In singlet carbene, the carbon atom has a lone pair of electrons and forms two covalent bonds. It has a total of three electron domains (two bonds and one lone pair). According to VSEPR theory, this arrangement leads to a bent geometry and requires $sp^2$ hybridization. One $sp^2$ orbital accommodates the lone pair, two $sp^2$ orbitals form sigma bonds with other atoms, and the remaining unhybridized p orbital is empty.
In triplet carbene, the carbon atom has two unpaired electrons (one in each of two different orbitals) and forms two covalent bonds. With two bonds and two separate unpaired electrons, the carbon atom effectively uses two hybrid orbitals for bonding and has two unhybridized orbitals. This requires $sp$ hybridization. The two $sp$ orbitals form sigma bonds, while the two perpendicular, unhybridized p orbitals contain the single electrons, following Hund's rule.
Therefore, the carbon atom in singlet carbene is $sp^2$ hybridized, and the carbon atom in triplet carbene is $sp$ hybridized.
The correct option is C.
The Miller indices of the shown lattice plane in a simple cubic Bravais lattice are :

The R/S configuration of C - 2 and C - 3 in the given molecule will be :
Predict the % product formation in the given reaction :