Question

Which ion is larger in each pair$?$
$C{a^{2 + }},M{g^{2 + }};\,\,\,\,\,C{l^ - },{P^{3 - }};\,\,\,\,\,C{u^ + },C{u^{2 + }}$.
(i) $C{a^{2 + }};\,\,C{l^ - };\,\,C{u^{2 + }}$
(ii) $C{a^{2 + }};\,\,{P^{3 - }};\,\,C{u^ + }$
(iii) $M{g^{2 + }};\,\,{P^{3 - }};\,\,C{u^{2 + }}$
(iv) $M{g^{2 + }};\,\,C{l^ - };\,\,C{u^ + }$

Answer 1

The size of a cation is smaller than its corresponding neutral atom. Also the size of an ion depends upon the nuclear charge i.e. the number of protons present in the nucleus and the number of electrons present in the orbits surrounding the nucleus. Use these facts to determine the larger ion in each pair.

Complete step-by-step answer:
i) $C{a^{2 + }},M{g^{2 + }}$:The corresponding neutral atoms are $Ca,Mg$ respectively. $Ca$ is present in group $2$ of period $4$ while $Mg$ is present in group $2$ of period $3$. Hence both of them belong to the same group. As we go down in a group the electrons are added to subsequent shells leading to decrease in nuclear charge hence the size of elements increases as we go down the group. The cations are smaller in size compared to the neutral form but cations of different elements follow the same trend as the neutral element. Hence the size of$C{a^{2 + }} > M{g^{2 + }}$.

ii) $C{l^ - },{P^{3 - }}$: The corresponding neutral atoms are $Cl,P$ respectively. The atomic number of $Cl$is $17$, hence it has $17$ electrons and $17$ protons. $Cl$ accepts one electron for the formation of $C{l^ - }$hence the total number of electrons becomes $18$. The atomic number of $P$is $15$, hence it has $15$ electrons and $15$ protons. $P$ accepts three electrons for the formation of ${P^{3 - }}$hence the total number of electrons becomes $18$. So $C{l^ - },{P^{3 - }}$ are isoelectronic species having equal number of electrons. Since the number of protons present in $C{l^ - }$is more compared to ${P^{3 - }}$, it has more nuclear charge and hence can pull the orbital electrons more closer to the nucleus leading to the decrease in size. Hence size of $C{l^ - } < {P^{3 - }}$.

iii) $C{u^ + },C{u^{2 + }}$: The neutral atom for both of the cations is $Cu$. The atomic number of $Cu$is $29$, hence it has $29$ electrons and $29$ protons. $Cu$loses one electron for the formation of $C{u^ + }$hence the total number of electrons becomes $28$. Also $Cu$loses two electrons for the formation of $C{u^{2 + }}$hence the total number of electrons becomes $27$. Hence the number of protons present in the nucleus in both cases are the same hence the nuclear charge is the same. But since in $C{u^{2 + }}$the number of electrons present is lesser than that of $C{u^ + }$ the protons can pull the electrons more strongly leading to the decrease In size. Hence the size of $C{u^ + } > C{u^{2 + }}$.

Hence the correct answer is (ii) $C{a^{2 + }};\,\,{P^{3 - }};\,\,C{u^ + }$.

Note: Make sure to consider both the facts to determine the size of the ions. Also for these kinds of questions it is always better to know the periodic table and the placement of the different elements in the periodic table as periodic tables can help in easy determination of the atomic number.