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CHXII09:COORDINATION COMPOUNDS

322320 Read Assertion and Reason carefully and mark correct option.
Assertion :
The spin only magnetic moment value for ${[Fe (CN)_{6}]}^{3 -}$ is 1.74 B.M., whereas for ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ is 5.92 B.M.
Reason :
In both complexes, Fe is present in $+ 3$ oxidation state.

1 Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

2 Assertion is false but Reason is true.

3 Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

4 Assertion is true but Reason is false.

Explanation:

${[Fe {(H_{2} O)}_{6}]}^{3 +}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration
supporting img

$H_{2} O$ is a weak field ligand.

Thus, ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ has 5 unpaired electrons.
Hence, $n = 5$
$μ = \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.92 B . M$ .
${[Fe (CN)_{6}]}^{3 -}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration

${CN}^{-}$ is a strong field ligand. Pairing of unpaired electrons will takes place.
supporting img

Thus, ${[Fe (CN)_{6}]}^{3 -}$ has only one unpaired electron, $n = 1$
$μ = \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, the values of magnetic moment are due to the number of unpaired electrons present and not because of the oxidation states of Fe. Hence, both assertion and reason are true, but reason is not the correct explanation.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322321 Which of the following complexes is diamagnetic and square planar?

1

{[{CoF}_{6}]}^{3 -}

2

{[Co {({NH}_{3})}_{6}]}^{3 +}

3

{[{NiCl}_{4}]}^{2 -}

4

{[Ni (CN)_{4}]}^{2 -}

Explanation:

${[{CoF}_{6}]}^{3 -} \to$ Paramagnetic and octahedral $({sp}^{3} d^{2})$ (4 unpaired electrons)
${[Co {({NH}_{3})}_{6}]}^{3 +} \to$ Diamagnetic and octahedral $(d^{2} {sp}^{3})$ (zero unpaired electrons) ${[{NiCl}_{4}]}^{2 -} \to$ Paramagnetic and tetrahedral $({sp}^{3})$ (2 unpaired electrons)
${[Ni (CN)_{4}]}^{2 -} \to$ Diamagnetic and square planar (dsp $^{2}$ ) (zero unpaired electrons)

MHTCET - 2021

CHXII09:COORDINATION COMPOUNDS

322322 Which one of the following complexes will exhibit the least paramagnetic behavior?

1

{[Mn {(H_{2} O)}_{6}]}^{2 +}

2

{[Fe {(H_{2} O)}_{6}]}^{2 +}

3

{[Co {(H_{2} O)}_{6}]}^{2 +}

4

{[Cr {(H_{2} O)}_{6}]}^{2 +}

Explanation:

More no. of unpaired electrons, higher is the paramagnetism
supporting img
Least paramagnetic behaviour is exhibited by $[Co {(H_{2} O_{6}]}^{2 +}$ as it has least no. of unpaired electrons. So, the correct option is (3).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322323 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ is square planar and diamagnetic.
Reason :
It has no unpaired electrons due to presence of strong field ligand.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$In {[Ni (CN)_{4}]}^{- 2}$ , oxidation state of $Ni$ is $+ 2$
$Ni : 3 d^{8} 4 s^{2}$ and ${Ni}^{2 +} : 3 d^{8} 4 s^{0}$
supporting img
The compound is diamagnetic as evident from the absence of unpaired electron. It is a square planar complexes, the hybridisation involved is ${dsp}^{2}$ .Therefore, option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322320 Read Assertion and Reason carefully and mark correct option.
Assertion :
The spin only magnetic moment value for ${[Fe (CN)_{6}]}^{3 -}$ is 1.74 B.M., whereas for ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ is 5.92 B.M.
Reason :
In both complexes, Fe is present in $+ 3$ oxidation state.

1 Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

2 Assertion is false but Reason is true.

3 Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

4 Assertion is true but Reason is false.

Explanation:

${[Fe {(H_{2} O)}_{6}]}^{3 +}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration
supporting img

$H_{2} O$ is a weak field ligand.

Thus, ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ has 5 unpaired electrons.
Hence, $n = 5$
$μ = \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.92 B . M$ .
${[Fe (CN)_{6}]}^{3 -}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration

${CN}^{-}$ is a strong field ligand. Pairing of unpaired electrons will takes place.
supporting img

Thus, ${[Fe (CN)_{6}]}^{3 -}$ has only one unpaired electron, $n = 1$
$μ = \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, the values of magnetic moment are due to the number of unpaired electrons present and not because of the oxidation states of Fe. Hence, both assertion and reason are true, but reason is not the correct explanation.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322321 Which of the following complexes is diamagnetic and square planar?

