The colour of iodine is different in different conditions which of the following statements is/are correct for iodine

In gaseous  $I_2$ colour arises due to electronic transition from  $\stackrel{*}{\pi }MOto\stackrel{*}{\sigma }MO$, for which energy is absorbed in yellow region. So it appear violet in colour when  $I_2$ is dissolved in solvents such as  $C{S}_2,CHC{l}_3,CC{l}_4$ which do not contain LP, the electronic transition is from up  $\stackrel{*}{\pi p}MOto\stackrel{*}{\sigma p}MO$  so violet. But when dissolved in solvents such as pyridine, alcohol etc. While contain LP, that can donate LP to  $\stackrel{*}{\sigma }{p}_z.$  So electronic transition between  $\stackrel{*}{\pi p}MOto\stackrel{*}{\sigma }2pMO$ is not possible. Due to charge transfer from solvent to  $I_2$ molecule brown colour arises. With increase in size of halogen atom in the group the energy difference between  $\stackrel{*}{\pi 2p}and\stackrel{*}{\sigma }2p$ orbitals goes on decreasing. This causes increase in the intensity of colour.