Determine the standard enthalpy of formation of SrI2(s) using a Born–Haber cycle. Enthalpy of sublimation of Sr(s) = 164 kJ/mol 1st ionization energy of Sr(g) = 549 kJ/mol 2nd ionization energy of Sr(g) = 1064 kJ/mol Enthalpy of sublimation of I2(s) = 62.4 kJ/mol Bond dissociation energy of I2(g) = 152.55 kJ/mol 1st electron affinity of I(g) = –295.15 kJ/molLattice energy of SrI2(s) = –1959.75 kJ/mol

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.