Some Basic Principles of Organic Chemistry MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Carbon has atomic number \(6\), so its six electrons occupy the orbitals according to the ground-state configuration \(\mathrm{1s^2\,2s^2\,2p^2}\). The outermost shell is the second shell, containing two electrons in the \(2s\) orbital and two electrons in the \(2p\) orbitals. Thus, carbon possesses four valence electrons. A stable octet requires eight electrons around carbon in its ordinary neutral covalent compounds. Instead of completely losing or gaining four electrons, carbon generally achieves this arrangement through electron sharing. Each ordinary covalent bond represents a shared electron pair, and carbon contributes one electron to each bond in the simplest four-bond description. Four covalent bond units therefore allow carbon to complete its octet. The four bond units need not connect carbon to four different atoms because a double bond contributes two bond units and a triple bond contributes three. For example, carbon satisfies tetravalency in methane through four single bonds and in ethene through two C–H single bonds and one C=C double bond. Hence, tetravalency describes the total covalent bonding capacity of carbon rather than merely the number of neighbouring atoms.

Explanation: \(\textbf{Identify the attachments:}\) The carbon atom in methanal is bonded directly to two hydrogen atoms and one oxygen atom. Therefore, the number of atoms directly attached to carbon is three. However, the number of attached atoms is not necessarily equal to the number of covalent bond units. \(\textbf{Count the bond units:}\) Each C–H single bond contributes one bond unit. The C=O double bond contributes two bond units because it consists of one sigma bond and one pi bond. Consequently, the total number of covalent bond units around carbon is \[ 2(1)+1(2)=4 \] The carbon atom therefore satisfies its ordinary tetravalency despite being attached to only three atoms. The two C–H bonds and the C=O bond provide the required shared-electron contribution for an octet around carbon. Counting the double bond as only one bond unit would incorrectly give a valency of three. The correct distinction is between the number of neighbouring atoms and the sum of the bond orders.

Explanation: A neutral carbon atom in an ordinary closed-shell organic structure must have a total covalent bond order of four. Hydrogen and chlorine ordinarily form one covalent bond each in these structures. A single bond contributes one bond unit, whereas a double bond contributes two. In option A, the central carbon forms one single bond to the methyl carbon, one single bond to chlorine and one double bond to the terminal carbon. Its total bond order is therefore \(1+1+2=4\), making the proposed connectivity valid. In option B, the terminal \(\mathrm{CH_2Cl}\) carbon has two C–H bonds, one C–Cl bond and one C–C bond, giving four bond units. The two alkene carbons in that structure also satisfy tetravalency. In option C, the middle carbon forms four single bonds: one to hydrogen, one to chlorine and two to neighbouring carbons. In option D, the middle carbon is shown with a C–H bond, a C–Cl bond, a C–C single bond and a C=C double bond. Its total bond order would be \(1+1+1+2=5\). That would require five ordinary covalent bond units around a neutral carbon atom, violating the usual octet and tetravalency requirements. The error is therefore not the presence of chlorine or the double bond individually, but the excessive combined bonding assigned to the central carbon.

Explanation: \(\textbf{Identify the carbon framework:}\) The structure contains two carbons in the C=C bond, one methyl-group carbon and two carbons in the C≡C bond. Therefore, the total number of carbon atoms is \[ N_C=2+1+2=5 \] \(\textbf{Apply carbon tetravalency:}\) Each carbon must possess four covalent bond units. The terminal \(\mathrm{CH_2}\) carbon has two C–H bonds and one C=C double bond. The carbon attached to the methyl group has one C=C double bond and two C–C single bonds, so it carries no hydrogen. The methyl carbon has three hydrogen atoms. The carbon between the methyl-substituted alkene carbon and the terminal alkyne carbon forms one single bond and one triple bond, so it also carries no hydrogen. The terminal alkyne carbon has one hydrogen atom. \(\textbf{Count the hydrogen atoms:}\) \[ N_H=2+0+3+0+1 \] \[ N_H=6 \] \(\textbf{Check the degree of unsaturation:}\) A saturated open-chain hydrocarbon containing five carbon atoms would have \[ H_{\mathrm{sat}}=2(5)+2=12 \] One C=C bond reduces the hydrogen count by two relative to saturation, and one C≡C bond reduces it by four. \[ H=12-2-4=6 \] Both independent approaches give the same molecular formula. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_5H_6}} \]

