Masterclass: 25 Ultra-Challenging JEE Advanced Problems on p-Block Elements (Groups 17 & 18)
From the anomalous electron gain enthalpy of Fluorine and the explosive power of Interhalogens, to the forced hybridization of Xenon in its exotic fluorides. Master the Halogens and Noble Gases.
Group 17 (Halogens) contains the most fiercely reactive and electronegative elements known, while Group 18 (Noble Gases) was long thought to be perfectly inert. To conquer JEE Advanced, you must bridge this gap: understanding how the hyper-oxidizing power of Fluorine physically forces the "inert" Xenon to expand its octet into empty $d$-orbitals, creating molecules that rewrite bonding theory.
View Solution
Step 1: Define the Contradiction
Fluorine pulls electrons better than Chlorine in a covalent bond ($\chi_F = 4.0, \chi_{Cl} = 3.0$). However, when adding a free electron, Chlorine releases more energy ($\Delta_{eg}H^{\circ} = -349 \text{ kJ/mol}$) than Fluorine ($\Delta_{eg}H^{\circ} = -333 \text{ kJ/mol}$).
Step 2: Analyze Orbital Size and Sterics
In Fluorine, the incoming electron must enter the $2p$ subshell. The $2p$ orbital is extremely compact and physically tiny. It already contains 5 electrons tightly packed together. Adding a 6th electron into this tiny volume creates massive interelectronic repulsion. This repulsion counteracts the strong pull of the nucleus, significantly reducing the net energy released.
Step 3: The Chlorine Advantage
In Chlorine, the incoming electron enters the $3p$ subshell. The $3p$ orbital is significantly larger and spatially more diffuse. The 5 existing electrons are spread over a much larger volume, so the incoming 6th electron experiences very little interelectronic repulsion. The nucleus can pull the electron in efficiently, releasing maximum energy.
View Solution
Step 1: The Expected Trend
Based purely on atomic radius, the $F-F$ bond is the shortest, so it should be the strongest. The expected order is $F_2 > Cl_2 > Br_2 > I_2$.
Step 2: The Actual Experimental Trend
The true order of bond dissociation enthalpy is: $Cl_2 > Br_2 > F_2 > I_2$. The $F-F$ bond is astonishingly weak ($\approx 158.8 \text{ kJ/mol}$), even weaker than the $Br-Br$ bond ($\approx 192.8 \text{ kJ/mol}$).
Step 3: The Cause of the Anomaly (Lone Pair Repulsion)
The Fluorine atom is incredibly small. In the $F_2$ molecule, the two extremely tiny atoms are forced to sit very close to each other to form the covalent bond. Each Fluorine atom possesses 3 highly compressed lone pairs in the $2p$ orbitals.
Because the internuclear distance is so short, these dense lone pairs on adjacent atoms physically clash, resulting in ferocious lone pair-lone pair electrostatic repulsion. This intense internal repulsion actively forces the two atoms apart, severely weakening the $F-F$ bond.
Step 4: Heavier Halogens
In $Cl_2$ and $Br_2$, the atoms are larger, the bond length is longer, and the lone pairs (in $3p$ and $4p$ orbitals) are far enough apart that this repulsion is negligible, restoring normal bond strengths.
View Solution
Step 1: Evaluating the Inductive Effect Trap
Because Fluorine is the most electronegative, one might assume it polarizes the $H-F$ bond the most, making the proton highly positive and easily lost. However, bond polarity is vastly overshadowed by Bond Strength.
Step 2: Orbital Overlap and Bond Strength
- In $HF$, the bond is formed by the overlap of the Hydrogen $1s$ orbital and the Fluorine $2p$ orbital. Because $1s$ and $2p$ are somewhat comparable in size, the overlap is extremely efficient, resulting in a phenomenally strong $H-F$ bond ($\approx 562 \text{ kJ/mol}$).
- Moving down the group, Hydrogen ($1s$) must overlap with much larger, more diffuse orbitals ($3p$ for Cl, $4p$ for Br, $5p$ for I). The size mismatch becomes enormous. The $1s-5p$ overlap in $HI$ is incredibly poor, resulting in a very long, very weak $H-I$ bond ($\approx 299 \text{ kJ/mol}$).
