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Hint: Polarising power of cations is directly proportional to the cation’s charge density. We can say that higher the charge density of cations, higher will be its polarising power.
Complete step by step answer:
Let us first establish what the polarising power of a cation really is before going on to try and determine the correct order of the same from the given cations.
> Polarising power of any cation describes its ability to distort any anion. We can say that it also describes the power of cation to attract the electron density. It has been observed that polarising power of a cation is directly proportional to their charge density. So, for cations with the same radius, its polarising power will increase as its charge increases. For cations with the same charge, its polarising power will increase as its radius decreases.
> Therefore, we can easily observe that smaller cations that have higher positive charges will have better polarizing power because the positive charge is distributed on a relatively small area resulting in higher charge density.
> Na, Mg, Al and Si all are elements of the same period. So, they will have their valence electrons in the same shell. Thus we can predict that their radius will not have such a difference.
> The ions given to us are $N{{a}^{+}},M{{g}^{2+}},A{{l}^{3+}}\text{and S}{{\text{i}}^{4+}}$ . We can see that positive charge on cations are different and that will be the deciding factor in their charge density and hence polarising power.
> We can say that $S{{i}^{4+}}$ will have the highest polarising power as its charge density will be highest. Then $A{{l}^{3+}},M{{g}^{2+}}\text{ and N}{{\text{a}}^{+}}$ will come in decreasing order.
Thus, we can conclude that the answer to this question is option (b).
Note:Cations which possess pseudo-noble gas configurations have greater polarising power due to poor shielding of the inner 10 d-electrons, which results in greater effective nuclear charge. For e.g. $Ag^+$,$Cu^+$ etc.
Complete step by step answer:
Let us first establish what the polarising power of a cation really is before going on to try and determine the correct order of the same from the given cations.
> Polarising power of any cation describes its ability to distort any anion. We can say that it also describes the power of cation to attract the electron density. It has been observed that polarising power of a cation is directly proportional to their charge density. So, for cations with the same radius, its polarising power will increase as its charge increases. For cations with the same charge, its polarising power will increase as its radius decreases.
> Therefore, we can easily observe that smaller cations that have higher positive charges will have better polarizing power because the positive charge is distributed on a relatively small area resulting in higher charge density.
> Na, Mg, Al and Si all are elements of the same period. So, they will have their valence electrons in the same shell. Thus we can predict that their radius will not have such a difference.
> The ions given to us are $N{{a}^{+}},M{{g}^{2+}},A{{l}^{3+}}\text{and S}{{\text{i}}^{4+}}$ . We can see that positive charge on cations are different and that will be the deciding factor in their charge density and hence polarising power.
> We can say that $S{{i}^{4+}}$ will have the highest polarising power as its charge density will be highest. Then $A{{l}^{3+}},M{{g}^{2+}}\text{ and N}{{\text{a}}^{+}}$ will come in decreasing order.
Thus, we can conclude that the answer to this question is option (b).
Note:Cations which possess pseudo-noble gas configurations have greater polarising power due to poor shielding of the inner 10 d-electrons, which results in greater effective nuclear charge. For e.g. $Ag^+$,$Cu^+$ etc.
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