HBr
(C) HCI
(D) HF
Answer: (D) HF
In simple words: Hydrogen fluoride exists as a liquid at room temperature due to strong intermolecular hydrogen bonding, which causes its molecules to associate.

🎯 Exam Tip: Hydrogen bonding is a crucial factor influencing the physical states of compounds, particularly for hydrides of highly electronegative elements like fluorine.

Question vii. In pyrosulfurous acid oxidation state of sulfur is
(A) Only +2
(B) Only +4
(C) +2 and +6
(D) Only +6
Answer: (B) Only + 4
In simple words: In pyrosulfurous acid (\( H_2S_2O_5 \)), the sulfur atoms exhibit an oxidation state of +4.

🎯 Exam Tip: To determine the oxidation state, remember that hydrogen is typically +1 and oxygen is -2; use the sum of oxidation states in a neutral compound equal to zero.

Question viii. Stability of interhalogen compounds follows the order
(A) BrF > IBr > ICI > CIF > BrCI
(B) IBr > BeF > ICI > CIF > BrCI
(C) CIF > ICI > IBr > BrCI > BrF
(D) ICI > CIF > BrCI > IBr > BrF
Answer: (C) CIF > ICI > IBr > BrCI > BrF
In simple words: The stability of interhalogen compounds generally decreases as the size difference between the constituent halogen atoms increases, and also depends on the bond strength.

🎯 Exam Tip: Factors affecting interhalogen compound stability include bond polarity, bond strength, and steric hindrance, often correlating with the size difference between halogens.

Question ix. BrCl reacts with water to form
(A) HBr
(B) Br2 + Cl2
(C) HOBr
(D) HOBr + HCI
Answer: (D) HOBr + HCI
In simple words: When bromine monochloride (BrCl) reacts with water, it undergoes hydrolysis to produce hypobromous acid (HOBr) and hydrochloric acid (HCl).

🎯 Exam Tip: Interhalogen compounds often undergo hydrolysis, where the more electronegative halogen forms the acid with hydrogen, and the less electronegative halogen forms a oxyacid (or an acid if its oxidation state allows).

Question x. Chlorine reacts with excess of fluorine to form.
(A) CIF
(B) CIF3
(C) CIF2
(D) CI2F3
Answer: (B) CIF3
In simple words: When chlorine reacts with an excess of fluorine, the larger amount of fluorine allows for the formation of a higher oxidation state interhalogen compound, specifically chlorine trifluoride (\( ClF_3 \)).

🎯 Exam Tip: The stoichiometry of reactants (e.g., excess of one halogen) dictates the type and oxidation state of the interhalogen compound formed.

Question xi. In interhalogen compounds, which of the following halogens is never the central atom.
(A) I
(B) CI
(C) Br
(D) F
Answer: (D) F
In simple words: Fluorine, being the most electronegative element and having the smallest atomic size, always acts as the terminal atom, never the central atom, in interhalogen compounds.

🎯 Exam Tip: The central atom in interhalogen compounds is always the less electronegative and larger halogen, allowing it to accommodate multiple surrounding halogen atoms.

Question xii. Which of the following has one lone pair of electrons?
(A) IF3
(B) ICI
(C) IF5
(D) CIF3
Answer: (C) IF5
In simple words: In iodine pentafluoride (\( IF_5 \)), iodine has five bond pairs with fluorine atoms and one remaining lone pair of electrons, resulting in a total of six electron domains around the central iodine atom.

🎯 Exam Tip: To determine lone pairs, calculate the total valence electrons for the central atom minus electrons used in bonding, then divide by two. This is crucial for VSEPR theory predictions.

Question xiii. In which of the following pairs, molecules are paired with their correct shapes?
(A) [13] : bent
(B) BrF5 : trigonal bipyramid
(C) CIF3: trigonal planar
(D) [BrF4]: square planar
Answer: (A) [13]: bent
In simple words: The question asks to identify the correctly matched molecule and its shape. While the triiodide ion (\( I_3^- \)) is typically linear, the given answer indicates that \( [I_3] \) (possibly referring to a different species or a specific interpretation) is considered bent in this context.

🎯 Exam Tip: When determining molecular shapes using VSEPR theory, correctly count bond pairs and lone pairs around the central atom. Be aware of common polyatomic ions like \( I_3^- \) which are linear, but always refer to specific curriculum context if an unconventional answer is provided.

Question xiv. Among the known interhalogen compounds, the maximum number of atoms is
(A) 3
(B) 6
(C) 7
(D) 8
Answer: (D) 8
In simple words: The largest known interhalogen compound is iodine heptafluoride (\( IF_7 \)), which consists of one iodine atom and seven fluorine atoms, totaling eight atoms.

🎯 Exam Tip: Recall the common formulas for interhalogen compounds, \( XY_n \), and the maximum value of 'n' (typically 1, 3, 5, or 7) is found with fluorine as the peripheral atom due to its small size and high electronegativity, allowing the central atom to accommodate more. The maximum number of atoms for \( IF_7 \) is 1+7 = 8.

2. Answer The Following.

Question i. Write the order of the thermal stability of the hydrides of group 16 elements.
Answer: The thermal stability of the hydrides of group 16 elements decreases in the order of \( H_2O > H_2S > H_2Se > H_2Te \).
In simple words: Thermal stability of Group 16 hydrides decreases down the group because the bond strength between hydrogen and the central atom weakens as the central atom's size increases.

🎯 Exam Tip: Thermal stability is inversely related to bond length; longer bonds (with larger central atoms) are weaker and break more easily upon heating.

Question ii. What is the oxidation state of Te in TeO2?
Answer: The oxidation state of Te in \( TeO_2 \) is + 4.
In simple words: In tellurium dioxide (\( TeO_2 \)), oxygen typically has an oxidation state of -2, so for a neutral molecule with two oxygen atoms, tellurium must have an oxidation state of +4 to balance the charges.

🎯 Exam Tip: To calculate oxidation states, assign known values (like -2 for oxygen, +1 for hydrogen) and then solve for the unknown element, ensuring the sum of oxidation states equals the charge of the compound or ion.

Question iii. Name two gases which deplete ozone layer.
Answer: Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.
In simple words: Nitrogen oxides and chlorofluorocarbons are two major types of gases that catalyze the breakdown of ozone molecules in the stratosphere, leading to ozone layer depletion.

🎯 Exam Tip: Focus on understanding the chemical reactions (e.g., catalytic cycles) by which these pollutants interact with and destroy ozone molecules.

Question iv. Give two uses of CIO2
Answer:
(i) CIO2 is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.
In simple words: Chlorine dioxide (\( ClO_2 \)) is a strong oxidizing agent primarily used for bleaching in industries like paper and textile, and for disinfecting water.

🎯 Exam Tip: When listing uses, categorize them (e.g., industrial, environmental) to help recall and provide clear, distinct applications for chemical compounds.

Question v. What is the action of bromine on magnesium metal?
Answer: Bromine reacts instantly with magnesium metal to give magnesium bromide.
\( Mg_{(s)} + Br_{2(l)} \rightarrow MgBr_{2(s)} \)
Magnesium bromide
In simple words: Magnesium metal reacts readily with bromine liquid in a direct combination reaction to form solid magnesium bromide.

🎯 Exam Tip: Remember that halogens readily react with active metals like magnesium to form ionic halides, often in a vigorous reaction.

Question vi. Write the names of allotropic forms of selenium.
Answer: Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form
In simple words: Selenium exists in different physical forms, including a red form which is non-metallic and a grey form that exhibits metallic properties.

🎯 Exam Tip: Allotropy refers to the property of some chemical elements to exist in two or more different forms, often with distinct physical and chemical properties, like carbon (diamond, graphite) or oxygen (\( O_2, O_3 \)).

Question vii. What is the oxidation state of S in H2SO4.
Answer: The oxidation state of S in \( H_2SO_4 \) is + 6.
\( \overset{+2}{(H_2)} \overset{+6}{(S)} \overset{-8}{(O_4)} \)
In simple words: In sulfuric acid (\( H_2SO_4 \)), the sulfur atom has an oxidation state of +6, which is its maximum oxidation state, resulting from the sum of charges from hydrogen (+1 each) and oxygen (-2 each).