Step 3: The Acidity Trend
Because the $H-I$ bond is so weak, it breaks almost effortlessly in water to yield massive amounts of $H^+$, making $HI$ the strongest acid. The $H-F$ bond is so tightly held that it only partially dissociates, making $HF$ a weak acid.
View Solution
Step 1: The Thermodynamic Cycle
To go from gaseous $X_2$ molecules to aqueous $X^-$ ions, three energy steps are involved:
1. Bond Dissociation Enthalpy ($\frac{1}{2} \Delta_{\text{diss}}H^{\circ}$): Breaking $1/2$ mole of $X_2$ into $X_{(g)}$ atoms. (Requires energy, endothermic).
2. Electron Gain Enthalpy ($\Delta_{eg}H^{\circ}$): Adding an electron to the gaseous atom to form $X^-_{(g)}$. (Releases energy, exothermic).
3. Hydration Enthalpy ($\Delta_{\text{hyd}}H^{\circ}$): Plunging the gaseous anion into water to form $X^-_{(aq)}$. (Releases energy, exothermic).
Step 2: Analyzing the Terms for Fluorine
We know Fluorine has a lower electron gain enthalpy than Chlorine, which should make it a weaker oxidizer.
However, the $F-F$ bond dissociation energy is anomalously low (meaning very little energy is required in Step 1).
Step 3: The Dominant Factor (Hydration)
The crucial deciding factor is Hydration Enthalpy. The resulting Fluoride ion ($F^-$) is exceptionally tiny. It has an immense charge density. When placed in water, it attracts a massive, tightly bound hydration shell of water molecules. The heat released during this intense hydration is colossal (highly exothermic).
This massive release of hydration energy (combined with the low bond dissociation energy) overwhelmingly compensates for the slightly lower electron gain enthalpy, making the overall $\Delta G^{\circ}$ massively negative.
View Solution
Part (a): $ClF_3$ (Chlorine trifluoride)
- Central atom: Chlorine (7 valence $e^-$).
- It forms 3 single bonds with 3 Fluorines, using 3 $e^-$.
- Remaining $e^- = 4$, which equals 2 Lone Pairs.
- Steric Number = 3 BP + 2 LP = 5 $\rightarrow$ $sp^3d$ hybridization.
- In a trigonal bipyramidal base geometry, lone pairs strictly occupy the equatorial positions to minimize $90^{\circ}$ repulsions. The 3 Fluorines occupy the two axial and one equatorial positions.
- Shape: Bent T-Shape.
Part (b): $BrF_5$ (Bromine pentafluoride)
- Central atom: Bromine (7 valence $e^-$).
- It forms 5 single bonds with 5 Fluorines, using 5 $e^-$.
- Remaining $e^- = 2$, which equals 1 Lone Pair.
- Steric Number = 5 BP + 1 LP = 6 $\rightarrow$ $sp^3d^2$ hybridization.
- In an octahedral base geometry, the single lone pair occupies any position, forcing the 5 Fluorines into the remaining slots.
- Shape: Square Pyramidal.
Part (c): $IF_7$ (Iodine heptafluoride)
- Central atom: Iodine (7 valence $e^-$).
- It forms 7 single bonds with 7 Fluorines, using all 7 $e^-$.
- Remaining $e^- = 0$, which equals 0 Lone Pairs.
- Steric Number = 7 BP + 0 LP = 7 $\rightarrow$ $sp^3d^3$ hybridization.
- Shape: Pentagonal Bipyramidal.
View Solution
Step 1: Pure Halogens vs Interhalogens
In a pure halogen molecule ($X-X$), the two atoms are identical in size and electronegativity. The orbital overlap ($np-np$) is perfectly symmetrical and highly efficient, resulting in a relatively strong, non-polar covalent bond (e.g., the $Cl-Cl$ bond is $242 \text{ kJ/mol}$).
Step 2: The Weakness of the $X-X'$ Bond
In an interhalogen compound ($X-X'$), the two atoms have different sizes and different electronegativities.
