Match List I with List II .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top LIST I (Complex ion) LIST II (Spin only magnetic moment in B.M.) A. [Cr(NH_3)_6]^3+ I. 4.90 B. [NiCl_4]^2- II. 3.87 C. [CoF_6]^3- III. 0.0 D. [Ni(CN)_4]^2- IV. 2.83 Choose the correct answer from the options given below :

(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Coordination Compounds PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q4 with Solution

Match List I with List II

	LIST I (Complex ion)		LIST II (Spin only magnetic moment in B.M.)
A.	$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$	I.	4.90
B.	$\left[\mathrm{NiCl}_4\right]^{2-}$	II.	3.87
C.	$\left[\mathrm{CoF}_6\right]^{3-}$	III.	0.0
D.	$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$	IV.	2.83

Choose the correct answer from the options given below :

A
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
B
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
C
(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
D
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

View written solutionFree

Correct answer: D

Use spin-only magnetic moment formula

For a complex with $n$ unpaired electrons,

\mu = \sqrt{n(n+2)} \text{ B.M.}

Given values correspond to:

$n=4 \Rightarrow \mu = \sqrt{24} \approx 4.90$ B.M.
$n=3 \Rightarrow \mu = \sqrt{15} \approx 3.87$ B.M.
$n=2 \Rightarrow \mu = \sqrt{8} \approx 2.83$ B.M.
$n=0 \Rightarrow \mu = 0.0$ B.M.

So:

I $\to 4$ unpaired electrons
II $\to 3$ unpaired electrons
III $\to 0$ unpaired electrons
IV $\to 2$ unpaired electrons

Analyze each complex

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

Oxidation state of Cr = $+3$
Cr: $[Ar]3d^5 4s^1$
$\mathrm{Cr}^{3+} \Rightarrow 3d^3$
Octahedral complex with $d^3$ configuration always has 3 unpaired electrons.

Thus,

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87

So, A $\to$ II.

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

Oxidation state of Ni = $+2$
$\mathrm{Ni}^{2+}: 3d^8$
$\mathrm{Cl}^-$ is a weak-field ligand.
Four-coordinate $\mathrm{Ni}^{2+}$ with weak ligand forms tetrahedral complex.
Tetrahedral $d^8$ has 2 unpaired electrons.

Thus,

\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83

So, B $\to$ IV.

C. $\left[\mathrm{CoF}_6\right]^{3-}$

Oxidation state of Co = $+3$
Co: $[Ar]3d^7 4s^2$
$\mathrm{Co}^{3+} \Rightarrow 3d^6$
$\mathrm{F}^-$ is a weak-field ligand, so this is high-spin octahedral.
High-spin $d^6$ configuration has 4 unpaired electrons.

Thus,

\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90

So, C $\to$ I.

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

Oxidation state of Ni = $+2$
$\mathrm{Ni}^{2+}: 3d^8$
$\mathrm{CN}^-$ is a strong-field ligand.
Four-coordinate $d^8$ with strong ligand forms square planar complex.
Square planar $d^8$ is diamagnetic, so unpaired electrons = 0.

Thus,

\mu = 0.0

So, D $\to$ III.

Final matching

A $\to$ II
B $\to$ IV
C $\to$ I
D $\to$ III

This corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, the derived answer agrees with the stored answer.

More from Coordination Compounds

Match List I with List II

	LIST I (Complex ion)		LIST II (Spin only magnetic moment in B.M.)
A.	$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$	I.	4.90
B.	$\left[\mathrm{NiCl}_4\right]^{2-}$	II.	3.87
C.	$\left[\mathrm{CoF}_6\right]^{3-}$	III.	0.0
D.	$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$	IV.	2.83

Choose the correct answer from the options given below :

A
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
B
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
C
(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
D
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

View written solutionFree

Correct answer: D

Use spin-only magnetic moment formula

For a complex with $n$ unpaired electrons,

\mu = \sqrt{n(n+2)} \text{ B.M.}

Given values correspond to:

$n=4 \Rightarrow \mu = \sqrt{24} \approx 4.90$ B.M.
$n=3 \Rightarrow \mu = \sqrt{15} \approx 3.87$ B.M.
$n=2 \Rightarrow \mu = \sqrt{8} \approx 2.83$ B.M.
$n=0 \Rightarrow \mu = 0.0$ B.M.

So:

I $\to 4$ unpaired electrons
II $\to 3$ unpaired electrons
III $\to 0$ unpaired electrons
IV $\to 2$ unpaired electrons

Analyze each complex

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

Oxidation state of Cr = $+3$
Cr: $[Ar]3d^5 4s^1$
$\mathrm{Cr}^{3+} \Rightarrow 3d^3$
Octahedral complex with $d^3$ configuration always has 3 unpaired electrons.

Thus,

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87

So, A $\to$ II.

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

Oxidation state of Ni = $+2$
$\mathrm{Ni}^{2+}: 3d^8$
$\mathrm{Cl}^-$ is a weak-field ligand.
Four-coordinate $\mathrm{Ni}^{2+}$ with weak ligand forms tetrahedral complex.
Tetrahedral $d^8$ has 2 unpaired electrons.

Thus,

\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83

So, B $\to$ IV.

C. $\left[\mathrm{CoF}_6\right]^{3-}$

Oxidation state of Co = $+3$
Co: $[Ar]3d^7 4s^2$
$\mathrm{Co}^{3+} \Rightarrow 3d^6$
$\mathrm{F}^-$ is a weak-field ligand, so this is high-spin octahedral.
High-spin $d^6$ configuration has 4 unpaired electrons.

Thus,

\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90

So, C $\to$ I.

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

Oxidation state of Ni = $+2$
$\mathrm{Ni}^{2+}: 3d^8$
$\mathrm{CN}^-$ is a strong-field ligand.
Four-coordinate $d^8$ with strong ligand forms square planar complex.
Square planar $d^8$ is diamagnetic, so unpaired electrons = 0.

Thus,

\mu = 0.0

So, D $\to$ III.

Final matching

A $\to$ II
B $\to$ IV
C $\to$ I
D $\to$ III

This corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, the derived answer agrees with the stored answer.

Coordination Compounds question

Coordination Compounds question

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

C. $\left[\mathrm{CoF}_6\right]^{3-}$

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

C. $\left[\mathrm{CoF}_6\right]^{3-}$

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

More from Coordination Compounds

A. [Cr(NH3)6]3+\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}[Cr(NH3​)6​]3+

B. [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-}[NiCl4​]2−

C. [CoF6]3−\left[\mathrm{CoF}_6\right]^{3-}[CoF6​]3−

D. [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}[Ni(CN)4​]2−

More from Coordination Compounds

A. [Cr(NH3)6]3+\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}[Cr(NH3​)6​]3+

B. [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-}[NiCl4​]2−

C. [CoF6]3−\left[\mathrm{CoF}_6\right]^{3-}[CoF6​]3−

D. [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}[Ni(CN)4​]2−

More from Coordination Compounds

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

C. $\left[\mathrm{CoF}_6\right]^{3-}$

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

A. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$

B. $\left[\mathrm{NiCl}_4\right]^{2-}$

C. $\left[\mathrm{CoF}_6\right]^{3-}$

D. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$