Identify the incorrect order of the given par | vedclass

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(D) The melting point of $3d$ transition elements depends on the number of unpaired electrons and the strength of metallic bonding. $Mn$ has a low mel

Vedclass · Accepted Answer

$CuS > ZnS > Na_2S$ (order of solubility in aqueous medium)

Vedclass · Answer

(D) The melting point of $3d$ transition elements depends on the number of unpaired electrons and the strength of metallic bonding. $Mn$ has a low melting point due to its stable $d^5$ configuration,which leads to weak metallic bonding. $V$ has a high melting point due to a large number of unpaired electrons. The order of melting points for $Co, Ni, Cu, Zn$ is $Co > Ni > Cu > Zn$ because the metallic bond strength decreases as the number of unpaired electrons decreases.Regarding solubility,$Na_2S$ is an ionic compound and is highly soluble in water. Among metal sulfides,the solubility is determined by the solubility product constant $(K_{sp})$. The $K_{sp}$ of $ZnS$ is higher than that of $CuS$,so $ZnS$ is more soluble than $CuS$. Therefore,the correct order of solubility is $Na_2S > ZnS > CuS$. Option $D$ states $CuS > ZnS > Na_2S$,which is the incorrect order.

Column - $I$	Column - $II$
$(a)$ Copper	$(i)$ Non-metal
$(b)$ Fluorine	$(ii)$ Transition Metal
$(c)$ Silicon	$(iii)$ Lanthanoid
$(d)$ Cerium	$(iv)$ Metalloid

Identify the $incorrect$ order of the given parameter.

Similar Questions

Match the element in column $I$ with that in column $II$.

Column - $I$ Column - $II$

$(a)$ Copper $(i)$ Non-metal

$(b)$ Fluorine $(ii)$ Transition Metal

$(c)$ Silicon $(iii)$ Lanthanoid

$(d)$ Cerium $(iv)$ Metalloid

Identify the correct match:

Which of the following transition metals is used as a catalyst?

The transition metals have a less tendency to form ions due to

Which of the following is not arranged in the correct sequence?

The number of statement$(s)$ correct from the following for copper (atomic number $29$) is/are $...$
$(A) \ Cu(II)$ complexes are always paramagnetic
$(B) \ Cu(I)$ complexes are generally colourless
$(C) \ Cu(I)$ is easily oxidized
$(D)$ In Fehling solution,the active reagent has $Cu(II)$

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