1. The size difference causes poor, inefficient orbital overlap (e.g., overlapping a $5p$ orbital of Iodine with a $3p$ orbital of Chlorine in $ICl$).
2. The electronegativity difference makes the bond highly polar ($X^{\delta+} - X'^{\delta-}$), rendering it susceptible to nucleophilic and electrophilic attack.
Step 3: The Resulting Reactivity
Because the $X-X'$ bond is fundamentally weaker than a standard $X-X$ bond, interhalogens have a much lower activation energy for bond cleavage, making them far more reactive.
Step 4: The Fluorine Exception
The $F_2$ molecule is the sole exception. As established in Problem 2, the $F-F$ bond is anomalously weak due to intense lone pair repulsion. Therefore, $F_2$ remains more reactive than almost all interhalogen compounds.
View Solution
Step 1: The Reactants
Dry slaked lime is Calcium Hydroxide, $Ca(OH)_2$.
Step 2: The Reaction
When Chlorine gas ($Cl_2$) is passed over the dry powder, a disproportionation-like reaction occurs with the base.
$2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O$
Step 3: The True Composition
While often simplified in elementary textbooks as $CaOCl_2$ (Calcium chlorohypochlorite), the actual solid isolated is a complex mixture containing Calcium hypochlorite, Calcium chloride, Calcium hydroxide, and water of crystallization.
The true representation of bleaching powder is:
$Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O$
Step 4: The Active Ingredient
The bleaching action and oxidizing power are derived exclusively from the Hypochlorite ion ($OCl^-$), which easily releases nascent oxygen $[O]$ or reacts with acids to liberate Chlorine gas ("available chlorine").
View Solution
Step 1: Reaction with Cold, Dilute $NaOH$
In a mild, cold alkaline environment, Chlorine ($O.S. = 0$) disproportionates to the $-1$ state and the mildly oxidized $+1$ state.
Reaction: $Cl_2 + 2NaOH \text{ (cold, dilute)} \rightarrow \mathbf{NaCl + NaOCl + H_2O}$
Products: Sodium chloride (Cl is $-1$) and Sodium hypochlorite (Cl is $+1$).
Step 2: Reaction with Hot, Concentrated $NaOH$
Under harsh, hot alkaline conditions, the hypochlorite ion ($OCl^-$) is unstable and further disproportionates. The overall reaction drives the Chlorine deeper into the highly oxidized, thermodynamically stable $+5$ state.
Reaction: $3Cl_2 + 6NaOH \text{ (hot, conc.)} \rightarrow \mathbf{5NaCl + NaClO_3 + 3H_2O}$
Products: Sodium chloride (Cl is $-1$) and Sodium chlorate (Cl is $+5$).
(a) Arrange them in strictly increasing order of Acidic Strength.
(b) Arrange them in strictly increasing order of Oxidizing Power.
Explain the opposing thermodynamic principles governing these two trends.
View Solution
Part (a): Acidic Strength Trend
Acidity increases as the number of terminal oxygen atoms ( $=O$ ) increases. In Perchloric acid ($HClO_4$), the resulting perchlorate anion ($ClO_4^-$) can delocalize its negative charge symmetrically over four highly electronegative oxygen atoms via intense $p\pi-d\pi$ resonance. This massive stabilization of the conjugate base makes it eager to release a proton.
Order: $HClO < HClO_2 < HClO_3 < HClO_4$ (Strongest acid).
Part (b): Oxidizing Power Trend
Paradoxically, the oxidizing power is exactly the reverse. As the number of oxygen atoms increases, the central Chlorine atom becomes sterically shielded by an impenetrable sphere of oxygens. More importantly, the intense $p\pi-d\pi$ resonance stabilizes the intact molecule, making it kinetically very sluggish to break apart and release its highly oxidized oxygen atoms. Hypochlorous acid ($HClO$) has almost no resonance stabilization and breaks apart effortlessly, making it a ferocious, rapid oxidizer.
Order: $HClO_4 < HClO_3 < HClO_2 < HClO$ (Strongest oxidizer).
