Question

Which one of the following statements is false

1. double bond is shorter and stronger than a single bond
2. a resonance hybrid is a mixture of all the resonance structure for a molecule
3.atom which is sp2 hybridized will have two un-hybridized p orbitals
4. only atoms which have d-orbitals in their valence shells can form sp3d hybrid orbitals

Answer 1

The false statement is that 'an atom which is sp2 hybridized will have two un-hybridized p orbitals', as it only has one un-hybridized p orbital. The correct hybridization for sp2 involves having just one un-hybridized p orbital that participates in pi (π) bonding.

Step-by-step explanation:

The statement which is false among the given options is: "Atom which is sp2 hybridized will have two un-hybridized p orbitals." An atom that is sp2 hybridized will have one un-hybridized p orbital, not two. For instance, in the molecule benzene, each carbon atom is sp2 hybridized, bonded to three other atoms with no lone pairs, resulting in one un-hybridized p orbital that forms a delocalized π (pi) bond throughout the ring.

Resonance structures do not represent the actual electron placement but illustrate the delocalization. Resonance does not affect hybridization since it involves only the unhybridized orbitals and the structure of π bonds. The statement "Only atoms which have d-orbitals in their valence shells can form sp3d hybrid orbitals" is also potentially misleading as atoms from period 3 onwards, such as phosphorus and sulfur, can utilize d-orbitals to form expanded octets, thus achieving sp3d or sp3d2 hybridization.