MCQ
Among the following, the compound that is both paramagnetic and coloured is
- A$K_2Cr_2O_7$
- B$(NH_4)_2[TiCl_6]$
- ✓$VOSO_4$
- D$K_3[Cu(CN)_4]$
$Cr ^{-6}=[ Ar ] 3 d ^0$
In $\left( NH _4\right)_2\left[ TiCl _6\right]$, the oxidation state of $Ti$ is $+4$.
$Ti ^{+4}=[ Ar ] 3 d ^0$
In $VOSO _4$, the oxidation state of $V$ is $+4$.
$V ^{+4}=[ Ar ] 3 d ^1$
In $K _3\left[ Cu ( CN )_4\right]$, the oxidation state of $Cu$ is $+1$.
$Cu ^{+1}=[ Ar ] 3 d ^{10}$
In $VOSO _4, V ^{+4}$ configuration is $d ^1$, so it has one unpaired d-electron so it is paramagnetic and colored. All other compounds don't have any unpaired $d$ electrons, so they are diamagnetic in nature.
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$L(1s^2, 2s^2 2p^4); Q(1s^2, 2s^2 2p^6, 3s^2 3p^5)$
$P(1s^2, 2s^2 2p^6, 3s^1); R(1s^2, 2s^2 2p^6, 3s^2)$
The formulae of ionic compounds that can be formed between these elements are