Explanation: The two structures contain five carbon atoms each and have only carbon–carbon single bonds. Both are saturated, acyclic hydrocarbons, so they follow the general molecular formula \(\mathrm{C_nH_{2n+2}}\). For five carbon atoms, each structure therefore contains twelve hydrogen atoms. This can also be verified directly by expanding the condensed structural formulas. In P, the two terminal carbons are \(\mathrm{CH_3}\) groups, while the three internal carbons are \(\mathrm{CH_2}\) groups. Thus, P contains two carbon atoms directly attached to only one other carbon atom. In Q, the three \(\mathrm{CH_3}\) groups are each attached to just one other carbon atom. The branching carbon in Q is attached to three other carbon atoms, while the remaining \(\mathrm{CH_2}\) carbon is attached to two. Consequently, Q contains three carbon atoms directly attached to only one other carbon atom. Both structures have four carbon–carbon sigma bonds and twelve carbon–hydrogen sigma bonds. Changing the branching pattern therefore changes the carbon connectivity without changing the molecular formula or total number of sigma bonds. The decisive distinction is the distribution of carbon neighbours, not the total carbon or hydrogen count. This illustrates how carbon tetravalency permits different valid carbon frameworks with the same molecular composition.

Explanation: \(\textbf{Interpret the molecular formula:}\) An open-chain saturated hydrocarbon with four carbon atoms would have the formula \(\mathrm{C_4H_{10}}\). The given formula has two fewer hydrogen atoms. This difference can arise either from introducing one C=C double bond or from forming one ring. Therefore, the molecular formula alone does not distinguish the two structural possibilities. \(\textbf{Examine the ring structure:}\) A four-carbon single ring contains four carbon–carbon connections. Because all its carbon–carbon bonds are single bonds, all four carbon centres are ordinarily \(sp^3\)-hybridised. Each connection contains one sigma bond, giving four C–C sigma bonds. \(\textbf{Examine the open-chain alkene:}\) An open chain containing four carbon atoms has three carbon–carbon connections. Introducing one double bond changes one of these connections from a single bond to a double bond. The double bond still contributes only one C–C sigma bond, together with one pi bond. Its two doubly bonded carbons are \(sp^2\)-hybridised, while the remaining saturated carbons are \(sp^3\)-hybridised. The open-chain alkene therefore has three C–C sigma bonds and two \(sp^2\) carbon centres. \(\textbf{Distinguishing conclusion:}\) Four \(sp^3\) carbon centres together with four C–C sigma bonds identify the single-ring alternative. The important reasoning is that ring formation adds a carbon–carbon connection, whereas formation of a double bond adds a pi bond without increasing the number of carbon–carbon connections.

Explanation: Methane contains one carbon atom bonded to four hydrogen atoms through four equivalent C–H sigma bonds. Carbon is \(sp^3\)-hybridised in methane, producing four equivalent hybrid orbitals directed towards tetrahedral positions. Each hybrid orbital overlaps with a hydrogen \(1s\) orbital to form a sigma bond. The ideal angle between any two C–H bonds is approximately \(109.5^\circ\). This arrangement gives methane a tetrahedral geometry rather than a square-planar arrangement. Consequently, the four hydrogen nuclei occupy the vertices of a tetrahedron and are not all situated in one plane. The Assertion is therefore true. The Reason is also true because the tetrahedral arrangement of the bonding directions is precisely what produces the non-coplanar geometry. A flat Lewis structure showing four C–H bonds around carbon is only a two-dimensional representation of connectivity. Such a drawing does not establish the actual three-dimensional positions of the hydrogen atoms. A hypothetical square-planar arrangement would give different bond-angle relationships and would not represent the normal geometry of methane. The Reason therefore correctly explains the Assertion.