View Solution
Step 1: The Initial Discovery
Bartlett was working with Platinum hexafluoride ($PtF_6$), an incredibly powerful oxidizing agent (a red volatile solid). He discovered that it was strong enough to rip an electron directly out of a stable Oxygen molecule ($O_2$), forming the ionic salt Dioxygenyl hexafluoroplatinate: $O_2^+ [PtF_6]^-$.
Step 2: The Mathematical Rationale
He realized that the first ionization enthalpy of molecular Oxygen ($O_2 \rightarrow O_2^+ + e^-$) is exactly $1175 \text{ kJ/mol}$.
Looking at the periodic table, the first ionization enthalpy of Xenon ($Xe$) is remarkably identical, at $1170 \text{ kJ/mol}$. Furthermore, their atomic/molecular sizes are very similar.
Step 3: The Groundbreaking Reaction
Deducing that if $PtF_6$ can oxidize $O_2$, it must also have the thermodynamic power to oxidize Xenon, he mixed Xenon gas with $PtF_6$ vapor. Instantly, a deep red solid precipitated out.
Reaction: $Xe_{(g)} + PtF_{6(g)} \rightarrow Xe^+ [PtF_6]^-_{(s)}$
View Solution
Step 1: Synthesis of $XeF_2$
To stop the reaction at the lowest oxidation state, Xenon must be maintained in a massive excess compared to Fluorine to prevent further fluorination of the initial product.
Ratio: $Xe$ in excess (typically 2:1 ratio).
Conditions: Heated at $673 \text{ K}$ at $1 \text{ bar}$ pressure.
$Xe_{(g)} + F_{2(g)} \rightarrow XeF_{2(s)}$
Step 2: Synthesis of $XeF_4$
To push the oxidation further, the ratio must heavily favor Fluorine.
Ratio: $1 : 5$ ratio of $Xe : F_2$.
Conditions: Heated at $873 \text{ K}$ at $7 \text{ bar}$ pressure.
$Xe_{(g)} + 2F_{2(g)} \rightarrow XeF_{4(s)}$
Step 3: Synthesis of $XeF_6$
To force Xenon into the $+6$ oxidation state, an overwhelming, massive excess of Fluorine and extreme pressure are required.
Ratio: $1 : 20$ ratio of $Xe : F_2$.
Conditions: Heated at $573 \text{ K}$ at high pressure ($60-70 \text{ bar}$).
$Xe_{(g)} + 3F_{2(g)} \rightarrow XeF_{6(s)}$
1. $XeF_2$: Excess $Xe$, $673\text{K}$, $1\text{bar}$.
2. $XeF_4$: $1:5$ ratio, $873\text{K}$, $7\text{bar}$.
3. $XeF_6$: $1:20$ ratio, $573\text{K}$, $60-70\text{bar}$.
View Solution
1. $XeF_2$:
- Valence $e^-$ on Xe = 8. It uses 2 $e^-$ for bonding with 2 F atoms.
- Remaining $e^- = 6$, which equals 3 Lone Pairs.
- Steric Number = 2 BP + 3 LP = 5 $\rightarrow$ $sp^3d$ hybridization (Trigonal bipyramidal base).
- To minimize $90^{\circ}$ lone-pair repulsions, all 3 lone pairs must occupy the equatorial plane. The 2 Fluorines occupy the axial positions.
- Final Shape: Perfectly Linear.
2. $XeF_4$:
- Valence $e^-$ on Xe = 8. It uses 4 $e^-$ for bonding with 4 F atoms.
- Remaining $e^- = 4$, which equals 2 Lone Pairs.
- Steric Number = 4 BP + 2 LP = 6 $\rightarrow$ $sp^3d^2$ hybridization (Octahedral base).
- To minimize repulsion, the 2 lone pairs locate exactly $180^{\circ}$ apart on the axial positions. The 4 Fluorines form a flat plane along the equator.
- Final Shape: Square Planar.
3. $XeF_6$:
- Valence $e^-$ on Xe = 8. It uses 6 $e^-$ for bonding.
- Remaining $e^- = 2$, which equals 1 Lone Pair.
