How to calculate magnetic moment?

Using the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

What is the hybridization of Ni(CO)4?

It is $sp^3$ hybridized, resulting in a Tetrahedral shape.

What is the magnetic moment of Mn2+ in [MnBr4]2-?

Mn2+ has 5 unpaired electrons, so $\mu = \sqrt{5(5+2)} \approx 5.9$ BM.

Is CO a strong field ligand?

Yes, CO is a very strong field ligand and causes maximum pairing of electrons.

What shape does dsp2 hybridization form?

It forms a Square Planar geometry.

What shape does sp3 hybridization form?

It forms a Tetrahedral geometry.

Why is [NiCl4]2- paramagnetic?

Cl- is a weak field ligand, so electrons do not pair up, leaving 2 unpaired electrons.

Match the LIST-I (Complex/Species) with LIST-II (Shape & magnetic moment)

Match the LIST-I (Complex/Species) with LIST-II (Shape & magnetic moment) | JEE Main Chemistry

JEERANKERS

ChemistryCoordination

QMatching

Match the LIST-I (Complex/Species) with LIST-II (Shape & magnetic moment) and choose the correct answer:

LIST-I:
A. [Ni(CO)$_4$]
B. [Ni(CN)$_4$]$^{2-}$
C. [NiCl$_4$]$^{2-}$
D. [MnBr$_4$]$^{2-}$

LIST-II:
I. Tetrahedral, 2.8 BM
II. Square planar, 0 BM
III. Tetrahedral, 0 BM
IV. Tetrahedral, 5.9 BM

✅ Correct Answer

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

✓

Short Solution

A. [Ni(CO)$_4$]

Ni is in 0 oxidation state ($3d^8 4s^2$). CO is a strong field ligand. It causes pairing, making it $3d^{10}$. Hybridization is $sp^3$ (Tetrahedral) and Unpaired electrons ($n$) = 0. So, $\mu = 0$ BM. Matches with III.

B. [Ni(CN)$_4$]$^{2-}$

Ni is in +2 state ($3d^8$). CN$^-$ is a strong field ligand. It pairs the 8 electrons into 4 orbitals, leaving one $d$-orbital empty for $dsp^2$ (Square Planar). $n = 0$, so $\mu = 0$ BM. Matches with II.

C. [NiCl$_4$]$^{2-}$

Ni is in +2 state ($3d^8$). Cl$^-$ is a weak field ligand. No pairing occurs. Two unpaired electrons remain ($n=2$). Hybridization is $sp^3$ (Tetrahedral). $\mu = \sqrt{2(2+2)} \approx 2.8$ BM. Matches with I.

D. [MnBr$_4$]$^{2-}$

Mn is in +2 state ($3d^5$). Br$^-$ is a weak field ligand. All 5 electrons remain unpaired ($n=5$). Hybridization is $sp^3$ (Tetrahedral). $\mu = \sqrt{5(5+2)} \approx 5.9$ BM. Matches with IV.

Final Combination

A $\rightarrow$ III, B $\rightarrow$ II, C $\rightarrow$ I, D $\rightarrow$ IV.

Final Answer: Option C

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Theory

1. Valence Bond Theory (VBT)

Valence Bond Theory explains the bonding in coordination compounds through the hybridization of atomic orbitals. Metal ions provide empty orbitals to accept electron pairs from ligands. If the inner $d$-orbitals are used, it’s called an inner orbital complex (often square planar or octahedral); if outer orbitals are used, it’s an outer orbital complex (often tetrahedral). The choice depends heavily on the “strength” of the ligand. Strong field ligands like CO and CN$^-$ force electrons to pair up in order to vacate inner orbitals, whereas weak field ligands like Cl$^-$ and Br$^-$ do not affect the electron distribution of the metal ion significantly.

2. Spectrochemical Series

Ligands are arranged in a series based on their ability to split the $d$-orbitals, known as the spectrochemical series. Common order: I$^-$ < Br$^-$ < Cl$^-$ < F$^-$ < OH$^-$ < H$_2$O < NH$_3$ < en < CN$^-$ < CO. Ligands on the left (like Cl$^-$ and Br$^-$) are weak field ligands that usually lead to high-spin complexes with many unpaired electrons. Ligands on the right (like CN$^-$ and CO) are strong field ligands that lead to low-spin complexes. This series is the primary tool for predicting whether a complex will be paramagnetic (contains unpaired electrons) or diamagnetic (all electrons paired).

3. Magnetic Moment Calculation

The magnetic properties of a complex are determined by the number of unpaired electrons ($n$) in the metal’s $d$-subshell. The “spin-only” magnetic moment ($\mu$) is calculated using the formula $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM). For example, if $n=1$, $\mu \approx 1.73$ BM; if $n=2$, $\mu \approx 2.83$ BM; and if $n=5$ (like in Mn$^{2+}$ weak field), $\mu \approx 5.92$ BM. Measuring this value experimentally helps chemists verify the hybridization and electronic configuration of the central metal atom in a coordination compound.

4. Geometry and Hybridization

The coordination number (number of ligand attachments) determines the available geometries. For coordination number 4, two hybridizations are common: $sp^3$ and $dsp^2$. An $sp^3$ hybridization leads to a Tetrahedral geometry, which is typical for weak field ligands or $d^{10}$ systems. A $dsp^2$ hybridization results in a Square Planar geometry, which is very common for $d^8$ metal ions (like Ni$^{2+}$, Pt$^{2+}$) when paired with strong field ligands. Understanding this relationship is critical for solving matching-type questions in JEE Main, as geometry and magnetic behavior are intrinsically linked.

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FAQs

Why is [Ni(CN)4]2- square planar?

Because CN- is a strong field ligand, it forces the 3d8 electrons to pair up, vacating one 3d orbital. This allows for dsp2 hybridization, which is square planar.

What does 0 BM indicate?

It indicates that all electrons are paired (diamagnetic), resulting in no net magnetic moment.

Is there a shortcut for magnetic moment?

Yes, if there are ‘n’ unpaired electrons, the value is always ‘n.something’ BM. For n=5, it’s 5.9; for n=2, it’s 2.8.

Why is Ni(CO)4 tetrahedral and not square planar?

In Ni(0), the configuration becomes 3d10 after pairing. Since no 3d orbitals are empty, it must use 4s and 4p orbitals, leading to sp3 (Tetrahedral).

What is the color of [NiCl4]2-?

It is typically bright green, whereas [Ni(CN)4]2- is yellow/colorless.

Can dsp2 be paramagnetic?

Usually, dsp2 complexes are diamagnetic because the pairing of electrons is what creates the empty d-orbital needed.

Is Br- always a weak field ligand?

Yes, in almost all standard coordination chemistry problems, halogens are considered weak field ligands.

What is ‘BM’ unit?

BM stands for Bohr Magneton, the physical constant used to express the magnetic moment of an electron.

Why does Mn2+ have 5 unpaired electrons?

Mn is [Ar] 3d5 4s2. Mn2+ is 3d5. In a weak field, all 5 electrons occupy separate d-orbitals according to Hund’s rule.

How do I distinguish between sp3 and dsp2 quickly?

Check the ligand. Strong field + d8 often = dsp2 (Square Planar). Weak field or d10 = sp3 (Tetrahedral).

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