1.

The pair having the same magnetic moment is [At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27]

Answer»

`[Cr(H_(2)O)_(6)]^(2+)`and `[CoCl_(4)]^(2-)`
`[Cr(H_(2)O)_(6)]^(2+) and [Fe(H_(2)O)_(6)]^(2+)`
`[Mn(H_(2)O)_(6)]^(2+) and [Cr(H_(2)O)_(6)]^(2+)`
`[CoCl_(4)]^(2-) and [Fe(H_(2)O)_(6)]^(2+)`

Solution :`[Ci(H_(2)O)_(6)]^(2+):Cr^(2+)=[Ar]3D^(4)4s^(0)`

Thus, `[Cr(H_(2)O)_(6)]^(2+)` and `[Fe(H_(2)O)_(6)]^(2+)` have the same number of UNPAIRED electrons and HENCE have same magnetic MOMENT.


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