- Steric Number = 6 BP + 1 LP = 7 $\rightarrow$ $sp^3d^3$ hybridization.
- The presence of 1 lone pair distorts the perfect octahedral geometry of the 6 Fluorines. The lone pair is stereochemically active and pushes through the faces of the octahedron, creating a highly fluxional, dynamic molecule.
- Final Shape: Distorted Octahedral (or Capped Octahedral).
View Solution
Step 1: Hydrolysis of $XeF_2$ (Redox)
In $XeF_2$, Xenon is in the $+2$ oxidation state. When dropped in water, it cannot form a stable $+2$ oxide. Instead, it aggressively oxidizes the oxygen in water from $-2$ to $0$ ($O_2$ gas), while reducing itself completely back to elemental Xenon gas ($0$).
Reaction: $2XeF_2 + 2H_2O \rightarrow 2Xe \uparrow + 4HF + O_2 \uparrow$
Because Xe goes from $+2 \rightarrow 0$, this is definitively a Redox reaction.
Step 2: Hydrolysis of $XeF_6$ (Non-Redox)
In $XeF_6$, Xenon is in the highly stable $+6$ oxidation state. When exposed to excess water, the highly electronegative Fluorine atoms are simply swapped out for Oxygen atoms via a nucleophilic substitution-like process. It forms Xenon trioxide.
Reaction: $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
In Xenon trioxide ($XeO_3$), the Xenon atom is still perfectly in the $+6$ oxidation state. No electrons were transferred; it is strictly a hydrolysis/substitution, making it a Non-redox reaction.
View Solution
Step 1: First Partial Hydrolysis
Adding just $1 \text{ mole}$ of $H_2O$ replaces 2 Fluorines.
Reaction: $XeF_6 + H_2O \rightarrow \mathbf{XeOF_4} + 2HF$
Structure of $XeOF_4$: Central Xe ($8 e^-$). 1 double bond to O ($2 e^-$), 4 single bonds to F ($4 e^-$). Remaining $e^- = 2$ (1 Lone Pair). Steric number 6 ($sp^3d^2$). The O and the lone pair occupy axial positions to minimize repulsion. Shape: Square Pyramidal.
Step 2: Second Partial Hydrolysis
Adding a second mole of $H_2O$ replaces 2 more Fluorines.
Reaction: $XeOF_4 + H_2O \rightarrow \mathbf{XeO_2F_2} + 2HF$
Structure of $XeO_2F_2$: Central Xe ($8 e^-$). 2 double bonds to O ($4 e^-$), 2 single bonds to F ($2 e^-$). Remaining $e^- = 2$ (1 Lone Pair). Steric number 5 ($sp^3d$). To minimize repulsion, the lone pair and the two bulky $=O$ groups must occupy the equatorial plane. The two Fluorines occupy the axial positions. Shape: See-Saw (or Teeter-Totter).
(a) $XeF_2$ reacts with Phosphorus pentafluoride ($PF_5$).
(b) $XeF_6$ reacts with Rubidium fluoride ($RbF$).
View Solution
Part (a): Acting as a Fluoride Donor
$PF_5$ is a highly electron-deficient Lewis Acid (with empty d-orbitals). $XeF_2$ donates a fluoride ion ($F^-$) to it.
- $XeF_2$ loses $F^-$ to become the linear cation: $[XeF]^+$
- $PF_5$ accepts $F^-$ to become the octahedral anion: $[PF_6]^-$
Reaction: $XeF_2 + PF_5 \rightarrow \mathbf{[XeF]^+ [PF_6]^-}$
Part (b): Acting as a Fluoride Acceptor
$RbF$ is an ionic salt, providing a massive source of free $F^-$ ions (a Lewis base). $XeF_6$ acts as a Lewis acid and eagerly accepts the fluoride ion into its coordination sphere, expanding its coordination number.
- $RbF$ provides the cation: $Rb^+$
- $XeF_6$ accepts $F^-$ to become the massive heptacoordinate anion: $[XeF_7]^-$
Reaction: $XeF_6 + RbF \rightarrow \mathbf{Rb^+ [XeF_7]^-}$
View Solution
Step 1: Valence Electron Count
- Central atom: Xenon (starts with 8 valence $e^-$).
