Given below are two statements:Statement-I: H | vedclass

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(C) In $K_{3}[Co(CO_{3})_{3}]$, $Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $CO_{3}^{2-}$ is a weak field ligand, so it forms an outer or

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Both statement-$I$ and statement-$II$ are true

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(C) In $K_{3}[Co(CO_{3})_{3}]$, $Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $CO_{3}^{2-}$ is a weak field ligand, so it forms an outer orbital complex: $sp^{3}d^{2}$ hybridisation, octahedral geometry, and $4$ unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{4(6)} = \sqrt{24} \approx 4.9 \ BM$. Statement-$I$ is true.For $[Ni(CN)_{4}]^{2-}$, $Ni^{2+}$ is $d^{8}$. $CN^{-}$ is a strong field ligand, so $dsp^{2}$ hybridisation, square planar, $0$ unpaired electrons, $\mu = 0 \ BM$.For $[MnBr_{4}]^{2-}$, $Mn^{2+}$ is $d^{5}$. $Br^{-}$ is a weak field ligand, so $sp^{3}$ hybridisation, tetrahedral, $5$ unpaired electrons, $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$.For $[CoF_{6}]^{3-}$, $Co^{3+}$ is $d^{6}$. $F^{-}$ is a weak field ligand, so $sp^{3}d^{2}$ hybridisation, octahedral, $4$ unpaired electrons, $\mu = 4.9 \ BM$.All values in Statement-$II$ are correct. Thus, both statements are true.

List-$I$	List-$II$
$(A). [Ag(CN)_2]^-$	$1. \text{Square planar, } 1.73 \, B.M.$
$(B). [Cu(CN)_4]^{3-}$	$2. \text{Linear, } 0 \, B.M.$
$(C). [Cu(CN)_6]^{4-}$	$3. \text{Octahedral, } 0 \, B.M.$
$(D). [Cu(NH_3)_4]^{2+}$	$4. \text{Tetrahedral, } 0 \, B.M.$
$(E). [Fe(CN)_6]^{4-}$	$5. \text{Octahedral, } 1.73 \, B.M.$

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