Select the incorrect order(s) of electronegat | vedclass

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Question

(C) Electronegativity $(E.N.)$ depends on the oxidation state and hybridization of the atom.$A$. $F$ $(4.0)$ is more electronegative than $N$ $(3.04)$

Vedclass · Accepted Answer

$E.N.$ of $C$ atom in $CH_4$ $>$ $E.N.$ of $C$ atom in $CO_2$

Vedclass · Answer

(C) Electronegativity $(E.N.)$ depends on the oxidation state and hybridization of the atom.$A$. $F$ $(4.0)$ is more electronegative than $N$ $(3.04)$ on the Pauling scale. This is correct.$B$. As the oxidation state of an atom increases,its electronegativity increases. In $Cl_2O_7$,the oxidation state of $Cl$ is $+7$,while in $Cl_2O_5$,it is $+5$. Thus,$E.N.$ of $Cl$ in $Cl_2O_7$ $>$ $E.N.$ of $Cl$ in $Cl_2O_5$. This is correct.$C$. In $CH_4$,$C$ is $sp^3$ hybridized ($25\% \ s$-character),while in $CO_2$,$C$ is $sp$ hybridized ($50\% \ s$-character). Higher $s$-character leads to higher electronegativity. Therefore,$E.N.$ of $C$ in $CO_2$ $>$ $E.N.$ of $C$ in $CH_4$. The given statement is incorrect.$D$. Higher positive charge on an ion increases its electronegativity. $Cu^{2+}$ has a higher positive charge than $Cu^+$,so $E.N.$ of $Cu^{2+}$ $>$ $E.N.$ of $Cu^+$. This is correct.Therefore,the incorrect statement is $C$.

Select the incorrect order$(s)$ of electronegativity of the element$(s)$:

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