🎯 Exam Tip: When calculating oxidation states, remember the common assignments: +1 for hydrogen, -2 for oxygen (unless in peroxides or with fluorine), and the sum must equal the overall charge of the compound.

Question viii. The pKa values of HCl is -7.0 and that of Hl is -10.0. Which is the stronger acid?
Answer: For HCI, pKa = -7.0, hence its dissoClation constant is, \( K_a = 1 \times 10^{-7} \).
For HI pKa = - 10.0, hence its dissoClation constant is \( K_a = 1 \times 10^{-7} \). Hence HCI dissoClates more than HI.
Therefore HCI is a stronger acid than HI.
In simple words: A lower pKa value indicates a stronger acid, meaning it dissociates more readily. Based on the provided pKa values, HI is a stronger acid than HCl.

🎯 Exam Tip: Remember that the stronger the acid, the lower its pKa value (more negative) and the larger its dissociation constant (Ka). If pKa(HI) is -10.0 and pKa(HCl) is -7.0, then \( K_a(HI) = 10^{10} \) and \( K_a(HCl) = 10^7 \), indicating HI is stronger. There seems to be a discrepancy in the provided Ka values or conclusion for the given pKa values.

Question ix. Give one example showing reducing property of ozone.
Answer: Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. \( O_{3(g)} \rightarrow O_{2(g)} + O \)
For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO4).
\( PbS_{(s)} + 4O_{3(g)} \rightarrow PbSO_{(s)} + 4O_{2(g)} \)
(ii) Potassium iodide, Kl is oxidised to iodine, \( I_2 \) in the solution.
\( 2KI_{(aq)} + H_2O_{(l)} + O_{3(g)} \rightarrow 2KOH_{(aq)} + I_{2(s)} + O_{2(g)} \)
In simple words: Although the question asks for the reducing property of ozone, the provided answer describes ozone's powerful oxidizing property, illustrating it by the oxidation of lead sulfide and potassium iodide due to the release of nascent oxygen. Ozone is primarily an oxidizing agent.

🎯 Exam Tip: Be careful to distinguish between oxidizing and reducing properties. Ozone is a strong oxidizing agent; it oxidizes other substances while being reduced itself. If asked for a reducing property, an example where ozone donates electrons or reduces another species would be required (which is rare for ozone).

Question x. Write the reaction of conc. H2SO4 with sugar.
Answer: Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.
\( C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O \)
The carbon that is left behind is called sugar charcoal and the process is called char.
In simple words: Concentrated sulfuric acid acts as a dehydrating agent, removing water from sugar (sucrose) to leave behind black carbon, a process known as charring.

🎯 Exam Tip: Remember concentrated sulfuric acid's strong dehydrating property, often demonstrated by its reaction with carbohydrates like sugar, producing carbon and water.

Question xi. Give two uses of chlorine.
Answer: Chlorine is used for :
• for sterilization of drinking water.
• bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
• in the manufacture of organic compounds like \( CHCl_3, CCl_4 \), DDT, dyes and drugs.
• in the extraction of metals like gold and platinum.
• in the manufacture of refrigerant like Freon (i.e., \( CCl_2F_2 \)).
• in the manufacture of several poisonous gases like mustard gas (Cl-C2H4-S-C2H4-Cl), phosgene (\( COCl_2 \)) used in warfare.
• in the manufacture of tear gas (\( CCl_3NO_2 \)).
In simple words: Chlorine is widely used as a disinfectant for water purification and as a powerful bleaching agent in the paper and textile industries, as well as in the synthesis of various organic and inorganic compounds.

🎯 Exam Tip: When listing uses of elements or compounds, focus on broad categories like disinfection, bleaching, or synthesis of other important chemicals. Providing multiple distinct uses showcases comprehensive knowledge.

Question xii. Complete the following.
1. \( ICI_3 + H_2O \rightarrow \) ___________ \( + ICI \)
2. \( I_2 + KClO_3 \rightarrow \) ___________ \( + KIO_2 \)
3. \( BrCl + H_2O \rightarrow \) ___________ \( + HCl \)
4. \( Cl_2 + ClF_3 \rightarrow \) ___________
5. \( H_2C=CH_2 + ICl \rightarrow \) ___________
6. \( XeF_4 + SiO_2 \rightarrow \) ___________ \( + SiF_4 \)
7. \( XeF_6 + 6H_2O \rightarrow \) ___________ \( + HF \)
8. \( XeOF_4 + H_2O \rightarrow \) ___________ \( + HF \)
Answer:
1. \( 2ICl_3 + 3H_2O \rightarrow 5HCl + HIO_3 + ICl \)
2. \( I_2 + KClO_3 \rightarrow ICl + KIO_3 \)
3. \( BrCl + H_2O \rightarrow HOBr + HCl \)
4. \( Cl_2 + ClF_3 \rightarrow 3ClF \)
5. \( CH_2 = CH_2 + ICl \rightarrow CH_2I - CH_2Cl \)
6. \( 2XeF_6 + SiO_2 \rightarrow 2XeOF_4 + SiF_4 \)
7. \( XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF \)
8. \( XeOF_4 + H_2O \rightarrow XeO_2F_2 + 2HF \)
In simple words: These reactions demonstrate various chemical transformations involving interhalogen compounds, xenon fluorides, and their reactions with water or other compounds, often leading to hydrolysis or substitution products.

🎯 Exam Tip: For completing reactions, identify the reactants and apply knowledge of common reaction types (hydrolysis, redox, addition) and balancing to determine the products. Pay attention to stoichiometry and reaction conditions.

Question xiii. Match the following
A-B
XeOF2 – Xenon trioxydifluoride
XeO2F2 – Xenon monooxydifluoride
XeO3F2 - Xenon dioxytetrafluoride
XeO2F4 - Xenon dioxydifluoride
Answer:
XeOF2 - Xenon monooxydifluoride
XeO2F2 - Xenon dioxydifluoride
XeO3F2 - Xenon trioxydifluoride
XeO2F4 - Xenon dioxytetrafluoride
In simple words: This question matches the chemical formulas of xenon oxyfluorides with their correct systematic names, which describe the number of oxygen and fluorine atoms present.

🎯 Exam Tip: Systematically name xenon oxyfluorides by counting the number of oxygen ("oxo") and fluorine ("fluoro") atoms, and ensure the prefixes (mono-, di-, tri-, tetra-) correspond to the correct count for each element.

Question xiv. What is the oxidation state of xenon in the following compounds?
XeOF4, XeO3, XeF5, XeF4, XeF2.
Answer:

Compound	Oxidation state of Xe
XeOF4	+ 6
XeO3	+ 6
XeF6	+ 6
XeF4	+ 4
XeF2	+ 2

In simple words: The oxidation state of xenon in its compounds is determined by balancing the charges of the other elements (like oxygen at -2 and fluorine at -1). Note that while XeF5 was asked, its oxidation state is not provided in the answer table, which includes XeF6 instead.

🎯 Exam Tip: Carefully list all compounds mentioned in the question and ensure their oxidation states are derived correctly. If there's a mismatch between the question's list and the answer's list, be prepared to address the compounds given in both contexts.

3. Answer The Following.

Question i. The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 \( kJ\ mol^{-1} \), respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, \( ^{16}S < ^{17}Cl < ^{18}Ar \).
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 \( kJ\ mol^{-1} \)) is higher than that of S (1000 \( kJ\ mol^{-1} \)).
(iv) Ar has electronic configuration \( 3s^23p^6 \). Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 \( kJ\ mol^{-1} \)).
In simple words: Across a period, ionization enthalpy generally increases due to increasing effective nuclear charge and decreasing atomic size. Argon has the highest ionization enthalpy because its stable, full octet electron configuration makes it very difficult to remove an electron.

🎯 Exam Tip: Ionization enthalpy trends across a period are influenced by increasing effective nuclear charge and decreasing atomic radius, with noble gases showing exceptionally high values due to their stable electron configurations.

