COMEDK 2025 Afternoon Shift | Coordination Compounds Question 4 | Chemistry

S.No.	Structure		IUPAC name
A.	$\left[\mathrm{CoBr}_2(\mathrm{en})_2\right] \mathrm{Cl}$	P.	Dibromidodi-( ethan-1, 2 - diamine)cobalt (III) chloride
B.	$\mathrm{Na}\left[\mathrm{PtBrCl}\left(\mathrm{NO}_2\right)\left(\mathrm{NH}_3\right)\right]$	Q.	Sodium amminebromidochloridonitrito- N - palatinate(II)
C.	$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\right] \mathrm{SO}_4$	R.	Tetraamminediaquachloridochromium (III) sulphate.
D.	$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{PtCl}_4\right]$	S.	Tetraammineplatinum (II) tetrachloridoplatinate (II)

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

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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)

-0

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

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

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

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

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

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