Explanation: The C=C double bond in propene consists of one sigma bond and one pi bond. Each carbon participating in this double bond is \(sp^2\)-hybridised. The three sigma-bond directions at an \(sp^2\)-hybridised carbon are approximately trigonal planar. The carbon–carbon single bond connecting the alkene region to the methyl carbon lies along one of these sigma-bond directions. Consequently, all three carbon nuclei can lie in the same plane defined by the alkene framework. However, the carbon of the methyl group forms four sigma bonds and is \(sp^3\)-hybridised. Its three C–H bonds and one C–C bond have an approximately tetrahedral spatial arrangement. All four bond directions around this methyl carbon cannot lie in a single plane. Since its C–C bond connects the methyl carbon to the alkene plane, the remaining three C–H bond directions cannot all remain within that plane. The student's mistake is to extend local trigonal-planar geometry at the double bond to every atom in the molecule. Hybridisation and geometry must be assigned separately to each relevant carbon atom. A molecule can therefore contain a planar unsaturated region and a tetrahedral saturated region simultaneously. The presence of one double bond is insufficient to establish complete molecular planarity.

Explanation: Each carbon atom in ethyne is bonded directly to one hydrogen atom and the other carbon atom. The C≡C triple bond contributes three bond units, and the C–H single bond contributes one. Thus, each carbon satisfies tetravalency. Each carbon has two sigma-bonding directions, so it undergoes \(sp\) hybridisation. One \(s\) orbital mixes with one \(p\) orbital to form two \(sp\) hybrid orbitals. These hybrid orbitals point in opposite directions and produce an approximately \(180^\circ\) bond angle. One hybrid orbital on each carbon participates in forming the C–C sigma bond, while the other forms a C–H sigma bond. Two unhybridised \(p\) orbitals remain on each carbon. The corresponding unhybridised orbitals overlap sideways in two mutually perpendicular planes to form two pi bonds. Therefore, the C≡C bond contains one sigma bond and two pi bonds. The two pi bonds do not imply that the carbon atoms are trigonal planar or that their hybridisation is \(sp^2\). The linear geometry follows from the two \(sp\)-hybridised sigma-bond directions, while the unhybridised orbitals account for the additional pi bonding. This separates the roles of hybrid and unhybridised orbitals in a triple bond.

Explanation: A carbon–carbon double bond contains a sigma bond and a pi bond. The pi bond involves sideways overlap of unhybridised \(p\) orbitals on the two doubly bonded carbon atoms. Free rotation about the C=C bond would disrupt this overlap, so the double bond restricts rotation. However, restricted rotation alone does not guarantee geometrical isomerism. For two distinct geometrical arrangements to exist in a simple alkene, each doubly bonded carbon must be attached to two different substituents. In ethene, each doubly bonded carbon carries two hydrogen atoms, so exchanging their apparent positions does not create a different isomer. In but-\(1\)-ene, the terminal doubly bonded carbon is \(\mathrm{CH_2}\), again carrying two identical hydrogen substituents. In \(2\)-methylprop-\(1\)-ene, one doubly bonded carbon carries two hydrogen atoms, while the other carries two methyl groups. It therefore also lacks the required substituent distinction. In but-\(2\)-ene, each doubly bonded carbon carries one hydrogen atom and one methyl group. The methyl groups can occupy the same side or opposite sides of the double-bond framework. These arrangements represent distinct geometrical isomers because ordinary rotation about the double bond is restricted. Thus, geometrical isomerism depends jointly on the presence of a double bond, restricted rotation and the substituents attached to both doubly bonded carbons.

