Question
Match the species in List I with their geometry in List IIList I List II A. $PCl_5$ I. Tetrahedral B. $BrF_5$ II. Square Planar C. $BF_4^-$ III. Trigonal bipyramidal D. $[Ni(CN)_4]^{2-}$ IV. Square pyramidal
Choose the correct answer from the options given below:
This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
A-III, B-IV, C-I, D-II
Matching Chemical Species with Molecular Geometry
This question requires matching chemical species with their corresponding molecular geometries based on VSEPR (Valence Shell Electron Pair Repulsion) theory.
Geometry Determination
- A. $PCl_5$: The central atom Phosphorus (P) has 5 valence electrons and forms 5 bonds with 5 Chlorine (Cl) atoms. There are 5 electron domains and no lone pairs. This results in a Trigonal bipyramidal geometry.
- B. $BrF_5$: The central atom Bromine (Br) has 7 valence electrons. It forms 5 bonds with 5 Fluorine (F) atoms and has 1 lone pair. The electron geometry is octahedral, but the molecular geometry, considering only bond pairs, is Square pyramidal.
- C. $BF_4^-$: The central atom Boron (B) has 3 valence electrons. The negative charge adds one electron, making it 4. It forms 4 bonds with 4 Fluorine (F) atoms. There are 4 electron domains and no lone pairs. This results in a Tetrahedral geometry.
- D. $[Ni(CN)_4]^{2-}$: The central atom Nickel (Ni) is in the +2 oxidation state ($Ni^{2+}$), which has a $d^8$ configuration. Cyanide (CN) is a strong field ligand. The complex has a coordination number of 4, and $dsp^2$ hybridization leads to a Square planar geometry.
Correct Matches
Based on the analysis:
- A matches with III (Trigonal bipyramidal)
- B matches with IV (Square pyramidal)
- C matches with I (Tetrahedral)
- D matches with II (Square planar)
Final Answer
The correct matching is A-III, B-IV, C-I, D-II.
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