Masterclass: 25 Ultra-Challenging JEE Advanced Problems on p-Block Elements (Groups 13 & 14)
From the quantum anomaly of $p\pi-p\pi$ back-bonding in Boron halides to the intricate 3D topologies of Silicates and Silicones. Master the chemistry of the Boron and Carbon families.
Groups 13 and 14 are notorious for breaking standard octet rules and periodic trends. To conquer JEE Advanced, you must perfectly understand electron-deficient multi-center bonds (banana bonds), the thermodynamic stabilization of the Inert Pair Effect, and the hydrolytic vulnerabilities created by empty d-orbitals.
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Step 1: The Inductive Expectation
Based purely on the $-I$ inductive effect, Fluorine should drain the most electron density from Boron, making $BF_3$ the most electron-deficient and strongest Lewis acid. However, this is chemically false.
Step 2: $p\pi-p\pi$ Back-Bonding
Boron is $sp^2$ hybridized with an empty $2p$ orbital. The attached halogen atoms have full $p$-orbitals containing lone pairs. These halogen lone pairs can coordinate internally with the empty Boron orbital to form a partial double bond (back-bonding).
Step 3: Orbital Overlap Efficiency
- In $BF_3$, the back-bonding is between a $2p$ (Fluorine) and a $2p$ (Boron) orbital. Because the orbitals are of identical size and energy, the overlap is extremely highly efficient. The internal electron deficiency of Boron is heavily satisfied, making it a very weak Lewis acid.
- In $BCl_3$, the overlap is $3p-2p$, which is less efficient due to a size mismatch.
- In $BBr_3$ ($4p-2p$) and $BI_3$ ($5p-2p$), the size mismatch is so severe that back-bonding is practically negligible. The Boron atom remains fiercely electron-deficient.
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Step 1: Terminal Bonds
There are four terminal $B-H$ bonds. These are standard, localized, two-center two-electron ($2c-2e$) bonds. They consume 8 electrons.
Step 2: Bridge Bonds (Banana Bonds)
This leaves exactly 4 electrons to bind the two Boron atoms and the two central bridging Hydrogen atoms. Boron uses $sp^3$ hybrid orbitals. A single molecular orbital is formed by the overlap of an $sp^3$ orbital from Boron 1, the $1s$ orbital of a bridging Hydrogen, and an $sp^3$ orbital from Boron 2. Two electrons drop into this massive delocalized orbital, forming a three-center two-electron ($3c-2e$) bond. There is one bridge above the plane and one below it.
Step 3: Geometry and Hybridization
Because Boron forms four bonds (two terminal, two bridging halves), its steric number is 4, dictating $sp^3$ hybridization. The two Boron atoms and the four terminal Hydrogens lie perfectly flat in a single plane. The two bridging Hydrogens sit in a plane strictly perpendicular ($90^{\circ}$) to this flat plane.
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Step 1: Reaction with Trimethylamine (Symmetrical Cleavage)
Trimethylamine ($NMe_3$) is an extremely bulky Lewis base. If two $NMe_3$ molecules tried to attach to the same Boron atom, the steric hindrance would be catastrophic. Therefore, the base attacks symmetrically, splitting the dimer perfectly in half.
$B_2H_6 + 2NMe_3 \rightarrow 2[BH_3 \cdot NMe_3]$ (A neutral adduct).
Step 2: Reaction with Ammonia (Unsymmetrical Cleavage)
Ammonia ($NH_3$) is a small, hard Lewis base. It suffers very little steric hindrance. Two $NH_3$ molecules attack a single Boron atom in the diborane structure simultaneously. This forces the remaining $BH_2$ unit to tear away with the bridging hydrides, yielding an ionic salt.
$B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+ [BH_4]^-$
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Step 1: Mechanism of Acidity
Boron in $B(OH)_3$ is electron-deficient (only 6 valence electrons). It acts as a Lewis acid by accepting a lone pair from the oxygen atom of a water molecule. This forms a complex that immediately kicks out the water's proton.
$B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+_{(aq)}$
Step 2: Enhancement by cis-Diols
This equilibrium lies far to the left ($pK_a \approx 9.2$), making it a very weak acid. However, when compounds containing cis-diols (like glycerol, catechol, or mannitol) are added, they rapidly react with the $[B(OH)_4]^-$ anion.
