Identify the incorrect order of the given par | vedclass

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Question

(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-$II$
$(1)$ General electronic configuration of $d$-block elements	$(A)$ Transition elements
$(2)$ $d$-block elements	$(B)$ $(n-1)d^{1-10}ns^{1-2}$
$(3)$ Groups $3$ to $12$	$(C)$ $Z = 58$ to $71$
$(4)$ Lanthanoid elements	$(D)$ Last electron enters $d$-orbital

List-$I$	List-$II$
$A$. Oxidation state of $V$ in $VOCl_2$	$I$. $0$
$B$. Number of unpaired electrons in $MnO_4^{2-}$ ion	$II$. $1$
$C$. Number of unpaired electrons in $[NiCl_4]^{2-}$ ion	$III$. $5$
$D$. This oxidation state is exhibited by all lanthanide ions	$IV$. $3$
	$V$. $4$
	$VI$. $2$

Identify the $incorrect$ order of the given parameter.

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