JEE Advanced 2021 Paper 2 Q05 Chemistry Coordination Chemistry Nomenclature & Isomerism Medium

JEE Advanced 2021 Paper 2 · Q05 · Nomenclature & Isomerism

The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are)

(Note: py = pyridine. Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively.)

  1. A. [FeCl$_4$]$^-$ and [Fe(CO)$_4$]$^{2-}$
  2. B. [Co(CO)$_4$]$^-$ and [CoCl$_4$]$^{2-}$
  3. C. [Ni(CO)$_4$] and [Ni(CN)$_4$]$^{2-}$
  4. D. [Cu(py)$_4$]$^+$ and [Cu(CN)$_4$]$^{3-}$
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Correct answer:A, B, D

Solution

Identify the d-electron count and the field strength of the ligand. Tetrahedral $\Rightarrow$ $sp^3$ (or $sd^3$); square planar $\Rightarrow$ $dsp^2$ (typical for d$^8$ with strong-field ligand).

(A) [FeCl$_4$]$^-$: Fe$^{3+}$, d$^5$, Cl$^-$ weak field $\Rightarrow$ $sp^3$, tetrahedral. [Fe(CO)$_4$]$^{2-}$: Fe$^{2-}$ (d$^{10}$), CO is strong field; only $sp^3$ is possible (no d-electrons to pair beyond a closed shell) $\Rightarrow$ tetrahedral. CORRECT.

(B) [Co(CO)$_4$]$^-$: Co($-$1) d$^{10}$, $sp^3$ tetrahedral. [CoCl$_4$]$^{2-}$: Co$^{2+}$ d$^7$, Cl$^-$ weak field, $sp^3$ tetrahedral. CORRECT.

(C) [Ni(CO)$_4$]: Ni(0) d$^{10}$, $sp^3$ tetrahedral. [Ni(CN)$_4$]$^{2-}$: Ni$^{2+}$ d$^8$, CN$^-$ strong field $\Rightarrow$ $dsp^2$ square planar. Pair is NOT both tetrahedral. WRONG.

(D) [Cu(py)$_4$]$^+$: Cu$^+$ d$^{10}$, $sp^3$ tetrahedral. [Cu(CN)$_4$]$^{3-}$: Cu$^+$ d$^{10}$, $sp^3$ tetrahedral. CORRECT.

Hence A, B, D are correct.

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