Step 3: Chelate Formation (Le Chatelier's Principle)
The cis-diols form highly stable, 5-membered spirane-like chelate complexes with the borate anion. This continuous removal of the $[B(OH)_4]^-$ product heavily drives the original equilibrium forward (to the right) according to Le Chatelier's Principle, releasing massive amounts of $H^+$ and making the solution strongly acidic (capable of being titrated with $NaOH$ using phenolphthalein).
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Step 1: Aromaticity and Polarity
Borazine has 6 $\pi$-electrons delocalized over the ring (donated by Nitrogen's lone pairs into Boron's empty p-orbitals). While somewhat aromatic, the $B-N$ bonds are highly polar because Nitrogen is significantly more electronegative than Boron.
The formal charges from the dative bond place negative charge on Boron, but actual electronegativity dictates that Nitrogen is $\delta-$ and Boron is $\delta+$.
Step 2: The Attack Trajectory
$HCl$ is a polar molecule ($H^{\delta+} - Cl^{\delta-}$). Because Borazine is highly polarized, the addition is remarkably facile compared to benzene.
- The electrophilic proton ($H^+$) is strongly attracted to the electron-rich, more electronegative Nitrogen atom.
- The nucleophilic chloride ion ($Cl^-$) is strongly attracted to the electron-deficient Boron atom.
Step 3: The Resulting Product
The addition occurs across all three $B=N$ bonds, completely destroying the weak aromaticity of the ring and yielding a saturated cyclohexane-like structure where every Nitrogen gains an $H$ and every Boron gains a $Cl$.
View Solution
Step 1: Dehydration and Swelling
Heating first drives off the 10 molecules of water of crystallization. The sudden steam generation causes the mass to swell massively into anhydrous borax.
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 + 10H_2O \uparrow$
Step 2: Formation of the Glassy Bead
Further intense red-heat causes the anhydrous borax to decompose into a transparent, molten liquid consisting of Sodium Metaborate and Boric Anhydride. Upon cooling, this solidifies into a glassy bead.
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + \mathbf{B_2O_3}$ (Boric Anhydride)
Step 3: The Colored Complex Formation
The $B_2O_3$ component acts as an aggressive acidic flux. When heated with a basic transition metal oxide (like $CoO$, which forms when the cobalt salt is heated), it dissolves the metal oxide to form a highly characteristic, colored metal metaborate salt.
$CoO + B_2O_3 \xrightarrow{\Delta} \mathbf{Co(BO_2)_2}$ (Cobalt Metaborate, Brilliant Blue)
View Solution
Step 1: Analyze $B_2H_6$ (Banana Bonding)
Boron has an empty orbital. Hydrogen, however, has ONLY one electron, which is already used in the single $B-H$ bond. There are strictly zero lone pairs available anywhere in the molecule. It is physically impossible to form a dative bond.
To dimerize, the two electrons already making up the $B-H$ bond must delocalize and stretch to cover a second Boron atom simultaneously. This creates a highly unusual, electron-deficient $3c-2e$ (Three-center, two-electron) Banana bond.
Step 2: Analyze $Al_2Cl_6$ (Dative Bonding)
Aluminium has an empty $3p$ (and $3d$) orbital. Chlorine, unlike Hydrogen, possesses three full, non-bonding lone pairs. To dimerize, a Chlorine atom from one $AlCl_3$ molecule simply takes one of its lone pairs and donates it entirely into the empty orbital of the Aluminium atom on the other molecule.
This forms a classic Coordinate Covalent (Dative) Bond. It is a standard two-electron bond, just formed by one atom donating both electrons. Because the bridge involves 3 atoms ($Al-Cl-Al$) held together by 4 electrons (one normal covalent bond, one dative bond), it is classed as a $3c-4e$ bond.
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Step 1: The Inert Pair Effect
Thallium ($Z=81$) has the valence configuration $6s^2 6p^1$. The inner $4f$ and $5d$ subshells shield the nucleus extremely poorly. Thus, the effective nuclear charge ($Z_{eff}$) heavily pulls on the highly penetrating $6s^2$ electrons, binding them so tightly to the nucleus that they refuse to participate in bonding (they become "inert"). Removing all three electrons to form $Tl^{3+}$ requires an impossibly high promotional/ionization energy.
