NCERT Solutions · Class 11 Chemistry Chemical Bonding and Molecular Structure

A geometry claim about \(\displaystyle \mathrm{CH_4}\) can only be tested by looking at what it predicts for a molecule derived from \(\displaystyle CH_4\) and here square planar and tetrahedral geometry make different, checkable predictions.Notice first that this cannot be settled by looking at \(\displaystyle CH_4\) itself. All four atoms attached to carbon are identical (H), so both candidate shapes are highly symmetric: In either case, the four identical C–H bond dipoles are arranged symmetrically enough that their vector sum is zero. So "does \(\displaystyle CH_4\) have a dipole moment?" gives the same answer (no) whichever shape is correct — it cannot distinguish them. The aside people miss: you have to break the symmetry by substitution before dipole moment becomes a useful test.Break the symmetry: replace two H atoms by two Cl atoms, giving \(\displaystyle CH_2Cl_2\).Ask how many different ways there are to place $\displaystyle 2$ Cl and $\displaystyle 2$ H on each candidate shape.If \(\displaystyle CH_4\) were square planar (H at the four corners of a square), the two Cl atoms could occupy: These are genuinely different molecules, not the same structure viewed two ways, because a square has two geometrically distinct pairs of corners (adjacent vs. diagonal). So square planar geometry predicts two isomers of \(\displaystyle CH_2Cl_2\), with different properties. In particular, for the diagonal (trans) form the two \(\displaystyle C-Cl\) bond-dipole vectors point in exactly opposite directions and cancel: \[\vec{\mu}_{net} = \vec{\mu}_{C-Cl,1} + \vec{\mu}_{C-Cl,2} = 0 \quad (\text{trans, } \angle Cl-C-Cl = 180^\circ) \] while for the adjacent (cis) form the two dipoles are at \(\displaystyle 90^\circ\) to each other and do not cancel, giving a nonzero resultant. So square planar geometry predicts one isomer with zero dipole moment and one isomer with a nonzero dipole moment.If \(\displaystyle CH_4\) is tetrahedral, look at the four corners of a regular tetrahedron. Every pair of corners is related to every other pair by a symmetry operation of the tetrahedron — a regular tetrahedron has no "adjacent" versus "diagonal" corners the way a square does; all six edges (and hence all \(\displaystyle \binom{4}{2}=6\) ways of picking $\displaystyle 2$ vertices out of $\displaystyle 4$) are equivalent. So however you choose $\displaystyle 2$ of the $\displaystyle 4$ H positions to replace with Cl, you get the same single structure. Tetrahedral geometry predicts exactly one isomer of \(\displaystyle CH_2Cl_2\), with a \(\displaystyle Cl-C-Cl\) angle close to \(\displaystyle 109.5^\circ\). Since \(\displaystyle 109.5^\circ \neq 180^\circ\), the two C–Cl dipoles do not cancel, so this one isomer must have a nonzero net dipole moment.Compare with experiment. Only one compound called dichloromethane (\(\displaystyle CH_2Cl_2\)) is known — chemists have never isolated two distinct isomers with different melting points, boiling points, or spectra. And that single compound has a measured, nonzero dipole moment (\(\displaystyle \mu \approx 1.60\ D\)).This matches the tetrahedral prediction exactly: one isomer, nonzero dipole moment. It flatly contradicts the square planar prediction, which requires two isomers to exist, one of them (the trans form) with zero dipole moment. Since only one form is ever found, and it is polar, the square planar hypothesis is falsified by this evidence, while the tetrahedral hypothesis survives it.This same reasoning is why chemists trust \(\displaystyle sp^3\) tetrahedral carbon generally: replacing any two of methane's four equivalent hydrogens must give a single, unique, chemically identical product, and every disubstituted methane derivative ever isolated confirms exactly that pattern.**Answer: \(\displaystyle CH_4\) is tetrahedral and not square planar because the two geometries make different, testable predictions about the disubstituted derivative \(\displaystyle CH_2Cl_2\): square planar geometry requires two isomers (a cis form with \(\displaystyle Cl-C-Cl=90^\circ\) and nonzero dipole moment, and a trans form with \(\displaystyle Cl-C-Cl=180^\circ\) and zero dipole moment), whereas tetrahedral geometry — since all pairs of vertices of a regular tetrahedron are equivalent — predicts only one isomer, necessarily with a nonzero dipole moment (as \(\displaystyle 109.5^\circ \neq 180^\circ\)). Experimentally, only one form of \(\displaystyle CH_2Cl_2\) exists and it does have a nonzero dipole moment (\(\displaystyle \approx 1.60\ D\)), which matches the tetrahedral prediction and rules out the square planar one.

