If element 25X+Y has spin magnetic moment 1.732 B.M then

= 1.732Number of unpaired electrons, n = 1.25X : [Ar] 4s23d5For having one unpaired electron, 6 electrons are to be removed (2 from 4s & 4 from 3d).∴ Y = 6.

Atomic StructureHard

Question

If element ₂₅X^+Y has spin magnetic moment 1.732 B.M then

Options

A.number of unpaired electron = 1

B.number of unpaired electron = 2

C.Y = 4

D.Y = 6

Solution

= 1.732
Number of unpaired electrons, n = 1.
₂₅X : [Ar] 4s²3d⁵
For having one unpaired electron, 6 electrons are to be removed (2 from 4s & 4 from 3d).
∴ Y = 6.

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