RI 2024 Chemical Bonding 2 Tutorial Suggested Ans

1 RAFFLES INSTITUTION YEAR 5 H2 CHEMISTRY 2024 Tutorial 8: Chemical Bonding II Suggested Answers There is no 2sp − 2sp2 overlap since the sp and sp2 C are not next to each other. At the C=C and C≡C bonds, the  bonds are formed by the side-on overlap between the 2p orbitals of C. The 1s and 2sp overlap occur between the triple bonded C and the H atom which uses its 1s orbitals for  bond with the 2sp hybrid orbital of C. Answer: C 2. The terminal carbon atoms (1 and 4) are sp3 and not sp2 hybridised. Hence option C is not correct and so is the answer. Answer: C 3. (a) (b) A –bond is formed when valence orbitals overlap head–on while a –bond is formed when valence orbitals overlap side–on. In ethene, the two carbon atoms form a –bond via the head–on overlapping of two sp2 hybrid orbitals i.e. In ethene, the two carbon atoms also form a –bond via the side–on overlapping of the two unhybridised p orbitals i.e. (c) sp2 hybridisation 120o 120o 120o 120o 2sp 2sp 2sp3 2sp2 2sp2 1. sp3 sp sp sp3 sp3

2 (d) 3 sp2 orbitals (each with one small and one large lobe): 4. (a) (b) BeF2 sp hybridisation [BeF4]2− sp3 hybridisation 5. (a) (b) Shape: Bent O-N-O Bond angle: 115 (Any answer between 109.5o and 120o is acceptable) (c) sp2. (d) The unhybridised p orbital of N and p orbitals of the O atoms are perpendicular to the plane of the ion (containing  bonds). These p orbitals overlap side-on continuously, so that the  electrons are delocalised over the N and O atoms. Thus, NO2− exhibits resonance and the two N–O bonds are equivalent, with the same bond length. 6. (a) B N BN B N B N BN N B N B N B N B N N BN B N B N B 109.5o N − Both B and N are sp2 hybridised and each B and N atom has an unhybridised p orbital perpendicular to the plane of atoms. Each B atom forms 3  bonds with 3 N atoms and each N atom also forms 3  bonds with 3 B atoms. B is from Group 13 and its unhybridised p orbital is empty while N is from Group 15 and its unhybridised p orbital has 2 electrons. The unhybridised p orbitals overlap side -on continuously with its neighbours resulting in the delocalisation of  electrons over the whole layer.

3 (b) It is expected that BN is an electrical conductor due to the delocalistion of  electrons from the continuous side -on overlap of p orbitals . However, since N is more electronegative than B, the  electrons tend to stay with N rather than delocalise throughout the -electron cloud . Hence, BN is a poorer electrical conductor than graphite.