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JEE MainChemistry

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Coordination Compounds: JEE Main Chemistry Question with Solution

Match List I with List II :-

\begin{array}{|l|l|l|l|} \hline & \text{List I} & & \text{List II} \\ & \text{Coordination Complex} & & \text{Number of unpaired electrons} \\ \hline \text{A.} & [\text{Cr}(\text{CN})_6]^{3-} & \text{I.} & 0 \\ \hline \text{B.} & [\text{Fe}(\text{H}_2\text{O})_6]^{2+} & \text{II.} & 3 \\ \hline \text{C.} & [\text{Co}(\text{NH}_3)_6]^{3+} & \text{III.} & 2 \\ \hline \text{D.} & [\text{Ni}(\text{NH}_3)_6]^{2+} & \text{IV.} & 4 \\ \hline \end{array}

Choose the correct answer from the options given below :-

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A

\text{A}-\text{II},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{III}

B

\text{A}-\text{IV},\ \text{B}-\text{III},\ \text{C}-\text{II},\ \text{D}-\text{I}

C

\text{A}-\text{III},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{II}

D

\text{A}-\text{II},\ \text{B}-\text{I},\ \text{C}-\text{IV},\ \text{D}-\text{III}

Step-by-step solutionView

Correct answer

By determining the central metal ion's oxidation state, d-electron count, and the ligand field strength (weak vs. strong field), we find $\text{Cr}^{3+}$ ( $d^3$ ) has 3, $\text{Fe}^{2+}$ ( $d^6$ , weak field) has 4, $\text{Co}^{3+}$ ( $d^6$ , strong field) has 0, and $\text{Ni}^{2+}$ ( $d^8$ ) has 2 unpaired electrons, giving $\text{A}-\text{II},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{III}$ .

Option analysis

Why each option works or fails

A ·

\text{A}-\text{II},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{III}

None. The configuration and crystal field splitting for each metal complex are evaluated correctly. Correctly determined oxidation states, electronic configurations, and ligand splitting behavior for each complex.

B ·

\text{A}-\text{IV},\ \text{B}-\text{III},\ \text{C}-\text{II},\ \text{D}-\text{I}

Treating the number of valence electrons from neutral atoms or misinterpreting d-counts without accounting for the formal oxidation states. First determine the oxidation state of the transition metal, remove electrons starting from the $4s$ orbital, and then populate the split $d$ -orbitals according to ligand field strength.

C ·

\text{A}-\text{III},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{II}

Assigning 2 unpaired electrons to $[ ext{Cr}( ext{CN})_6]^{3-}$ by mistakenly pairing its 3 electrons in the $t_{2g}$ set under strong-field assumption. Remember that for $d^1$ , $d^2$ , and $d^3$ systems, Hund's rule applies regardless of ligand strength because the three $t_{2g}$ orbitals each hold one unpaired electron before pairing can occur.

D ·

\text{A}-\text{II},\ \text{B}-\text{I},\ \text{C}-\text{IV},\ \text{D}-\text{III}

Assuming that ligands behave identically across different metal ions, or reversing the strong/weak field pairing behavior for $\text{Fe}^{2+}$ and $\text{Co}^{3+}$ . Identify $\text{H}_2\text{O}$ as a weak-field ligand causing high-spin configuration ( $t_{2g}^4 e_g^2$ , 4 unpaired electrons) for $\text{Fe}^{2+}$ , and $\text{NH}_3$ as a strong-field ligand causing low-spin configuration ( $t_{2g}^6 e_g^0$ , 0 unpaired electrons) for $\text{Co}^{3+}$ .

Reviewed route

Solution

StepWorking

01concept

Determine the oxidation state of the central metal ion, write its $3d^n$ configuration, and apply Crystal Field Theory considering whether the ligand is strong-field (pairing occurs) or weak-field (high spin).

02option_verdict

A: $\text{Cr}^{3+} (3d^3)$ has $t_{2g}^3 e_g^0 \implies 3$ unpaired electrons (II). B: $\text{Fe}^{2+} (3d^6)$ with weak field $\text{H}_2\text{O}$ has $t_{2g}^4 e_g^2 \implies 4$ unpaired electrons (IV). C: $\text{Co}^{3+} (3d^6)$ with strong field $\text{NH}_3$ has $t_{2g}^6 e_g^0 \implies 0$ unpaired electrons (I). D: $\text{Ni}^{2+} (3d^8)$ has $t_{2g}^6 e_g^2 \implies 2$ unpaired electrons (III). Matching gives A-II, B-IV, C-I, D-III.

03option_verdict

Incorrectly matches A to IV and D to I, ignoring proper d-electron counting and CFT splitting.

04option_verdict

Matches A to III and D to II, miscounting $\text{Cr}^{3+}$ as $d^2$ instead of $d^3$ .

05option_verdict

Swaps the electron configurations of $\text{Fe}^{2+}$ (weak field) and $\text{Co}^{3+}$ (strong field), assuming both pair up or both stay high-spin.

