Question

For what min. value of x the compound VClx is having the value 2.89 BM

Answer 1

Answer:

The ‘value of x’ is 4.

Solution:

The number of unpaired electrons can be found from the magnetic moment using the below relation,

\bold{\text {Magnetic moment}=\sqrt{n(n+2)}}Magnetic moment=n(n+2)

Here, n is the number of unpaired electrons in an element.

So, 1.73=\sqrt{n(n+2)}\quad \rightarrow(1)1.73=n(n+2)→(1)

Squaring on both sides of equation (1), we get

1.73 \times 1.73=n^{2}+2 n \quad \rightarrow(2)1.73×1.73=n2+2n→(2)

(2) \Rightarrow n^{2}+2 n-3=0 \quad \rightarrow(3)(2)⇒n2+2n−3=0→(3)

The quadratic equation (3) is solved as follows

\Rightarrow n^{2}+3 n-n-3=0⇒n2+3n−n−3=0

\Rightarrow n(n+3)-(n+3)=0⇒n(n+3)−(n+3)=0

\Rightarrow(n+3)(n-1)=0⇒(n+3)(n−1)=0

So the positive value of n will be 1.

Thus, it is found that the number of unpaired electrons in the vanadium with magnetic moment of 1.73 BM should be 1.

The electronic configuration of Vanadium shows that the last shells has configuration of \bold{3 \mathrm{d}^{3} 4 \mathrm{s}^{2}}3d34s2 .

So in order to have one unpaired electrons in this element, four electrons should be given away that means the oxidation state of Vanadium will be in +4 state.