MCQ
Identify the correct order of solubility in aqueous medium.
- A$Na_2S > CuS > ZnS$
- ✓$Na_2S > ZnS > CuS$
- C$CuS > ZnS > Na_2S$
- D$ZnS > Na_2S > CuS$
✓
Answer
Correct option: B.
$Na_2S > ZnS > CuS$
b
lonic compounds are more soluble in water or in aqueous medium.
lonic compounds are more soluble in water or in aqueous medium.
According to Fajans' rule,
Size of the cation increases the ionic character also increases.
lonic character oc size of cation (if anion is same)
The order of size of cation is
$N a^{+}>Z n^{2+}>C u^{2+}$
$\therefore$ The order of ionic character and hence, of solubility in water is as
$N a_{2} S>Z n S>C u S$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Product $Z$ of above reaction is
→If the bidentate ligand ethane $1, 2-$ diamine $(en)$ in progressively added in green aqueous solution of $[Ni(H_2O)_6]Cl_2$ in the molar ratios $en : Ni$ as $1 : 1, 2 : 1, 3 : 1$ then their associated spin only magnetic moment (in $B.M.\, unit$) are respectively
→The equilibrium constant $(K_c)$ for the reaction,
→$CaS{O_4}.5{H_2}{O_{(s)}}\, \rightleftharpoons \,CaS{O_4}.3{H_2}{O_{(s)}} + 2{H_2}{O_{(g)}}$ is equal to
Aniline in a set of reactions yielded a product $D$.The structure of product $D$ would be
→image$\mathop {\xrightarrow{{NaN{O_2}}}}\limits_{HCl} A\,\,\xrightarrow{{CuCN}}B\,\,\mathop {\xrightarrow{{{H_2}}}}\limits_{Ni} C\,\,\xrightarrow{{HN{O_2}}}D$
The reaction, $X + 2Y + Z \to N$ occurs by the following mechanism
→$(i)$ $X + Y \rightleftharpoons M$ very rapid equilibrium
$(ii)$ $M + Z \to P$ slow
$(iii)$ $O + Y \to N$ very fast
What is the rate law for this reaction
Which of the following outer orbital complex has maximum number of unpaired electron
How many electrons can fit in the orbital for which $n = 3$ and $l = 1\ ?$
→For a reaction, the rate constant is expressed as, $K = A.e^{-40000/T}$ . The energy of the activation is ....... $cal$
→Figure $\xrightarrow{{{H^ + }/{H_2}O}}P$
→Figure $\xrightarrow[{(ii)\,\,NaB{D_4}}]{{(i)\,\,Hg{{(OAc)}_2}\,{H_2}O}}Q$
Figure $\xrightarrow[{(ii)\,\,{H_2}{O_2}\,/\,O{H^ - }}]{{(i)\,\,B{D_6}\,,\,THF}}R$
Products $P, Q$ and $R$ are
The electronic configuration of three elements $A,B$ and $C$ are given below. The molecular formula of the compound formed from $B$ and $C$ will be
→$A : 1s^2\, 2s^2\, 2p^6$
$B : 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^3$
$C : 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$