A molecule's net dipole moment is the vector sum of its individual bond dipoles, not just a list of them — direction matters as much as size.Each Be–H bond in \(\displaystyle \mathrm{BeH_2}\) is polar because beryllium and hydrogen have different electronegativities, so each bond carries its own small dipole moment, with the negative end pointing toward the more electronegative atom (H is more electronegative than Be, so each dipole points from Be towards H).Step $\displaystyle 1$: Find the shape of the molecule.Beryllium has the configuration \(\displaystyle 1s^2 2s^2\). To form two bonds it promotes one \(\displaystyle 2s\) electron to \(\displaystyle 2p\), giving \(\displaystyle 2s^1 2p^1\), and these two orbitals mix to form two equivalent \(\displaystyle sp\) hybrid orbitals. Two \(\displaystyle sp\) hybrid orbitals always point in exactly opposite directions, \(\displaystyle 180^\circ\) apart, because that is the arrangement that keeps them as far apart as possible. So \(\displaystyle \mathrm{BeH_2}\) is a linear molecule: \[\mathrm{H-Be-H} \] with a bond angle of \(\displaystyle 180^\circ\).This is the step students skip: you cannot decide whether a molecule is polar just by looking at whether its bonds are polar — you must first fix the geometry, because the dipole moment depends on the angle between the bonds.Step $\displaystyle 2$: Add the two bond-dipole vectors.Let each Be–H bond dipole have magnitude \(\displaystyle \mu\), directed from Be towards each H atom. Since the two Be–H bonds point in exactly opposite directions (\(\displaystyle 180^\circ\) apart), the two dipole vectors are equal in magnitude and opposite in sense along the same line: \[\vec{\mu}_1 = -\vec{\mu}_2 \] The resultant dipole moment of the molecule is the vector sum: \[\vec{\mu}_{\text{net}} = \vec{\mu}_1 + \vec{\mu}_2 = \vec{\mu}_1 + (-\vec{\mu}_1) = 0 \]So the two bond dipoles are equal in magnitude and opposite in direction, and they cancel each other exactly.Why this matters: a "polar bond" is a property of one bond in isolation (it depends only on the electronegativity difference between the two bonded atoms). A "polar molecule" is a property of the whole molecule and depends on the geometry too. \(\displaystyle \mathrm{BeH_2}\) has polar bonds but a symmetric linear geometry, so the bond dipoles are positioned to cancel exactly, making the molecule as a whole nonpolar — its net dipole moment is zero.Answer: \(\displaystyle \mathrm{BeH_2}\) is linear (Be is \(\displaystyle sp\) hybridized, bond angle \(\displaystyle 180^\circ\)), so its two equal and polar Be–H bond dipoles point in exactly opposite directions along the same line. Being equal in magnitude and opposite in direction, they cancel each other vectorially, giving the molecule a net dipole moment of zero even though each individual Be–H bond is polar.

The net dipole moment of a pyramidal AX₃E molecule is the vector sum of the three bond dipoles AND the dipole due to the lone pair — and in NH₃ these two contributions add, while in NF₃ they oppose each other. That is why NH₃ ($\displaystyle 1.47$ D) beats NF₃ ($\displaystyle 0.24$ D), even though fluorine is the far more electronegative atom.Setting up the geometryBoth NH₃ and NF₃ have nitrogen as sp³ hybridised, with three bond pairs and one lone pair, giving a pyramidal shape (point group \(\displaystyle C_{3v}\)). By this symmetry, whatever net dipole the molecule has must lie along the \(\displaystyle C_3\) axis — the axis running through N and the lone pair, straight through the centre of the base formed by the three H (or F) atoms. So the total dipole moment is\[\vec{\mu}_{net} = \vec{\mu}_{bonds} + \vec{\mu}_{lone\ pair} \]where \(\displaystyle \vec{\mu}_{bonds}\) is the vector sum of the three individual N–H (or N–F) bond dipoles, and \(\displaystyle \vec{\mu}_{lone\ pair}\) is the dipole contributed by the lone pair sitting on nitrogen.A dipole moment vector is drawn pointing from the less electronegative (partially positive, \(\displaystyle \delta^+\)) atom toward the more electronegative (partially negative, \(\displaystyle \delta^-\)) atom.The lone pair sits in a