Hybridization = no. of sigma bond + no.of lone pairH2c=o has 3 sigma bond and zero lone pairans sp2

Hybridisation of Carbon in H2C=O

To determine the hybridisation state of carbon in H2C=O, we need to consider the bonding and electron arrangement around the carbon atom.

Bonding in H2C=O:
In the given molecule, H2C=O, carbon is bonded to two hydrogen atoms (H) and one oxygen atom (O).

Electron arrangement of Carbon:
Carbon has an atomic number of 6, which means it has 6 electrons. The electron configuration of carbon is 1s2 2s2 2p2.

When forming bonds, carbon can use its 2s and 2p orbitals. The 2s orbital can hold a maximum of 2 electrons, and the 2p orbital can hold a maximum of 6 electrons (3 orbitals x 2 electrons each).

Hybridisation Theory:
According to the hybridisation theory, carbon can undergo hybridisation, where it mixes its 2s and 2p orbitals to form a new set of hybrid orbitals. The number of hybrid orbitals formed is equal to the number of atomic orbitals involved in hybridisation.

The possible hybridisation states for carbon are:

- sp3 hybridisation: Carbon mixes one 2s orbital and three 2p orbitals to form four sp3 hybrid orbitals. The resulting arrangement is tetrahedral.

- sp2 hybridisation: Carbon mixes one 2s orbital and two 2p orbitals to form three sp2 hybrid orbitals. The remaining p orbital remains unhybridised. The resulting arrangement is trigonal planar.

- sp hybridisation: Carbon mixes one 2s orbital and one 2p orbital to form two sp hybrid orbitals. The remaining two p orbitals remain unhybridised. The resulting arrangement is linear.

Explanation:
In the case of H2C=O, the carbon atom is double bonded to the oxygen atom. This means that the carbon atom is involved in three sigma bonds (two with hydrogen and one with oxygen) and one pi bond with oxygen.

The presence of a double bond indicates that the carbon atom must be sp2 hybridised. This is because in order to form a double bond, carbon must have three orbitals available for bonding. These three orbitals are the three sp2 hybrid orbitals formed by mixing one 2s and two 2p orbitals.

The remaining unhybridised p orbital on carbon forms the pi bond with oxygen.

Therefore, the correct answer is option 'B': sp2 hybridised carbon, trigonal planar.