JEE Advanced 2021 · Paper 2 · Chemistry Q5

ChemistryMultiple correct+4 / -2Question 24 of 57

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)

\ce{[FeCl4]^-}

and

\ce{[Fe(CO)4]^{2-}}

\ce{[Co(CO)4]^-}

and

\ce{[CoCl4]^{2-}}

\ce{[Ni(CO)4]}

and

\ce{[Ni(CN)4]^{2-}}

\ce{[Cu(py)4]^+}

and

\ce{[Cu(CN)4]^{3-}}

Answer and worked solution

Official answer: A, B, D

Geometry of each complex.

(A) $\ce{[FeCl4]^-}$ ( $\ce{Fe^{3+}}$ , $d^5$ , weak-field ligand) and $\ce{[Fe(CO)4]^{2-}}$ (Fe(−II), $d^{10}$ ): both tetrahedral.
(B) $\ce{[Co(CO)4]^-}$ (Co(−I), $d^{10}$ ) and $\ce{[CoCl4]^{2-}}$ : both tetrahedral.
(C) $\ce{[Ni(CO)4]}$ is tetrahedral, but $\ce{[Ni(CN)4]^{2-}}$ ( $d^8$ , strong-field ligand) is square planar.
(D) $\ce{[Cu(py)4]^+}$ and $\ce{[Cu(CN)4]^{3-}}$ ( $\ce{Cu+}$ , $d^{10}$ ): both tetrahedral.

(A), (B), (D)

Topics:Valence bond theory Metal carbonyls and EAN rule

JEE Advanced 2021 · Paper 2, Chemistry question 5