hybrid orbital that points away from the three bonding pairs, along the \(\displaystyle C_3\) axis, out of the top of the pyramid (with N at the apex). It is a concentration of electron density with no nucleus at the far end to pull it back, so it acts like a dipole pointing outward from N, away from the three bonds — and this direction is fixed by the geometry alone; it does not depend on whether the atoms at the base are H or F.Case $\displaystyle 1$: NH₃Nitrogen (electronegativity ≈ $\displaystyle 3.0$) is more electronegative than hydrogen (≈ $\displaystyle 2.1$). So each N–H bond dipole points from H toward N — that is, from the base of the pyramid up toward the apex, the same general direction as the lone pair's dipole.Here \(\displaystyle \vec{\mu}_{bonds}\) and \(\displaystyle \vec{\mu}_{lone\ pair}\) point the same way along the \(\displaystyle C_3\) axis, so they reinforce:\[\mu_{NH_3} = \mu_{bonds} + \mu_{lone\ pair}\ \text{(magnitudes add)} \]This gives NH₃ a comparatively large observed dipole moment, \(\displaystyle \mu_{NH_3} = 1.47\ \text{D}\).Case $\displaystyle 2$: NF₃Now the electronegativity order is reversed: fluorine (≈ $\displaystyle 4.0$) is more electronegative than nitrogen (≈ $\displaystyle 3.0$). So each N–F bond dipole points from N toward F — from the apex down toward the base — which is the opposite direction to the lone pair's dipole (still pointing outward from N, away from the base, because that direction is fixed by geometry, not electronegativity).This is the step people miss: they assume "F is more electronegative, so NF₃ should have the bigger dipole moment." It is exactly the opposite, because the lone pair's contribution does not flip with the substituent — only the bond dipoles do, and in NF₃ they now fight the lone pair instead of helping it.Here \(\displaystyle \vec{\mu}_{bonds}\) and \(\displaystyle \vec{\mu}_{lone\ pair}\) point in opposite directions along the \(\displaystyle C_3\) axis, so they partly cancel:\[\mu_{NF_3} = |\mu_{lone\ pair} - \mu_{bonds}|\ \text{(magnitudes subtract)} \]Even though each individual N–F bond is more polar than an N–H bond, this cancellation leaves NF₃ with a small net dipole moment, \(\displaystyle \mu_{NF_3} = 0.24\ \text{D}\).Answer: NH₃ has the higher dipole moment ($\displaystyle 1.47$ D vs. $\displaystyle 0.24$ D for NF₃). In both molecules the lone pair on N contributes a dipole pointing away from the three bonds along the pyramidal axis. In NH₃, N is more electronegative than H, so the N–H bond dipoles point toward N — the same direction as the lone pair — and they add. In NF₃, F is more electronegative than N, so the N–F bond dipoles point toward F — opposite to the lone pair — and they partly cancel. That cancellation, not the individual bond polarities, is why NF₃ ends up with the smaller net dipole moment.

Hybridisation is atomic orbitals of an atom mixing together and re‑emerging as a new set of orbitals — same number, same total energy, but now all identical in shape and energy, and pointing in specific directions in space.An atom's pure orbitals (one \(\displaystyle s \), three \(\displaystyle p \)) do not, by themselves, explain the bond angles seen in real molecules. Take methane, \(\displaystyle \mathrm{CH_4} \): carbon's ground-state configuration is \(\displaystyle 1s^2\,2s^2\,2p_x^1\,2p_y^1 \), which offers only two half-filled orbitals — not enough to form four bonds, and nowhere near the observed \(\displaystyle 109.5^\circ \) angle if you tried to use unmixed \(\displaystyle s \) and \(\displaystyle p \) orbitals directly. Hybridisation resolves this: one electron is promoted from \(\displaystyle 2s \) to the empty \(\displaystyle 2p_z \), giving four half-filled orbitals, and then these four orbitals — one \(\displaystyle s \) and three \(\displaystyle p \) — blend into four new, identical orbitals called hybrid orbitals. Each carries a fixed share of \(\displaystyle s \) and \(\displaystyle p \) character, and this share is what fixes the geometry.Rules that govern hybridisation (why the shapes come out fixed, not arbitrary): The type of hybridisation is named after which