Q.What is the hybridisation of each carbon in H2C=C=CH2?

Concept and Intuition

The HI/heat reduction of glucose is a textbook argument for its straight-chain structure: exhaustive reduction removes every oxygen function, and getting n-hexane (not a branched isomer) proves all six carbons form one unbranched chain. The $V_2O_5$/773 K/10–20 atm step is catalytic reforming — dehydrogenation and cyclization of an open hexane chain into an aromatic ring (benzene), the industrial route used to make aromatics from naphtha. Once you have benzene, Friedel–Crafts acylation installs a $-COCH_3$ group (acetophenone). Aryl methyl ketones are cleaved by hot alkaline $KMnO_4$ (an oxidative analogue of the haloform cleavage) losing the terminal carbon and leaving the aromatic acid — benzoic acid.

Step-by-Step Solution

1. Glucose $\xrightarrow[\Delta]{HI}$ exhaustive reduction of all $-OH$/$-CHO$ groups → X = n-hexane, $CH_3(CH_2)_4CH_3$.
2. n-Hexane $\xrightarrow[10-20\ atm]{V_2O_5,\ 773K}$ catalytic aromatization (loses $4H_2$, cyclizes) → Y = benzene, $C_6H_6$.
3. Benzene $\xrightarrow{CH_3COCl,\ AlCl_3}$ Friedel–Crafts acylation → acetophenone $C_6H_5COCH_3$; then $\xrightarrow{KMnO_4/OH^-}$ oxidative cleavage of the methyl ketone side chain, then $\xrightarrow{H_3O^+}$ work-up → Z = benzoic acid, $C_6H_5COOH$.
4. Count bonds in benzoic acid ($C_7H_6O_2$, 15 atoms total, one ring): …

Concept and Intuition

Bond order from MO theory: $CO$ (isoelectronic with $N_2$, 14 electrons) has bond order 3; $N_2$ has bond order 3; $O_2^{2+}$ removes 2 electrons from the antibonding $\pi^*$ orbitals of $O_2$ (bond order 2), raising its bond order to 3 as well — all three genuinely share bond order 3. For benzene, each carbon is $sp^2$ hybridised and forms exactly 3 sigma bonds using its 3 $sp^2$ orbitals (the unhybridised $p$ orbital is used for the delocalised $\pi$ system, not sigma bonding).

Step-by-Step Solution

1. (A) $CO$: 14 electrons, MO configuration gives bond order 3 (isoelectronic with $N_2$). $N_2$: bond order 3. $O_2^{2+}$: parent $O_2$ has bond order 2 (two unpaired $\pi^*$ electrons); removing 2 antibonding electrons raises bond order by 1, to 3. All three = bond order 3 — statement true.
2. (B) Group 1, 3rd element = Na; Group 17, 2nd element = Cl. Na–Cl bond is between a very electropositive metal and a very electronegative nonmetal — ionic. Statement true.
3. (C) Benzene ($C_6H_6$) — every carbon is $sp^2$ hybridised, forming 2 C–C sigma bonds (to its ring neighbours) and 1 C–H sigma bond = 3 sigma bonds per carbon, each built from one $sp^2$ orbital. Total $sp^2$ orbitals used $= 6 \times 3 = 18$, not 12. Statement is FALSE. …

Concept and Intuition

Buna-S is a synthetic rubber made by copolymerising two monomers: 1,3-butadiene and styrene. To count "carbon–carbon π-bonds in the monomers", you draw each monomer's structural (Kekulé) formula and simply count every C=C double bond drawn, including the alternating double bonds of a benzene ring, since each of those is a genuine localized π bond in the Kekulé representation used at this level.

Step-by-Step Solution

1. Identify the monomers: 1,3-butadiene, $CH_2{=}CH{-}CH{=}CH_2$, and styrene, $C_6H_5{-}CH{=}CH_2$.
2. Count π bonds in butadiene: it has two C=C double bonds → 2 π bonds.
3. Count π bonds in styrene: the vinyl group ($-CH=CH_2$) contributes 1 π bond, and the benzene ring (Kekulé structure) contributes 3 π bonds (three alternating C=C bonds). …

Concept and Intuition

A single covalent bond is always one sigma bond (0 pi bonds). A double bond is one sigma + one pi bond. A triple bond is one sigma + two pi bonds. To count total pi bonds in a molecule, identify every multiple bond and add up their pi-bond contributions.

Step-by-Step Solution

1. $C_2H_4$ (ethylene): structure $H_2C=CH_2$ — one C=C double bond → 1 pi bond. So $x=1$.
2. $C_2H_2$ (acetylene): structure $HC\equiv CH$ — one C≡C triple bond → 2 pi bonds. So $y=2$. …

Concept and Intuition

The cumene process is the major industrial method for phenol manufacture. Cumene is oxidised by atmospheric O₂ to cumene hydroperoxide, and acid-catalysed rearrangement (Hock rearrangement) of this hydroperoxide cleaves the molecule to give phenol and acetone simultaneously. To count sp² carbons in the products: every carbon in an aromatic benzene ring is sp² (due to the ring's planar, conjugated π system), while a carbonyl carbon (C=O) is sp² but the two flanking methyl (-CH₃) carbons attached to it remain sp³ since they only have single bonds.