Question ii. "Acidic character of hydrides of group 16 elements increases from H2O to H2Te" Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from \( H_2O \) to \( H_2Te \). This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from \( H_2O \) to \( H_2Te \).
In simple words: As you move down Group 16, the size of the central atom increases, leading to weaker H-E bonds. Weaker bonds mean the hydrogen can be released more easily as a proton, thus increasing the acidic character of the hydrides.

🎯 Exam Tip: For hydrides within a group, acidic strength is primarily determined by bond strength; weaker bonds (due to larger atomic size) lead to greater acidity because protons are more easily released.

Question iii. How is dioxygen prepared in laboratory from KClO3?
Answer: By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.
\( 2KClO_3 \xrightarrow{\Delta, MnO_2} 2KCl + 3O_{2(g)} \)
In simple words: Dioxygen can be prepared in the lab by heating potassium chlorate with manganese dioxide as a catalyst, which speeds up its decomposition into potassium chloride and oxygen gas.

🎯 Exam Tip: Remember that catalysts like manganese dioxide are used to lower the activation energy of a reaction, allowing it to proceed at a faster rate or lower temperature without being consumed itself.

Question iv. What happens when
a. Lead sulfide reacts with ozone (\( O_3 \)).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO4) changing the oxidation state of S from - 2 to +6.
\( PbS_{(s)} + 4O_{3(g)} \rightarrow PbSO_{(s)} + 4O_{2(g)} \)
(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
\( NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)} \)
In simple words: Ozone is a strong oxidizing agent; it oxidizes lead sulfide to lead sulfate and nitric oxide to nitrogen dioxide in these reactions.

🎯 Exam Tip: Ozone's strong oxidizing power is evident in its reactions, where it typically oxidizes other compounds while being reduced to oxygen itself.

Question v. Give two chemical reactions to explain oxidizing property of concentrated H2SO4.
Answer: Hot and concentrated \( H_2SO_4 \) acts as an oxidising agent, since it gives nascent oxygen on heating.
\( H_2SO_4 \xrightarrow{\Delta, \text{conc.}} H_2O + SO_2 + [O] \)
It oxidises metals and non-metals. For example,
\( Cu_{(s)} + 2H_2SO_4 \xrightarrow{\text{conc.}} CuSO_4 + 2H_2O + SO_2 \)
(metal)
\( C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O \)
(non-metal)
In simple words: Concentrated sulfuric acid is a potent oxidizing agent because it releases nascent oxygen when heated, allowing it to oxidize both metals (like copper) and non-metals (like carbon).

🎯 Exam Tip: Recognize concentrated sulfuric acid as a strong dehydrating and oxidizing agent; its oxidizing property is often demonstrated by its reactions where it is reduced to \( SO_2 \) while oxidizing another substance.

Question vi. Discuss the structure of sulfur dioxide.
Answer:
(i) \( SO_2 \) molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 \( kJ\ mol^{-1} \).
(ii) Sulphur in \( SO_2 \) is \( sp^2 \) hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In \( SO_2 \), each oxygen atom is bonded to sulphur by \( \sigma \) and a \( \pi \) bond.
(iv) A bond between S and O are formed by \( sp^2 \)-p overlapping.
(v) One of \( \pi \) bonds is formed by p\( \pi \) - p\( \pi \) overlapping while other \( \pi \) bond is formed by p\( \pi \) - d\( \pi \) overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सल्फर डाइऑक्साइड (\( SO_2 \)) की संरचना को दर्शाता है। इसमें सल्फर के ग्राउंड स्टेट, फर्स्ट एक्साइटेड स्टेट और \( sp^2 \) हाइब्रिडाइज्ड स्टेट के इलेक्ट्रॉनिक कॉन्फ़िगरेशन को दर्शाया गया है, जिसमें ऑर्बिटल्स में इलेक्ट्रॉनों का वितरण और उनके युग्मन को दिखाया गया है। साथ ही, \( SO_2 \) के अनुनाद संरचनाओं को दर्शाया गया है, जिसमें सल्फर और ऑक्सीजन के बीच \( \sigma \), \( p\pi-p\pi \) और \( p\pi-d\pi \) बंधों का वर्णन है, जो यह दर्शाता है कि बांड की लंबाई कैसे समान होती है।
In simple words: Sulfur dioxide has a bent V-shaped structure with \( sp^2 \) hybridized sulfur. Its bonds are identical due to resonance, involving both \( p\pi-p\pi \) and \( p\pi-d\pi \) overlap, and it has a bond angle slightly less than 120° due to lone pair repulsion.

🎯 Exam Tip: When describing molecular structures, always mention hybridization, bond angles, molecular geometry (VSEPR), and the role of resonance in equalizing bond lengths and strengths.

Question vii. Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:
• Halogens have outer electronic configuration \( ns^2 np^5 \).
• Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration \( ns^2np^6 \).
• Hence they are monovalent and show oxidation state - 1.
• Since fluorine does not have vacant d-orbital, it shows only one oxidation state of - 1 while all other halogens show variable oxidation states from - 1 to +7.
• These oxidation states are, - 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.
In simple words: Fluorine's inability to expand its octet due to the absence of d-orbitals restricts it to only a -1 oxidation state, whereas other halogens can utilize their vacant d-orbitals to exhibit higher positive oxidation states.

🎯 Exam Tip: The presence or absence of vacant d-orbitals is critical for explaining the variable oxidation states of elements. Second-period elements like fluorine cannot expand their octet, limiting their chemical behavior.

Question viii. What is the action of chlorine on the following
a. Fe
b. Excess of NH3
Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
\( 2Fe + 3Cl_2 \rightarrow 2FeCl_3 \)
(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, \( NH_4Cl \) and nitrogen.
\( 3Cl_2 + 8NH_3 \xrightarrow{\text{excess}} 6NH_4Cl + N_2 \)
In simple words: Chlorine reacts with iron to form iron(III) chloride and with excess ammonia to produce ammonium chloride and nitrogen gas.

🎯 Exam Tip: Remember specific reaction conditions (like "excess of NH3") as they often dictate the products formed; for example, if chlorine is in excess with ammonia, a different reaction pathway and products would occur.

Question ix. How is hydrogen chloride prepared from sodium chloride?
Answer:
1. In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated \( H_2SO_4 \).
\( NaCl + H_2SO_4 \xrightarrow{420\ K} NaHSO_4 + HCl \)
\( NaHSO_4 + NaCl \xrightarrow{420\ K} Na_2SO_4 + HCl \)
2. Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated \( H_2SO_4 \) and then collected by upward displacement of air.
In simple words: Hydrogen chloride is prepared by reacting sodium chloride with concentrated sulfuric acid, initially producing sodium bisulfate and HCl, which can further react with more NaCl at higher temperatures to yield sodium sulfate and more HCl. The gas is then dried and collected.

🎯 Exam Tip: Laboratory preparation reactions often involve specific temperature conditions and the use of drying agents. Remember the two-step process for HCl generation from NaCl and concentrated \( H_2SO_4 \).

Question x. Draw structures of XeF6, XeO3, XeOF4, XeF2.
Answer: The structures of the compounds are shown below:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विभिन्न क्सीनन यौगिकों की आणविक संरचनाओं को दर्शाता है: क्सीनन हेक्साफ्लोराइड (XeF₆) एक विकृत अष्टफलकीय आकार में, क्सीनन ट्राइऑक्साइड (XeO₃) एक त्रिकोणीय पिरामिड के आकार में एक एकाकी युग्म के साथ, क्सीनन ऑक्सिटेट्राफ्लोराइड (XeOF₄) एक वर्गाकार पिरामिड के आकार में, और क्सीनन डाइफ्लोराइड (XeF₂) एक रेखीय अणु के रूप में, जिसमें क्सीनन पर तीन एकाकी युग्म होते हैं।
In simple words: The diagram illustrates the molecular geometries of xenon compounds: XeF₆ as distorted octahedral, XeO₃ as trigonal pyramidal, XeOF₄ as square pyramidal, and XeF₂ as linear, showcasing how lone pairs and bonding atoms determine their shapes.

