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$K_3Cu (CN)_4$
✓
Answer
Correct option: C.
$VOSO_4$
c
$(d)$ In $K_3[Cu(CN)_4]$ $Cu$ is in $+1$ oxidation state hence has no unpaired electron hence colourless and diamagnetic.
$(d)$ In $K_3[Cu(CN)_4]$ $Cu$ is in $+1$ oxidation state hence has no unpaired electron hence colourless and diamagnetic.
$(b)$ In ${(N{H_4})_2}[TiC{l_6}]\,Ti$ is in $+4$ oxidation state, hence has no unpaired electron hence colourless and diamagnetic.
$(c)$ In $VOSO_4$, is in $+4$ oxidation state, hence has no unpaired electron ,thus it is coloured and paramagnetic.
$(a)$ In $K_2Cr_2O_7$ $Cr$ is in $+6$ oxidation state, hence has no unpaired electron and thus it is diamagnetic. Though $K_2Cr_2O_7$ has no unpaired electron but it is coloured. this is due to charge transfer.
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$LIST-I$ contains metal species and $LIST-II$ contains their properties.
→| $LIST-I$ | $LIST-II$ |
| $(I)$ $\left[ Cr ( CN )_6\right]^{4-}$ | $(P)$ $t_{2 g }$ orbitals contain 4 electrons |
| $(II)$ $\left[ RuCl _6\right]^{2-}$ | $(Q)$ $\mu$ (spin-only) $=4.9 BM$ |
| $(III)$ $\left[ Cr \left( H _2 O \right)_6\right]^{2+}$ | $(R)$ low spin complex ion |
| $(IV)$ $\left[ Fe \left( H _2 O \right)_6\right]^{2+}$ | $(S)$ metal ion in $4$+ oxidation state |
| $(T)$ $d^4$ species |
[Given : Atomic number of $Cr =24, Ru =44, Fe =26$ ] Metal each metal species in $LIST-I$ with their properties in $LIST-II$, and choose the correct option
"A" obtained by Ostwald's method involving air oxidation of $NH _3$, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test
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