The structure of IF₆^– is distorted octahedral This is due to presence of a “weak” lone pair.

The shape of IF6– is Tetragonally distorted octahedron.

Explanation:
The IF6– ion consists of a central iodine atom bonded to six fluorine atoms. To determine the shape of the ion, we need to consider its electronic geometry and the presence of any lone pairs on the central atom.

Electronic Geometry:
The iodine atom in IF6– has six valence electrons, and each fluorine atom contributes one valence electron. Therefore, the total number of valence electrons in the ion is 7 (iodine) + 6(7 fluorine) + 1 (negative charge) = 48.

According to the VSEPR theory (Valence Shell Electron Pair Repulsion theory), the electron pairs around the central atom will arrange themselves in a way that minimizes repulsion. This will determine the electronic geometry of the ion.

In the case of IF6–, the iodine atom has six bonding pairs and no lone pairs of electrons. Therefore, the electronic geometry of the ion is octahedral.

Presence of Lone Pairs:
However, the presence of lone pairs on the central atom can distort the shape of the molecule/ion. In the case of IF6–, the iodine atom has one lone pair of electrons.

Distortion of Octahedron:
The presence of a lone pair causes electron-electron repulsion, which leads to a distortion of the octahedral shape. In the case of IF6–, the lone pair occupies more space compared to the bonding pairs, resulting in a tetragonal distortion of the octahedron.

In a tetragonally distorted octahedron, the lone pair occupies one of the axial positions, causing the molecule/ion to have a slightly elongated shape. This distortion is due to the repulsion between the lone pair and the bonding pairs, which pushes the fluorine atoms closer together.

Conclusion:
In conclusion, the shape of IF6– is a tetragonally distorted octahedron. This distortion is caused by the presence of a lone pair on the central iodine atom, which pushes the fluorine atoms closer together.