Among the following,the compound that is both | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $K_3[Cu(CN)_4]$: $Cu$ is in $+1$ oxidation state $(3d^{10})$,so it has no unpaired electrons,making it diamagnetic and colourless.$(NH_4)_2[TiCl_6

Vedclass · Accepted Answer

$VOSO_4$

Vedclass · Answer

(C) $K_3[Cu(CN)_4]$: $Cu$ is in $+1$ oxidation state $(3d^{10})$,so it has no unpaired electrons,making it diamagnetic and colourless.$(NH_4)_2[TiCl_6]$: $Ti$ is in $+4$ oxidation state $(3d^0)$,so it has no unpaired electrons,making it diamagnetic and colourless.$VOSO_4$: $V$ is in $+4$ oxidation state $(3d^1)$,so it has one unpaired electron,making it paramagnetic and coloured.$K_2Cr_2O_7$: $Cr$ is in $+6$ oxidation state $(3d^0)$,so it has no unpaired electrons. It is diamagnetic,but appears coloured due to ligand-to-metal charge transfer $(LMCT)$.

Among the following,the compound that is both paramagnetic and coloured is

Similar Questions

The metal that does not give the borax bead test is

Write two uses of $KMnO_4$ and $K_2Cr_2O_7$.

Which of the following does not give oxygen on heating?

Assertion : Coloured cations can be identified by borax bead test.
Reason : Transparent bead $(NaBO_2 + B_2O_3)$ forms coloured bead with coloured cation.

Cigarette or gas lighter is made up of

Vedclass Test Series

Exam Paper Generator

Online Exam Module