The 'spin only' magnetic moment value of MO42−MO42− is _______ BM. (Where M is a metal having least metallic radii. among Sc,Ti,V,Cr,MnSc,Ti,V,Cr,Mn and ZnZn ).(Given atomic number: Sc=21,Ti=22, V=23,Cr=24,Mn=25Sc=21,Ti=22, V=23,Cr=24,Mn=25 and Zn=30Zn=30)
Question
The 'spin only' magnetic moment value of MO42−MO42− is _______ BM. (Where M is a metal having least metallic radii. among Sc,Ti,V,Cr,MnSc,Ti,V,Cr,Mn and ZnZn ).(Given atomic number: Sc=21,Ti=22, V=23,Cr=24,Mn=25Sc=21,Ti=22, V=23,Cr=24,Mn=25 and Zn=30Zn=30)
Solution
The metal with the least metallic radii among Sc, Ti, V, Cr, Mn, and Zn is Sc (Scandium).
The atomic number of Scandium is 21, which means it has 21 electrons. In the ground state, these electrons fill up to the 3d1 4s2 orbitals.
However, in the MO4^2- ion, Scandium loses its 3 outermost electrons to become Sc^3+. This means it loses the 2 electrons in the 4s orbital and 1 electron in the 3d orbital, leaving it with no unpaired electrons.
The formula for calculating the spin-only magnetic moment is √n(n+2) BM, where n is the number of unpaired electrons.
Since Scandium in MO4^2- has no unpaired electrons, the spin-only magnetic moment is √0(0+2) = 0 BM.
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