and how many orbitals were mixed, and the aside people miss is this: the percentage of \(\displaystyle s \)-character in a hybrid orbital, not the label "sp/sp²/sp³" by itself, is what determines the bond angle — more \(\displaystyle s \)-character pulls the electron density closer to the nucleus and pushes the hybrid orbitals farther apart.sp hybridisation ($\displaystyle 1$ s + $\displaystyle 1$ p → $\displaystyle 2$ sp orbitals): linear, \(\displaystyle 180^\circ \). One \(\displaystyle s \) orbital mixes with one \(\displaystyle p \) orbital. Two sp orbitals result, each with \(\displaystyle 50\% \) \(\displaystyle s \)-character and \(\displaystyle 50\% \) \(\displaystyle p \)-character. To stay as far apart as possible, two orbitals can only point in exactly opposite directions along one axis, so the shape is a straight line: \[\text{sp orbitals: } 180^\circ \text{ apart, linear} \] Example: in \(\displaystyle \mathrm{BeCl_2} \), beryllium's two sp hybrid orbitals overlap with chlorine orbitals on either side, giving a linear \(\displaystyle \mathrm{Cl - Be - Cl} \) molecule. The same hybridisation gives the linear \(\displaystyle \mathrm{H-C \equiv C-H} \) skeleton in acetylene.sp² hybridisation ($\displaystyle 1$ s + $\displaystyle 2$ p → $\displaystyle 3$ sp² orbitals): trigonal planar, \(\displaystyle 120^\circ \). One \(\displaystyle s \) orbital mixes with two \(\displaystyle p \) orbitals. Three equivalent sp² orbitals result, each with \(\displaystyle 33.3\% \) \(\displaystyle s \)-character and \(\displaystyle 66.7\% \) \(\displaystyle p \)-character (less \(\displaystyle s \)-character than sp, so a smaller angle than \(\displaystyle 180^\circ \)). Three orbitals maximise their mutual separation by lying in one plane, pointing to the corners of an equilateral triangle: \[\text{sp}^2 \text{ orbitals: } 120^\circ \text{ apart, trigonal planar} \] Example: boron in \(\displaystyle \mathrm{BCl_3} \) is sp² hybridised, giving a flat, triangular molecule with all three \(\displaystyle \mathrm{Cl-B-Cl} \) angles equal to \(\displaystyle 120^\circ \). The same hybridisation gives the flat \(\displaystyle \mathrm{C_2H_4} \) (ethylene) skeleton around each carbon.sp³ hybridisation ($\displaystyle 1$ s + $\displaystyle 3$ p → $\displaystyle 4$ sp³ orbitals): tetrahedral, \(\displaystyle 109.5^\circ \). One \(\displaystyle s \) orbital mixes with all three \(\displaystyle p \) orbitals. Four equivalent sp³ orbitals result, each with \(\displaystyle 25\% \) \(\displaystyle s \)-character and \(\displaystyle 75\% \) \(\displaystyle p \)-character — the least \(\displaystyle s \)-character of the three, hence the largest departure from a small angle and the characteristic tetrahedral value. Four orbitals maximise their separation only by pointing to the four corners of a regular tetrahedron, with the central atom at its centre: \[\text{sp}^3 \text{ orbitals: } 109^\circ 28' \,(\approx 109.5^\circ) \text{ apart, tetrahedral} \] Example: carbon in \(\displaystyle \mathrm{CH_4} \) is sp³ hybridised; all four \(\displaystyle \mathrm{H-C-H} \) angles are \(\displaystyle 109.5^\circ \), reproducing methane's familiar tetrahedral shape.The place students slip is treating hybridisation as something that happens to isolated atoms for its own sake — it is really a bookkeeping device that predicts the geometry an atom's bonds will take once you know how many hybrid orbitals it forms and how many pure \(\displaystyle p \) orbitals (versus \(\displaystyle s \)) went into each one.Answer: Hybridisation is the mixing of atomic orbitals of comparable energy on one atom to give an equal number of new, identical hybrid orbitals with fixed directional geometry. sp hybrid orbitals (from $\displaystyle 1$ s + $\displaystyle 1$ p) are linear, $\displaystyle 180$° apart; sp² hybrid orbitals (from $\displaystyle 1$ s + $\displaystyle 2$ p) are trigonal planar, $\displaystyle 120$° apart; sp³ hybrid orbitals (from $\displaystyle 1$ s + $\displaystyle 3$ p) are tetrahedral, $\displaystyle 109.5$° apart — each shape following from the orbitals arranging themselves as far apart as possible in space.

