JEE Challenger
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ChemMedMain 2026 • 5th Apr - Shift 2

Geometry of Interhalogen and Polyatomic Anions and Aluminium Hydroxide Formation

Given below are two statements :

Statement I : Aluminium upon reaction with NaOH\text{NaOH} forms [Al(OH)6]3−[\text{Al}(\text{OH})_6]^{3-} ion.

Statement II : The geometry of ICl4−\text{ICl}_4^-, ClO3−\text{ClO}_3^- and IBr2−\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 :

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Correct

Topics & Concepts

p-Block Elementsp-Block ElementsVSEPR Theory

Step-by-Step Solution

To evaluate the statements, we analyze each one using basic inorganic chemistry principles and the Valence Shell Electron Pair Repulsion (VSEPR) theory.

Analysis of Statement I:

When aluminium metal reacts with aqueous sodium hydroxide (NaOH\text{NaOH}), it dissolves to form sodium tetrahydroxoaluminate(III) along with the evolution of hydrogen gas: 2Al(s)+2NaOH(aq)+6H2O(l)⟶2Na+[Al(OH)4]−(aq)+3H2(g)2\text{Al}(s) + 2\text{NaOH}(aq) + 6\text{H}_2\text{O}(l) \longrightarrow 2\text{Na}^+[\text{Al}(\text{OH})_4]^-(aq) + 3\text{H}_2(g)

The species formed in solution is the tetrahydroxoaluminate(III) ion, [Al(OH)4]−[\text{Al}(\text{OH})_4]^-, where aluminium is four-coordinate, rather than the hexacoordinate [Al(OH)6]3−[\text{Al}(\text{OH})_6]^{3-} ion.

Thus, Statement I is false.

Analysis of Statement II:

Using VSEPR theory, we determine the hybridization, total electron pairs, and molecular geometries of the given species:

  1. ICl4−\text{ICl}_4^- ion:

    • Central atom (I\text{I}): 7 valence electrons +1+ 1 extra electron from negative charge =8= 8 valence electrons.
    • Number of bond pairs (σ\sigma-bonds with Cl\text{Cl}) =4= 4
    • Number of lone pairs =8−42=2= \frac{8 - 4}{2} = 2
    • Steric number =4+2=6  ⟹  sp3d2= 4 + 2 = 6 \implies sp^3d^2 hybridization.
    • With 4 bond pairs and 2 axial lone pairs, the molecular geometry is square planar.

  2. ClO3−\text{ClO}_3^- ion:

    • Central atom (Cl\text{Cl}): 7 valence electrons +1+ 1 extra electron from negative charge =8= 8 valence electrons.
    • Number of σ\sigma-bond pairs with oxygen atoms =3= 3
    • Number of lone pairs =8−62=1= \frac{8 - 6}{2} = 1 (since 6 electrons are involved in bonding with three oxygen atoms).
    • Steric number =3+1=4  ⟹  sp3= 3 + 1 = 4 \implies sp^3 hybridization.
    • With 3 bond pairs and 1 lone pair, the molecular geometry is pyramidal (trigonal pyramidal).

  3. IBr2−\text{IBr}_2^- ion:

    • Central atom (I\text{I}): 7 valence electrons +1+ 1 extra electron from negative charge =8= 8 valence electrons.
    • Number of bond pairs (σ\sigma-bonds with Br\text{Br}) =2= 2
    • Number of lone pairs =8−22=3= \frac{8 - 2}{2} = 3
    • Steric number =2+3=5  ⟹  sp3d= 2 + 3 = 5 \implies sp^3d hybridization.
    • With 2 axial bond pairs and 3 equatorial lone pairs, the molecular geometry is linear.

Thus, the geometries of ICl4−\text{ICl}_4^-, ClO3−\text{ClO}_3^-, and IBr2−\text{IBr}_2^- are indeed square planar, pyramidal, and linear, respectively.

Therefore, Statement II is true.

Conclusion:

  • Statement I is false
  • Statement II is true

Correct Option: D

Geometry of Interhalogen and Polyatomic Anions and Aluminium Hydroxide Formation | Chemistry PYQ Solution - JEE Challenger