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Match Coordination Complex Shapes and Magnetic Moments

Learn how to match coordination complexes with their shapes and magnetic moments using hybridization and ligand field concepts. This helps solve JEE..

❓ Question

Match the LIST-I with LIST-II

LIST-I (Complex Species)LIST-II (Shape & Magnetic Moment)
A. [Ni(CO)4][Ni(CO)_4]I. Tetrahedral, 2.8 BM2.8\ BM
B. [Ni(CN)4]2[Ni(CN)_4]^{2-}II. Square planar, 0 BM0\ BM
C. [NiCl4]2[NiCl_4]^{2-}III. Tetrahedral, 5.9 BM5.9\ BM
D. [MnBr4]2[MnBr_4]^{2-}IV. Tetrahedral, 0 BM0\ BM

Choose correct match.


đź–Ľ Question Image

Match Coordination Complex Shapes and Magnetic Moments


✍️ Short Explanation

This problem is based on:

👉 Coordination compounds
👉 Hybridization and geometry
👉 Magnetic moment and unpaired electrons.

Main idea:

ÎĽ=n(n+2)\boxed{ \mu=\sqrt{n(n+2)} }

Where:

  • nn = number of unpaired electrons
Match Coordination Complex Shapes and Magnetic Moments

đź”· Step 1 — Analyse [Ni(CO)4][Ni(CO)_4] đź’Ż

Oxidation state of Ni:

00

Electronic configuration:

Ni:3d84s2Ni:3d^84s^2

With CO ligand:

Ni03d10Ni^0\rightarrow3d^{10}

All electrons paired.

Geometry:

Tetrahedral\boxed{\text{Tetrahedral}}

Magnetic moment:

0 BM\boxed{0\ BM}

Thus:

AIVA\rightarrow IV

đź”· Step 2 — Analyse [Ni(CN)4]2[Ni(CN)_4]^{2-}

Oxidation state:

Ni2+Ni^{2+}

Configuration:

3d83d^8

CN^{-} is strong field ligand.

Hence pairing occurs.

Hybridization:

dsp2dsp^2

Geometry:

Square planar\boxed{\text{Square planar}}

All electrons paired.

ÎĽ=0 BM\mu=0\ BM

Thus:

BIIB\rightarrow II

đź”· Step 3 — Analyse [NiCl4]2[NiCl_4]^{2-}

Oxidation state:

Ni2+Ni^{2+}

Configuration:

3d83d^8

Cl^{-} is weak field ligand.

No pairing occurs.

Hybridization:

sp3sp^3

Geometry:

Tetrahedral\boxed{\text{Tetrahedral}}

Number of unpaired electrons:

n=2n=2

Magnetic moment:

ÎĽ=2(2+2)\mu=\sqrt{2(2+2)}
=82.8 BM=\sqrt8 \approx2.8\ BM

Thus:

CIC\rightarrow I

đź”· Step 4 — Analyse [MnBr4]2[MnBr_4]^{2-}

Oxidation state:

Mn2+Mn^{2+}

Configuration:

3d53d^5

Br^{-} is weak field ligand.

No pairing.

Geometry:

Tetrahedral\boxed{\text{Tetrahedral}}

Unpaired electrons:

n=5n=5

Magnetic moment:

ÎĽ=5(5+2)\mu=\sqrt{5(5+2)}
=355.9 BM=\sqrt{35} \approx5.9\ BM

Thus:

DIIID\rightarrow III

đź”· Step 5 — Final Matching

AIV, BII, CI, DIII\boxed{ A-IV,\ B-II,\ C-I,\ D-III }

đź”· Step 6 — JEE Trap Alert 🚨

❌ Strong field aur weak field ligand confuse kar dena

❌ Square planar vs tetrahedral mix kar dena

❌ Magnetic moment formula bhool jaana

Remember:

ÎĽ=n(n+2)\boxed{ \mu=\sqrt{n(n+2)} }

and:

✔ Strong field → pairing

✔ Weak field → no pairing


✅ Final Answer

AIV, BII, CI, DIII\boxed{ A-IV,\ B-II,\ C-I,\ D-III }


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