🎯 Exam Tip: For drawing structures, apply VSEPR theory to predict electron geometry and molecular geometry, paying attention to lone pairs on the central atom and their effect on bond angles and overall shape. Clearly labeling each structure is essential.

Question xi. What are interhalogen compounds? Give two examples.
Answer: Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example CIF, \( BrF_3 \), ICI
In simple words: Interhalogen compounds are binary compounds formed between two different halogens, where the less electronegative halogen acts as the central atom, and examples include chlorine monofluoride (ClF) and bromine trifluoride (BrF₃).

🎯 Exam Tip: Remember that in interhalogen compounds, the central atom is always the larger, less electronegative halogen, while the surrounding atoms are the smaller, more electronegative ones, predominantly fluorine.

Question xii. What is the action of hydrochloric acid on the following?
a. NH3
b. Na2CO3
Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
\( NH_3 + HCl \rightarrow NH_4Cl \)
b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
\( Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2 \)
In simple words: Hydrochloric acid reacts with ammonia to form ammonium chloride and reacts with sodium carbonate to produce sodium chloride, water, and carbon dioxide.

🎯 Exam Tip: Recognize HCl as a strong acid that neutralizes bases like ammonia and reacts with carbonates to produce carbon dioxide gas, a common test for carbonates.

Question xiii. Give two uses of HCI.
Answer: Hydrogen chloride (OR hydrochloric acid) is used :
• in the manufacture of chlorine and ammonium chloride,
• to manufacture glucose from corn, starch
• to manufacture dye
• in mediCIne and galvanising
• as an important reagent in the laboratory
• to extract glue from bones and for the purification of bone black.
• for dissolving metals, \( Fe + 2HCl_{(aq)} \rightarrow FeCl_2 + H_{2(g)} \)
In simple words: Hydrochloric acid is a versatile chemical used in various industrial processes, including manufacturing, food production, metallurgy, and as a laboratory reagent.

🎯 Exam Tip: When listing uses, provide diverse applications, such as its role in industrial synthesis, food processing, and analytical chemistry, to demonstrate a broad understanding of its importance.

Question xiv. Write the names and structural formulae of oxoacids of chlorine.
Answer: The names and structural formulae of the oxoacids of chlorine are provided in the table below.
In simple words: The question asks for the names and structural representations of chlorine's oxoacids, which are compounds containing chlorine, oxygen, and hydrogen, showcasing varying oxidation states of chlorine.

🎯 Exam Tip: When dealing with oxoacids of a single element, pay attention to the varying oxidation states of the central atom, as this directly influences the number of oxygen atoms and, consequently, the acid's name and structure.

 

Question xv.What happens when a. Cl2 reacts with F2 in equal volume at 437 K. b. Br2 reacts with excess of F2.
Answer:(a) Cl2 reacts with F2 in equal volumes at 437 K to give chlorine monofluoride CIF. Cl2 + F2 \( \xrightarrow{437 K} \) 2CIF (equal volumes)
In simple words: When chlorine and fluorine react in equal amounts at 437 K, they form chlorine monofluoride.

🎯 Exam Tip: Pay attention to reaction conditions (temperature, ratios) as they influence the product formed in interhalogen reactions.

 

(b) Br2 reacts with excess of F2 to give bromine trifluoride. Br2 + 3F2 \( \xrightarrow{\Delta} \) 2BrF3 (excess)
In simple words: When bromine reacts with an excess of fluorine, it forms bromine trifluoride.

🎯 Exam Tip: Note how excess reactant can determine the stoichiometry and product formation, leading to higher halogen fluorides.

 

Question xvi.How are xenon fluorides XeF2, XeF4 and XeF6 obtained ? Give suitable reactions.
Answer:Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction. (i) Preparation of XeF2 : Xe + F2 \( \xrightarrow{\text{sealed Ni tube, } 400 \text{ °C}} \) XeF2 (ii) Preparation of XeF4 : Xe + F2 \( \xrightarrow{\text{Ni tube, } 400 \text{ °C}} \) XeF4 (1:5) 5-6 atm
Xe + 2F2 \( \xrightarrow{\text{electric discharge}} \) XeF4 80 °C (iii) Preparation of XeF6: Xe + 3F2 \( \xrightarrow{\text{electric discharge}} \) XeF6 low temperature
In simple words: Xenon fluorides are synthesized by reacting xenon directly with fluorine, adjusting the stoichiometry, temperature, and pressure to obtain different fluorides like XeF2, XeF4, and XeF6.

🎯 Exam Tip: Memorize the specific reaction conditions (temperature, pressure, ratio of reactants) for preparing each xenon fluoride, as these are critical for distinguishing their synthesis methods.

 

Question xvii.How are XeO3 and XeOF4 prepared?
Answer:Preparation of XeO3 : Xenon trioxide (XeO3) is prepared by the hydrolysis of XeF4 or XeF6. • By hydrolysis of XeF4 : 3XeF4 + 6H20
\( \implies \) 2Xe + XeO3 + 12 HF + \( \frac{1}{2} \)O2 • By hydrolysis of XeF6 : XeF6 + 3H2O
\( \implies \) XeO3 + 6HF • Preparation of XeOF4 : Xenon oxytetrafluoride (XeOF4) is prepared by the partial hydrolysis of XeF6. XeF6 + H2O
\( \implies \) XeOF4 + 2HF
In simple words: Xenon trioxide (XeO3) is made by fully hydrolyzing XeF4 or XeF6, while xenon oxytetrafluoride (XeOF4) results from the partial hydrolysis of XeF6.

🎯 Exam Tip: Understand that the extent of hydrolysis (partial vs. complete) of xenon fluorides dictates the formation of different xenon-oxygen compounds like XeO3 and XeOF4.

 

Question xviii.Give two uses of neon and argon.
Answer:Uses of neon (Ne) : • Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes. • Neon signs are visible from a long distance and also have high penetrating power in mist or fog. • A mixture of neon and helium is used in voltage stabilizers and current rectifiers. • Neon is also used in the production of lasers and fluorescent tubes. Uses of argon (Ar) : • Argon is used to fill fluorescent tubes and radio valves. • It is used to provide inert atmosphere for welding and production of steel. • It is used along with neon in neon sign lamps to obtain different colours. • A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.
In simple words: Neon is commonly used for signs and lighting due to its bright glow and penetrating power, while argon is valued for creating inert atmospheres in welding, metallurgy, and in certain types of lighting.

🎯 Exam Tip: Focus on linking specific properties of noble gases (e.g., inertness, specific light emission) to their practical applications in industry and daily life.

 

Question xix.Describe the structure of Ozone. Give two uses of ozone.
Answer:(A) • Ozone has molecular formula O3. • The lewis dot and dash structures for O3 are :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ओजोन (O3) की अनुनाद संरचनाओं को दर्शाता है। इसमें एक केंद्रीय ऑक्सीजन परमाणु दो अन्य ऑक्सीजन परमाणुओं से जुड़ा होता है। एक संरचना में, एक ऑक्सीजन परमाणु केंद्रीय परमाणु से डबल बॉन्ड और दूसरा सिंगल बॉन्ड द्वारा जुड़ा होता है, जबकि दूसरी अनुनाद संरचना में बॉन्ड की स्थिति उलट जाती है। प्रत्येक ऑक्सीजन परमाणु पर उचित औपचारिक आवेश और lone pairs दिखाए गए हैं। • Infrared and electron diffraction spectra show that O3 molecule is angular with 0-0-0 bond angle 117°. • Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds. (B) Uses of Ozone : • Ozone sterilises drinking water by oxidising germs and bacteria present in it. • It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk. • Ozone is used to purify the air in crowded places like Cinema halls, railways, tunnels, etc. • In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.
In simple words: Ozone (O3) has an angular shape with a 117° bond angle, exhibiting resonance where both oxygen-oxygen bonds are identical and intermediate in length between single and double bonds. It is used for water purification, bleaching, and in chemical manufacturing due to its strong oxidizing property.

🎯 Exam Tip: When describing ozone's structure, emphasize its angular geometry, resonance, and the intermediate nature of its bond lengths. For uses, remember its strong oxidizing property is the basis for its applications in sterilization, bleaching, and industrial synthesis.

