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

Chemistry
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\[Magnetic \space moment = \sqrt{n(n+2)}\] where n-->number of unpaired electron :)
n=3
its given that u=1.73

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Other answers:

i m confuse
@Ruchi. \[1.73 = \sqrt{n(n+2)}\] Therefore n = 1 Therefore number of unpaired electron = 1 Now try to finish it off :)
its answer=[Ar]3d1
hw u gt n=1
@Ruchi. , substitute n=1 and see in the formula :P
wait
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.
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 :)
@shivam_bhalla i m nt getting its electrnic configration
1s2 2s2 2p6 3s2 3p6 4s2 3d3.
@Ruchi. , This question is more intersesting now :P. Let me have a check and let you know the electron configuration.
@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 ?
p
Typo ; I meant 3s orbital or 3d orbital ??
V+4=1S2 2s2 2p6 3s2 3p6 3d1.
am i right.
what r u telling i m nt understanding.
Ruchi, The configuration you gave for V4+ is wrong. Think why ?? (tip:check the number of electrons is 4s orbital )
You should always mention 4s 0 while writing the V4+ configuration. So you are partially right @Ruchi.
@Ruchi. V+4=1S2 2s2 2p6 3s2 3p6 4s0 3d1
0 ka mention torina nakariga.
kariga
Revise aufbau principle @ruchi. Remove 2 electrons from 4s orbital, then 2 electrons from 3d orbital to get V4+

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