A central atom's hybridisation is fixed by how many sigma-bond pairs (plus lone pairs) sit around it — count those first, on each side of the reaction.The reaction is \[\text{AlCl}_3 + \text{Cl}^- \longrightarrow \text{AlCl}_4^- \]Step $\displaystyle 1$: Look at Al in \(\displaystyle \mathrm{AlCl_{3}}\).Aluminium has $\displaystyle 3$ valence electrons \(\displaystyle (3s^2 3p^1)\). Each Cl atom takes one of these in a shared pair, so Al forms exactly $\displaystyle 3$ Al–Cl sigma bonds and is left with no lone pair of its own.Three bond pairs and zero lone pairs around the central atom means $\displaystyle 3$ electron domains, arranged in a plane at \(\displaystyle 120^\circ\) — a trigonal planar shape. That geometry is built from sp2 hybrid orbitals (one s + two p orbitals mixed to give $\displaystyle 3$ equivalent orbitals).So in \(\displaystyle \mathrm{AlCl_{3}}\), Al is sp2 hybridised, and because it has an empty, unhybridised \(\displaystyle p\) orbital left over, it is electron-deficient — that empty orbital is exactly why \(\displaystyle \mathrm{AlCl_{3}}\) is a strong Lewis acid.Step $\displaystyle 2$: Look at what the incoming \(\displaystyle \mathrm{Cl^{-}}\) does.A chloride ion carries a lone pair it is not using in any bond. This lone pair is donated into that empty p orbital on Al — a coordinate (dative) bond. Al does not lose or gain a proton or rearrange its own electrons; it simply accepts a new electron pair.Al now has $\displaystyle 4$ shared pairs around it (the original $\displaystyle 3$ Al–Cl sigma bonds plus the new Al←Cl coordinate bond), still with no lone pair. Four electron domains with no lone pair means a tetrahedral shape at \(\displaystyle 109.5^\circ\) angles, which requires mixing one s and three p orbitals — sp3 hybridisation.A step people skip here: don't try to explain the change by looking at oxidation state or charge — Al's oxidation state stays \(\displaystyle +3\) throughout. The hybridisation change is purely a count of electron domains going from $\displaystyle 3$ to $\displaystyle 4$, nothing else.Step $\displaystyle 3$: State the change.\[\text{Al (sp}^2\text{, trigonal planar, in AlCl}_3) \;\xrightarrow{\text{+ Cl}^-}\; \text{Al (sp}^3\text{, tetrahedral, in AlCl}_4^-) \]The extra electron pair donated by \(\displaystyle \mathrm{Cl^{-}}\) forces one more hybrid orbital to be made available, converting the vacant \(\displaystyle p\) orbital's role into a fourth \(\displaystyle sp^3\) hybrid orbital that houses the new bond pair.**Answer: Yes, the hybridisation of Al changes. In \(\displaystyle \mathrm{AlCl_{3}}\), Al is sp2 hybridised ($\displaystyle 3$ bond pairs, trigonal planar, with an empty p orbital that makes it electron-deficient). When \(\displaystyle \mathrm{Cl^{-}}\) donates its lone pair into that empty orbital to form \(\displaystyle \mathrm{AlCl_{4}}\)−, Al becomes sp3 hybridised ($\displaystyle 4$ bond pairs, tetrahedral).

Hybridisation is fixed by how many electron pairs sit around the atom, not by how many bonds it started with — so you have to recount the pairs on B and on N after the reaction, not just look at the reactants.The reaction referred to is boron trifluoride accepting the lone pair of ammonia to form a coordinate (dative) adduct:\[\text{BF}_3 \;+\; \text{NH}_3 \;\longrightarrow\; \text{F}_3\text{B}\!\leftarrow\!\text{NH}_3 \]Here the arrow \(\displaystyle \leftarrow\) stands for a coordinate bond: the shared pair of electrons is supplied entirely by nitrogen, not by boron.Step $\displaystyle 1$ — hybridisation of B and N before the reaction.The hybridisation of a central atom is decided by its steric number, \[\text{steric number} = (\text{number of }\sigma\text{ bonds}) + (\text{number of lone pairs}), \] with steric number $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$ corresponding to \(\displaystyle sp\), \(\displaystyle sp^2\), \(\displaystyle sp^3\). Step $\displaystyle 2$ — what the coordinate bond does to each atom's count.Nitrogen's lone pair, which is already sitting in one of its \(\displaystyle sp^3\) orbitals, is donated straight into boron's empty 2p orbital. No bonds break and no new orbitals need to be built on nitrogen — a pair that was non-bonding simply becomes bonding. On boron, though, a brand-new fourth electron pair has appeared where there was none before.A common mistake here is to assume that because nitrogen is the one donating, it must be nitrogen whose hybridisation changes. It is the opposite: donating a lone pair that already occupies a hybrid orbital changes nothing about that atom's orbital count; it is receiving a new pair into a previously empty orbital that forces a re-hybridisation.Step $\displaystyle 3$ — recount the steric numbers in the product \(\displaystyle \text{F}_3\text{B}\!\leftarrow\!\text{NH}_3\). So the change is one-sided: boron is forced to rehybridise to accommodate the new bond it accepts, while nitrogen's hybridisation was already set up to supply that bond without any change to itself.Answer: Yes, but only for boron. In \(\displaystyle \text{BF}_3\), boron is \(\displaystyle sp^2\) hybridised (steric number $\displaystyle 3$, trigonal planar); on forming the coordinate bond with \(\displaystyle \text{NH}_3\) it becomes \(\displaystyle sp^3\) hybridised (steric number $\displaystyle 4$, tetrahedral). Nitrogen remains \(\displaystyle sp^3\) hybridised throughout — it already had steric number $\displaystyle 4$ in \(\displaystyle \text{NH}_3\), and donating its lone pair as the new N→B bond does not change that count, so its hybridisation is unchanged.

