Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is $25$
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is $25$
With atomic number $25$ , the divalent ion in aqueous solution will have $d^{5}$ configuration (five unpaired electrons). The magnetic moment, $\mu$ is
$\mu=\sqrt{5(5+2)}=5.92 \,BM$
Similar Questions
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula $\mathrm{M}_2 \mathrm{O}_{\mathrm{n}}$ (where $\mathrm{n}=3,4,5$ ). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is. . . . . . ..$\mathrm{BM}$ (near integer)
(Given atomic number: $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23$, $\mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26, \mathrm{Co}: 27, \mathrm{Ni}: 28, \mathrm{Cu}: 29$, $\mathrm{Zn}: 30)$
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula $\mathrm{M}_2 \mathrm{O}_{\mathrm{n}}$ (where $\mathrm{n}=3,4,5$ ). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is. . . . . . ..$\mathrm{BM}$ (near integer)
(Given atomic number: $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23$, $\mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26, \mathrm{Co}: 27, \mathrm{Ni}: 28, \mathrm{Cu}: 29$, $\mathrm{Zn}: 30)$
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