1

{[{CoF}_{6}]}^{3 -}

2

{[Co {({NH}_{3})}_{6}]}^{3 +}

3

{[{NiCl}_{4}]}^{2 -}

4

{[Ni (CN)_{4}]}^{2 -}

Explanation:

${[{CoF}_{6}]}^{3 -} \to$ Paramagnetic and octahedral $({sp}^{3} d^{2})$ (4 unpaired electrons)
${[Co {({NH}_{3})}_{6}]}^{3 +} \to$ Diamagnetic and octahedral $(d^{2} {sp}^{3})$ (zero unpaired electrons) ${[{NiCl}_{4}]}^{2 -} \to$ Paramagnetic and tetrahedral $({sp}^{3})$ (2 unpaired electrons)
${[Ni (CN)_{4}]}^{2 -} \to$ Diamagnetic and square planar (dsp $^{2}$ ) (zero unpaired electrons)

MHTCET - 2021

CHXII09:COORDINATION COMPOUNDS

322322 Which one of the following complexes will exhibit the least paramagnetic behavior?

1

{[Mn {(H_{2} O)}_{6}]}^{2 +}

2

{[Fe {(H_{2} O)}_{6}]}^{2 +}

3

{[Co {(H_{2} O)}_{6}]}^{2 +}

4

{[Cr {(H_{2} O)}_{6}]}^{2 +}

Explanation:

More no. of unpaired electrons, higher is the paramagnetism
supporting img
Least paramagnetic behaviour is exhibited by $[Co {(H_{2} O_{6}]}^{2 +}$ as it has least no. of unpaired electrons. So, the correct option is (3).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322323 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ is square planar and diamagnetic.
Reason :
It has no unpaired electrons due to presence of strong field ligand.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$In {[Ni (CN)_{4}]}^{- 2}$ , oxidation state of $Ni$ is $+ 2$
$Ni : 3 d^{8} 4 s^{2}$ and ${Ni}^{2 +} : 3 d^{8} 4 s^{0}$
supporting img
The compound is diamagnetic as evident from the absence of unpaired electron. It is a square planar complexes, the hybridisation involved is ${dsp}^{2}$ .Therefore, option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322320 Read Assertion and Reason carefully and mark correct option.
Assertion :
The spin only magnetic moment value for ${[Fe (CN)_{6}]}^{3 -}$ is 1.74 B.M., whereas for ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ is 5.92 B.M.
Reason :
In both complexes, Fe is present in $+ 3$ oxidation state.

1 Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

2 Assertion is false but Reason is true.

3 Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

4 Assertion is true but Reason is false.

Explanation:

${[Fe {(H_{2} O)}_{6}]}^{3 +}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration
supporting img

$H_{2} O$ is a weak field ligand.

Thus, ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ has 5 unpaired electrons.
Hence, $n = 5$
$μ = \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.92 B . M$ .
${[Fe (CN)_{6}]}^{3 -}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration

${CN}^{-}$ is a strong field ligand. Pairing of unpaired electrons will takes place.
supporting img

Thus, ${[Fe (CN)_{6}]}^{3 -}$ has only one unpaired electron, $n = 1$
$μ = \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, the values of magnetic moment are due to the number of unpaired electrons present and not because of the oxidation states of Fe. Hence, both assertion and reason are true, but reason is not the correct explanation.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322321 Which of the following complexes is diamagnetic and square planar?

1

{[{CoF}_{6}]}^{3 -}

2

{[Co {({NH}_{3})}_{6}]}^{3 +}

3

{[{NiCl}_{4}]}^{2 -}

4

{[Ni (CN)_{4}]}^{2 -}

Explanation:

${[{CoF}_{6}]}^{3 -} \to$ Paramagnetic and octahedral $({sp}^{3} d^{2})$ (4 unpaired electrons)
${[Co {({NH}_{3})}_{6}]}^{3 +} \to$ Diamagnetic and octahedral $(d^{2} {sp}^{3})$ (zero unpaired electrons) ${[{NiCl}_{4}]}^{2 -} \to$ Paramagnetic and tetrahedral $({sp}^{3})$ (2 unpaired electrons)
${[Ni (CN)_{4}]}^{2 -} \to$ Diamagnetic and square planar (dsp $^{2}$ ) (zero unpaired electrons)

