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(A) $(I)$ $I.P.$ of $O_{(g)}$ is less than $I.P.$ of $O^{-}_{(g)}$ is $False$ because removing an electron from a negatively charged ion is easier tha

Vedclass · Accepted Answer

$F, F, T, F$

Vedclass · Answer

(A) $(I)$ $I.P.$ of $O_{(g)}$ is less than $I.P.$ of $O^{-}_{(g)}$ is $False$ because removing an electron from a negatively charged ion is easier than from a neutral atom due to electron-electron repulsion.$(II)$ $I.P.$ of $Ne_{(g)}$ is less than $I.P.$ of $Ne^{+}_{(g)}$ because removing an electron from a cation requires more energy than from a neutral atom. Thus,the statement is $False$.$(III)$ $E.A.$ of $O^{+}_{(g)}$ is greater than $E.A.$ of $O_{(g)}$ is $True$ because $O^{+}$ has a higher effective nuclear charge and is more electron-attracting than neutral $O$.$(IV)$ $I.P.$ of $N_{(g)}$ is less than $I.P.$ of $N^{+}_{(g)}$ because removing an electron from a cation is harder. Thus,the statement is $False$.Therefore,the sequence is $F, F, T, F$.

List-$I$	List-$II$
$(1)$ Active non-metal with high electron gain enthalpy.	$(A)$ $Na$
$(2)$ Soft metal with low ionization enthalpy.	$(B)$ $Sb$
$(3)$ Metalloid that forms ${M_2}{O_3}$ type oxide.	$(C)$ $He$
$(4)$ Chemically inert gas.	$(D)$ $Cl$

List-$I$	List-$II$
$A$. $Al^{3+} < Mg^{2+} < Na^{+} < F^{-}$	$I$. Ionisation Enthalpy
$B$. $B < C < O < N$	$II$. Metallic character
$C$. $B < Al < Mg < K$	$III$. Electronegativity
$D$. $Si < P < S < Cl$	$IV$. Ionic radii

Similar Questions

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