A double bond is one sigma (σ) bond plus one pi (π) bond, and a triple bond is one sigma bond plus two pi bonds — the σ bond comes from head-on overlap of hybrid orbitals, and each π bond comes from sideways overlap of unhybridized p orbitals. To "draw" this, you have to first fix the hybridization at each carbon, because that tells you how many p orbitals are left over to form π bonds.Ethylene, \(\displaystyle \text{C}_2\text{H}_4 \) — the double bondEach carbon here is \(\displaystyle sp^2 \) hybridized. One \(\displaystyle s \) orbital and two \(\displaystyle p \) orbitals mix to give three \(\displaystyle sp^2 \) hybrid orbitals, lying in one plane at \(\displaystyle 120^\circ \) to each other. The third \(\displaystyle p \) orbital of carbon (say \(\displaystyle 2p_z \)) is left unhybridized, standing perpendicular to that plane.Picture one carbon with its three \(\displaystyle sp^2 \) lobes pointing into a flat triangle, and a single unhybridized \(\displaystyle p_z \) orbital sticking up and down through the centre of that triangle (a dumb-bell perpendicular to the plane of the three lobes). The second carbon is built the same way.Now bring the two carbons together: So the C=C double bond in ethylene is: $\displaystyle 1$ σ bond (head-on \(\displaystyle sp^2\)–\(\displaystyle sp^2 \) overlap, along the C–C axis) + $\displaystyle 1$ π bond (sideways \(\displaystyle p_z\)–\(\displaystyle p_z \) overlap, above and below the plane).A step people skip: the π overlap is only possible because the two \(\displaystyle p_z \) orbitals stay parallel — this is exactly why there is no free rotation about a C=C bond (rotating one CH₂ end would break the sideways overlap).Acetylene, \(\displaystyle \text{C}_2\text{H}_2 \) — the triple bondEach carbon here is \(\displaystyle sp \) hybridized. One \(\displaystyle s \) orbital and one \(\displaystyle p \) orbital mix to give two \(\displaystyle sp \) hybrid orbitals, oriented at \(\displaystyle 180^\circ \) to each other (a straight line). That leaves two unhybridized \(\displaystyle p \) orbitals on each carbon — say \(\displaystyle 2p_y \) and \(\displaystyle 2p_z \) — mutually perpendicular to each other and both perpendicular to the \(\displaystyle sp \) axis.Picture one carbon with its two \(\displaystyle sp \) lobes pointing left and right along a straight line, and two p-orbital dumb-bells, one vertical and one horizontal, both centred on the carbon and both at right angles to that line (like a plus-sign cross-section sitting on the axis).Bringing the two carbons together, with hydrogens on the outside: So the C≡C triple bond in acetylene is: $\displaystyle 1$ σ bond (head-on \(\displaystyle sp\)–\(\displaystyle sp\) overlap) + $\displaystyle 2$ π bonds (two sideways \(\displaystyle p\)–\(\displaystyle p\) overlaps in mutually perpendicular planes). The combined π electron density from these two π bonds forms a cylindrical sheath wrapped around the C–C σ bond, which is why the whole \(\displaystyle \text{HC}{\equiv}\text{CH} \) molecule is linear and symmetric about that axis.The pattern to remember: hybridize the σ-framework first (\(\displaystyle sp^2 \) for a double bond, \(\displaystyle sp \) for a triple bond), let whatever \(\displaystyle p \) orbitals are left over sit perpendicular to that framework, and every extra bond beyond the first σ bond is one more sideways \(\displaystyle p\)–\(\displaystyle p\) overlap.**Answer: In \(\displaystyle \text{C}_2\text{H}_4 \), each carbon is \(\displaystyle sp^2 \) hybridized; the C=C double bond is one σ bond from head-on \(\displaystyle sp^2\)–\(\displaystyle sp^2 \) overlap plus one π bond from sideways overlap of the unhybridized \(\displaystyle p_z \) orbitals (above and below the planar molecule). In \(\displaystyle \text{C}_2\text{H}_2 \), each carbon is \(\displaystyle sp \) hybridized; the C≡C triple bond is one σ bond from head-on \(\displaystyle sp\)–\(\displaystyle sp \) overlap plus two π bonds from sideways overlap of the two pairs of unhybridized \(\displaystyle p \) orbitals lying in mutually perpendicular planes, giving the linear H–C≡C–H molecule a cylindrical π-electron sheath around the σ bond.

