JEE Main 2026 Chemistry: Tetrahedral Geometry and Incomplete Octet Analysis
Q. Given below are two statements:

Statement I: The number of species among SF4, NH4+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3.

Statement II: In the set [NO2, BeH2, BF3, AlCl3], all the molecules have incomplete octet around central atom.

In the light of the above statements, choose the correct answer:
Correct Answer: C

(Statement I is false but Statement II is true)

Step-by-Step Analysis

Analysis of Statement I (Tetrahedral Species):
We need to check the geometry of each listed species:
SpeciesSteric No. / HybridizationGeometry
SF45 (sp3d + 1 LP)See-saw
NH4+4 (sp3)Tetrahedral
[NiCl4]2-4 (sp3)Tetrahedral
XeF46 (sp3d2 + 2 LP)Square Planar
[PtCl4]2-4 (dsp2)Square Planar
SeF45 (sp3d + 1 LP)See-saw
[Ni(CN)4]2-4 (dsp2)Square Planar
Only 2 species (NH4+ and [NiCl4]2-) are tetrahedral. Statement I says 3, so it is False.
Analysis of Statement II (Incomplete Octet):
An incomplete octet means the central atom has fewer than 8 electrons in its valence shell.
  • NO2: Odd electron molecule. Nitrogen has 7 electrons in valence shell (incomplete).
  • BeH2: Beryllium has only 4 electrons around it (incomplete).
  • BF3: Boron has only 6 electrons around it (incomplete).
  • AlCl3: Aluminum has only 6 electrons around it (incomplete).
All molecules have an incomplete octet. Statement II is True.

Related Theory

1. Determining Geometry (VSEPR & Hybridization)
To find the geometry, we calculate the Steric Number (SN): \[ \text{SN} = \frac{1}{2} [V + M - C + A] \] Where \(V\) = valence electrons of central atom, \(M\) = monovalent atoms, \(C\) = cationic charge, and \(A\) = anionic charge. - SF4/SeF4: SN = 5. Geometry is See-saw because of one lone pair. - XeF4: SN = 6. Geometry is Square Planar because of two lone pairs.

2. Coordination Compounds Geometry
- [NiCl4]2-: \(Cl^-\) is a weak field ligand. Ni2+ (\(3d^8\)) remains high spin with \(sp^3\) hybridization (Tetrahedral). - [Ni(CN)4]2-: \(CN^-\) is a strong field ligand. Electrons pair up, leading to \(dsp^2\) hybridization (Square Planar). - [PtCl4]2-: For 5d series elements like Pt, all ligands (even \(Cl^-\)) act as strong field ligands. Hence, it is Square Planar.

3. Octet Rule Exceptions
The octet rule is not universal. Exceptions include: - Incomplete Octet: Common in Be, B, and Al compounds (Electron-deficient). - Expanded Octet: Common in 3rd period onwards (e.g., PCl5, SF6). - Odd-Electron Molecules: Species like NO and NO2.

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Related Covered Topics

jee mains 2026 chemical bonding VSEPR theory tetrahedral geometry incomplete octet coordination chemistry square planar geometry hybridization nickel complexes boron trifluoride

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