Given below are two statements:

$$\textbf{Statement I :}$$ Aluminium upon reaction with $$\text{NaOH}$$ forms $$[\text{Al(OH)}_6]^{3-}$$ ion.

$$\textbf{Statement II :}$$ The geometry of $$\text{ICl}_4^{-}$$, $$\text{ClO}_3^{-}$$ and $$\text{IBr}_2^{-}$$ is square planar, pyramidal and linear respectively.

In the light of the above statements, choose the correct answer from the options given below:

Statement I: Aluminium upon reaction with $$\text{NaOH}$$ forms $$[\text{Al}(\text{OH})_6]^{3-}$$ ion. 

When Aluminium ($$\text{Al}$$) reacts with an aqueous solution of sodium hydroxide ($$\text{NaOH}$$), it forms a soluble tetrahydroxoaluminate complex along with the evolution of hydrogen gas.

$$2\text{Al}(s) + 2\text{NaOH}(aq) + 6\text{H}_2\text{O}(l) \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4](aq) + 3\text{H}_2(g)$$

The coordination complex formed in this reaction contains the tetrahydroxoaluminate(III) ion, which is $$[\text{Al}(\text{OH})_4]^-$$, not the hexahydroxoaluminate ion ($$[\text{Al}(\text{OH})_6]^{3-}$$).

Hence, Statement 1 is False.

Statement II: The geometry of $$\text{ICl}_4^-$$, $$\text{ClO}_3^-$$ and $$\text{IBr}_2^-$$ is square planar, pyramidal and linear respectively.

We can determine the geometry of each chemical species using VSEPR theory:

1. $$\text{ICl}_4^-$$ (Iodine tetrachloride ion)

• Steric Number: Iodine (central atom) has 7 valence electrons + 1 negative charge = 8 electrons.
  It forms 4 single bonds with Chlorine atoms, leaving 4 electrons (2 lone pairs).
• Hybridization: $$sp^3d^2$$ (Steric Number = $$4 \text{ bond pairs} + 2 \text{ lone pairs} = 6$$).
• Geometry/Shape: To minimize repulsion, the 2 lone pairs occupy axial positions, giving the molecule a square planar molecular shape.

2. $$\text{ClO}_3^-$$ (Chlorate ion)

• Steric Number: Chlorine (central atom) has 7 valence electrons + 1 negative charge = 8 electrons.
  It forms 2 double bonds with two Oxygens and 1 single bond with one Oxygen (using up 5 electrons for bonding), leaving 2 electrons (1 lone pair).
• Hybridization: $$sp^3$$ (Steric Number = $$3\text{ }\sigma\ \text{}+1\text{ lone pair}=4$$).
• Geometry/Shape: An asymmetrical tetrahedral arrangement with one lone pair results in a trigonal pyramidal shape.

3. $$\text{IBr}_2^-$$ (Iodobromite ion)

• Steric Number: Iodine (central atom) has 7 valence electrons + 1 negative charge = 8 electrons. It forms 2 single bonds with Bromine atoms, leaving 6 electrons (3 lone pairs).
• Hybridization: $$sp^3d$$ (Steric Number = $$2 \text{ bond pairs} + 3 \text{ lone pairs} = 5$$).
• Geometry/Shape: The 3 lone pairs occupy the equatorial positions of a trigonal bipyramidal arrangement, resulting in a linear molecular shape.
  Hence, Statement II is true.