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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Assertion $A$: Alcohols react both as nucleophiles and electrophiles.
Reason $R$: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen.
In the light of the above statements, choose the correct answer from the options given below:
$F{e^{3 + }}(aq) + {e^ - } \to F{e^{ - 1}}(aq);\,{E^o} = + 0.77\,V$
$A{l^{3 + }}(aq) + 3{e^ - } \to Al(s);\,{E^o} = - 1.66\,V$
$B{r_2}(aq) + 2{e^ - } \to 2B{r^ - }(aq);\,{E^o} = + 1.08\,V$
Based on the data given above, reducing power of $F{e^{2 + }},\,Al$ and $B{r^ - }$ will increase in the order