Explanation: Hybridisation involves the mixing of atomic orbitals of comparable energy belonging to the same atom. The number of hybrid orbitals formed must equal the number of atomic orbitals participating in the mixing. A carbon atom has one valence \(s\) orbital and three valence \(p\) orbitals available for the conventional \(sp^3\), \(sp^2\) and \(sp\) descriptions. In \(sp^3\) hybridisation, one \(s\) orbital and three \(p\) orbitals combine to produce four hybrid orbitals. No unhybridised \(p\) orbital remains, so record P is incorrect despite giving the correct number of hybrid orbitals. In \(sp^2\) hybridisation, one \(s\) orbital and two \(p\) orbitals mix to form three hybrid orbitals. The third \(p\) orbital remains unhybridised and can participate in pi bonding. Both entries in record Q are therefore correct. In \(sp\) hybridisation, one \(s\) orbital and one \(p\) orbital produce only two hybrid orbitals, not three. The remaining two \(p\) orbitals remain unhybridised, so record R is partly correct but internally inconsistent. Record S assigns two hybrid orbitals and two unhybridised \(p\) orbitals to \(sp^2\), which actually describes the orbital count for \(sp\) hybridisation. The key conservation rule is that hybridisation redistributes the participating atomic orbitals without creating or destroying the number of orbitals. Correct orbital accounting must simultaneously satisfy the number of mixed orbitals, the number of hybrid orbitals and the number of unhybridised orbitals.

Explanation: \(\textbf{Identify the local bonding environments:}\) Hybridisation is assigned independently to each carbon atom according to its local sigma-bonding arrangement. The first carbon, \(C_1\), belongs to a methyl group. It forms three C–H sigma bonds and one C–C sigma bond, giving four sigma-bonding directions. Therefore, \(C_1\) is \(sp^3\)-hybridised. The second carbon, \(C_2\), participates in a C=C double bond. It forms three sigma bonds and participates in one pi bond through its remaining unhybridised \(p\) orbital. Therefore, \(C_2\) is \(sp^2\)-hybridised. The third carbon, \(C_3\), also participates in the same double bond and has three sigma-bonding directions. It is consequently \(sp^2\)-hybridised. \(\textbf{Examine the triple-bonded region:}\) The fourth carbon, \(C_4\), is attached through a single bond to \(C_3\) and through a triple bond to \(C_5\). It has two sigma-bonding directions and uses two unhybridised \(p\) orbitals for pi bonding. Thus, \(C_4\) is \(sp\)-hybridised. The terminal carbon, \(C_5\), forms a triple bond to \(C_4\) and a single bond to hydrogen. It also has two sigma-bonding directions and is \(sp\)-hybridised. \(\textbf{Check the complete sequence:}\) The structure contains one saturated carbon, two alkene carbons and two alkyne carbons. Their corresponding hybridisations are \(sp^3\), \(sp^2\), \(sp^2\), \(sp\) and \(sp\). All five carbon atoms satisfy tetravalency even though they have different local geometries. The molecule therefore cannot be assigned a single uniform carbon hybridisation. The correct sequence follows from counting sigma-bond directions rather than simply counting the total bonds attached to each carbon.

Explanation: \(\textbf{Identify the bonding framework:}\) The given molecule contains six carbon atoms arranged in one open chain. There are five carbon–carbon connections between successive carbon atoms. Each carbon–carbon connection contributes exactly one sigma bond, irrespective of whether it is single, double or triple. Therefore, the number of carbon–carbon sigma bonds is \[ N_{\sigma(\mathrm{C-C})}=5 \] \(\textbf{Count the carbon–hydrogen sigma bonds:}\) The hydrogen counts on the six carbons, reading from left to right, are \(3,2,0,0,1,2\). Each carbon–hydrogen bond is a single sigma bond. \[ N_{\sigma(\mathrm{C-H})}=3+2+0+0+1+2 \] \[ N_{\sigma(\mathrm{C-H})}=8 \] \(\textbf{Calculate total sigma bonds:}\) \[ N_{\sigma}=N_{\sigma(\mathrm{C-C})}+N_{\sigma(\mathrm{C-H})} \] \[ N_{\sigma}=5+8=13 \] \(\textbf{Count the pi bonds:}\) The C≡C triple bond contributes two pi bonds. The C=C double bond contributes one pi bond. All remaining bonds are single sigma bonds. \[ N_{\pi}=2+1=3 \] \(\textbf{Check carbon tetravalency:}\) Each carbon in the structure has a total bond order of four. The two triple-bonded carbons have two sigma-bonding directions, whereas the doubly bonded carbons have three. The molecular structure is therefore consistent with the stated sigma and pi counts. Counting a triple bond as three sigma bonds would be incorrect because only one of its three bonds is sigma. \(\textbf{Final answer:}\) \[ \boxed{(N_{\sigma},N_{\pi})=(13,3)} \]

