Polarisation and Fajan’s rule Questions in English

(A) According to Fajan's rule,the covalent character of an ionic bond increases with the polarizing power of the cation.
$Ag^{+}$ has a pseudo-inert gas configuration ($18$ electrons in the valence shell),which results in higher polarizing power compared to alkali and alkaline earth metal cations like $K^{+}$,$Ba^{2+}$,and $Ca^{2+}$.
Greater polarization of the $Cl^{-}$ anion leads to a higher covalent character.
Therefore,$AgCl$ exhibits the most covalent character and is the least ionic among the given compounds.

(A) According to $Fajan's$ rule and the concept of electronegativity,a bond with $100\%$ covalent character is formed between identical atoms.
This is because there is no electronegativity difference between identical atoms,resulting in an equal sharing of the electron pair.

(C) According to Fajan's rule,the covalent character of a bond increases as the electronegativity difference between the bonded atoms decreases.
Electronegativity values on the Pauling scale are: $C = 2.55$,$F = 3.98$,$O = 3.44$,$S = 2.58$,and $Br = 2.96$.
The electronegativity differences are:
$|C - F| = |2.55 - 3.98| = 1.43$
$|C - O| = |2.55 - 3.44| = 0.89$
$|C - Br| = |2.55 - 2.96| = 0.41$
$|C - S| = |2.55 - 2.58| = 0.03$
Since the electronegativity difference is the smallest for the $C - S$ bond,it possesses the most covalent character.

(D) According to $Fajan's$ rule,the covalent character of an ionic bond increases with the increase in the size of the anion.
As we move down the group in the periodic table,the size of the halide ion increases in the order: $F^- < Cl^- < Br^- < I^-$.
Therefore,the polarizability of the anion increases,leading to an increase in the covalent character of the bond.
Thus,$CaI_2$ has the highest covalent character among the given options.

(A) According to $Fajan's$ rule,the covalent character of an ionic bond increases with an increase in the polarizing power of the cation.
Polarizing power is directly proportional to the charge density of the cation (charge/size ratio).
Comparing the cations: $Al^{3+}$,$Mg^{2+}$,$Cs^+$,and $La^{3+}$.
$Al^{3+}$ has the smallest ionic radius and a high charge $(+3)$,resulting in the highest charge density.
Therefore,$AlCl_3$ exhibits the maximum covalent character among the given options.

(B) According to Fajan's rule,the covalent character in an ionic compound increases with a decrease in the size of the cation and an increase in its charge.
$Al^{3+}$ has a smaller size and higher charge compared to $Na^{+}$,$Mg^{2+}$,and $H^{-}$.
Due to the high polarizing power of the $Al^{3+}$ ion,it polarizes the electron cloud of the $Cl^{-}$ ion to a greater extent,leading to significant covalent character in the $Al-Cl$ bond.
Therefore,$AlCl_3$ is considered to have significant covalent character.

(A) According to Fajan's rule,the polarization of an anion is influenced by the charge and size of the cation.
Greater charge on the cation leads to greater polarization of the anion.
Smaller size of the cation leads to greater polarization of the anion.
Therefore,a cation with high charge and small size causes maximum polarization.
Thus,the correct statement is that maximum polarization is brought about by a cation of high charge.

(B) According to Fajan's rule,the polarising power of a cation is directly proportional to its charge density.
Greater the charge on the cation and smaller its size,the higher will be its polarising power.
Comparing the given ions:
$Na^{+}$ ($+1$ charge,larger size)
$Mg^{2+}$ ($+2$ charge,smaller than $Na^{+}$)
$Ca^{2+}$ ($+2$ charge,larger than $Mg^{2+}$)
$Al^{3+}$ ($+3$ charge,smallest size among the given options).
Therefore,$Al^{3+}$ has the highest charge-to-size ratio,resulting in the highest polarising power.

(D) According to Fajan's rule,covalent character increases with an increase in the polarising power of the cation and the polarisability of the anion.
$1$. The polarising power of the cation increases as the charge density increases (higher charge and smaller size). Here,$Al^{3+}$ has the highest charge among $Na^+$,$Mg^{2+}$,and $Al^{3+}$.
$2$. The polarisability of the anion increases with its size. Among $Cl^-$ and $I^-$,the iodide ion $(I^-)$ is larger and more polarisable.
$3$. Therefore,$AlI_3$ has the cation with the highest charge and the anion with the highest polarisability,resulting in the maximum covalent character.

(A) According to $Fajan's$ rule,the polarisability of an anion depends on its size.
As the size of the anion increases,the valence electrons are held less tightly by the nucleus,making them more easily distorted by a cation.
Therefore,the order of polarisability for halide ions is $F^{-} < Cl^{-} < Br^{-} < I^{-}$.

(C) According to Fajan's rule,covalent character in an ionic bond is favoured by:
$1$. Small size of the cation.
$2$. Large size of the anion.
$3$. High charge on the cation or anion.
$4$. Cation with pseudo-noble gas configuration (e.g.,$Cu^+$,$Zn^{2+}$).
Therefore,a small cation and a large anion favour covalent bond formation.

(A) According to Fajan's rule,the polarizing power of a cation increases with its charge and decreases with its size.
Therefore,a highly charged and small cation has the maximum polarizing power.
Conversely,the polarizability of an anion increases with its size.
Thus,statement $A$ is correct as polarization is more pronounced by highly charged cations.

