JEE Main 2026 — p Block Elements (Group 13 & 14) Question with Solution

Statement I:
Aluminium reacts with aqueous sodium hydroxide to form sodium tetrahydroxoaluminate(III) and liberates hydrogen gas. The reaction is given by:
$2Al+2NaOH+6H_{2}O\to 2N{a}^{+}[Al(OH{)}_4{]}^{-}+3H_2$
The complex ion formed is $[Al(OH{)}_4{]}^{-}$, not $[Al(OH{)}_6{]}^{3-}$. Thus, Statement I is false.

Statement II:
For $IC{l}_4^{-}$, 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 $s{p}^3{d}^2$ and the shape is square planar.
For $Cl{O}_3^{-}$, 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 $s{p}^3$ and the shape is trigonal pyramidal.
For $IB{r}_2^{-}$, 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 $s{p}^{3}d$ 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