Explanation: A carbon–carbon single bond in ethane consists of one sigma bond formed between two \(sp^3\)-hybridised carbon atoms. The carbon–carbon double bond in ethene consists of one sigma bond and one pi bond between two \(sp^2\)-hybridised carbon atoms. The carbon–carbon triple bond in ethyne consists of one sigma bond and two pi bonds between two \(sp\)-hybridised carbon atoms. Increasing bond order generally increases the electron density holding the bonded carbon nuclei together and produces a shorter, stronger carbon–carbon bond. Accordingly, the triple bond is shorter than the double bond, and the double bond is shorter than the single bond in these simple comparable compounds. Hybridisation provides an additional consistent explanation of the bonding differences. An \(sp^3\) hybrid orbital has \(25\%\) s-character, an \(sp^2\) orbital has approximately \(33.3\%\), and an \(sp\) orbital has \(50\%\). Greater s-character places more of the bonding electron density closer to the carbon nucleus. This contributes to the shorter sigma-bond framework associated with \(sp\)-hybridised carbon compared with \(sp^2\)- and \(sp^3\)-hybridised carbon in these examples. Both the increase in bond order and the relevant hybridisation trend support the stated ordering. The triple bond does not contain three sigma bonds; its additional bonding arises from two pi bonds. Likewise, a double bond does not have twice the number of sigma bonds present in a single bond. The comparison is specifically between the carbon–carbon bonds of ethane, ethene and ethyne, not an unrestricted rule for every bond in every substituted organic compound.

Explanation: \(\textbf{Determine the saturated reference formula:}\) A neutral saturated open-chain hydrocarbon with six carbon atoms follows the relation \[ \mathrm{C_nH_{2n+2}} \] Therefore, its saturated reference hydrogen count is \[ H_{\mathrm{sat}}=2(6)+2=14 \] \(\textbf{Account for the multiple bonds:}\) Introducing one C=C double bond into the corresponding acyclic carbon framework reduces the hydrogen count by two. Introducing one C≡C triple bond reduces the hydrogen count by four relative to the saturated framework. The molecule contains both types of multiple bonds. \[ H=14-2-4 \] \[ H=8 \] Thus, the molecular formula is \(\mathrm{C_6H_8}\). \(\textbf{Identify the hybridised carbon centres:}\) The two carbons forming the C=C bond are \(sp^2\)-hybridised. Each \(sp^2\) carbon retains one unhybridised \(p\) orbital for pi bonding. Therefore, the double-bonded carbons contribute \[ N_{p,\mathrm{double}}=2(1)=2 \] The two carbons forming the C≡C bond are \(sp\)-hybridised. Each \(sp\) carbon retains two unhybridised \(p\) orbitals. Therefore, the triple-bonded carbons contribute \[ N_{p,\mathrm{triple}}=2(2)=4 \] \(\textbf{Combine the orbital contributions:}\) \[ N_p=2+4=6 \] The remaining two carbon atoms are saturated \(sp^3\) centres and do not supply additional unhybridised \(p\) orbitals in this conventional bonding description. \(\textbf{Consistency check:}\) Six unhybridised carbon \(p\) orbitals form three pi bonds through three pairs of sideways-overlapping orbitals. One pi bond belongs to the C=C bond, and two belong to the C≡C bond. This agrees with the specified molecular bonding and independently checks the orbital count. \(\textbf{Final answer:}\) \[ \boxed{(N_H,N_p)=(8,6)} \]