Step 2: The Structural Reality of $TlI_3$
Because $Tl^{3+}$ is massively unstable, it is actually a phenomenal oxidizing agent. If it were to exist near an Iodide ion ($I^-$), it would instantly rip electrons from the iodide, oxidizing it to $I_2$ while reducing itself to the highly stable $Tl^+$ state.
Therefore, $TlI_3$ is not Thallium(III) Iodide. It is actually Thallium(I) Triiodide.
Step 3: The Triiodide Anion
The true structure is the ionic salt $Tl^+ [I_3]^-$. The central iodine in the triiodide anion expands its octet (using empty $5d$ orbitals) to accommodate the extra atoms, forming a linear $sp^3d$ hybridized anion. This makes $TlI_3$ structurally isomorphous to Potassium triiodide ($KI_3$).
View Solution
Step 1: Analyze Silicon Tetrachloride ($SiCl_4$)
Silicon belongs to the 3rd period. Its valence shell is $n=3$, meaning it strictly possesses empty $3d$ orbitals.
When a water molecule approaches, it can effortlessly donate its oxygen lone pair into one of these empty $3d$ orbitals. This forms a 5-coordinate intermediate, lowering the activation energy significantly and allowing the $Si-Cl$ bonds to eventually break and be replaced by $-OH$ groups.
Step 2: Analyze Carbon Tetrachloride ($CCl_4$)
Carbon belongs to the 2nd period. Its valence shell is $n=2$, which only contains $2s$ and $2p$ orbitals. It has absolutely no d-orbitals available. Furthermore, carbon is extremely small, and the four massive chlorine atoms create an impenetrable steric shield around it.
Because there is no empty orbital for the incoming water molecule to coordinate with, the required transition state physically cannot form. The activation energy is insurmountable, rendering $CCl_4$ completely inert to hydrolysis.
View Solution
Step 1: Analyze Trimethylamine ($N(CH_3)_3$)
Nitrogen is $sp^3$ hybridized, giving a pyramidal shape. Carbon is in the second period and has no empty d-orbitals. The lone pair remains entirely localized on the Nitrogen, causing lone-pair/bond-pair repulsion (pyramidal shape) and making the molecule highly basic.
Step 2: Analyze Trisilylamine ($N(SiH_3)_3$)
Silicon is in the third period and possesses empty $3d$ orbitals. The lone pair on Nitrogen resides in a $2p$ orbital. Because these orbitals are adjacent, the Nitrogen lone pair delocalizes and donates electron density into the empty $3d$ orbitals of the three Silicon atoms. This is called $p\pi-d\pi$ back bonding.
Step 3: Geometric Consequence
To maximize this orbital overlap, the Nitrogen atom rehybridizes from $sp^3$ to $sp^2$. The three $N-Si$ sigma bonds form a perfect trigonal planar geometry, and the unhybridized p-orbital (now shared via $\pi$-bonding) lies perpendicular to the plane. Because the lone pair is completely delocalized into the Silicon atoms, it is unavailable for protonation, rendering the molecule completely non-basic.
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Step 1: Pyrosilicates (Sorosilicates)
Two tetrahedra join together by sharing exactly one corner oxygen.
Formula derivation: $2 \times (SiO_4) - 1(O) = Si_2O_7$. Each terminal oxygen carries a $-1$ charge (6 total).
Ion formula: $Si_2O_7^{6-}$.
Step 2: Cyclic Silicates and Chain Silicates
To form a closed ring or an infinite linear chain, every tetrahedron must hold hands with two neighbors. Therefore, they share exactly two corner oxygens per tetrahedron.
Each tetrahedron has 1 Silicon, 2 unshared Oxygens (charge $-2$), and 2 shared halves of Oxygen (which equal 1 full neutral Oxygen).
Net formula per unit: $[SiO_3^{2-}]_n$.
Step 3: Three-Dimensional Framework Silicates
In a full 3D network like Quartz ($SiO_2$), every single corner is shared with another tetrahedron. Therefore, they share four corner oxygens.
Net formula per unit: $Si$ + 4(halves of O) = $SiO_2$ (neutral).
View Solution
Step 1: Linear Polymerization
Hydrolysis of $R_2SiCl_2$ yields $R_2Si(OH)_2$ (a diol). Because it has two $-OH$ groups, it can link to another molecule on its left and another on its right, propagating an infinite linear chain: $-O-Si(R_2)-O-Si(R_2)-O-$.
