Question

An ion M^+a has the magnetic moment equal to √24 BM. The value of a is( At.No. of M=24). I want fully solved solution of this question​

Answer 1

Answer:

4 unpaired, +2charge

Explanation:

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Answer 2

Answer:

The value of "a" is 2.

Explanation:

First, let's see how the magnetic moment is calculated.

$\mathrm{M=\sqrt{n(n+2)} }$           (1)

Where,

M=magnetic moment of the given ion or element

n=number of unpaired electron

From the question, we have that the magnetic moment is √24 BM.

By inserting the value of 'M' in equation (1) we get;

$\mathrm{\sqrt{24} =\sqrt{n(n+2)} }$

$\mathrm{24 =n(n+2) }$

$\mathrm{24 =n^2+2n }$

$\mathrm{n^2+2n-24=0 }$

$\mathrm{n^2+6n-4n-24=0 }$

$\mathrm{n(n+6)-4(n+6)=0 }$

$\mathrm{(n+6)(n-4)=0 }$

$\mathrm{n=-6,4}$           (we will take the positive value of "n" as cannot be negative)

That means there are 4 unpaired electrons.

The electronic configuration of the element having atomic number 24 is

1s²2s²2p⁶3s²3p⁶4s¹3d⁵.             (2)

Since it has 4 unpaired electrons then the electronic configuration will become, 1s²2s²2p⁶3s²3p⁶4s⁰3d⁴.        (3)

By comparing equations (2) and (3) we can see that when the element loses 2 electrons then it has 4 unpaired electrons. So, the value of "a" is 2.

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