Explanation: \(\textbf{Identify the initial bonding:}\) The starting compound is but-\(1\)-ene, containing one C=C double bond. The two doubly bonded carbon atoms are \(sp^2\)-hybridised and have approximately trigonal-planar local geometry. The double bond consists of one sigma bond and one pi bond. The molecular formula of the starting compound is \(\mathrm{C_4H_8}\). \(\textbf{Examine the structural change:}\) In the product, the same two carbon atoms are connected through a C≡C triple bond. The carbon skeleton remains unchanged, so no carbon–carbon connection is added or removed. However, the triple bond requires three covalent bond units between the two carbon atoms rather than two. To preserve tetravalency, the initially doubly bonded carbons must collectively lose two hydrogen attachments. The product therefore has molecular formula \(\mathrm{C_4H_6}\). \(\textbf{Determine the hybridisation change:}\) Each carbon of the new triple bond now has two sigma-bonding directions. Its hybridisation is therefore \(sp\), rather than the original \(sp^2\). The corresponding local geometry changes from approximately trigonal planar to linear. Each \(sp\)-hybridised carbon retains two unhybridised \(p\) orbitals, enabling formation of the two pi bonds of the triple bond. \(\textbf{Check the bond-count changes:}\) Initially, the C=C bond contributes one sigma and one pi bond. Finally, the C≡C bond contributes one sigma and two pi bonds. Therefore, the number of carbon–carbon sigma bonds remains unchanged while the total pi-bond count increases by one. The two removed C–H bonds also decrease the total sigma-bond count of the molecule by two. This is consistent with the change from \(\mathrm{C_4H_8}\) to \(\mathrm{C_4H_6}\). \(\textbf{Concluding inference:}\) The transformation requires two fewer hydrogen atoms, a change from \(sp^2\) to \(sp\) at both affected carbon atoms and one additional pi bond. The structural change is therefore consistent with carbon tetravalency, the local hybridisation assignments and the sigma–pi bonding model.

Explanation: Propadiene contains two consecutive carbon–carbon double bonds sharing a common central carbon atom. The central carbon forms one sigma bond with each terminal carbon and participates in two pi bonds. It therefore has two sigma-bonding directions, which correspond to \(sp\) hybridisation. The two \(sp\) hybrid orbitals are directed oppositely, giving an approximately linear carbon framework. Two mutually perpendicular unhybridised \(p\) orbitals remain on the central carbon. One of these orbitals overlaps sideways with an unhybridised \(p\) orbital on the first terminal carbon. The other overlaps sideways with an unhybridised \(p\) orbital on the second terminal carbon. Since the central carbon's two unhybridised \(p\) orbitals are mutually perpendicular, the two pi-bonding systems occupy perpendicular planes. Each terminal carbon forms three sigma bonds: two C–H bonds and one C–C bond. Consequently, each terminal carbon is \(sp^2\)-hybridised and has approximately trigonal-planar local geometry. The plane containing the two hydrogen atoms and the sigma-bond directions at one terminal carbon is therefore perpendicular to the corresponding plane at the other terminal carbon. The three carbon nuclei lie along the same straight line, but the complete molecule is not planar. This distinction demonstrates that a linear carbon skeleton does not necessarily imply that all atoms of the molecule occupy one plane. The correct geometry follows from the simultaneous requirements of carbon hybridisation, two perpendicular pi-bonding interactions and local trigonal planarity at the terminal carbon atoms.

