MCQ
Main reason for brown ring test of $NO_3^-$ is
- ANegligible tendency of $FeSO_4$ to form complex
- ✓Reducing power of $FeSO_4$ to convert $NO_3^-$ to $NO$
- CDecomposition of $FeSO_4$ to give $SO_2$ gas
- DOxidising property of $FeSO_4$
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|
List$-I$ (Parameter) |
List$-II$ (Unit) |
| $(a)$ Cell constant | $(i)$ $\mathrm{S}\, \mathrm{cm}^{2} \,\mathrm{~mol}^{-1}$ |
| $(b)$ Molar conductivity | $(ii)$ Dimensionless |
| $(c)$ Conductivity | $(iii)$ $\mathrm{m}^{-1}$ |
| $(d)$ Degree of dissociation of electrolyte | $(iv)$ $\Omega^{-1} \,\mathrm{~m}^{-1}$ |
Choose the most appropriate answer from the options given below :
$(A)$ $\left[ Cr \left( NH _3\right)_5 Cl ^2 Cl _2\right.$ and $\left[ Cr \left( NH _3\right)_4 Cl _2\right) Cl$
$(B)$ $\left[ Co \left( NH _3\right)_4 Cl _2\right]^{+}$and $\left[ Pt \left( NH _3\right)_2\left( H _2 O \right) Cl \right]^{+}$
$(C)$ $\left[ CoBr _2 Cl _2\right]^{2-}$ and $\left[ PtBr _2 Cl _2\right]^{2-}$
$(D)$ $\left[ Pt \left( NH _3\right)_3\left( NO _3\right) Cl\right.$ and $\left[ Pt \left( NH _3\right)_3 Cl \right] Br$
| List I (Complex) | List $II$ (Type of isomerism) |
| $A$. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}_2$ | $I$. Solvate isomerism |
| $B$. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{Br}$ | $II$. Linkage isomerism |
| $C$. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]$ | $III$. Ionization isomerism |
| $D$. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ | $IV$. Coordination isomerism |
Choose the correct answer from the options given below:
$V^{+4},\,Ni^{+2},\, Ti^{+3}, \,Co^{+2}, \,Fe^{+3}, \,Cu^{+2}$