Given below are two statements :
Statement I: Aluminium upon reaction with NaOH forms $[Al(OH)_6]^{3-}$ ion.
Statement II: The geometry of $ICl_4^-$, $ClO_3^-$ and $IBr_2^-$ 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 reacts with aqueous sodium hydroxide to form sodium tetrahydroxoaluminate(III) and liberates hydrogen gas. The reaction is given by:
$2Al + 2NaOH + 6H_2O \rightarrow 2Na^+[Al(OH)_4]^- + 3H_2$
The complex ion formed is $[Al(OH)_4]^-$, not $[Al(OH)_6]^{3-}$. Thus, Statement I is false.

Statement II:
For $ICl_4^-$, the central iodine atom has $7$ valence electrons $+ 1$ (from negative charge) $= 8$ electrons. It forms $4$ sigma bonds with Cl atoms and has $2$ lone pairs. The hybridization is $sp^3d^2$ and the shape is square planar.
For $ClO_3^-$, the central chlorine atom has $7$ valence electrons $+ 1 = 8$ electrons. It forms $3$ sigma bonds with O atoms and has $1$ lone pair. The hybridization is $sp^3$ and the shape is trigonal pyramidal.
For $IBr_2^-$, the central iodine atom has $7$ valence electrons $+ 1 = 8$ electrons. It forms $2$ sigma bonds with Br atoms and has $3$ lone pairs. The hybridization is $sp^3d$ and the shape is linear.
Thus, Statement II is true.

Therefore, Statement I is false but Statement II is true.

Answer: Statement I is false but Statement II is true