Chemistry Q59 – JEE-MAIN 2026

Home Chemistry Chemical BondingQuestion #226

Given below are two statements : Statement I: Aluminium upon reaction with NaOH forms

[A l (O H)_{6}]^{3 -}

ion. Statement II: The geometry of

I C l_{4}^{-}

C l O_{3}^{-}

and

I B r_{2}^{-}

is square planar, pyramidal and linear respectively. In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: Chemical Reactivity and Complex Formation

Recall the typical products formed when amphoteric metals like Aluminium react with strong bases like NaOH.

🧠 Full Solution Path

Step 1: Evaluate Statement I: Reaction of Aluminium with NaOH✦ Active

Aluminium is an amphoteric metal. It reacts with strong bases like NaOH to form sodium aluminate. The most common and stable complex ion formed in aqueous solution is the tetrahydroxoaluminate(III) ion, $[A l (O H)_{4}]^{-}$ . While the hexahydroxoaluminate(III) ion, $[A l (O H)_{6}]^{3 -}$ , is a known complex, it is not the primary or most commonly formed product under typical reaction conditions. Therefore, Statement I is considered false.

Step 2: Evaluate Statement II: Geometries of

I C l_{4}^{-}

C l O_{3}^{-}

, and

I B r_{2}^{-}

using VSEPR Theory○ Expand

1. For $I C l_{4}^{-}$ : The central atom is I. Valence electrons of I = 7. Four Cl atoms contribute 4 electrons. The negative charge adds 1 electron. Total valence electrons = $7 + 4 + 1 = 12$ electrons, or 6 electron pairs. There are 4 bond pairs (I-Cl) and $6 - 4 = 2$ lone pairs. With a steric number of 6 and 2 lone pairs, the geometry is **square planar**.

2. For $C l O_{3}^{-}$ : The central atom is Cl. Valence electrons of Cl = 7. The negative charge adds 1 electron. For VSEPR, oxygen is considered divalent, so it doesn't contribute electrons to the central atom's count for steric number calculation in this simplified method. Total electrons for steric number = $7 + 1 = 8$ , or 4 electron pairs. There are 3 bond pairs (Cl-O) and $4 - 3 = 1$ lone pair. With a steric number of 4 and 1 lone pair, the geometry is **pyramidal**.

3. For $I B r_{2}^{-}$ : The central atom is I. Valence electrons of I = 7. Two Br atoms contribute 2 electrons. The negative charge adds 1 electron. Total valence electrons = $7 + 2 + 1 = 10$ electrons, or 5 electron pairs. There are 2 bond pairs (I-Br) and $5 - 2 = 3$ lone pairs. With a steric number of 5 and 3 lone pairs, the geometry is **linear**.

All geometries mentioned in Statement II are correct. Therefore, Statement II is true.

💡 Teacher's Secret Hint

Remember to correctly count valence electrons, account for charges, and identify bond pairs and lone pairs for accurate VSEPR predictions.

Step 3: Conclusion○ Expand

Based on the analysis, Statement I is false, and Statement II is true.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 59 – JEE-MAIN 2026

Clue 1 of 3: Chemical Reactivity and Complex Formation

🧠 Full Solution Path