A single bond between two atoms is always one sigma \(\displaystyle (\sigma)\) bond; a double bond is one \(\displaystyle \sigma\) plus one pi \(\displaystyle (\pi)\); a triple bond is one \(\displaystyle \sigma\) plus two \(\displaystyle \pi\). Every other bond in the molecule — to hydrogen, or from one multiply-bonded carbon further along the chain — is a plain \(\displaystyle \sigma\) bond formed by head-on (axial) overlap. A \(\displaystyle \pi\) bond, by contrast, comes only from the sideways overlap of unhybridized p-orbitals left over once a carbon has hybridized. So to count bonds correctly, first work out how each carbon is hybridized, because that tells you how many p-orbitals are left over to form \(\displaystyle \pi\) bonds.(a) \(\displaystyle \mathrm{C_2H_2}\) — ethyne (acetylene), \(\displaystyle H-C\equiv C-H\)Each carbon here is bonded to only two other atoms (one H and one C), and the molecule is linear, so each carbon is sp hybridized. Two sp hybrid orbitals point in opposite directions along the molecular axis on each carbon; the two p-orbitals not used in hybridization (say \(\displaystyle p_y\) and \(\displaystyle p_z\)) stay unhybridized on each carbon.Building the bonds: So the triple bond between the carbons is itself \(\displaystyle 1\,\sigma + 2\,\pi\), on top of the two C–H sigma bonds:\[\sigma \text{ bonds} = 2\,(\mathrm{C-H}) + 1\,(\mathrm{C-C}) = 3, \qquad \pi \text{ bonds} = 2 \]Where people slip: it is tempting to say a triple bond is "three bonds" and stop there — but only one of those three is \(\displaystyle \sigma\); the other two are \(\displaystyle \pi\), and they must be counted separately because they come from a different kind of overlap.(b) \(\displaystyle \mathrm{C_2H_4}\) — ethene (ethylene), \(\displaystyle H_2C=CH_2\)Each carbon here is bonded to three other atoms (two H and one C), and the geometry around each carbon is trigonal planar, so each carbon is sp2 hybridized. Three sp2 orbitals lie in a plane on each carbon; one p-orbital (perpendicular to that plane) is left unhybridized on each carbon.Building the bonds: \[\sigma \text{ bonds} = 4\,(\mathrm{C-H}) + 1\,(\mathrm{C-C}) = 5, \qquad \pi \text{ bonds} = 1 \]Where people slip: the double bond \(\displaystyle \mathrm{C=C}\) is \(\displaystyle 1\,\sigma + 1\,\pi\), not two \(\displaystyle \sigma\) bonds — writing "$\displaystyle 5$ sigma, $\displaystyle 1$ pi" only comes out right if you first hybridize each carbon correctly (sp2, not sp3) and remember that only one p-orbital per carbon is left over to make the \(\displaystyle \pi\) bond.Answer: (a) \(\displaystyle \mathrm{C_2H_2}\): $\displaystyle 3$ sigma bonds and $\displaystyle 2$ pi bonds (total $\displaystyle 5$). (b) \(\displaystyle \mathrm{C_2H_4}\): $\displaystyle 5$ sigma bonds and $\displaystyle 1$ pi bond (total $\displaystyle 6$).

A sigma \(\displaystyle (\sigma)\) bond forms by head-on (end-to-end) overlap of orbitals along the line joining the two nuclei — the internuclear axis. If the orbitals overlap sideways instead, with their lobes lying parallel to each other above and below the axis, the bond is a pi \(\displaystyle (\pi)\) bond, not a sigma bond. So the test for each pair is: does the overlap happen along the x-axis, or across it?Go through each case, taking the internuclear axis as the x-axis.(a) 1s and 1s. An s orbital is spherically symmetric — it has no preferred direction, so it always presents a lobe pointing straight along whatever axis joins the two nuclei. Two 1s orbitals overlap head-on along the x-axis. \(\displaystyle \Rightarrow\) forms a \(\displaystyle \sigma\) bond.(b) 1s and \(\displaystyle 2p_x\). The \(\displaystyle 2p_x\) orbital's own axis of orientation is the x-axis. Since the x-axis is also the internuclear axis here, the \(\displaystyle 2p_x\) lobe points directly at the 1s orbital, and they overlap head-on. \(\displaystyle \Rightarrow\) forms a \(\displaystyle \sigma\) bond.