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A compound of vanadium has a magnetic mo...

A compound of vanadium has a magnetic moment of 1.73 BM. The electronic configuration of vanadium ion in the compound is:

A

`[Ar]3d^(2)`

B

`[Ar]3d^(1)4s^(0)`

C

`[Ar]3d^(3)`

D

`[Ar]3d^(0)4s^(1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Magnetic moment`=sqrt(n(n+1))BM`
`1.73=sqrt(n(n+2))`
`sqrt(3)=sqrt(n(n+2))`
n=1 (Number of unparied electrons)
`V_(22)to3d^(2)4s^(2)`
`V^(3+)to3d^(1)4s^(0)` (has one unpaired electron)
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Knowledge Check

  • Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes places only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. A compound of vanadium has a magnetic moment of 1.73 BM . Electronic configuration of the vanadium ion in the compound is :

    A
    `[Ar]4s^(0)3d^(1)`
    B
    `[Ar]4s^(2)3d^(3)`
    C
    `[Ar]4s^(1)3d^(0)`
    D
    `[Ar]4s^(0)3d^(5)`

  • Electronic configuration of vanadium is

    A
    `3d^(4)4s^(1)`
    B
    `3d^(3)4s^(2)`
    C
    `3d^(4)4s^(2)`
    D
    `3d^(2)4s^(2)`

  • A compound of vanadium has a magnetic moment of 1.73 BM . What will be the electronic configurations:

    A
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(1)`
    B
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(2)`
    C
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
    D
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)`

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