Question

1. What irreducible representations do the four H 1s orbitals on CH4 span? Are there s and p orbitals on the central C atom that may form molecular orbitals with them? In SiH4, where 3d orbitals might be available, could these orbitals play a role in forming molecular orbitals by overlapping with the H 1s orbitals?

Answer 1

Methane (CH4) molecule:
(a) On the basis of hybridisation.
Atomic number of C=6
Its electronic configuration in the ground
state is 1s² 2s² 2px¹ 2py¹
The electronic configuration in the excited
state is 1s² 2s¹ 2px¹ 2py¹ 2pz¹
One 2s and three 2p orbitals undergo sp3
hybridisation to form four sp hybrid orbitals which are arranged tetrahedrally at an angle of 109° 28' to each other.
The four sp3 hybrid orbitals overlap with the
half- filled 1s orbitals of four H-atoms, forming
CH4 . Thus, CH4, is a tetrahedral molecule with each H-C- H angle equal to 109° 28'.

(b) On the basis of VSEPR theory.

Central atom in CH4 is C. It has four valence electrons which are shared with four hydrogen atoms. Thus,there are four bond pairs of electrons present around the central atom carbon. Now according to vsepr theory these four electron pairs will repel each other and does not have so far apart that there are no for the repulsion between them the only shape possible is tetrahedral.

Thus clearly from this picture we can say s and p orbitals on central C atom will form molecular orbitals with 1s orbital of H

ht9928 O e & 2s 2py 2px 21. sp3 HYBRID ORBITALS I H-ATOM 1s. H sp3 sp3 ਹੈ। 109028/ Sp3 199°28 + 1s H-ATOM H sp3 H-ATOM s H H-ATOM

In SiH4, where 3d orbitals might be available but still it follows the same fashion as ch4 in bonding and undergo SP3 hybridization to form tetrahedral geometry with its 4 hybrid molecular orbitals in bonding with 1s of H. (No use of 3d orbitals)