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