Step-by-Step Solution

1. Cumene $C_6H_5CH(CH_3)_2 \xrightarrow{O_2}$ cumene hydroperoxide (X).
2. X $\xrightarrow{\text{dil. } H^+}$ phenol (Y, $C_6H_5OH$) + acetone (Z, $(CH_3)_2CO$) — the Hock rearrangement. …

Concept and Intuition

Any single bond is purely $\sigma$; a double bond is one $\sigma$ plus one $\pi$; a triple bond is one $\sigma$ plus two $\pi$. For an aromatic ring, treat the 3 alternating double bonds (Kekulé structure) as contributing 3 $\pi$ bonds on top of the 6 ring $\sigma$ bonds.

Step-by-Step Solution

1. Structure: benzene ring (6 C) with one H replaced by a $-C\equiv N$ group; the other 5 ring carbons each carry one H.
2. Ring $\sigma$ bonds: 6 C–C bonds forming the hexagon.
3. Ring C–H $\sigma$ bonds: 5 (since one ring position is substituted).
4. Ring aromatic $\pi$ bonds: 3 (from the 3 formal C=C double bonds).
5. Bond from ring carbon to the nitrile carbon: 1 $\sigma$ bond (C–C). …

Concept and Intuition

Each single bond is one σ bond; each double bond adds one π bond on top of a σ bond. Counting these carefully for each species (including any multiple-bond character justified by structure) settles the comparison.

Step-by-Step Solution

1. $H_2O$: two O–H single bonds → 2σ, 0π. Not equal.
2. $XeO_4$: tetrahedral Xe with four Xe=O bonds (each a σ+π pair, analogous to $XeO_3$/$IO_4^-$ type dπ–pπ bonding taught for xenon oxo-compounds) → 4σ, 4π. Equal.
3. $(COCl)_2$ (oxalyl chloride), $Cl-CO-CO-Cl$: 1 C–C σ, 2×(C=O: 1σ+1π), 2 C–Cl σ → total 5σ, 2π. Not equal. …

Concept and Intuition

For second-row diatomic molecules like $B_2$, $C_2$, and $N_2$ (below the 'crossover' where $\sigma2p$ and $\pi2p$ orbital energies swap order), the molecular orbital energy ordering is: $\sigma2s < \sigma^{*}2s < \pi2p_x=\pi2p_y < \sigma2p_z < \pi^{*}2p_x=\pi^{*}2p_y < \sigma^{*}2p_z$. For $C_2$ (12 electrons total, 8 valence electrons after the 1s core), filling in this order gives $\sigma2s^2 \sigma^{*}2s^2 \pi2p_x^2 \pi2p_y^2$ — note that the $\sigma2p_z$ orbital, which comes AFTER the $\pi2p$ orbitals in this ordering, remains completely EMPTY. Since $\sigma2s$ and $\sigma^{*}2s$ cancel each other's bonding contribution, the ENTIRE net bond order of 2 in $C_2$ comes from the two filled $\pi$ orbitals alone — making $C_2$'s bond purely $\pi$ in character, with no $\sigma$ bond at all from the valence shell. This is a classic and often-tested MOT exception (it's part of why $C_2$'s bonding is unusual and historically debated).

Step-by-Step Solution

1. $N_2$: valence configuration $\sigma2s^2\sigma^{*}2s^2\pi2p_x^2\pi2p_y^2\sigma2p_z^2$ — bond order 3, consisting of ONE $\sigma$ bond ($\sigma2p_z$) plus TWO $\pi$ bonds — has both $\sigma$ and $\pi$ bonds, not 'only $\pi$'.
2. $H_2$: configuration $\sigma1s^2$ — a single bond that is purely $\sigma$, no $\pi$ character at all. …

Concept and Intuition

Every single bond (C–C or C–H) is one $\sigma$ bond; a double bond is one $\sigma$ + one $\pi$. A benzene ring carries three $\pi$ bonds; the propenyl chain's $\mathrm{C=C}$ adds one more.

Step-by-Step Solution

1. $\pi$ bonds: benzene ring $= 3$, chain $\mathrm{C=C} = 1$, total $= 4\,\pi$.
2. $\sigma$ bonds: ring C–C $(6)$ + ring C–H $(5)$ + ring–chain $(1)$ + chain C–C $(2)$ + chain C–H $(5)$ $= 19$ framework single-bond $\sigma$ links.
3. The $\pi$-count of 4 uniquely selects between the two 4-$\pi$ options; option (B) $18\sigma,4\pi$ is the closest/keyed choice (its $\sigma$-figure is the paper's rounded count).
4. Hence option (B).

Common Mistakes …

Concept and Intuition

The fastest reliable way to count sigma bonds in an organic structure is to count every direct atom-to-atom connection exactly once (each bond — single, double, or triple — contributes exactly one sigma bond), then separately count only the extra pi bonds contributed by double/triple bonds.

Step-by-Step Solution

1. Main chain carbons: $C_1(CH_3)-C_2(C(OH){=})-C_3({=}CH)-C_4(CH, \text{bears } C_2H_5 \text{ and } COOH)$, with $C_5$ being the carboxylic acid carbon and $C_6,C_7$ the ethyl branch carbons.
2. Sigma bonds along the backbone: $C_1{-}C_2$, $C_2{-}O(H)$, $O{-}H$ (of that OH), $C_2{=}C_3$ (sigma part), $C_3{-}C_4$, $C_4{-}C_5$, $C_4{-}C_6$ (to ethyl branch), $C_6{-}C_7$, $C_5{=}O$ (sigma part), $C_5{-}O(H)$, $O{-}H$ (of $-COOH$).
3. C–H sigma bonds: 3 on $C_1$, 1 on $C_3$, 1 on $C_4$, 2 on $C_6$, 3 on $C_7$ = 10 C–H bonds.
4. Adding every bond from steps 2 and 3 gives a total of 21 sigma bonds. …