Question

A compound of vanadium has a magnetic moment of 1.73 BM.WORK out the electronic configuration of the vanadium ion in the compound.

Answer 1

\[Magnetic \space moment = \sqrt{n(n+2)}\]
where n-->number of unpaired electron :)

Answer 2

n=3

Answer 3

its given that u=1.73

Answer 4

i m confuse

Answer 5

@Ruchi.

\[1.73 = \sqrt{n(n+2)}\]
Therefore n = 1

Therefore number of unpaired electron = 1

Now try to finish it off :)

Answer 6

its answer=[Ar]3d1

Answer 7

hw u gt n=1

Answer 8

@Ruchi. , substitute n=1 and see in the formula :P

Answer 9

wait

Answer 10

1.73 = sqrt(3) = sqrt(n(n+2))

So, n(n+2) = 3

or n = 1

For V, Z = 23, [Ar]4s23d3
am i right.

Answer 11

Yes. @Ruchi, now tell me if Vanadium has 3 unpaired electrons, and you know that the vanadium ion is having 1 unpaired electron, then now can you figure out the charge on vanadium ion :)

Answer 12

@shivam_bhalla i m nt getting its electrnic configration

Answer 13

1s2 2s2 2p6 3s2 3p6 4s2 3d3.

Answer 14

@Ruchi. , This question is more intersesting now :P. Let me have a check and let you know the electron configuration.

Answer 15

@Ruchi. Your task is to get a vanadium ion with 1 unpaired electron . Now a simple question to you , first where will you remove the electron from vanadium first ?? s orbital or p orbital ?

Answer 16

p

Answer 17

Typo ; I meant 3s orbital or 3d orbital ??

Answer 18

V+4=1S2 2s2 2p6 3s2 3p6 3d1.

Answer 19

am i right.

Answer 20

what r u telling i m nt understanding.

Answer 21

Ruchi, The configuration you gave for V4+ is wrong. Think why ?? (tip:check the number of electrons is 4s orbital )

Answer 22

You should always mention 4s 0 while writing the V4+ configuration. So you are partially right @Ruchi.

Answer 23

@Ruchi.
V+4=1S2 2s2 2p6 3s2 3p6 4s0 3d1

Answer 24

0 ka mention torina nakariga.

Answer 25

kariga

Answer 26

Revise aufbau principle @ruchi. Remove 2 electrons from 4s orbital, then 2 electrons from 3d orbital to get V4+