Explanation: \(\textbf{Identify the common molecular composition:}\) All three hydrocarbons contain four carbon atoms and six hydrogen atoms. Therefore, each contains six C–H sigma bonds. Their differences arise from the number of carbon–carbon connections and the distribution of multiple bonds. \(\textbf{Analyse P — buta-1,3-diene:}\) Its structure is \(\mathrm{CH_2=CH-CH=CH_2}\). The open chain contains three carbon–carbon connections, each contributing one sigma bond. The two C=C bonds contribute one pi bond each. \[ N_{\sigma}(P)=6+3=9 \] \[ N_{\pi}(P)=1+1=2 \] \(\textbf{Analyse Q — cyclobutene:}\) Cyclobutene contains a four-membered carbon ring with one C=C double bond. The ring contains four carbon–carbon connections, giving four C–C sigma bonds. The double bond contributes one pi bond. \[ N_{\sigma}(Q)=6+4=10 \] \[ N_{\pi}(Q)=1 \] \(\textbf{Analyse R — but-1-yne:}\) Its structure is \(\mathrm{HC{\equiv}C-CH_2-CH_3}\). The four-carbon open chain has three carbon–carbon connections. Its C≡C triple bond contributes two pi bonds. \[ N_{\sigma}(R)=6+3=9 \] \[ N_{\pi}(R)=2 \] \(\textbf{Compare the results:}\) \[ N_{\sigma}(P)=N_{\sigma}(R)=9 \] \[ N_{\sigma}(Q)=10 \] \[ N_{\pi}(P)=N_{\pi}(R)=2 \] \[ N_{\pi}(Q)=1 \] The cyclic structure has one additional carbon–carbon sigma connection compared with either open-chain structure. The diene and alkyne contain the same total number of pi bonds despite having different types of multiple bonds. The important conclusion is that identical molecular formulas do not require identical sigma–pi distributions. Ring formation and multiple-bond formation can produce the same hydrogen deficiency through different bonding arrangements.

Explanation: The molecule is pent-\(2\)-yne and contains a triple bond between \(C_2\) and \(C_3\). Each triple-bonded carbon forms two sigma bonds and participates in two pi bonds. Therefore, both \(C_2\) and \(C_3\) are \(sp\)-hybridised. The two \(sp\) hybrid orbitals on each carbon point in opposite directions. At \(C_2\), the sigma bonds towards \(C_1\) and \(C_3\) form an angle of approximately \(180^\circ\). Consequently, \(C_1,C_2,C_3\) lie along a straight line. At \(C_3\), the sigma bonds towards \(C_2\) and \(C_4\) also form an angle of approximately \(180^\circ\). Since the \(C_2-C_3\) bond is common to both arrangements, \(C_4\) lies on the same straight line as \(C_1,C_2,C_3\). The first four carbon nuclei are therefore collinear. However, \(C_4\) is a \(\mathrm{CH_2}\) carbon forming four sigma bonds. It is \(sp^3\)-hybridised and possesses approximately tetrahedral local geometry. The \(C_3-C_4-C_5\) bond angle is consequently close to the tetrahedral value rather than \(180^\circ\). The \(C_4-C_5\) bond therefore departs from the line established by the triple-bonded region. The presence of a triple bond creates local linearity, but this linearity does not automatically extend through an adjacent \(sp^3\)-hybridised carbon. This illustrates why molecular geometry must be inferred by combining the local hybridisations of successive atoms rather than assigning one shape to the entire carbon chain.

Explanation: The molecule contains three carbon atoms: two methyl-group carbons and one carbonyl carbon. The first methyl carbon forms three C–H sigma bonds and one C–C sigma bond. It therefore has four sigma-bonding directions and is \(sp^3\)-hybridised. The second methyl carbon is attached to oxygen and three hydrogen atoms. It also forms four sigma bonds and is \(sp^3\)-hybridised. The carbonyl carbon has a different bonding environment. It forms a single sigma bond with the first methyl carbon, a sigma bond with the singly bonded oxygen and a sigma bond as part of the C=O double bond. Thus, it has three sigma-bonding directions and is \(sp^2\)-hybridised. The carbonyl carbon retains one unhybridised \(p\) orbital. Sideways overlap involving this orbital and an appropriate oxygen \(p\) orbital produces the pi component of the C=O double bond. The C=O bond therefore contributes one pi bond, even though no carbon–carbon multiple bond is present. The carbonyl carbon has approximately trigonal-planar local geometry, whereas the two methyl carbons have approximately tetrahedral local geometry. The three carbon atoms need not possess the same hybridisation merely because they belong to one molecule. The decisive distinction is that pi bonding may occur between carbon and a heteroatom, not only between two carbon atoms. Accordingly, the molecule contains one \(sp^2\) carbon, two \(sp^3\) carbons and one pi bond.