 

Question xx.Explain the trend in following atomic properties of group 16 elements. i. Atomic radii ii. Ionisation enthalpy iii. Electronegativity.
Answer:(1) Atomic and ionic radii : 1. As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge. 2. The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po (2) Ionisation enthalpy : • The ionisation enthalpy of group 16 elements has quite high values. • Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group. • The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations. Group 15 : (valence shell) ns2 npx¹ npy¹ npz¹ Group 16 : (valence shell) ns2 npx² npy¹ npz¹ Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals. Hence group 16 elements have lower first ionisation enthalpy than group 15 elements. (3) Electronegativity : • The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods. • Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4) • On moving down the group electronegativity decreases from oxygen to polonium. • On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

Elements	O	S	Se	Te	Po
Electronegativity	3.5	2.44	2.48	2.01	1.76

In simple words: Group 16 elements show increasing atomic radii down the group, decreasing ionization enthalpy down the group, and generally high electronegativity (decreasing down the group), with these trends influenced by nuclear charge, electron shells, and electronic configuration.

🎯 Exam Tip: For trends in atomic properties, always relate the changes to fundamental principles like increasing atomic number, addition of new electron shells, and effective nuclear charge. Pay attention to exceptions or specific comparisons, like Group 15 vs. Group 16 ionization enthalpies.

 

4. Answer The Following.

 

Question i.Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur	Monoclinic sulphur
1. It is pale yellow.	1. It is bright yellow.
2. Orthorhombic crystals	2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K	3. Melting point, 393 K
4. Density, 2.069 g/cm³	4. Density: 1.989 g/cm³
5. Insoluble in water, but soluble in CS2	5. Soluble in CS2
6. It is stable below 369 K and transforms to α-sulphur above this temperature.	6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S8 molecules with a structure of a puckered ring.	7. It exists as S8 molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS2	8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

In simple words: Rhombic and monoclinic sulfur are two allotropic forms differing in color, crystal structure, melting points, densities, and preparation methods, with rhombic being stable below 369 K and monoclinic above.

🎯 Exam Tip: When comparing allotropes, focus on key differentiating properties like crystal structure, physical appearance, melting point, density, and specific preparation methods.

 

Question ii.Give two reactions showing oxidizing property of concentrated H2SO4.
Answer:Hot and concentrated H2SO4 acts as an oxidising agent, since it gives nascent oxygen on heating. H2SO4 \( \xrightarrow{\text{conc. } \Delta} \) H2O + SO2 + [O] It oxidises metals and non-metals. For example, Cu(s) + 2H2SO4
\( \implies \) CuSO4 + 2H2O + SO2 (metal) (conc.) C + 2H2SO4
\( \implies \) CO2 + 2SO2 + 2H2O (non-metal)
In simple words: Hot and concentrated sulfuric acid acts as a strong oxidizing agent by releasing nascent oxygen upon heating, which can then oxidize both metals like copper and non-metals like carbon.

🎯 Exam Tip: Remember that concentrated H2SO4's oxidizing power stems from its ability to produce nascent oxygen; illustrate this with reactions involving a metal and a non-metal.

 

Question iii.How is SO2 prepared in the laboratory from sodium sulfite? Give two physical properties of SO2.
Answer:(A) Laboratory method (From sulphite) : • Sodium sulphite on treating with dilute H2SO4 forms SO2. Na2SO3 + H2SO4(aq)
\( \implies \) Na2SO4 + H2O(l) + SO2(g) • Sodium sulphite, Na2SO3 on reaction with dilute hydrochloric acid solution forms SO2. Na2SO3(aq) + 2HCl(aq)
\( \implies \) 2NaCl(aq) + H2O(l) + SO2(g) (B) Physical properties of SO2 • It is a colourless gas with a pungent smell. • It is highly soluble in water and forms sulphurous acid, H2SO3. SO2(g) + H2O(l)
\( \implies \) H2SO3(aq) • It is poisonous in nature. • At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.
In simple words: Sulfur dioxide can be prepared in the lab by reacting sodium sulfite with either dilute sulfuric acid or hydrochloric acid, and it is a colorless, pungent, poisonous gas that is highly soluble in water.

🎯 Exam Tip: For laboratory preparations, remember the common reagents and conditions. For physical properties, focus on easily observable characteristics like color, odor, and solubility.

 

Question iv.Describe the manufacturing of H2SO4 by contact process.
Answer:Contact process of the manufacture of sulphuric acid involves following steps : (1) Preparation of SO2 : Sulphur or pyrite ores like iron pyrites, FeS2 on burning in excess of air, form SO2. S(g) + O2(g) \( \xrightarrow{\Delta} \) SO2(g) 4FeS2(s) + 11O2(g) \( \xrightarrow{\Delta} \) 2Fe2O3(s) + 8SO2(g) (2) Oxidation of SO2 to SO3 : SO2 is oxidised to SO3 in the presence of a heterogeneous catalyst V2O5 and atmospheric oxygen. This oxidation reaction is reversible. 2SO2(g) + O2(g) \( \xrightarrow{V2O5} \) 2SO3(g) \( \text{ \(\Delta\)H = -196.6 kJ} \) The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO2 and O2 are taken in the ratio 2:3. To avoid the poisoning of a costly catalyst, it is necessary to make SO2 free from the impurities like dust, moisture, As2O3 poison, etc. (3) Dissolution of SO3 : SO3 obtained from catalytic converter is absorbed in 98% H2SO4 to obtain H2S2O7, oleum or fuming sulphuric acid. SO3(g) + H2SO4
\( \implies \) H2S2O7 98% H2S2O7 + H2O
\( \implies \) 2H2SO4
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र संपर्क विधि द्वारा सल्फ्यूरिक एसिड के निर्माण की प्रक्रिया को दर्शाता है। इसमें सल्फर से सल्फर डाइऑक्साइड का उत्पादन, फिर V2O5 उत्प्रेरक का उपयोग करके SO2 का SO3 में ऑक्सीकरण और अंत में 98% H2SO4 में SO3 को अवशोषित करके ओलियम (H2S2O7) बनाना, जिसे बाद में जल मिलाकर सल्फ्यूरिक एसिड में परिवर्तित किया जाता है। विभिन्न चरण जैसे डस्ट प्रेसिपिटेटर, वाशिंग और कूलिंग टावर, आर्सेनिक प्यूरीफायर और उत्प्रेरक कनवर्टर दिखाए गए हैं।
In simple words: The Contact Process for sulfuric acid manufacturing involves three main steps: first, producing sulfur dioxide from sulfur or pyrite; second, oxidizing SO2 to SO3 using a V2O5 catalyst; and third, absorbing SO3 in concentrated H2SO4 to form oleum, which is then diluted to yield sulfuric acid.

🎯 Exam Tip: For the Contact Process, remember the three key stages (SO2 production, SO3 oxidation, oleum formation), the catalyst (V2O5), and the optimal conditions (temperature, pressure) for efficient conversion.

 

Question 7.1 (Textbook Page No 141)

 

12th Chemistry Digest Chapter 7 Elements Of Groups 16, 17 And 18 Intext Questions And Answers

 

Question 1.Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns2np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns2np4) of group 16 elements of the corresponding period.
In simple words: Group 16 elements have lower first ionization enthalpy than Group 15 elements in the same period because Group 15 has a stable half-filled p-orbital configuration, making it harder to remove an electron.

🎯 Exam Tip: The exceptional stability of half-filled and completely filled orbitals is a crucial concept; recognize how it affects ionization enthalpy trends for elements like Group 15.

 

Question 7.2 (Textbook Page No 141)

 

Question 1.The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol-¹, respectively. Explain the observed trend.
Answer:The elements S and Cl belong to second period of the periodic table. Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.
In simple words: Ionization enthalpy increases across a period because the effective nuclear charge increases and atomic size decreases, making it harder to remove electrons, hence Cl has a higher value than S.

🎯 Exam Tip: Remember that increasing effective nuclear charge and decreasing atomic size across a period are the primary reasons for the increase in ionization enthalpy.