- Add 1 electron for the negative charge $= 9$ total valence $e^-$.
- It forms 5 single bonds with 5 Fluorines, consuming 5 $e^-$.
- Remaining $e^- = 4$, which pairs up perfectly to form 2 Lone Pairs.
Step 2: Hybridization
- Steric Number = 5 Bond Pairs + 2 Lone Pairs = 7.
- A steric number of 7 demands $sp^3d^3$ hybridization, based on a pentagonal bipyramidal geometric framework.
Step 3: Geometry and Shape
In a pentagonal bipyramid, the axial positions are at $90^{\circ}$ to the equatorial plane, but the equatorial positions are spread $72^{\circ}$ apart. To minimize the massive repulsion between the two stereochemically active lone pairs, they locate strictly opposite each other at the axial positions.
This forces all 5 Fluorine atoms to occupy the equatorial plane.
Because the 5 bonds lie perfectly flat in a single plane arranged as a pentagon, the molecular shape is uniquely and completely Pentagonal Planar.
View Solution
Step 1: The Decomposition Reaction
When shocked or heated, solid $XeO_3$ decomposes instantly into its constituent elements:
$2XeO_{3(s)} \rightarrow 2Xe_{(g)} + 3O_{2(g)}$
Step 2: The Enthalpy Drive ($\Delta H \ll 0$)
The $Xe-O$ bonds are relatively weak because Xenon is a noble gas and highly resists giving up electron density to oxygen to maintain the $+6$ oxidation state. Conversely, the $O=O$ double bond in the $O_2$ product is phenomenally strong ($\approx 498 \text{ kJ/mol}$). The massive thermodynamic payback from forming three incredibly stable oxygen molecules makes the decomposition violently exothermic.
Step 3: The Entropy Drive ($\Delta S \gg 0$)
The reaction starts with 2 moles of a highly ordered solid lattice ($XeO_3$). It instantly transforms into 5 moles of expanding gas ($2Xe + 3O_2$). This phase change from solid to gas creates a colossal, instantaneous increase in entropy (disorder).
Because both $\Delta H$ is negative and $\Delta S$ is massively positive, the Gibbs free energy equation ($\Delta G = \Delta H - T\Delta S$) becomes phenomenally negative, driving the explosive detonation.
View Solution
Proof 1: Addition across a Double Bond
Just like Chlorine ($Cl_2$) adds across an alkene to form a vicinal dichloride, Cyanogen gas adds across an alkene (often under thermal or catalytic conditions) to form a vicinal dicyanide.
$CH_2=CH_2 + (CN)_2 \rightarrow NC-CH_2-CH_2-CN$
Proof 2: Formation of Insoluble Silver Salts
Just as the Chloride ion ($Cl^-$) reacts with Silver to form a highly insoluble white precipitate ($AgCl$), the Cyanide pseudohalide ion ($CN^-$) reacts with $Ag^+$ to form a highly insoluble white precipitate of Silver Cyanide.
$Ag^+_{(aq)} + CN^-_{(aq)} \rightarrow \mathbf{AgCN_{(s)} \downarrow}$
Proof 3: Disproportionation in Alkalies
Just as Chlorine disproportionates in cold, dilute $NaOH$ to yield chloride ($-1$) and hypochlorite ($+1$), Cyanogen disproportionates perfectly in $NaOH$ to yield the cyanide ion and the cyanate ion.
$Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$
$(CN)_2 + 2NaOH \rightarrow \mathbf{NaCN + NaOCN + H_2O}$
View Solution
Step 1: The Architecture of the Host
When water freezes into ice, it forms an extensive 3D tetrahedral network of hydrogen bonds. Because hydrogen bonds are highly directional, this network forces the water molecules apart, creating massive, empty, cage-like cavities (voids) within the crystal lattice.