✓discriminator

$\text{Cr}^{3+}$ is always $d^3$ (3 unpaired electrons $\implies \text{A-II}$ ) and $\text{Co}^{3+}$ with $\text{NH}_3$ is low-spin diamagnetic (0 unpaired electrons $\implies \text{C-I}$ ), which uniquely identifies option (0).

✓ Source and academic review

Question type: Match the following
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does

[ ext{Cr}( ext{CN})_6]^{3-}

have 3 unpaired electrons even though

ext{CN}^-

is a strong field ligand?

Because $ext{Cr}^{3+}$ is a $d^3$ ion, and all 3 electrons singly occupy the lower-energy $t_{2g}$ orbitals ( $t_{2g}^3$ ) regardless of ligand field strength.

Answer

By determining the central metal ion's oxidation state, d-electron count, and the ligand field strength (weak vs. strong field), we find $\text{Cr}^{3+}$ ( $d^3$ ) has 3, $\text{Fe}^{2+}$ ( $d^6$ , weak field) has 4, $\text{Co}^{3+}$ ( $d^6$ , strong field) has 0, and $\text{Ni}^{2+}$ ( $d^8$ ) has 2 unpaired electrons, giving $\text{A}-\text{II},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{III}$ .

Why each option works or fails

A · correct: $\text{A}-\text{II},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{III}$ - None. The configuration and crystal field splitting for each metal complex are evaluated correctly. Correctly determined oxidation states, electronic configurations, and ligand splitting behavior for each complex.
B: $\text{A}-\text{IV},\ \text{B}-\text{III},\ \text{C}-\text{II},\ \text{D}-\text{I}$ - Treating the number of valence electrons from neutral atoms or misinterpreting d-counts without accounting for the formal oxidation states. First determine the oxidation state of the transition metal, remove electrons starting from the $4s$ orbital, and then populate the split $d$ -orbitals according to ligand field strength.
C: $\text{A}-\text{III},\ \text{B}-\text{IV},\ \text{C}-\text{I},\ \text{D}-\text{II}$ - Assigning 2 unpaired electrons to $[ ext{Cr}( ext{CN})_6]^{3-}$ by mistakenly pairing its 3 electrons in the $t_{2g}$ set under strong-field assumption. Remember that for $d^1$ , $d^2$ , and $d^3$ systems, Hund's rule applies regardless of ligand strength because the three $t_{2g}$ orbitals each hold one unpaired electron before pairing can occur.
D: $\text{A}-\text{II},\ \text{B}-\text{I},\ \text{C}-\text{IV},\ \text{D}-\text{III}$ - Assuming that ligands behave identically across different metal ions, or reversing the strong/weak field pairing behavior for $\text{Fe}^{2+}$ and $\text{Co}^{3+}$ . Identify $\text{H}_2\text{O}$ as a weak-field ligand causing high-spin configuration ( $t_{2g}^4 e_g^2$ , 4 unpaired electrons) for $\text{Fe}^{2+}$ , and $\text{NH}_3$ as a strong-field ligand causing low-spin configuration ( $t_{2g}^6 e_g^0$ , 0 unpaired electrons) for $\text{Co}^{3+}$ .

Step-by-step solution

concept: Determine the oxidation state of the central metal ion, write its $3d^n$ configuration, and apply Crystal Field Theory considering whether the ligand is strong-field (pairing occurs) or weak-field (high spin).
option_verdict: A: $\text{Cr}^{3+} (3d^3)$ has $t_{2g}^3 e_g^0 \implies 3$ unpaired electrons (II). B: $\text{Fe}^{2+} (3d^6)$ with weak field $\text{H}_2\text{O}$ has $t_{2g}^4 e_g^2 \implies 4$ unpaired electrons (IV). C: $\text{Co}^{3+} (3d^6)$ with strong field $\text{NH}_3$ has $t_{2g}^6 e_g^0 \implies 0$ unpaired electrons (I). D: $\text{Ni}^{2+} (3d^8)$ has $t_{2g}^6 e_g^2 \implies 2$ unpaired electrons (III). Matching gives A-II, B-IV, C-I, D-III.
option_verdict: Incorrectly matches A to IV and D to I, ignoring proper d-electron counting and CFT splitting.
option_verdict: Matches A to III and D to II, miscounting $\text{Cr}^{3+}$ as $d^2$ instead of $d^3$ .
option_verdict: Swaps the electron configurations of $\text{Fe}^{2+}$ (weak field) and $\text{Co}^{3+}$ (strong field), assuming both pair up or both stay high-spin.
discriminator: $\text{Cr}^{3+}$ is always $d^3$ (3 unpaired electrons $\implies \text{A-II}$ ) and $\text{Co}^{3+}$ with $\text{NH}_3$ is low-spin diamagnetic (0 unpaired electrons $\implies \text{C-I}$ ), which uniquely identifies option (0).