(c) \(\displaystyle 2p_y\) and \(\displaystyle 2p_y\). The \(\displaystyle 2p_y\) orbital is oriented along the y-axis, which is perpendicular to the internuclear (x) axis. Two \(\displaystyle 2p_y\) orbitals on neighbouring atoms therefore lie side by side, both pointing up-down along y while sitting apart along x. Their lobes overlap sideways (above and below the internuclear axis), not head-on. \(\displaystyle \Rightarrow\) this is exactly how a \(\displaystyle \pi\) bond is defined — lateral overlap of parallel p orbitals whose axes are perpendicular to the internuclear axis. This pair does NOT form a sigma bond; it would give a \(\displaystyle \pi\) bond instead.(d) 1s and 2s. Like 1s, the 2s orbital is spherically symmetric, so it also always overlaps head-on along whatever axis connects the two nuclei. \(\displaystyle \Rightarrow\) forms a \(\displaystyle \sigma\) bond.A common mix-up here is treating "any two p orbitals" as automatically sigma-forming. What matters is not that the orbitals are p-type, but whether each orbital's own lobe is directed along the internuclear axis (\(\displaystyle p_x\) is, when x is the internuclear axis) or perpendicular to it (\(\displaystyle p_y\) and \(\displaystyle p_z\) are not).Answer: (c) \(\displaystyle 2p_y\) and \(\displaystyle 2p_y\) will not form a sigma bond, because a \(\displaystyle \sigma\) bond needs head-on overlap along the internuclear (x) axis, and the \(\displaystyle 2p_y\) orbitals are oriented perpendicular to that axis — they can only overlap sideways, which gives a \(\displaystyle \pi\) bond, not a \(\displaystyle \sigma\) bond.

The hybridisation of a carbon atom is decided by how many other atoms are directly (sigma) bonded to it, not by how many total bonds (sigma + pi) it has. A carbon joined to four different atoms by four single (sigma) bonds is \(\displaystyle sp^3 \). A carbon that has one double bond (one sigma + one pi, to the same neighbour) has only three sigma-bonded neighbours, so it is \(\displaystyle sp^2 \). A carbon with a triple bond, or two double bonds, has only two sigma-bonded neighbours, so it is \(\displaystyle sp \). The double or triple bond's pi part is built from the leftover, unhybridised p orbital(s) — it plays no role in fixing the hybridisation itself.So the working, for each molecule, is simply: draw every carbon, count how many atoms are sigma-bonded to it, and read off the hybridisation from that count.(a) \(\displaystyle CH_3-CH_3 \) (ethane)Each carbon is bonded to three H atoms and the other carbon — four sigma bonds, no multiple bond anywhere in the molecule.Both carbon atoms: \(\displaystyle sp^3 \).(b) \(\displaystyle CH_3-CH=CH_2 \) (propene) So: \(\displaystyle CH_3 \) carbon is \(\displaystyle sp^3 \); both carbons of the \(\displaystyle C=C \) double bond are \(\displaystyle sp^2 \).(c) \(\displaystyle CH_3-CH_2-OH \) (ethanol) Here's the aside worth flagging: the oxygen having lone pairs does not change the carbon's hybridisation — only what is sigma-bonded to the carbon itself counts. Both carbons are \(\displaystyle sp^3 \).(d) \(\displaystyle CH_3-CHO \) (acetaldehyde) So: \(\displaystyle CH_3 \) carbon is \(\displaystyle sp^3 \); the carbonyl (\(\displaystyle CHO \)) carbon is \(\displaystyle sp^2 \).(e) \(\displaystyle CH_3COOH \) (acetic acid) The step people miss here: the carboxyl carbon has two different oxygens attached (one by a single bond, one by a double bond), but it is still only one pi bond, so it stays $\displaystyle 3$ sigma-neighbours → \(\displaystyle sp^2 \), not something "extra" because there are two C–O connections.So: \(\displaystyle CH_3 \) carbon is \(\displaystyle sp^3 \); the \(\displaystyle COOH \) carbon is \(\displaystyle sp^2 \).Answer: (a) both C atoms \(\displaystyle sp^3 \); (b) the \(\displaystyle CH_3 \) carbon \(\displaystyle sp^3 \), the two double-bonded carbons (\(\displaystyle CH= \) and \(\displaystyle =CH_2 \)) both \(\displaystyle sp^2 \); (c) both C atoms \(\displaystyle sp^3 \); (d) the \(\displaystyle CH_3 \) carbon \(\displaystyle sp^3 \), the \(\displaystyle CHO \) (carbonyl) carbon \(\displaystyle sp^2 \); (e) the \(\displaystyle CH_3 \) carbon \(\displaystyle sp^3 \), the \(\displaystyle COOH \) carbon \(\displaystyle sp^2 \). In every case the rule is the same: a carbon with four single (sigma) bonds is \(\displaystyle sp^3 \); a carbon involved in one double bond (three sigma-bonded neighbours) is \(\displaystyle sp^2 \).