Step 2: The Chain Stopper Requirement
To terminate the chain, you must cap the ends with a unit that has no remaining $-OH$ groups to continue the sequence.
Step 3: The Trialkyl Reagent
You must add Trialkylchlorosilane ($R_3SiCl$). Upon hydrolysis, this yields $R_3SiOH$. This molecule only has one hydroxyl group. It will condense onto the end of the growing linear polymer chain, attaching a bulky $-SiR_3$ group. Because this group lacks any further $-OH$ sites, polymerization at that end permanently ceases.
Note: Conversely, adding $RSiCl_3$ allows 3D cross-linking because it provides three active condensation sites.
View Solution
Step 1: The Carbon Advantage
Carbon is extremely small (Period 2). The overlap between its tiny $2p$ and $2s$ hybrid orbitals is phenomenally efficient. Consequently, the Carbon-Carbon ($C-C$) single bond is incredibly strong, possessing a bond enthalpy of roughly $\mathbf{348 \text{ kJ/mol}}$. Because this bond is so thermodynamically stable, long carbon chains do not spontaneously break apart.
Step 2: The Silicon Weakness
Silicon is significantly larger (Period 3). The $3p-3p$ orbital overlap is much more diffuse and less efficient. The Silicon-Silicon ($Si-Si$) single bond is considerably weaker, with a bond enthalpy of only $\approx \mathbf{297 \text{ kJ/mol}}$.
Step 3: The Competing Oxygen Trap
Crucially, Silicon has a massive thermodynamic affinity for Oxygen. The $Si-O$ bond is extremely strong ($\approx 368 \text{ kJ/mol}$) compared to the $Si-Si$ bond. Therefore, any long chains of $Si-Si$ bonds are thermodynamically unstable in the environment and will eagerly react with oxygen or water to break the chains and form highly stable $Si-O-Si$ linkages (Silica/Silicates).
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Step 1: The Donor Atom
In the MO diagram of $CO$, the Highest Occupied Molecular Orbital (HOMO) is a slightly antibonding $\sigma$-orbital that has predominant Carbon character (because Carbon is less electronegative, its atomic orbitals are closer in energy to this HOMO). Thus, $CO$ always coordinates to the metal through the Carbon atom.
Step 2: Sigma Donation (Forward Bonding)
The carbon atom donates this lone pair of electrons from its $\sigma$-HOMO into an empty d-orbital (or hybrid orbital) on the transition metal, forming a standard $M \leftarrow C$ coordinate $\sigma$-bond.
Step 3: Pi-Back Donation (Synergic Effect)
Transition metals in low oxidation states have full d-orbitals packed with electron density. The $CO$ molecule possesses empty, low-lying $\pi^*$ (antibonding) orbitals (the LUMO).
The metal spontaneously pushes electron density from its full d-orbitals backward into the empty $\pi^*$ orbitals of the $CO$ ligand. This is called $\pi$-back bonding.
Step 4: The Synergic Loop
This creates a self-reinforcing loop: Sigma donation makes the metal more electron-rich, which increases its ability to push electrons back via $\pi$-bonding. The $\pi$-back bonding removes excess negative charge from the metal, making it a better $\sigma$-acceptor. This "Synergic" effect creates a phenomenally strong metal-carbon bond.
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Step 1: Thermodynamic Baselines
By convention in thermodynamics, the most stable allotrope of an element at standard state ($1 \text{ bar}$, $298 \text{ K}$) is assigned a formation enthalpy of zero.
For Carbon, $\Delta H_f^{\circ} (Graphite) = 0 \text{ kJ mol}^{-1}$.
We are given $\Delta H_f^{\circ} (Diamond) = +1.90 \text{ kJ mol}^{-1}$. This means diamond sits at a higher energy state than graphite.
Step 2: The Reaction Enthalpy
We are looking at the phase transition: $C_{(Diamond)} \rightarrow C_{(Graphite)}$.
$\Delta H_{\text{reaction}} = \sum \Delta H_f^{\circ} (\text{Products}) - \sum \Delta H_f^{\circ} (\text{Reactants})$
$\Delta H_{\text{reaction}} = [0] - [+1.90]$
$\Delta H_{\text{reaction}} = -1.90 \text{ kJ mol}^{-1}$.
