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

Question

A compound of vanadium has a magnetic moment 1.73BM. Work out the electronic configuration of vanadium ion in this compound.
I got unpaired electron as one and figured out that there should be 21 electrons but the answer says 19 electrons. Could any one explain why?

Answer 1

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

So, n(n+2) = 3

or n = 1

For V, Z = 23, [Ar]4s23d3

Now in V+ it should be [Ar]4s13d3 or 22 electrons

Answer 2

1.73 = sqrt(n[n+2])
1.73 X 1.73 = n[n+2]
2.99 = n^2 + 2n
n^2 + 2n - 3 = 0
this implies that, n = 1 or n = -3.
since n=-3 is not possible
hence no. of unpaired electrons is 1 [n]
or, total spin [S] = n/2
S = 1/2

electronic configuration of V is [Ar] 4s^2 3d^3
since S=1/2, so only 1 electron will be left in subshell 3d
Electron loss will start with 4s first, taking 2 electrons, then with 3d taking 2 electrons thus leaving only 1 electron in subshell 3d.
Now, 4 electrons have been lost so configuration of V^4+ is [Ar]3d^1