Question

Calculate Bohr Magneton (BM) of M
ion Z=26.​

Answer 1

Answer:

μs​=n(n+2) ​B. M n-number of unpaired electrons. 1 B.M (Bohr magneton)=2πmceh​if magnetic moment is zero the substance is diamagnetic. If an ion of 25​Mn has a magnetic moment of 3.873 B.M.

Explanation:

Answer 2

Complete question:

Calculate the ‘Spin only’ magnetic moment in Bohr Magnetons of M⁺² (aq) ion (Z = 26).

Answer:

the ‘Spin only’ magnetic moment in Bohr Magnetons of M⁺² (aq) ion (Z = 26) is 3.96 BM.

Explanation:

Given,

The atomic number of M⁺² ion is (Z) = 26.

To find,

The magnetic moment of M⁺² ion (μ) in Bohr magnetons.

Concept,

The magnetic moment of an atom or an ion is given by

$\mu = \sqrt{n(n +2)}BM$

Where n = no. of unpaired electrons present in an atom or an ion.

Calculation,

The electronic configuration of an atom M is [Ar] 3d⁶ 4s².

So, the electronic configuration of the M⁺² ion is [Ar] 3d⁶.

As in the d orbital, there are five subshells and the electrons fill in the subshells according to Hund's rule.

i.e. the paring of the electron is done after each subshell is filled with a single electron. So as there are 5 subshells, 5 electrons first fill in every subshell each then the remaining 1 electron pair in any one of the subshells (for simplicity let it be 1st subshell).

Hence there are 4 unpaired electrons present in the d orbital of the M⁺² ion.

i.e. n = 4

From equation (1):

$\mu = \sqrt{n(n +2)}BM$

$\implies \mu =\sqrt{4(4+2)} BM$

⇒ μ = √24 BM

⇒ μ = 3.96 BM.

Therefore, the ‘Spin only’ magnetic moment in Bohr Magnetons of M⁺² (aq) ion (Z = 26) is 3.96 BM.

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