The type of hybridisation and magnetic property of the complex [MnCl6]3-, respectively, are
(a) sp3d

Q: Why does Manganese lose electrons from the 4s orbital before the 3d orbital?

Even though the 4s orbital is filled before the 3d orbital (according to the Aufbau principle), once the orbitals are populated, the 4s orbital is actually higher in energy and further from the nucleus than the 3d orbital. Therefore, when ionizing, electrons are always removed from the outermost principal quantum shell first.

Q: How do I know if a ligand is strong or weak?

You need to memorize the basic order of the spectrochemical series. Halides (like $Cl^-$, $F^-$) and oxygen donors (like $H_2O$) are generally weak field ligands. Nitrogen donors (like $NH_3$) and carbon donors (like $CN^-$, $CO$) are strong field ligands.

Q: What would happen if we had a strong field ligand instead of chloride?

If a strong field ligand like $CN^-$ was present, the crystal field splitting energy would be large ($\Delta_o > P$). This would force the four $3d$ electrons to pair up into just two orbitals, leaving two $3d$ orbitals completely empty. The metal would then use these inner orbitals to form a $d^2sp^3$ (inner orbital) complex.

Q: Why is it called an 'outer orbital' complex?

It is called an outer orbital complex because the metal uses the $d$-orbitals from the outermost principal quantum shell (the $4d$ orbitals) for hybridization, rather than the inner $3d$ orbitals.

Q: How does the number of unpaired electrons relate to paramagnetism?

Every electron acts like a tiny magnet due to its spin. When electrons are paired, their opposite spins cancel out each other's magnetic fields (diamagnetic). When electrons are unpaired, their magnetic fields add up, causing the entire molecule to be attracted to an external magnetic field (paramagnetic).

Question

The type of hybridisation and magnetic property of the complex $[MnCl_6]^{3-}$, respectively, are

Options:

(A) $sp^3d^2$ and diamagnetic

(B) $d^2sp^3$ and diamagnetic

(C) $d^2sp^3$ and paramagnetic

(D) $sp^3d^2$ and paramagnetic

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to the fascinating world of Coordination Chemistry! Today, we are going to decode the geometry and magnetic behavior of the hexachloridomanganate(III) ion, mathematically written as $[MnCl_6]^{3-}$.

To understand how this complex is built, we must first look at the architect: the central Manganese atom. Our first mission is to determine its oxidation state. Let's assume the oxidation state of Manganese is $x$. We know that each chloride ligand carries a charge of $-1$, and there are six of them. The overall charge of the complex sphere is $-3$.

Setting up our master equation:

$x + 6(-1) = -3$

Solving for $x$, we get $x = +3$. This tells us that Manganese is sitting in a $+3$ oxidation state.

### The Electronic Blueprint

Manganese ($Mn$) has an atomic number of $25$. In its neutral ground state, its electronic configuration is $[Ar] 3d^5 4s^2$. However, our Manganese is $Mn^{3+}$, meaning it has lost three electrons. It will first lose the two outermost electrons from the $4s$ orbital, and then one electron from the $3d$ orbital.

This leaves us with the configuration:

$Mn^{3+}: [Ar] 3d^4$

Now, we have four electrons sitting in the $3d$ subshell. According to Hund's rule, they will occupy four separate orbitals singly.

### The Role of the Ligand

Here comes the crucial part: the ligands approaching the metal. We have six chloride ($Cl^-$) ions. According to the **spectrochemical series**, chloride is a **weak field ligand**.

What does a weak field ligand do? It produces a very small crystal field splitting energy ($\Delta_o < P$). Because the splitting is so small, it is not energetically favorable to force the electrons to pair up against their natural repulsion. Therefore, the four electrons in the $3d$ orbitals remain happily unpaired.

### Hybridisation and Final Calculation

Since the $3d$ orbitals are occupied and cannot be emptied by pairing, the $Mn^{3+}$ ion must look outward to find empty rooms for the six incoming chloride pairs. It utilizes its outer empty orbitals: one $4s$, three $4p$, and two $4d$ orbitals.

Mixing these together gives us **$sp^3d^2$ hybridisation**. Because it uses the outer $4d$ orbitals rather than the inner $3d$ orbitals, this is called an **outer orbital complex**.

Finally, let's check the magnetic property. We established earlier that there are four unpaired electrons ($n = 4$) in the $3d$ subshell. Any species with unpaired electrons is attracted to an external magnetic field, making it **paramagnetic**.

Thus, the complex is $sp^3d^2$ hybridized and paramagnetic!

Animated Solution for Chemistry - Coordination Compounds: The type of hybridisation and magnetic property of the complex $[M n C l_{6}]^{3 -}$ , respectively, are

Select Answer:

Visualized Solution

$Oxidation State of Mn$

$Electronic Configuration$

$Ligand Field Strength$

$Hybridisation$

$Magnetic Property$

$The Way Forward$

The Sigma Insight: Bonding and Crystal field

Analyzing the Setup

The Electronic Blueprint

The Role of the Ligand

Hybridisation and Final Calculation

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Animated Solution for Chemistry - Coordination Compounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The type of hybridisation and magnetic property of the complex [MnCl6​]3−, respectively, are