Step 2: The Trapping Mechanism
Under high pressure, noble gas atoms are physically forced into these empty cages just as the water freezes around them. Once the ice crystal is solid, the noble gas atom is physically trapped inside the cage. It forms absolutely zero chemical bonds with the water molecules; it is merely a prisoner held by weak Van der Waals dispersion forces. If the ice is melted, the cage collapses, and the gas instantly escapes.
Step 3: The Failure of Helium and Neon
For the cage to be stable, the guest atom must fit snugly. If the atom is too small, it can easily slip right through the gaps in the hydrogen-bonded water cage and escape.
Helium and Neon are exceptionally tiny atoms. They are physically small enough to diffuse straight through the walls of the ice cages, making it impossible to trap them. Argon, Krypton, and Xenon are large enough to be physically blocked from escaping.
View Solution
Step 1: Preparation of HOF
Passing Fluorine gas over ice at $-40^{\circ}\text{C}$ yields small quantities of the highly unstable HOF. It is isolated by trapping it at ultra-low temperatures.
Reaction: $F_2 + H_2O \text{ (ice)} \xrightarrow{-40^{\circ}\text{C}} \mathbf{HOF + HF}$
Step 2: The Oxidation State Limit
In all higher oxoacids of halogens (like $HClO_4$), the central halogen atom must achieve a highly positive oxidation state ($+3, +5, \text{ or } +7$).
Fluorine is the most electronegative element in the universe. It is physically impossible for oxygen (which is less electronegative) to force Fluorine to yield electron density and adopt a positive oxidation state. In HOF, the oxidation state of Fluorine is rigorously $-1$, and Oxygen is forced into the highly unusual $0$ state.
Step 3: The Orbital Limit
Furthermore, to form multiple bonds with numerous oxygen atoms (as in $HClO_4$), the central atom must expand its octet. Because Fluorine is in Period 2, it possesses absolutely no d-orbitals in its valence shell. It physically cannot form more than one covalent bond.
View Solution
Step 1: The Danger of Nitrogen at Depth
At deep ocean depths, the hydrostatic pressure is immense. The compressed air the diver breathes is at a very high pressure. According to Henry's Law, this high partial pressure forces massive amounts of Nitrogen gas to dissolve directly into the diver's blood and fatty tissues.
Step 2: Decompression Sickness (The Bends)
When the diver swims back to the surface, the external pressure drops rapidly. The solubility of Nitrogen plummets. The massive amount of dissolved Nitrogen suddenly bursts out of solution, forming large gas bubbles directly inside the blood vessels. These bubbles block capillaries (embolisms), causing excruciating pain, paralysis, and death.
Step 3: The Helium Solution
Helium is a noble gas with an exceptionally small, tightly held electron cloud. Its polarizability is near zero, meaning its intermolecular forces with solvent molecules (blood) are virtually non-existent.
Consequently, the Henry's Law constant ($K_H$) for Helium is astronomically high, meaning its solubility in blood is phenomenally low, even under crushing deep-sea pressures. Because very little Helium dissolves in the blood in the first place, no dangerous bubbles form when the diver surfaces.
View Solution
Step 1: Analyze $XeF_2$
- Central atom: Xenon. Starts with 8 valence $e^-$.
- Forms 2 single bonds with Fluorine, using 2 $e^-$.
- Remaining $e^- = 6$, which equals 3 Lone Pairs.
- Total valence electrons on central atom system: 8 (from Xe) + 2 (contributed by F bonds) = 10 (or 2 BP + 3 LP = 10 electrons). Wait, standard count: $8 + 2 \times 7 = 22$ total valence electrons.
- Steric Number = 2 BP + 3 LP = 5 $\rightarrow$ $sp^3d$ hybridization. The 3 LP go equatorial, yielding a Linear shape.
Step 2: Analyze $I_3^-$
- Central atom: Iodine. Starts with 7 valence $e^-$.
- Add 1 electron for the anionic charge = 8 valence $e^-$.
- Forms 2 single bonds with the two terminal Iodine atoms, using 2 $e^-$.
- Remaining $e^- = 6$, which equals 3 Lone Pairs.
- Total valence electrons: $3 \times 7 + 1 = 22$ total valence electrons.