 

Question 7.4 (Textbook Page No 141)

 

Question 1.Fluorine has less negative electron gain affinity than chlorine. Why?
Answer:The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.
In simple words: Fluorine has a less negative electron gain enthalpy than chlorine due to its very small size, which leads to strong electron-electron repulsions within its compact 2p orbitals, making it less receptive to an incoming electron.

🎯 Exam Tip: For electron gain enthalpy comparisons, always consider both atomic size and electron-electron repulsion, especially for small atoms like fluorine where electron density is high.

 

Try This... (Textbook Page No 140)

 

Question 1.Explain the trend in the following properties of group 17 elements. (1) Atomic size, (2) Ionisation enthalpy, (3) Electronegativity, (4) Electron gain enthalpy.
Answer:(1) Atomic size : • Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element. • The atomic radii increase in the order F < Cl < Br < I • Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest. (2) Ionisation enthalpy : • The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction. • The ionisation ethalpies decrease down the group since the atomic size increases. • The ionisation enthalpy decreases in the order F > Cl > Br > I. • Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

Element	F	Cl	Br	I
Ionisation enthalpy kJ/mol	1680	1256	1142	1008

(3) Electronegativity : • Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge. • Each halogen is the most electronegative element of its period. • Fluorine has the highest electronegativity as compared to any element in the periodic table. • The electronegativity decreases as, F > Cl > Br > I 4.0 3.2 3.0 2.7 (electronegativity) (4) Electron gain enthalpy (ΔegH) : • The halogens have the highest negative values for electron gain enthalpy. • Electron gain enthalpies of halogens are negative indicating release of energy. • Halogens liberate maximum heat by gain of electron as compared to other elements. • Since halogens have outer valence electronic configuration, ns2 np5, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements. • In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl. F(g) + e-
\( \implies \) F¯(g) ΔegH = -333 klmol-1 Cl(g) + e-
\( \implies \) Cl¯(g) ΔegH = -349 kJ mol-1 • The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.
In simple words: For Group 17 elements (halogens), atomic size increases down the group, ionization enthalpy decreases down the group (Fluorine highest), electronegativity decreases down the group (Fluorine highest), and electron gain enthalpy values are highly negative, with Chlorine having the most negative value due to fluorine's small size.

🎯 Exam Tip: Remember the general periodic trends (size, IE, EN decrease down a group), but also highlight the specific exceptions or nuances, like Fluorine's relatively lower electron gain enthalpy due to its compact size and electron-electron repulsion.

 

Question 2.Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:• Oxygen has a smaller atomic size than sulphur. • It is more electronegative than sulphur. • It has a larger electron density. • Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.
In simple words: Oxygen has a less negative electron gain enthalpy than sulfur because its smaller atomic size leads to higher electron density and stronger electron-electron repulsions, making it less favorable for an incoming electron despite being more electronegative.

🎯 Exam Tip: When comparing electron gain enthalpy for elements in the same group, especially second period elements with subsequent ones, remember to consider the effect of small atomic size leading to high electron density and inter-electronic repulsion.

 

Question 7.3 (Textbook Page No 141)

 

Question 1.Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer:Oxygen exists as diatomic molecule (O2) whereas sulphur exists as polyatomic molecule (S8). The van der Waals forces of attraction between O2 molecules are relatively weak owing to the much smaller size. The large van der Waals attractive forces in the S8 molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.
In simple words: Oxygen has much lower melting and boiling points than sulfur because oxygen exists as small O2 diatomic molecules with weak van der Waals forces, while sulfur forms larger S8 polyatomic molecules with significantly stronger intermolecular forces.

🎯 Exam Tip: Emphasize the difference in molecular structure (diatomic vs. polyatomic) and its direct impact on intermolecular forces and, consequently, melting and boiling points when explaining property differences within a group.

 

Question 7.5 (Textbook Page No 141)

 

Question 1.Bond dissociation enthalpy of F2 (158.8 kJ mol-¹) is lower than that of Cl2 (242.6 kJ mol-¹) Why?
Answer:Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F2 molecule are so close together that they strongly repel each other, and make the F - F bond weak. Thus, it requires less amount of energy to break the F - F bond. In Cl2 molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less. Thus Cl - Cl bond is comparatively stronger. Therefore bond dissociation enthalpy of F2 is lower than that of Cl2.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक F-F अणु में lone pair-lone pair प्रतिकर्षण को दर्शाता है। इसमें दो फ्लोरीन परमाणु एक दूसरे से सहसंयोजक बंधन से जुड़े हैं, और प्रत्येक परमाणु पर lone pairs दिखाए गए हैं। इन lone pairs के बीच का प्रतिकर्षण, फ्लोरीन के छोटे आकार के कारण, F-F बंधन को कमजोर कर देता है।
In simple words: The F-F bond in fluorine is weaker than the Cl-Cl bond in chlorine, leading to lower bond dissociation enthalpy, because fluorine's small atomic size causes strong lone pair-lone pair repulsions between the closely packed electron clouds of the two fluorine atoms.

🎯 Exam Tip: When explaining bond strengths, especially anomalies in Group 17, remember to consider the effect of atomic size on lone pair repulsion, as this can significantly weaken bonds in smaller atoms like fluorine.

 

Question 7.6 (Textbook Page No 142)

 

Question 1.Noble gases have very low melting and boiling points. Why?
Answer:Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.
In simple words: Noble gases have extremely low melting and boiling points because they exist as single atoms with only weak van der Waals forces of attraction between them, requiring very little energy to overcome these forces.

🎯 Exam Tip: The monoatomic nature of noble gases and the presence of only weak van der Waals forces are the key points to explain their low melting and boiling points.

 

Can You Tell? (Textbook Page No 142)

 

Question 1.The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:The first member of a group usually differs in properties from the rest of the members of the group for the following reasons : • Its small size • High electronegativity • Absence of vacant d-orbitals in its valence shell.
In simple words: The first element in any group often shows distinct properties compared to its heavier congeners due to its exceptionally small size, high electronegativity, and the lack of available d-orbitals for expanded valence.

🎯 Exam Tip: Focus on the three key factors-small size, high electronegativity, and absence of d-orbitals-to explain the anomalous behavior of the first element in a group.

 

Use Your Brain Power! (Textbook Page No 142)

 

Question 1.Oxygen forms only OF2 with fluorine while sulphur forms SF6. Explain. Why?
Answer:• Oxygen combines with the most electronegative element fluorine to form OF2 and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant d-orbitals it cannot exhibit higher oxidation states. • Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF6.
In simple words: Oxygen forms only OF2 because it lacks vacant d-orbitals to expand its valence, limiting its oxidation state to +2 with fluorine, whereas sulfur forms SF6 due to the availability of vacant d-orbitals, allowing it to expand its octet and achieve a +6 oxidation state.

🎯 Exam Tip: The presence or absence of vacant d-orbitals is a critical factor in explaining the variable oxidation states and maximum valency of elements, especially when comparing second period elements with heavier ones in the same group.

 

Question 2.Which of the following possesses hydrogen bonding? H2S, H2O, H2Se, H2Te
Answer:• Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H2O. • The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds. • Thus, hydrogen bonding is not present in the other hydrides H2S, H2Se and H2Te.
In simple words: Only H2O forms hydrogen bonds among the hydrides of Group 16 elements (H2S, H2Se, H2Te) because oxygen is significantly more electronegative than sulfur, selenium, or tellurium, allowing for strong intermolecular hydrogen bonding.

🎯 Exam Tip: Recall that hydrogen bonding requires hydrogen to be directly bonded to a highly electronegative atom like F, O, or N, which explains why H2O forms these bonds but H2S, H2Se, and H2Te do not.

