Which one of the following statements is not correct for benzene?

Correct — D, It has six isomers. Benzene is one definite compound, C6H6, not a set of six. Isomerism does not enter the picture until you start substituting on the ring, and even then the counts are small and specific: a monosubstituted benzene has only one form, because all six positions are equivalent, and a disubstituted benzene has three — ortho, meta and para. The two Kekule structures with alternating double bonds that every textbook draws are not isomers either; they are resonance structures, two ways of drawing the same molecule, which is why the ring is usually drawn with a circle instead. That circle is what the other three statements describe, and each is correct. Every carbon is sp2 hybridised and uses two of its sigma bonds on neighbouring carbons and the third on a hydrogen, giving the flat hexagon its 120 degree angles. The leftover p electron on each carbon joins a single delocalised cloud lying above and below the ring plane. And because that cloud spreads the bonding evenly, all six carbon-carbon links are identical at about 140 picometres — longer than a double bond, at about 135, and shorter than a single bond, at about 147.

• (a)Each carbon atom forms sigma bonds with two other carbon atoms with a bond angle of 120°. — True. Each carbon in benzene is sp2 hybridised, forming sigma bonds to its two neighbouring carbons and one to a hydrogen; the three sigma bonds lie in a plane at 120 degrees to one another, which is what makes the ring a regular flat hexagon.
• (b)Delocalized electrons create a symmetrical 'cloud' of electrons above and below the plane. — True, and it is the heart of benzene's chemistry. The six p electrons left over after the sigma framework is built are not tied to particular pairs of atoms; they form a symmetrical ring-shaped cloud above and below the plane, and that delocalisation is why benzene resists addition and undergoes substitution instead.
• (c)The length of C – C bonds is intermediate between single and double bonds. — True, and it is the experimental proof of statement (b). If benzene really alternated single and double bonds you would measure two different lengths; instead every carbon-carbon bond comes out the same, at a value lying between the two.

Benzene, C6H6, is the parent aromatic hydrocarbon: a flat six-carbon ring in which each carbon is sp2 hybridised and bonded to two carbons and one hydrogen. The unhybridised p orbital on every carbon overlaps sideways all the way round, so the six pi electrons are delocalised into a doughnut-shaped cloud above and below the ring rather than being localised into three double bonds. Delocalisation makes the molecule unusually stable, equalises all six carbon-carbon bond lengths, and dictates that benzene reacts by substitution rather than by the addition typical of alkenes.

Three of the statements are things you can check against a picture of the molecule; the fourth is a claim about isomer counting, which is a different kind of statement altogether, and that mismatch is the first clue. Beyond that, it helps to be precise about what an isomer is: two compounds with the same molecular formula but different structures. Molecules other than benzene do share the formula C6H6, but they are separate substances with their own names, not 'benzene's isomers', and there is no standard list of six. What students should hold instead are the substituted counts, since those are what examiners test — one monosubstituted form, three disubstituted forms. Note one printed detail: the paper sets option (c) as C – C with spaces around the dash, which is the booklet's own typography for a carbon-carbon bond and not a minus sign.

• Benzene is C6H6, a planar regular hexagon with all bond angles 120 degrees and every carbon sp2 hybridised.
• All six carbon-carbon bonds are identical, about 140 pm, between a C=C double bond at about 135 pm and a C-C single bond at about 147 pm.
• The six pi electrons are delocalised into a symmetrical cloud above and below the ring plane, which is why benzene is drawn with a circle inside the hexagon.
• The two Kekule structures are resonance forms of one molecule, not two isomers.
• Substituted benzenes give the isomer counts worth remembering: one monosubstituted form, and three disubstituted forms called ortho, meta and para.

• Treating the two Kekule structures as two different substances; they are two drawings of one molecule.
• Expecting benzene to show two different carbon-carbon bond lengths, as an alternating structure would demand.
• Assuming a not-correct stem must hide a subtle error; here the false statement is of a completely different kind from the other three.

Usually as a which-statement-is-not-correct item on benzene's structure, or as a direct question on its bond length, hybridisation or number of pi electrons.

• practice — not a real PYQ

  The hybridisation of each carbon atom in a benzene molecule is

  • (a)sp
  • (b)sp2
  • (c)sp3
  • (d)sp3d

  Answer(b) sp2 — three sigma bonds in a plane at 120 degrees, with the remaining p orbital contributing to the delocalised pi cloud above and below the ring.

• practice — not a real PYQ

  How many isomers are possible for a disubstituted benzene carrying two identical substituents?

  • (a)One
  • (b)Two
  • (c)Three
  • (d)Six

  Answer(c) Three — the ortho, meta and para arrangements. A monosubstituted benzene has only one form, because all six ring positions are equivalent.