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The magnetic moment of .25Mn in ionic st...

The magnetic moment of `._25Mn` in ionic state is `sqrt(15)B.M`, then Mn is in:

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
4
`25^(Mn)-[Ar]3d^(5)4s^(2)`

Given `sqrt(n(n+2))=sqrt(15)rArrn=3`
Hence to have `'3'` unpaired electrons `Mn` must be in `'+4'` state
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The magnetic moment of M^(+) (atomic number = 25 is sqt(15) BM .The number of unpaired electron and the value of x respectivekly are

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Knowledge Check

  • The magnetic moment of ._(25)Mn in ionic state is sqrt(15) B.M, then Mn is in:

    A
    `+2` state
    B
    `+3` state
    C
    `+4` state
    D
    `+5` state

  • The magneitc moment of ._(25)Mn in ionic state is 4.83 B.M., then Mn is in :

    A
    `+2` state
    B
    `+3` state
    C
    `+4` state
    D
    `+5` state

  • The magneitc moment of ._(25)Mn in ionic state is 4.83 B.M., then Mn is in :

    A
    `+2` state
    B
    `+3` state
    C
    `+4` state
    D
    `+5` state

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    Choose the most appropriate options. The magnetic moment of M^(x +) (atomic number = 25) is sqrt 15 BM. Then, the oxidation number x of M is :

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