$A$. $Al^{3+}$ $B$. $Cu^{2+}$ $C$. $Ba^{2+}$ $D$. $Co^{2+}$ $E$. $Mg^{2+}$
Choose the correct answer from the options given below.
| Group | Cations |
| Group $-II$ | $\mathrm{Cu}^{2+}$ |
| Group $-III$ | $\mathrm{Al}^{3+}$ |
| Group $-IV$ | $\mathrm{Co}^{2+}$ |
| Group $-V$ | $\mathrm{Ba}^{2+}$ |
| Group $-VI$ | $\mathrm{Mg}^{2+}$ |
The correct order of group number of ions is $\underset{\text { (B) }}{\mathrm{Cu}^{2+}} < \underset{\text { (A) }}{\mathrm{Al}^{3+}} < \underset{\text { (D) }}{\mathrm{Co}^{2+}} < \underset{\text { (C) }}{\mathrm{Ba}^{2+}} < \underset{\text { (E) }}{\mathrm{Mg}^{2+}}$
$\therefore$ The correct order is $\mathrm{B}, \mathrm{A}, \mathrm{D}, \mathrm{C}, \mathrm{E}$
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(Assume $100\%$ ionization)
$A.$ $0.500\,M\,C _2 H _5 OH ( aq )$ and $0.25\, M\, KBr ( aq )$
$B.$ $0.100\,M\,K _4\left[ Fe ( CN )_6\right]$ (aq) and $0.100\, M$ $FeSO _4\left( NH _4\right)_2 SO _4$ (aq)
$C.$ $0.05 \,M\, K _4\left[ Fe ( CN )_6\right]( aq )$ and $0.25\, M\, NaCl$ (aq)
$D.$ $0.15\, M\, NaCl ( aq )$ and $0.1\, M BaCl _2$ (aq)
$E.$ $0.02\, M\, KCl\, MgCl _{2 .} 6 H _2 O ( aq )$ and $0.05\, M$ $KCl ( aq )$