To determine the orbital in which the pair of electrons in the carbanion $CH_3C \equiv C^{-}$ is present, we need to analyze the hybridization of the carbon atom carrying the negative charge.$\\$The structure of the carbanion $CH_3C \equiv C^{-}$ can be broken down as follows:$\\$• The carbon atom with the negative charge is part of a triple bond with another carbon atom.$\\$• In a triple bond, one bond is a sigma bond and two are pi bonds.$\\$The hybridization of a carbon atom involved in a triple bond is $sp. \\$This is because:$\\$• One $s$ orbital and one $p$ orbital hybridize to form two $sp$ orbitals.$\\$• The remaining two $p$ orbitals are unhybridized and form the two pi bonds.$\\$In the carbanion $CH_3C \equiv C^{-},$ the lone pair of electrons is located in the $sp$ hybrid orbital.$\\$This is because the $sp$ hybrid orbitals are used for forming sigma bonds and accommodating lone pairs.$\\$Therefore, the pair of electrons in the carbanion $CH_3C \equiv C^{-}$ is present in an $sp$ orbital.$\\$Thus, the correct option is Option 2: $sp.$