The correct shape and I - I - I bond angles respectively in $$I_3^-$$ ion are:

The $$I_3^-$$ ion is formed when an iodide ion ($$I^-$$) donates a lone pair to an iodine molecule ($$I_2$$). The central iodine atom in $$I_3^-$$ has 7 valence electrons plus 1 from the negative charge, giving a total of 8 valence electrons around it. Of these, 2 electrons form bonds with the two terminal iodine atoms, and the remaining 6 electrons form 3 lone pairs.

According to VSEPR theory, the central iodine atom has 5 electron pairs (2 bonding pairs and 3 lone pairs), which corresponds to a trigonal bipyramidal electron geometry. The two bonding pairs occupy the axial positions (which are 180° apart) to minimise lone pair-bonding pair repulsions, while the three lone pairs occupy the equatorial positions.

With the bonding pairs in the axial positions, the molecular shape of $$I_3^-$$ is linear, and the $$I-I-I$$ bond angle is $$180°$$.

Therefore, the correct answer is Option (4): Linear; 180°.