
Out of the list I, select the cation which has less polarizing power than $C{{a}^{2+}}$, and from list II, select the anion having more polarizability:
List I: $M{{g}^{2+}},S{{c}^{3+}},{{K}^{+}}$
List II: ${{O}^{2-}},C{{l}^{-}},{{P}^{3-}}$
A. $M{{g}^{2+}},{{O}^{2-}}$
B. ${{K}^{+}},{{P}^{3-}}$
C. $S{{c}^{3+}},{{P}^{3-}}$
D. $M{{g}^{2+}},C{{l}^{-}}$
Answer
133.8k+ views
Hint: Think about the definition of polarizability and what rule governs the polarizing power of ions and atoms. Consider how the number of electrons or radius affect the polarizing power.
Complete step by step solution:
Before attempting to answer this question, we need to know what polarizability is. Polarizability is the measure of how easily a cloud of electrons around an atom or ion can be distorted in the presence of an electric field.
The polarizability depends on the charge and radius of the atoms or ions. In cations, the more the effective nuclear charge present on the electron, the harder it will be to distort the cloud. The effective nuclear charge is defined as the force with which the nucleus attracts each electron. So, the ion with the larger effective nuclear charge will have less polarizing power.
This is concisely put by the Fajan’s rule. It states that:
i) High charge and small size of a cation will increase the polarizing power
ii) High charge and large size of an anion will increase the polarizability
Amongst the list, I, the cation with a lower charge than $C{{a}^{2+}}$will have a lower polarizing power. The only such cation present is the ${{K}^{+}}$. So, ${{K}^{+}}$ will have a lower polarizing power than $C{{a}^{2+}}$.
Amongst list II, the anion with the highest charge will have more polarizability. The anion that has the highest charge is ${{P}^{3-}}$, so it has the most polarizability.
Hence, the answer to this question is ‘B. ${{K}^{+}},{{P}^{3-}}$’
Note: When considering list one, note that $M{{g}^{2+}}$ has the same charge as $C{{a}^{2+}}$. In such a case, we should look at the radii of both the ions to determine their polarizing power. Magnesium is present in the third period and calcium is present in the fourth. So, calcium will have a greater radius and thus lesser polarizing power than magnesium.
Complete step by step solution:
Before attempting to answer this question, we need to know what polarizability is. Polarizability is the measure of how easily a cloud of electrons around an atom or ion can be distorted in the presence of an electric field.
The polarizability depends on the charge and radius of the atoms or ions. In cations, the more the effective nuclear charge present on the electron, the harder it will be to distort the cloud. The effective nuclear charge is defined as the force with which the nucleus attracts each electron. So, the ion with the larger effective nuclear charge will have less polarizing power.
This is concisely put by the Fajan’s rule. It states that:
i) High charge and small size of a cation will increase the polarizing power
ii) High charge and large size of an anion will increase the polarizability
Amongst the list, I, the cation with a lower charge than $C{{a}^{2+}}$will have a lower polarizing power. The only such cation present is the ${{K}^{+}}$. So, ${{K}^{+}}$ will have a lower polarizing power than $C{{a}^{2+}}$.
Amongst list II, the anion with the highest charge will have more polarizability. The anion that has the highest charge is ${{P}^{3-}}$, so it has the most polarizability.
Hence, the answer to this question is ‘B. ${{K}^{+}},{{P}^{3-}}$’
Note: When considering list one, note that $M{{g}^{2+}}$ has the same charge as $C{{a}^{2+}}$. In such a case, we should look at the radii of both the ions to determine their polarizing power. Magnesium is present in the third period and calcium is present in the fourth. So, calcium will have a greater radius and thus lesser polarizing power than magnesium.
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