 

Question 3.Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र जल (H2O) के अणुओं के बीच हाइड्रोजन बॉन्डिंग को दर्शाता है। इसमें ऑक्सीजन परमाणु पर आंशिक नकारात्मक आवेश (δ-) और हाइड्रोजन परमाणुओं पर आंशिक सकारात्मक आवेश (δ+) होते हैं। एक अणु का हाइड्रोजन परमाणु दूसरे अणु के ऑक्सीजन परमाणु के lone pair के साथ एक इंटरमॉलिक्युलर हाइड्रोजन बॉन्ड बनाता है, जिससे अणुओं का एक नेटवर्क बनता है।
In simple words: Hydrogen bonding in water (H2O) occurs because the highly electronegative oxygen atom pulls electron density from hydrogen, creating partial positive charges on hydrogen and allowing them to form strong intermolecular attractions with the lone pairs of oxygen atoms in adjacent water molecules.

🎯 Exam Tip: When illustrating hydrogen bonding, clearly indicate the partial charges (δ+ and δ-) on the atoms and use dotted lines to represent the intermolecular hydrogen bonds between hydrogen of one molecule and a highly electronegative atom (like oxygen) of another.

 

Try This..... (Textbook Page No 143)

 

Question 1.Complete the following tables :

Element	O	O	S	F
Compound	H2O	OF2	H2S	HF
Oxidation state	-2	......	......	......
Element	Se	Se	Te	Cl
Compound	SeO2	SeO3	TeF6	HOCl
Oxidation state	......	+6	......	......

Answer:

Element	O	O	S	F
Compound	H2O	OF2	H2S	HF
Oxidation state	-2	+2	-2	-1
Element	Se	Se	Te	Cl
Compound	SeO2	SeO3	TeF6	HOCl
Oxidation state	+4	+6	+6	+1

In simple words: This table shows the typical oxidation states of elements (O, S, Se, Te, F, Cl) in various compounds like H2O, OF2, H2S, HF, SeO2, SeO3, TeF6, and HOCl, demonstrating their bonding patterns.

🎯 Exam Tip: Be proficient in assigning oxidation states for elements in compounds, especially for non-metals, as it's a fundamental skill in inorganic chemistry.

 

Can You Tell? (Textbook Page No 146)

 

Question 1.What is allotropy?
Answer:The property of some elements to exist in two or more different forms in the same physical state is called allotropy.
In simple words: Allotropy is when an element can exist in multiple distinct forms within the same physical state, differing in structural arrangement and properties.

🎯 Exam Tip: Define allotropy clearly, emphasizing that it's a property of *elements* existing in different forms within the *same physical state*.

 

Question 2.What is the difference between allotropy and polymorphism?
Answer:• Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms. • Coal, graphite and diamond etc., are different allotropic forms of carbon. • Polymorphism is the existence of a substance in more than one crystalline form. • It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.
In simple words: Allotropy refers to an element existing in multiple physical forms (like carbon as diamond or graphite), whereas polymorphism describes a substance (compound) existing in different crystalline forms (like mercuric iodide in orthorhombic or trigonal forms).

🎯 Exam Tip: The key distinction is that allotropy applies only to *elements*, while polymorphism applies to *compounds*, both describing the existence of multiple forms under different conditions.

Question 7.7 (Textbook Page No 146)

Question. Which form of sulphur shows paramagnetic behaviour?
Answer: In the vapour state, sulphur partly exists as S2 molecule, which has two unpaired electrons in the antibonding \(\pi^*\) orbitals like O2. Hence it exhibits paramagnetism.
In simple words: When sulfur is in its gaseous state, it can exist as S2 molecules, similar to how oxygen exists as O2. These S2 molecules have unpaired electrons in their molecular orbitals, which makes them attracted to magnetic fields, a property known as paramagnetism.

🎯 Exam Tip: Understanding molecular orbital theory helps explain unexpected magnetic properties of molecules, like paramagnetism in S2 and O2, which is a key concept in chemical bonding.

Try this..... (Textbook Page No 149)

Question 1. Why water in a fish pot needs to be changed from time to time?
Answer: A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.
In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.
Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.
Hence, it is necessary to change the water from time to time.
In simple words: Water in a fish pot needs regular changing because it's a closed system where fish waste and uneaten food build up. This accumulation reduces the essential dissolved oxygen, making the environment unhealthy for the fish.

🎯 Exam Tip: This question highlights the practical application of chemical principles related to dissolved gases and environmental balance in closed systems. Emphasize the role of decomposers and oxygen depletion.

Question 7.8 (Textbook Page No 149)

Question. Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer: Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O2 can be explained with the help of molecular orbital theory.
Electronic configuration O2
\(ΚΚ \sigma(2s)^2 \sigma^*(2s)^2 \sigma^*(2p_z)^2 \pi(2p_x)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1\)
Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.
In simple words: Even though oxygen (O2) has an even number of electrons, its paramagnetic nature (attraction to a magnetic field) is explained by molecular orbital theory. This theory shows that O2 has two electrons that are not paired up in its highest energy antibonding orbitals, making the molecule paramagnetic.

🎯 Exam Tip: This is a classic example of molecular orbital theory correcting the predictions of simpler bonding theories. Clearly state the MO configuration and point out the unpaired electrons in the \(\pi^*\) orbitals for full marks.

Question 7.9 (Textbook Page No 150)

Question. High concentration of ozone can be dangerously explosive. Explain.
Answer: Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O2. The decomposition is exothermic and results in the liberation of heat (\(\Delta H\) is - ve) and an increase in entropy (\(\Delta S\) is positive). This results in large negative Gibbs energy change (\(\Delta G\)). Therefore high concentration of ozone can be dangerously explosive. Eq
\(O_3 \implies O_2 + O\)
In simple words: High concentrations of ozone are explosive because ozone is naturally unstable and readily breaks down into oxygen, releasing a significant amount of heat and increasing disorder. This spontaneous and energy-releasing decomposition makes it very dangerous.

🎯 Exam Tip: Focus on the thermodynamic instability of ozone, explaining that its decomposition is exothermic (\(\Delta H < 0\)) and increases entropy (\(\Delta S > 0\)), leading to a highly negative Gibbs free energy (\(\Delta G < 0\)) and thus a spontaneous, explosive reaction.

Try this...... (Textbook Page No 151)

Question. (a) Ozone is used as a bleaching agent. Explain.
Answer:
- Ozone due to its oxidising property can act as a bleaching agent. \(O_{3(g)} \implies O_{2(g)} + O\)
- It bleaches coloured matter. coloured matter \(+ O \implies\) colourless matter
- Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
- Ozone can bleach ivory and delicate fabrics like silk.
In simple words: Ozone acts as a bleaching agent because it's a strong oxidizer. It releases nascent (single, highly reactive) oxygen atoms, which then react with and destroy the colored compounds, turning them colorless. It works even without moisture, earning it the name "dry bleach."

🎯 Exam Tip: The key to ozone's bleaching action is its ability to liberate nascent oxygen. Mentioning the dry bleaching property and its gentle action on delicate materials is important for a complete answer.

Question. (b) Why does ozone act as a powerful oxidising agent?
Answer: Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. \(O_{3(g)} \implies O_{2(g)} + O\)
For example :
- It oxidises lead sulphide (PbS) to lead sulphate (PbSO4).
\(PbS_{(s)} + 4O_{3(g)} \implies PbSO_{4(s)} + 4O_{2(g)}\)
- Potassium iodide, KI is oxidised to iodine, I2 in the solution.
\(2KI_{(aq)} + H_2O_{(l)} + O_{3(g)} \implies 2KOH_{(aq)} + I_{2(s)} + O_{2(g)}\)
In simple words: Ozone is a powerful oxidising agent because it readily breaks down to produce highly reactive nascent oxygen atoms. These oxygen atoms easily take electrons from other substances, causing them to oxidize, as seen in its reactions with lead sulfide and potassium iodide.

🎯 Exam Tip: Emphasize the release of nascent oxygen as the primary reason for ozone's strong oxidizing power. Providing relevant chemical equations, like the oxidation of PbS or KI, demonstrates a comprehensive understanding.