Step 3: Conclusion
Because $\Delta H$ is negative, the process is exothermic. Diamonds are kinetically stable (immense activation energy required to break the 3D covalent network), but thermodynamically unstable. Over billions of years, or if heated intensely without oxygen, diamond will spontaneously collapse into graphite, releasing heat.
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Step 1: The Aluminosilicate Framework
Pure 3D silica ($SiO_2$) is a neutral framework of $SiO_4$ tetrahedra. In a zeolite, a fraction of the Silicon(IV) atoms in the lattice are systematically replaced by Aluminium(III) atoms.
Step 2: The Origin of the Charge
Silicon has a $+4$ valency. Aluminium has a $+3$ valency. Every time an $Al^{3+}$ ion replaces an $Si^{4+}$ ion in the rigid tetrahedral framework, the structure becomes short by one unit of positive charge. This leaves a permanent net negative charge localized on the framework.
To maintain electrical neutrality, mobile cations (like $Na^+$, $K^+$, or $H^+$) sit loosely in the pores of the zeolite. When these cations are $H^+$, the interior of the zeolite acts as an incredibly potent, solid Bronsted acid, making it a spectacular catalyst.
Step 3: ZSM-5 and Shape Selectivity
The pores of zeolites are of exact, uniform molecular dimensions. Only reactant molecules of a specific size can enter, and only products of a specific size can exit. The catalyst ZSM-5 (Zeolite Socony Mobil-5) possesses pores uniquely sized to dehydrate alcohols (like methanol) and polymerize them directly into branched, high-octane hydrocarbon chains that perfectly constitute gasoline.
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Step 1: The Composition of $Pb_3O_4$
The fractional average oxidation state of Pb in $Pb_3O_4$ is $+8/3$. This implies it is a mixture of two distinct states. Lead favors the $+2$ state (due to the inert pair effect) and the $+4$ state.
Structurally, Red Lead is a stoichiometric mixture of exactly two moles of Lead(II) oxide and one mole of Lead(IV) oxide: $2PbO \cdot PbO_2$.
Step 2: Reaction with Dilute Nitric Acid
When $2PbO \cdot PbO_2$ is treated with dilute $HNO_3$:
- The basic $PbO$ component reacts readily with the acid in a simple neutralization to form soluble Lead(II) nitrate.
- The $PbO_2$ component is highly covalent and inert to dilute acids (it is a powerful oxidizing agent, but $HNO_3$ cannot be oxidized further). It remains completely unreacted.
Step 3: The Chemical Equation
$Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 \text{ (soluble)} + \mathbf{PbO_2 \downarrow \text{ (insoluble brown/black precipitate)}} + 2H_2O$
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Step 1: Water Gas (Syngas)
Passing superheated steam ($H_2O$) over red-hot coke ($1000^{\circ}\text{C}$) forces an endothermic reaction:
$C_{(s)} + H_2O_{(g)} \rightarrow \mathbf{CO_{(g)} + H_{2(g)}}$
Composition: An equimolar mixture of Carbon Monoxide and Hydrogen gas.
Step 2: Producer Gas
Passing air (approx. $78\% N_2, 21\% O_2$) over red-hot coke leads to the exothermic partial combustion of carbon. The massive amount of inert nitrogen simply flows through unaffected.
$2C_{(s)} + O_{2(g)} + 4N_{2(g)} \rightarrow \mathbf{2CO_{(g)} + 4N_{2(g)}}$
Composition: A mixture of predominantly Carbon Monoxide and Nitrogen gas.
Step 3: The Calorific Difference
In Water Gas ($CO + H_2$), both components are highly combustible fuels that release massive heat when burned in oxygen.
In Producer Gas ($CO + N_2$), the Nitrogen ($N_2$) is completely inert and incombustible. It acts as dead weight, absorbing heat during combustion and drastically diluting the calorific value of the $CO$ component.
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Step 1: First Heating Stage ($370 \text{ K}$)
Gentle heating causes orthoboric acid to lose one molecule of water, forming Metaboric acid.
$H_3BO_3 \xrightarrow{370 \text{ K}} HBO_2 + H_2O \uparrow$
Step 2: Second Heating Stage ($\approx 410 \text{ K}$)
Further heating forces four molecules of metaboric acid to condense, losing more water to form Tetraboric acid.