MHTCET - 2021

CHXII09:COORDINATION COMPOUNDS

322322 Which one of the following complexes will exhibit the least paramagnetic behavior?

1

{[Mn {(H_{2} O)}_{6}]}^{2 +}

2

{[Fe {(H_{2} O)}_{6}]}^{2 +}

3

{[Co {(H_{2} O)}_{6}]}^{2 +}

4

{[Cr {(H_{2} O)}_{6}]}^{2 +}

Explanation:

More no. of unpaired electrons, higher is the paramagnetism
supporting img
Least paramagnetic behaviour is exhibited by $[Co {(H_{2} O_{6}]}^{2 +}$ as it has least no. of unpaired electrons. So, the correct option is (3).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322323 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ is square planar and diamagnetic.
Reason :
It has no unpaired electrons due to presence of strong field ligand.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$In {[Ni (CN)_{4}]}^{- 2}$ , oxidation state of $Ni$ is $+ 2$
$Ni : 3 d^{8} 4 s^{2}$ and ${Ni}^{2 +} : 3 d^{8} 4 s^{0}$
supporting img
The compound is diamagnetic as evident from the absence of unpaired electron. It is a square planar complexes, the hybridisation involved is ${dsp}^{2}$ .Therefore, option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322320 Read Assertion and Reason carefully and mark correct option.
Assertion :
The spin only magnetic moment value for ${[Fe (CN)_{6}]}^{3 -}$ is 1.74 B.M., whereas for ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ is 5.92 B.M.
Reason :
In both complexes, Fe is present in $+ 3$ oxidation state.

1 Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

2 Assertion is false but Reason is true.

3 Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

4 Assertion is true but Reason is false.

Explanation:

${[Fe {(H_{2} O)}_{6}]}^{3 +}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration
supporting img

$H_{2} O$ is a weak field ligand.

Thus, ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ has 5 unpaired electrons.
Hence, $n = 5$
$μ = \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.92 B . M$ .
${[Fe (CN)_{6}]}^{3 -}$ contains ${Fe}^{3 +}$ with $3 d^{5}$ configuration

${CN}^{-}$ is a strong field ligand. Pairing of unpaired electrons will takes place.
supporting img

Thus, ${[Fe (CN)_{6}]}^{3 -}$ has only one unpaired electron, $n = 1$
$μ = \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, the values of magnetic moment are due to the number of unpaired electrons present and not because of the oxidation states of Fe. Hence, both assertion and reason are true, but reason is not the correct explanation.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322321 Which of the following complexes is diamagnetic and square planar?

1

{[{CoF}_{6}]}^{3 -}

2

{[Co {({NH}_{3})}_{6}]}^{3 +}

3

{[{NiCl}_{4}]}^{2 -}

4

{[Ni (CN)_{4}]}^{2 -}

Explanation:

${[{CoF}_{6}]}^{3 -} \to$ Paramagnetic and octahedral $({sp}^{3} d^{2})$ (4 unpaired electrons)
${[Co {({NH}_{3})}_{6}]}^{3 +} \to$ Diamagnetic and octahedral $(d^{2} {sp}^{3})$ (zero unpaired electrons) ${[{NiCl}_{4}]}^{2 -} \to$ Paramagnetic and tetrahedral $({sp}^{3})$ (2 unpaired electrons)
${[Ni (CN)_{4}]}^{2 -} \to$ Diamagnetic and square planar (dsp $^{2}$ ) (zero unpaired electrons)

MHTCET - 2021

CHXII09:COORDINATION COMPOUNDS

322322 Which one of the following complexes will exhibit the least paramagnetic behavior?

1

{[Mn {(H_{2} O)}_{6}]}^{2 +}

2

{[Fe {(H_{2} O)}_{6}]}^{2 +}

3

{[Co {(H_{2} O)}_{6}]}^{2 +}

4

{[Cr {(H_{2} O)}_{6}]}^{2 +}

Explanation:

More no. of unpaired electrons, higher is the paramagnetism
supporting img
Least paramagnetic behaviour is exhibited by $[Co {(H_{2} O_{6}]}^{2 +}$ as it has least no. of unpaired electrons. So, the correct option is (3).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322323 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ is square planar and diamagnetic.
Reason :
It has no unpaired electrons due to presence of strong field ligand.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$In {[Ni (CN)_{4}]}^{- 2}$ , oxidation state of $Ni$ is $+ 2$
$Ni : 3 d^{8} 4 s^{2}$ and ${Ni}^{2 +} : 3 d^{8} 4 s^{0}$
supporting img
The compound is diamagnetic as evident from the absence of unpaired electron. It is a square planar complexes, the hybridisation involved is ${dsp}^{2}$ .Therefore, option (1) is correct.