COMEDK 2024 Evening Shift | Coordination Compounds Question 15 | Chemistry

Based on Crystal Field theory, match the Complex ions listed in Column I with the electronic configuration in the d orbitals of the central metal ion listed in Column II.

No.	Complexion	No.	d orbital configuration of central metal ion.
(A)	$$ \left[M n(C N)_6\right]^{4-} $$	(P)	$$ e_g^2 t_{2 g}^3 $$
(B)	$$ \left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} $$	(Q)	$$ t_{2 g}^4 e_g^2 $$
(C)	$$ \left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} $$	(R)	$$ t_{2 g}^5 $$
(D)	$$ \left[\mathrm{MnCl}_4\right]^{2-} $$	(S)	$$ t_{2 g}^5 e_g^2 $$

$$ \mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q} $$

$$ \mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P} $$

$$ \mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{R} $$

$$ \mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q} $$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

On the basis of crystal field theory, electronic configuration of a low spin $$\mathrm{d}^4$$ complex is:

$$t_{2 g}^1 e_g^3$$

$$t_{2 g}^4 e_g$$

$$t_{2 g}^3 e_g^1$$

$$t_{2 g}^2 e_g^2$$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

The molar conductivity of the complex $$\mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}$$ is found to be the same as that of a $$1: 3$$ electrolyte. The structural formula of the compound is :

$$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_3 $$

$$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\left(\mathrm{H}_2 \mathrm{O}\right)\right] \mathrm{Cl}_2 \mathrm{H}_2 \mathrm{O} $$

$$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Cl}_2\right] \mathrm{Cl} \mathrm{H}_2 \mathrm{O} $$

$$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}\right] \mathrm{Cl}_2 $$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

The type of hybridisation and the spin only magnetic moment for the complex $$[\mathrm{Ni}(\mathrm{CO})_4]$$ are respectively:

$$\mathrm{sp}^3, \sqrt{8} \mathrm{~BM}$$

$$\mathrm{dsp}^2$$, zero

$$\mathrm{sp}^3$$, zero

$$\mathrm{dsp}^2, \sqrt{8} \mathrm{~BM}$$

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