JEE Main 2023 · 6 April · evening · Numerical answer · +4 / −0
The number of species having a square planar shape from the following is __________.
XeF₄, SF₄, SiF₄, BF₄⁻, BrF₄⁻,[Cu(NH₃)₄]²⁺,[FeCl₄]²⁻,[PtCl₄]²⁻
Answer and solution
Answer 4
Solution
Chemistry ma’am · 3 steps
Main-group species need lone pairs counted. For complexes, check whether the metal uses dsp² or sp³.
- Main-group rule. Count bond pairs and lone pairs on the central atom. Square planar needs 4 bond pairs and 2 lone pairs, sitting opposite each other in an octahedral arrangement.
- Complex rule. Four ligands give square planar with dsp² hybridisation and tetrahedral with sp³. A d⁸ ion such as Pt²⁺, or the d⁹ ion Cu²⁺, goes dsp². The d⁶ ion Fe²⁺ with weak-field Cl⁻ stays sp³.
Species Electron count or metal ion Hybridisation Shape XeF₄ 4 bond pairs, 2 lone pairs sp³d² square planar SF₄ 4 bond pairs, 1 lone pair sp³d see-saw SiF₄ 4 bond pairs, 0 lone pairs sp³ tetrahedral BF₄⁻ 4 bond pairs, 0 lone pairs sp³ tetrahedral BrF₄⁻ 7 + 1 = 8 electrons: 4 bond pairs, 2 lone pairs sp³d² square planar [Cu(NH₃)₄]²⁺ Cu²⁺, d⁹ dsp² square planar [FeCl₄]²⁻ Fe²⁺, d⁶, weak-field Cl⁻ sp³ tetrahedral [PtCl₄]²⁻ Pt²⁺, d⁸, 5d metal dsp² square planar - So. The square planar species are XeF₄, BrF₄⁻, [Cu(NH₃)₄]²⁺ and [PtCl₄]²⁻. That makes 4.
The common slip: 3. Calling [Cu(NH₃)₄]²⁺ tetrahedral just because it has four ligands gives 3. Cu²⁺ uses dsp² here, so it is square planar.
The idea to keep. Four ligands do not decide the shape. In main-group species, 4 bond pairs plus 2 lone pairs give square planar. In complexes, dsp² gives square planar and sp³ gives tetrahedral.
Answer from the NTA final answer key. Solution by PrepWiser.