- Steric Number = 2 BP + 3 LP = 5 $\rightarrow$ $sp^3d$ hybridization. The 3 LP go equatorial, yielding a Linear shape.
Step 3: Conclusion
Both species possess exactly 22 total valence electrons (isoelectronic). Both have a central atom with 2 Bond Pairs and 3 Lone Pairs (isostructural), dictating an identical $sp^3d$ linear geometry.
View Solution
Step 1: The First Abstraction
The Chlorine radical ($Cl^{\bullet}$) abstracts a hydrogen atom from the methyl group of Toluene. This leaves behind a Benzylic Radical ($C_6H_5-C^{\bullet}H_2$).
Step 2: Resonance Stabilization
This radical is exceptionally stable because the single unpaired electron delocalizes perfectly into the extensive $\pi$-cloud of the adjacent benzene ring. The radical character spreads to the ortho and para positions, drastically lowering the activation energy for this abstraction step compared to a normal alkane.
Step 3: The Exhaustive Chain
Once Benzyl chloride is formed, it still possesses two benzylic hydrogens. Because the benzylic position remains massively activated by resonance stabilization, the Chlorine radicals continue to preferentially attack this exact carbon atom, effortlessly abstracting the remaining hydrogens sequentially until the carbon is completely saturated with Chlorine atoms.
View Solution
Step 1: The Initial Attack
The exceptionally strong $Si-F$ bond ($\approx 565 \text{ kJ/mol}$) drives the reaction. $HF$ aggressively attacks the solid silica matrix, breaking the $Si-O$ bonds and replacing them with Fluorine atoms to form a volatile gas.
Reaction: $SiO_{2(s)} + 4HF_{(aq)} \rightarrow \mathbf{SiF_{4(g)} \uparrow} + 2H_2O_{(l)}$
Step 2: Formation of the Soluble Complex
Silicon tetrafluoride ($SiF_4$) is an electron-deficient Lewis acid (with empty $3d$ orbitals). In the presence of excess $HF$, it accepts two more Fluoride ions ($F^-$) to expand its octet and form a highly stable, water-soluble complex acid.
Reaction: $SiF_{4(g)} + 2HF_{(aq)} \rightarrow \mathbf{H_2[SiF_6]_{(aq)}}$
Conclusion:
By converting the rigid solid silica network into a gas and a soluble liquid complex, the $HF$ effectively eats completely through the walls of the glass container. It must be stored in specialized Teflon or wax-coated plastic bottles.
View Solution
Step 1: Identifying Gas A
A pale greenish-yellow, poisonous gas is the universal physical description of Chlorine gas ($Cl_2$). Thus, A = $Cl_2$.
Step 2: Cold/Dilute Disproportionation (B and C)
Chlorine in cold, dilute $NaOH$ yields Sodium chloride and Sodium hypochlorite.
$Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$
Since C is a bleaching agent, C is Sodium hypochlorite ($NaOCl$). Therefore, B is Sodium chloride ($NaCl$).
Step 3: Hot/Conc Disproportionation (D)
Chlorine in hot, concentrated $NaOH$ yields Sodium chloride and Sodium chlorate.
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
Therefore, highly oxidized salt D is Sodium chlorate ($NaClO_3$).
Step 4: Thermal Decomposition (E)
Heating Sodium chlorate (or Potassium chlorate) with $MnO_2$ catalyst decomposes it violently to yield Sodium chloride and Oxygen gas.
$2NaClO_3 \xrightarrow{MnO_2, \Delta} 2NaCl + 3O_2 \uparrow$
The life-sustaining gas E is Oxygen ($O_2$).
Mastering the Extremes
Congratulations on conquering these 25 ultra-challenging problems on Groups 17 and 18! You have successfully navigated the paradoxical electron gain enthalpy of Chlorine, the explosive thermodynamics of Xenon trioxide, and the intricate disproportionation cascades of the Halogens. Remember, in JEE Advanced, noble gases are never truly "inert" and Fluorine never obeys the rules of the rest of its group. Keep evaluating orbital availability and hybridization limits to predict these exotic molecules. Visit Chemca.in for more elite masterclasses!
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