The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is ........... .

Q: Why is the electronic configuration of Chromium $3d^5 4s^1$ and not $3d^4 4s^2$?

Chromium is an exception to the standard Aufbau filling order. A half-filled $d$-subshell ($3d^5$) is significantly more stable due to higher exchange energy and symmetrical charge distribution. Therefore, one electron from the $4s$ orbital gets excited to the $3d$ orbital.

Q: Why do we remove electrons from the $4s$ orbital before the $3d$ orbital when forming the $Cr^{3+}$ ion?

Even though the $4s$ orbital is filled before the $3d$ orbital, 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 (highest $n$) first.

Q: Does it matter if oxalate is a strong field or weak field ligand in this specific question?

No, it doesn't matter at all! For any $d^1$, $d^2$, or $d^3$ octahedral complex, the electrons will simply fill the lower energy $t_{2g}$ orbitals singly. The pairing energy and ligand strength only become relevant when you have 4 to 7 $d$-electrons.

Q: What is the hybridisation of the Chromium atom in this complex?

Since the $e_g$ orbitals (which are the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals) are completely empty, they are available for bonding. The complex will use these two inner $3d$ orbitals, along with one $4s$ and three $4p$ orbitals, resulting in $d^2sp^3$ hybridisation (an inner orbital complex).

Q: What is the coordination number of Chromium in this complex?

The coordination number is 6. Even though there are only 3 oxalate ligands, each oxalate is a bidentate ligand, meaning it forms two coordinate bonds with the central metal atom. $3 \times 2 = 6$.

Question

The total number of unpaired electrons present in the complex $K_3[Cr(oxalate)_3]$ is ........... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Let's embark on a journey to find the number of unpaired electrons in the coordination complex $K_3[Cr(C_2O_4)_3]$. To understand the magnetic properties and the electronic structure of this complex, our very first step is to determine the oxidation state of the central metal atom, which is Chromium ($Cr$).

We know that the overall charge on this neutral complex is zero. Potassium ($K$) is an alkali metal, so each potassium ion carries a $+1$ charge. The ligand here is oxalate ($C_2O_4^{2-}$), which is a bidentate ligand carrying a $-2$ charge.

Let's set up a simple algebraic equation to find the oxidation state of Chromium, let's call it $x$:

$$3(+1) + x + 3(-2) = 0$$

### The Master Equation
Now, let's solve this equation.

$$3 + x - 6 = 0$$

$$x - 3 = 0$$

$$x = +3$$

This tells us that Chromium is present in the $+3$ oxidation state in this complex.

Next, we need to look at the electronic configuration of a neutral Chromium atom. Chromium is a classic exception to the Aufbau principle because a half-filled $d$-subshell provides extra exchange energy and stability. Its ground state configuration is:

$$Cr \longrightarrow [Ar] 3d^5 4s^1$$

Since our Chromium is in the $+3$ state, it has lost three electrons. Electrons are always removed from the outermost shell first. So, we remove one electron from the $4s$ orbital and two electrons from the $3d$ orbitals. This leaves us with:

$$Cr^{3+} \longrightarrow [Ar] 3d^3 4s^0$$

### Crystal Field Splitting and Final Calculation
In an octahedral complex, the approach of the six ligand donor atoms creates an electrostatic field that breaks the degeneracy of the five $3d$ orbitals. They split into two distinct energy levels: a lower energy set of three orbitals called $t_{2g}$, and a higher energy set of two orbitals called $e_g$.

We have exactly three electrons in the $3d$ subshell of $Cr^{3+}$. According to Hund's Rule of Maximum Multiplicity, electrons will fill degenerate orbitals singly before any pairing occurs.

Because there are only three electrons, they will comfortably occupy the three $t_{2g}$ orbitals one by one.

**Notice a beautiful fact here:** For a $d^3$ configuration in an octahedral field, the strength of the ligand (whether it is a strong field or weak field ligand) does not matter at all! The first three electrons will always go into the $t_{2g}$ level unpaired.

Therefore, the electronic configuration in the crystal field is $t_{2g}^3 \ e_g^0$.

Counting them up, we can clearly see that there are exactly **3 unpaired electrons** present in the complex.

Animated Solution for Chemistry - Coordination Compounds: The total number of unpaired electrons present in the complex $K_{3} [C r (o x a l a t e)_{3}]$ is ........... .

Enter Numerical Value:

Visualized Solution

$K_{3} [C r (C_{2} O_{4})_{3}]$

$Oxidation State Setup$

$Solving for x$

$Electronic Configuration of C r$

$Electronic Configuration of C r^{3 +}$

$Crystal Field Splitting$

$Filling Electrons$

$Final Answer$

$The Way Forward$

The Sigma Insight: Bonding and Crystal field

Analyzing the Setup Let's embark on a journey to find the number of unpaired electrons in the coordination complex $K_{3} [C r (C_{2} O_{4})_{3}]$

The Master Equation

Crystal Field Splitting and Final Calculation In an octahedral complex, the approach of the six ligand donor atoms creates an electrostatic field that breaks the degeneracy of the five $3 d$ orbitals

Similar Questions

The total number of unpaired electrons present in $[C o (N H_{3})_{6}] C l_{2}$ and $[C o (N H_{3})_{6}] C l_{3}$ is ....... .

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Which of the following facts about the complex $[C r (N H_{3})_{6}] C l_{3}$ is wrong?

Two complexes $[C r (H_{2} O)_{6}] C l_{3}$ (A) and $[C r (N H_{3})_{6}] C l_{3}$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is

The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to $+ 3$ state is

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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 total number of unpaired electrons present in the complex K3​[Cr(oxalate)3​] is ........... .

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K3​[Cr(C2​O4​)3​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation State Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electronic Configuration of Cr

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electronic Configuration of Cr3+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Crystal Field Splitting

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Filling Electrons

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward