Which of the following statements is correct?

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Question

(A) Let us analyze each option:$A$. $[ZnBr_{4}]^{2-}$: $Br^{-}$ is a weak field ligand and does not cause pairing of electrons. $Zn^{2+}$ has a $d^{10

Vedclass · Accepted Answer

$[ZnBr_{4}]^{2-}$ is tetrahedral.

Vedclass · Answer

(A) Let us analyze each option:$A$. $[ZnBr_{4}]^{2-}$: $Br^{-}$ is a weak field ligand and does not cause pairing of electrons. $Zn^{2+}$ has a $d^{10}$ configuration. The hybridization is $sp^3$,resulting in a tetrahedral geometry. This statement is correct.$B$. $[Ni(NH_{3})_{6}]^{2+}$: $Ni^{2+}$ is $3d^8$. In the presence of $NH_{3}$,it forms an outer orbital complex with $sp^3d^2$ hybridization. It is not an inner orbital complex.$C$. $[Co(NH_{3})_{6}]^{2+}$: $Co^{2+}$ is $3d^7$. In the presence of $NH_{3}$,it forms a low-spin complex,but it is typically oxidized to $Co^{3+}$ in the presence of air/ligands. However,strictly considering $[Co(NH_{3})_{6}]^{2+}$,it has one unpaired electron and is paramagnetic. Note: The question asks for the correct statement,and $A$ is definitively correct.$D$. $[CoBr_{2}(en)_{2}]^{-}$: This complex does not contain ambidentate ligands (like $NO_{2}^{-}$ or $SCN^{-}$),so it cannot exhibit linkage isomerism.Therefore,the most accurate statement is $A$.

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