The magnetic moment of Mx+ (atomic number M=25) is root of 15 Bohr Magneton. The number of unpaired electrons and the value of X respectively are? Also please explain Magnetic Moment?

Dear student
Please find the solution of the asked query:
The magnetic moment of Mx+ (an element with atomic number 25 ) = 15 B.M
We know that magnetic moment is given by n(n+2) B.Mwhere n is the no.of unpaired electron in the element

So ,According to question:
n(n+2)  =15Squaring both sidesn(n+2) =15n2+2n -15 = 0Solving we get  n=3

Thus there are 3 unpaired electrons  in the element.
The configuration of an element with atomic number 25 is 1s22s22p63s23p64s23d5 or [Ar]4s23d5
This configuration has 5 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 


The magnetic moment of any species is equal to the sum of orbital and spin motion contribution of each unpaired electron. In case of transition metal ions, the orbital contribution is suppressed by the electrostatic filed of surrounding atoms, molecules or ions. So, the effective magnetic moment is spin only magnetic moment in case of transition ions. 
Magnetic moment, n(n+2)  B.M
 where, n is the number of unpaired electrons.


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  • 7
magnetic moment = root of ne * (ne + 2) BM where ne = no of umpaired electron.
    after squaring, ne(ne + 2) = 15
   so, ne = 3
now electronic configuration of M = 25 is 1s​22s22p63s23p64s23d5 
this configuration has 5 unpaired electron but we have 3 unpaired electron.
hence x = 5-3 = 2
  • 13
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