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

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$I_{2}$ dissolved in $C S_{2}, C H C l_{3} o r C C l_{4}$ is violet.

Solid iodine is blackish violet but gaseous iodine is violet in colour.

$I_{2}$ dissolved in n- hexane, benzene, ethanol and pyridine are violet, purple, brown and yellow respectively, but on adding pyridine to solutions of $I_{2}$ in n-hexane, benzene and ethanol also changes to yellow

In gaseous state form $F_{2} to I_{2}$ the intensity of colour of halogens increases

answer is A, B, C, D.

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Detailed Solution

In gaseous $I_{2}$ colour arises due to electronic transition from $\overset{*}{π} M O t o \overset{*}{σ} M O$ , for which energy is absorbed in yellow region. So it appear violet in colour when $I_{2}$ is dissolved in solvents such as $C S_{2}, C H C l_{3}, C C l_{4}$ which do not contain LP, the electronic transition is from up $\overset{*}{π p} M O t o \overset{*}{σ p} M O$ so violet. But when dissolved in solvents such as pyridine, alcohol etc. While contain LP, that can donate LP to $\overset{*}{σ} p_{z} .$ So electronic transition between $\overset{*}{π p} M O t o \overset{*}{σ} 2 p M O$ 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 $\overset{*}{π 2 p} a n d \overset{*}{σ} 2 p$ orbitals goes on decreasing. This causes increase in the intensity of colour.