Question

The magnetic moment of m( z = 25) is √15



b.M. The number of unpaired electrons and the value of x is

Answer 1

Explanation:

[n(n+2)]^1/2 = [15]^1/2

Squaring both sides

n(n+2)=15n(n+2)=15

n2+2n−15=0n2+2n−15=0

Solving we get n=3n=3

Thus there are 3 unpaired electrons in the element.

The configuration of an element with atomic number 25 is [Ar]4s23d5

This configuration has electrons in the valence shell.

We have been given that the symbol of the element is Mx∗ So the value of X becomes 5−3=2

Answer 2

"It seems this is what you are looking for"

The magnetic moment of Mˣ⁺(atomic number = 25) is √15 BM. The number of unpaired electrons and the value of x is:

Answer:

The number of unpaired electrons in Mˣ⁺ is equal to three.

The value of x is equal to 4.

Explanation:

Given: the atomic mass of M $=25$

Therefore the electronic configuration of M will be:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{2$

The number of unpaired electrons $=7$

The magnetic moment is given by:

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

$\mu=\sqrt{7(7+2)}=\sqrt{35} BM$

But the given magnetic moment is $\sqrt{15}BM$. It means the number of unpaired electrons are less than seven.

$\sqrt{n(n+2)}=\sqrt{15}\\ n(n+2)=15$

If we put n=2 in above equation:

$3(3+2)=15\\15=15$

Therefore the number of unpaired electrons in  Mˣ⁺ is equal three.

Then the oxidation state of M $=7-3=+4$

Therefore, M is present as M⁴⁺ and the value of x is 4.