$4HBO_2 \xrightarrow{410 \text{ K}} H_2B_4O_7 + H_2O \uparrow$
Step 3: Final Red Heat
Intense red heat completely dehydrates the molecule, stripping away all remaining hydrogen as water and leaving behind a hard, transparent, glass-like solid (the primary fluxing agent in the borax bead test).
$H_2B_4O_7 \xrightarrow{\text{Red Heat}} \mathbf{2B_2O_3} + H_2O \uparrow$
View Solution
Step 1: Reaction with Acid
Aluminium sits well above Hydrogen in the electrochemical series. It easily reduces the $H^+$ ions from hydrochloric acid to yield Aluminium chloride and Hydrogen gas.
$2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2AlCl_{3(aq)} + \mathbf{3H_{2(g)} \uparrow}$
Step 2: Reaction with Base
Aluminium dissolves in concentrated $NaOH$ because the hydroxide ions act as strong ligands. They coordinate with the electron-deficient aluminium to form a highly soluble, stable coordination complex (Sodium tetrahydroxoaluminate). Water acts as the oxidizing agent here, being reduced to Hydrogen gas.
$2Al_{(s)} + 2NaOH_{(aq)} + 6H_2O_{(l)} \rightarrow \mathbf{2Na[Al(OH)_4]_{(aq)}} + \mathbf{3H_{2(g)} \uparrow}$
Step 3: The Shared Danger
Both reactions aggressively evolve highly flammable, explosive Hydrogen gas. This is why drain cleaners containing $NaOH$ often have aluminium shards mixed in—the fierce generation of hot $H_2$ gas bubbles physically agitates and breaks apart clogs in pipes.
View Solution
Step 1: The Hydrolysis Equation
When dissolved in water, the tetraborate anion completely hydrolyzes. It breaks apart to yield Orthoboric acid (a very weak acid) and free Hydroxide ions (from the strong base $NaOH$).
$Na_2B_4O_7 + 7H_2O \rightleftharpoons 2NaOH + 4H_3BO_3$
Ionically: $B_4O_7^{2-} + 7H_2O \rightleftharpoons 2OH^- + 4H_3BO_3$
The overwhelming presence of $OH^-$ makes the solution strongly alkaline.
Step 2: The Titration
When you add $HCl$, it neutralizes the $NaOH$ component.
$Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$
Because 1 mole of Borax consumes 2 moles of $HCl$, its n-factor is 2.
Step 3: Choosing the Indicator
At the exact equivalence point, all the $NaOH$ has been neutralized. The flask now contains only $NaCl$ (neutral) and $H_3BO_3$ (Orthoboric acid). Because Orthoboric acid is a weak acid, the pH at the equivalence point will be distinctly acidic ($pH \approx 4-5$).
Phenolphthalein ($pH$ range 8-10) is useless here. You strictly require an indicator that changes color in the acidic range. Methyl Orange (transition range $3.1-4.4$) is the mandatory indicator.
View Solution
Step 1: Vertices, Edges, and Faces
Vertices ($V$) = 60 (since there are 60 Carbon atoms).
Because every carbon atom forms 3 bonds (edges), and every edge connects 2 atoms, Total Edges ($E$) = $\frac{60 \times 3}{2} = 90$.
Step 2: Euler's Formula
For any closed convex polyhedron: $V - E + F = 2$, where $F$ is the total number of faces (rings).
$60 - 90 + F = 2 \implies -30 + F = 2 \implies F = 32$.
There are exactly 32 total rings in the molecule.
Step 3: Separation into Pentagons and Hexagons
Let $x$ be the number of pentagons ($5$-sided) and $y$ be the number of hexagons ($6$-sided).
Eq 1: $x + y = 32$
To count the edges: each pentagon has 5 edges, each hexagon has 6. Since every edge is shared by exactly 2 faces:
$5x + 6y = 2E = 2(90) = 180$
Eq 2: $5x + 6y = 180$
Step 4: Solve the System
Multiply Eq 1 by 5: $5x + 5y = 160$
Subtract from Eq 2: $(5x + 6y) - (5x + 5y) = 180 - 160 \implies \mathbf{y = 20}$ (Hexagons).
Substitute back: $x + 20 = 32 \implies \mathbf{x = 12}$ (Pentagons).
