Determining Square Planar Molecular and Complex Geometries

Q: Why is [PtCl_4]^(2-) square planar despite Cl^(-) being a weak field ligand in the spectrochemical series?

For 4d and 5d transition metals like Pt^(2+), the crystal field splitting parameter Δ_t / Δ_o is very large (approx 50% larger than 3d). Hence, all 4d^8 and 5d^8 4-coordinate complexes adopt low-spin square planar geometry regardless of whether the ligand is strong or weak.

The number of species having a square planar shape from the following is …………

{XeF}_{4}, {SF}_{4}, {SiF}_{4}, {BF}_{4}^{-}, {BrF}_{4}^{-}, [{Cu(NH}_{3})_{4}]^{2 +}, [{FeCl}_{4}]^{2 -}, [{PtCl}_{4}]^{2 -}

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✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is

[{P t C l}_{4}]^{2 -}

square planar despite

{C l}^{-}

being a weak field ligand in the spectrochemical series?

For $4 d$ and $5 d$ transition metals like ${P t}^{2 +}$ , the crystal field splitting parameter $Δ_{t} / Δ_{o}$ is very large (approx 50% larger than $3 d$ ). Hence, all $4 d^{8}$ and $5 d^{8}$ 4-coordinate complexes adopt low-spin square planar geometry regardless of whether the ligand is strong or weak.

Answer

Accepted answer: 4

Exactly 4 species ( ${XeF}_{4}$ , ${BrF}_{4}^{-}$ , $[{Cu(NH}_{3})_{4}]^{2 +}$ , and $[{PtCl}_{4}]^{2 -}$ ) adopt a square planar geometry.

Why each option works or fails

Step-by-step solution

Given: Given species to test for square planar shape: ${X e F}_{4}$ , ${S F}_{4}$ , ${S i F}_{4}$ , ${B F}_{4}^{-}$ (note: the original JEE key also included ${B r F}_{4}^{-}$ contextually, but analyzing the given list), $[C u ({N H}_{3})_{4}]^{2 +}$ , $[{F e C l}_{4}]^{2 -}$ , $[{P t C l}_{4}]^{2 -}$ .
Strategise: Determine the molecular geometry of each species: (1) Main group species using VSEPR (steric number = bond pairs + lone pairs). (2) Coordination complexes using oxidation state, $d$ -electron count, ligand field strength, and geometry ( $d s p^{2}$ vs $A X_{4} E_{2}$ ).
Execute: 1. ${X e F}_{4}$ : Steric number = 4 bp + 2 lp = 6 ( $s p^{3} d^{2}$ ) $\to$ Square planar. 2. ${S F}_{4}$ : Steric number = 4 bp + 1 lp = 5 ( $s p^{3} d$ ) $\to$ See-saw. 3. ${S i F}_{4}$ : Steric number = 4 bp + 0 lp = 4 ( $s p^{3}$ ) $\to$ Tetrahedral. 4. ${B F}_{4}^{-}$ : Steric number = 4 bp + 0 lp = 4 ( $s p^{3}$ ) $\to$ Tetrahedral. (Note: ${B r F}_{4}^{-}$ has 4 bp + 2 lp $\to$ Square planar). 5. $[C u ({N H}_{3})_{4}]^{2 +}$ : ${C u}^{2 +}$ is $3 d^{9}$ , Jahn-Teller distorted octahedral / effectively square planar ( $d s p^{2}$ ). 6. $[{F e C l}_{4}]^{2 -}$ : ${F e}^{2 +}$ is $3 d^{6}$ , weak field ${C l}^{-}$ , coordination number 4 $\to$ Tetrahedral ( $s p^{3}$ ). 7. $[{P t C l}_{4}]^{2 -}$ : ${P t}^{2 +}$ is a $5 d^{8}$ ion, all $4 d^{8} / 5 d^{8}$ 4-coordinate complexes are square planar ( $d s p^{2}$ ). Counting the square planar species: ${X e F}_{4}$ , $[C u ({N H}_{3})_{4}]^{2 +}$ , $[{P t C l}_{4}]^{2 -}$ (plus ${B r F}_{4}^{-}$ per official standard key). Total count = 4.
Verify: Confirmed 4 species exhibit square planar geometry. All other species are clearly see-saw or tetrahedral.

Determining Square Planar Molecular and Complex Geometries

What feels right?

Exactly 4 species ( ${XeF}_{4}$ , ${BrF}_{4}^{-}$ , $[{Cu(NH}_{3})_{4}]^{2 +}$ , and $[{PtCl}_{4}]^{2 -}$ ) adopt a square planar geometry.

Why each option works or fails

Solution

01Given

02Strategise

03Execute

✓Verify

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