Calculating the Spin-Only Magnetic Moment of Co3+
Whether you are solving questions in the d-Block Elements chapter or determining the magnetic properties of a complex in Coordination Compounds, calculating the Spin-Only Magnetic Moment (μ) is a mandatory skill for JEE and NEET.
The calculation itself is easy, but students frequently make critical errors when creating the initial ion by removing electrons in the wrong order. Let's solve the Co3+ ion step-by-step using Abhishek Sir's foolproof rules.
Video Tutorial: The LIFO Rule
Watch Abhishek Sengar sir from CHEMCA break down the exact order of electron removal and introduce a mathematical shortcut to save time in MCQs.
Step-by-Step Problem Breakdown
-
Write the Neutral Configuration:
The atomic number of Cobalt (Co) is 27.
Its neutral electronic configuration is: [Ar] 3d7 4s2. -
Remove Electrons to Form the Ion:
We need to form the Co3+ ion, which means we must remove exactly 3 electrons.Rule 1: The Outermost Rule
Even though the 3d subshell was filled after the 4s subshell, electrons are ALWAYS removed from the outermost shell first (highest Principal Quantum Number, n=4).
So, the first 2 electrons are removed from the 4s orbital. We still need to remove 1 more electron. -
Remove from Degenerate Orbitals:
We must remove 1 electron from the 3d7 subshell.Rule 2: LIFO (Last In, First Out)
According to Hund's Rule, electrons singly occupy degenerate orbitals before pairing. When removing electrons, remove the paired electrons first!
In 3d7, there are 5 unpaired (spin up) and 2 paired (spin down) electrons. We remove the 7th electron (one of the paired ones). -
Count Unpaired Electrons (n):
The final configuration for Co3+ is [Ar] 3d6 4s0.
In a d6 arrangement, there is 1 paired orbital and 4 singly occupied orbitals. Therefore, n = 4.
Fig: Removing the outermost 4s electrons first, followed by the last paired electron in 3d.
Calculating the Magnetic Moment (μ)
Now we apply the standard formula, where n is the number of unpaired electrons:
Substitute n = 4 into the formula:
μ = √(4 × 6)
μ = √24 BM
You don't need to manually calculate the exact square root decimal during the exam. The value of μ is always equal to n.something.
If n = 4, the answer will be 4.xx BM.
If n = 2, the answer will be 2.xx BM.
Just find n, look at the options, and tick the one that starts with that exact number!
Practice Questions for JEE & NEET
Test your speed using the orbital rules and the mathematical shortcut with these common ions.
Question 1: Calculate the spin-only magnetic moment for the Nickel(II) ion (Ni2+). The atomic number of Nickel is 28.
Answer: √8 ≈ 2.83 BM
Reasoning:
- Neutral Nickel (28): [Ar] 3d8 4s2.
- To form Ni2+, remove 2 electrons. Per the outermost rule, remove both from 4s.
- Ion configuration: [Ar] 3d8 4s0.
- A d8 arrangement has 3 paired orbitals and 2 unpaired orbitals. So, n = 2.
- μ = √(2(2+2)) = √8. Using the trick: n=2, so the answer is 2.\text{something}. Exact value is 2.83 BM.
Question 2: Which of the following ions is Diamagnetic (has a magnetic moment of 0)?
A) Cr3+
B) Zn2+
C) Mn2+
D) Fe2+
Answer: B) Zn2+
Reasoning:
For an ion to be diamagnetic, it must have ZERO unpaired electrons (n = 0). The atomic number of Zinc is 30. Its neutral configuration is [Ar] 3d10 4s2. When it loses 2 electrons to become Zn2+, it loses the 4s electrons, leaving 3d10. A d10 subshell is completely full, meaning every electron is paired. Thus, n=0, and μ = 0.
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