View Solution
Step 1: The Substitution Sites
The terminal $B-H$ bonds are standard covalent bonds and are exposed on the periphery of the molecule. The methyl groups ($-CH_3$) easily replace these four terminal hydrogens.
Step 2: The Structural Consequence
The resulting structure is 1,1,2,2-tetramethyldiborane. The four bulky methyl groups sit on the outside, completely dominating the steric space. The two central bridging hydrogen atoms remain perfectly intact.
Step 3: The Bridge Blockade
Why can't the last two bridging hydrogens be replaced?
1. Steric Hindrance: A methyl group is massive compared to a hydrogen atom. If a methyl group attempted to occupy the central bridging position, it would catastrophically crash into the other bulky methyl groups already present on the terminal positions. The molecule would physically tear itself apart.
2. Bonding Mechanics: The bridge requires an electron-deficient $3c-2e$ banana bond. A methyl carbon (which requires a stable $sp^3$ tetrahedral geometry with full octets) strongly resists being forced into a bizarre 3-center bridging geometry, whereas the tiny spherical $1s$ orbital of Hydrogen is perfectly suited for it.
View Solution
Step 1: Analyze Tin ($Sn$)
Tin sits in Period 5. While the inert pair effect begins to manifest, the $+4$ oxidation state remains vastly more thermodynamically stable than the $+2$ state. Therefore, Tin in $SnCl_2$ ($Sn^{2+}$) is in an "uncomfortable" low oxidation state. It desperately wants to lose two more electrons to reach the stable $+4$ state ($SnCl_4$). Because it wants to be oxidized, $SnCl_2$ acts as a powerful Reducing Agent.
Step 2: Analyze Lead ($Pb$)
Lead sits at the absolute bottom in Period 6. The $6s^2$ electrons are deeply penetrating and poorly shielded by the massive $4f$ and $5d$ orbitals. The effective nuclear charge locks these $6s^2$ electrons tightly, refusing to let them bond (the severe Inert Pair Effect). Consequently, for Lead, the $+2$ oxidation state is phenomenally more stable than the $+4$ state.
Lead in $PbCl_4$ ($Pb^{4+}$) is in an energetically disastrous state. It violently wants to rip two electrons from anything nearby to drop down to the stable $Pb^{2+}$ state ($PbCl_2$). Because it desperately wants to be reduced, $PbCl_4$ acts as a powerful Oxidizing Agent.
View Solution
Step 1: The Dehydration Mechanism
Malonic acid: $HO-C(=O)-CH_2-C(=O)-OH$.
Aggressive dehydration removes $2H_2O$. The oxygen atoms from the $-OH$ groups depart with the two protons from the central $-CH_2-$ group. The remaining valence electrons collapse inward to form cumulative double bonds.
Step 2: Structural Geometry and Hybridization
The resulting structure is a cumulated polyene (a ketene derivative) with the formula: $O=C=C=C=O$.
Every single Carbon atom forms exactly two $\sigma$-bonds and two $\pi$-bonds. A steric number of 2 dictates that all three Carbon atoms are perfectly $sp$ hybridized. The entire 5-atom molecule is perfectly linear ($180^{\circ}$ bond angles).
Step 3: Oxidation States
- Central Carbon: It is double-bonded to two other Carbon atoms. Because the bonds are between identical elements (C=C), the electronegativity difference is zero. No electrons are "assigned" to or from it. Its absolute oxidation state is $0$.
- Terminal Carbons: Each is double-bonded to one Carbon (no charge effect) and double-bonded to one highly electronegative Oxygen atom. Oxygen takes the $-2$ state, meaning the terminal Carbon must take the $+2$ state to balance the local bond.
Note: The fractional average oxidation state often calculated algebraically ($\frac{4}{3}$ or $+1.33$) is merely the mathematical average of these distinct $+2, 0, +2$ states.
Mastering the Anomalies
Congratulations on conquering these 25 ultra-challenging problems! Groups 13 and 14 are the definitive transition zones of the periodic table. They bridge the gap between electropositive metals and highly electronegative non-metals. To truly dominate this section in JEE Advanced, you must shift your mindset away from rigid octets. Embrace the weirdness of 3c-2e banana bonds, the massive thermodynamic disruptions caused by the Inert Pair Effect, and the quantum tricks of back-bonding. Keep refining your inorganic logic, and visit Chemca.in for more world-class masterclasses!
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