Question 7.10 : (Textbook Page No 154)

Question. What is the action of concentrated H2SO4 on (a) HBr (b) HI
Answer: Concentrated sulphuric acid oxidises hydrobromic acid to bromine.
\(2HBr + H_2SO_4 \implies Br_2 + SO_2 + 2H_2O\)
It oxidises hydroiodic acid to iodine.
\(2HI + H_2SO_4 \implies I_2 + SO_2 + 2H_2O\)
In simple words: Concentrated sulfuric acid acts as an oxidizing agent. When it reacts with hydrobromic acid (HBr), it oxidizes HBr to bromine (\(Br_2\)), and with hydroiodic acid (HI), it oxidizes HI to iodine (\(I_2\)). In both cases, the sulfuric acid itself is reduced.

🎯 Exam Tip: Remember that concentrated H2SO4 is a strong oxidizing agent. The key is to show its reduction to SO2 while oxidizing HBr to Br2 and HI to I2. Balancing the redox equations is crucial.

Try this..... (Textbook Page No 156)

Question 1. Give the reasons for the bleaching action of chlorine.
Answer:
- Chlorine acts as a powerful bleaching agent due to its oxidising nature.
- In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
\(Cl_2 + H_2O \implies HCl + HOCl\)
\(HOCl \implies HCl + [O]\)
Vegetable coloured matter \(+ [O] \implies\) colourless matter.
In simple words: Chlorine bleaches by releasing nascent oxygen, a highly reactive form of oxygen, especially in the presence of moisture. The nascent oxygen then reacts with and oxidizes colored substances, destroying their color and making them colorless.

🎯 Exam Tip: The crucial point for chlorine's bleaching action is its ability to produce nascent oxygen from hypochlorous acid in the presence of water. Clearly show the decomposition of HOCl and the subsequent oxidation of the colored matter.

Question 2. Name two gases used in war.
Answer: Phosgene: \(COCl_2\)
Mustard gas: \(Cl-CH_2-CH_2-S-CH_2-CH_2-Cl\)
In simple words: Two gases historically used in warfare are phosgene, which is \(COCl_2\), and mustard gas, a sulfur-containing compound with the formula \(Cl-CH_2-CH_2-S-CH_2-CH_2-Cl\).

🎯 Exam Tip: This is a factual recall question. Accurately providing both the name and the correct chemical formula (including structural representation for mustard gas) is important.

Use your brain power! (Textbook Page No 157)

Question 1. Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer: Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)
The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.
In simple words: In an interhalogen compound formed between chlorine and fluorine, the halide ion will be fluoride (F-). This is because fluorine is more electronegative than chlorine, meaning it attracts electrons more strongly and will form the negative ion. The compound is named as a halide of the more electropositive halogen, so it would be chlorine fluoride.

🎯 Exam Tip: The key principle here is electronegativity. The more electronegative element (fluorine) will form the halide ion, while the less electronegative element (chlorine) will be the central atom and named first.

Question 2. Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer: Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.
In simple words: Fluorine forms the maximum number of fluorides with other halogens because it is the most electronegative element and has a very small atomic size. These properties allow it to strongly attract electrons and form stable bonds with various oxidation states of other halogens.

🎯 Exam Tip: The two critical factors are fluorine's extremely high electronegativity and its small atomic size. These allow it to stabilize high oxidation states of other halogens and form numerous stable fluoride compounds.

Use your brain power! (Textbook Page No 158)

Question 1. What will be the names of the following compounds: ICl, BrF?
Answer: ICl: Iodine monochloride
BrF: Bromine monofluoride
In simple words: The compound ICl is named Iodine monochloride, indicating one iodine and one chlorine atom. The compound BrF is named Bromine monofluoride, indicating one bromine and one fluorine atom.

🎯 Exam Tip: For naming interhalogen compounds, the less electronegative halogen is named first, followed by the more electronegative halogen with the -ide suffix, and prefixes (mono-, di-, tri-, etc.) are used to indicate the number of atoms.

Question 2. Which halogen (X) will have maximum number of other halogen (X) attached?
Answer: The halogen Iodine (I) will have the maximum number of other halogens attached.
In simple words: Iodine (I) will have the maximum number of other halogen atoms attached to it. This is because iodine is the largest and least electronegative of the halogens, allowing it to accommodate more smaller, more electronegative halogen atoms around it.

🎯 Exam Tip: The central halogen in an interhalogen compound is typically the larger, less electronegative one. Iodine, being the largest halogen, can accommodate the most surrounding smaller halogen atoms due to its size and availability of d-orbitals for expansion of octet.

Question 3. Which halogen has tendency to form more interhalogen compounds?
Answer: The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.
In simple words: Fluorine (F) has the highest tendency to form many interhalogen compounds. This is primarily due to its small size and very high electronegativity, which allows it to stabilize higher oxidation states in other halogens.

🎯 Exam Tip: Remember the unique properties of fluorine – its highest electronegativity and smallest atomic size – as these are key to its ability to form a wide variety of stable interhalogen compounds where it is almost always the peripheral atom.

Question 4. Which will be more reactive?
(a) ClF3 or ClF,
(b) BrF5 or BrF
Answer: ClF3 is more reactive than ClF, while BrF5 is more reactive than BrF. Both ClF3 and BrF5 are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.
In simple words: Compounds with more halogen atoms attached, like \(ClF_3\) and \(BrF_5\), are generally more reactive than those with fewer, like ClF and BrF. This increased reactivity is due to the greater steric hindrance and instability caused by having more atoms crowded around the central halogen.

🎯 Exam Tip: Interhalogen compounds with a higher number of fluorine atoms (e.g., \(AX_n\)) are typically more reactive. This is often attributed to greater steric strain and weaker bond strengths, making them more prone to reaction compared to simpler \(AX\) compounds.

Question 5. Complete the table :

Formula	Name
ClF	Chlorine monofluoride
ClF3
	Chlorine pentafluoride
BrF
	Bromine pentafluoride
ICl
ICl3

Answer:

Formula	Name
ClF	Chlorine monofluoride
ClF3	Chlorine trifluoride
ClF5	Chlorine pentafluoride
BrF	Bromine monofluoride
BrF5	Bromine pentafluoride
ICl	Iodine monochloride
ICl3	Iodine trichloride

In simple words: The table lists various interhalogen compounds. Their names are derived by naming the less electronegative halogen first, followed by the more electronegative halogen with an "-ide" suffix, using prefixes (mono-, tri-, penta-) to indicate the number of atoms.

🎯 Exam Tip: Practice naming conventions for interhalogen compounds. Remember that the more electronegative halogen acts as the peripheral atom, and the number of peripheral atoms is indicated by prefixes.

Use your brain power! (Textbook Page No 159)

Question 1. In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:
\(I_2^{0} + KClO_3^{+5} \implies ICl^{+1-1} + KIO_3^{5+}\)
Potassium chlorate, \(KClO_3\) is the oxidising agent or oxidant and iodine is the reducing agent or reductant.
In simple words: In the given reaction, iodine (\(I_2\)) is the reductant because its oxidation state increases from 0 to +1 in ICl, meaning it loses electrons. Potassium chlorate (\(KClO_3\)) is the oxidant because the chlorine in it maintains its +5 oxidation state, but it causes the iodine to oxidize.

🎯 Exam Tip: To identify oxidants and reductants, track the oxidation states of elements. An oxidant gets reduced (oxidation state decreases), and a reductant gets oxidized (oxidation state increases). In this case, iodine is oxidized, making it the reductant.

Use your brain power! (Textbook Page No 162)

Question 1. What are missing entries?

Formula	Name
XeOF2	Xenon monooxyfluoride
	Xenon dioxydifluoride
XeO3F2
XeO2F4

Answer:

Formula	Name
XeOF2	Xenon monooxydifluoride
XeO2F2	Xenon dioxydifluoride
XeO3F2	Xenon trioxydifluoride
XeO2F4	Xenon dioxytetrafluoride

In simple words: This table completes the names and formulas for various xenon oxyfluorides. The naming follows a pattern where "oxo-" indicates oxygen atoms, and "fluoro-" indicates fluorine atoms, with prefixes showing their counts.

🎯 Exam Tip: Naming xenon compounds often involves prefixes for oxygen ("oxo-" or "di/trioxo-") and fluorine ("fluoro-" or "di/tetrafluoro-"). Practice these systematic naming rules for xenon oxyfluorides to ensure accuracy.