Paramagnetism is a property due to the presence of unpaired electrons. In case of transition metals, as they contain unpaired contain unpaired electrons in the `(n-1)` d orbitals , most of the transition metal ions and their compounds are paramagnetic . Paramagnetism increases with increases in number of unpaired electrons. Magnetic moment is calculated from 'spin only formula' `Vz`
`mu=sqrt(n(n+2)) B.M n="number of unpaired electrons"`
Similarly the colour of the compounds of transition metals may be attributed to the presence of incomplete `(n-1)` d sub-shell. When an electron from a lower energy of d-orbitals is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to complementry colour of the light obserbed. The frequency of the light absorbed is determined by the nature of the ligand.
The colourless species is:

Options

A. `VCl_(3)`
B. `VOSO_(4)`
C. `Na_(3)VO_(4)`
D. `[V(H_(2)O_(6))SO_(4)]`

Answer

C

Explanation

(A) in `VCl_(3)` , the vanadium (III) has valence shell electron configuration `3d^(1)` . SO its aqueous solution will be coloured due d-d transition of electron.
(B) in `VOSO_(4)` , the vanadium (IV) has valence shell electron configuration `3d^(1)` . SO its aqueous solution will be coloured due d-d transition of electron.
(C)In `Na_(3)VO_(4)`, the vanadium (V) has valence shell electron configuration `3d^(3)`. So its aqueous solution will be colourless as there will be no. d-d transition.
In `[V(H_(2)O)_(5)SO_(4)]2H_(2)O,` The vanadium (II) ha valence shell electron configuration `3d^(3)` . SO its aqueous solution will be coloured due to d-d transition of electron.