Question

Calculate the spin only magnetic moment of [CoF6]3- and Calculate the spin only magnetic moment of [CoF6]3- and [Co(NH3)6]3-Co(NH3)6]3-

Answer 1

Answer:

Magnetic moment can be calculated using the formula √{(n)(n+2)} , where n= number of unpaired electrons.

Plugging in the value of n, we get the magnetic moment

M=√{(4)(6)}

M=√24= 4.89BM.

Answer 2

Answer:

[CoF₆]³⁻ and [Co(NH₃)₆]³⁻ have spin-only magnetic moments of 4.8 BM.

Let's talk about the magnetic moment and its formula, and then examine how we got the solution.

Explanation:

What exactly is the magnetic moment?

The magnetic strength and direction of a magnet or other item that generates a magnetic field are defined as magnetic moment.

• The magnetic moment is measured as a vector quantity.
• The items tend to align themselves such that the magnetic moment vector becomes parallel to the magnetic field lines.

Formula:

The formula for calculating the magnetic moment based only on spin is,

μ = √4s (s + 1)

here, s = spin magnetic moment

μ = √n (n + 2)

here, n denotes the number of unpaired electrons

The spin-only magnetic moment is  measured in BM (Bohr Magneton).

In the given question,

the [CoF₆]³⁻ complex has Co in +3 oxidation state.

(Co) = [Ar] 3d⁷ 4s²

(Co⁺³) = [Ar] 3d⁶

No. of Co⁺³ unpaired electrons is 4

So, spin only magnetic moment = √4 (4 + 2) = √24 = 4.8 BM

Similarly, in [Co(NH₃)₆]³⁻ complex, Co is present in +3 oxidation state.

(Co) = [Ar] 3d⁷ 4s²

(Co⁺³) = [Ar] 3d⁶

No. of Co⁺³ unpaired electrons is 4

So, spin only magnetic moment = √4 (4 + 2) = √24 = 4.8 BM

Hence, both the complex will have same spin only magnetic moment, i.e., 4.8 BM.

Hope the concept is clear.