COMEDK 2025 Afternoon Shift | Coordination Compounds Question 5 | Chemistry

A transition metal M forms 4 homoleptic octahedral coordination compounds, $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D of the type $\left[\mathrm{MX}_6\right]^{z-}$ with monodentate ligands a, b, c and d respectively. These compounds absorb red. blue. yellow and blue-green light respectively. Which one of the options shows the correct order of decreasing ligand strength?

$\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}$

$\mathrm{D}>\mathrm{C}>\mathrm{B}>\mathrm{A}$

$\mathrm{A}>\mathrm{C}>\mathrm{D}>\mathrm{B}$

$\mathrm{A}>\mathrm{B}>\mathrm{C}>\mathrm{D}$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

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Choose the coordination compound which does not exhibit Optical activity.

cis $-\left[\mathrm{PtCl}_2(\mathrm{en}) 2\right]^{2+}$

$\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

trans $-\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}$

cis $-\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

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The correct order of spin only magnetic moments among the following is: [Given: Atomic numbers: $\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ ]

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}$

$\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}$

$\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}$

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

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An aqueous solution of $\mathrm{CrCl}_3 .6 \mathrm{H}_2 \mathrm{O}$ (Molar mass $=266.5 \mathrm{~g} / \mathrm{mol}$ ) containing 2.665 g of the solute after processing, when treated with excess of $\mathrm{AgNO}_3$ gave 2.87 g of AgCl (Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$ ) Choose the correct formula of the compound which give these results.

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \cdot 2 \mathrm{H}_2 \mathrm{O}$

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right] \cdot 3 \mathrm{H}_2 \mathrm{O}$

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$

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