(C) The melting point of ionic halides depends on the lattice energy of the crystal.
According to Fajan's rule,as the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the covalent character increases and the ionic character decreases.
$NaF$ has the smallest anion,which results in the highest lattice energy and the strongest ionic character among the given options.
Therefore,$NaF$ has the highest melting point.

(D) The formation of an ionic bond is favored by factors that maximize lattice energy and minimize the energy required to form ions.
Specifically,a low ionization energy $(I.E.)$ for the metal,high electron affinity for the non-metal,and high lattice energy are required.
According to Fajan's rule,large cations and small anions minimize polarization,thereby favoring ionic character over covalent character.
Thus,a low positive charge,large cation,and small anion are helpful in the formation of an ionic bond.

(C) According to Fajan's rule,the covalent character of an ionic bond increases with the polarising power of the cation.
Polarising power is directly proportional to the charge density of the cation.
The charge and size of the cations are $Na^{+}$ ($+1$,large size),$Li^{+}$ ($+1$,smaller than $Na^{+}$),and $Be^{2+}$ ($+2$,smallest size).
The order of polarising power is $Be^{2+} > Li^{+} > Na^{+}$.
Therefore,the order of increasing covalent character is $NaCl < LiCl < BeCl_2$.

(A) According to Fajan's rule,the polarising power of a cation depends on its charge density.
$1$. Higher the charge on the cation,the greater is its polarising power.
$2$. Smaller the size of the cation,the greater is its polarising power.
Comparing the given ions: $Cs^{+}$ ($1+$ charge,large size),$K^{+}$ ($1+$ charge,smaller than $Cs^{+}$),$Mg^{2+}$ ($2+$ charge,smaller than $K^{+}$),and $Al^{3+}$ ($3+$ charge,smallest size).
Therefore,the correct order of polarising power is $Cs^{+} < K^{+} < Mg^{2+} < Al^{3+}$.

(A) According to Fajan's rule,the polarising power of a cation is directly proportional to its charge density.
$A$. $A$ small,highly positive ion (cation) has a high charge-to-size ratio,which allows it to strongly attract the electron cloud of an anion,resulting in the highest polarising ability.
Therefore,the correct option is $A$.

(D) The correct option is $D$.
$LiCl$ is more covalent in nature due to the high polarizing power of the small $Li^+$ ion (Fajans' rule).
Because of its covalent character,$LiCl$ has a lower melting point and lower electrical conductivity in the fused state compared to the more ionic $NaCl$.

(A) According to Fajan's rule,the polarizing power of the cation decreases as the size of the cation increases down the group $(Be^{2+} < Mg^{2+} < Ca^{2+} < Ba^{2+})$.
Consequently,the covalent character decreases and the ionic character increases as we move down the group from $Be$ to $Ba$.
Therefore,the correct order of increasing ionic character is $BeCl_2 < MgCl_2 < CaCl_2 < BaCl_2$.

(A) The ionic character of alkaline earth metal halides depends on the polarizing power of the cation.
As we move down the group from $Mg$ to $Ra$,the size of the cation increases and the ionization energy $(I.E.)$ decreases.
According to Fajan's rule,smaller cations have higher polarizing power,which leads to more covalent character.
Therefore,$Mg^{2+}$ has the smallest size and the highest polarizing power among the given options,making $MgCl_2$ the most covalent and least ionic.
The order of ionic character is $MgCl_2 < CaCl_2 < SrCl_2 < BaCl_2 < RaCl_2$.

(C) The correct answer is $(C)$.
$NaF$ has the highest melting point among the given halides.
According to Fajan's rule,as the size of the anion increases,the covalent character increases,and the ionic character decreases.
Since the fluoride ion $(F^-)$ is the smallest among the halides,$NaF$ has the maximum ionic character and the strongest electrostatic forces of attraction,leading to the highest melting point.

(B) The melting point of an ionic compound is directly proportional to its ionic character. According to Fajan's rule,the polarizing power of a cation is small and the polarizability of the anion is small when the electronegativity difference between the bonded atoms is large. In $NaF$,the electronegativity difference between $Na$ $(0.9)$ and $F$ $(4.0)$ is the largest among alkali metal halides,resulting in the maximum ionic character. Therefore,$NaF$ has the highest melting point.

(A) The correct answer is $A$.
$AgI$ (Silver iodide) is insoluble in water due to its high covalent character,which arises from the large size of the iodide ion $(I^-)$ and the polarizing power of the silver ion $(Ag^+)$ according to Fajan's rule.
In contrast,$KBr$,$CaCl_2$,and $AgF$ are ionic compounds and are generally soluble in water.

(B) According to Fajan's rule,ionic character is inversely proportional to the polarizing power of the cation.
Polarizing power increases as the size of the cation decreases and its charge increases.
Among the given cations ($Li^+$,$Rb^+$,$Be^{2+}$,$Mg^{2+}$),$Rb^+$ has the largest size and lowest charge,resulting in the lowest polarizing power and maximum ionic character.
$Be^{2+}$ has the smallest size and high charge,resulting in the highest polarizing power and minimum ionic character.
Therefore,$RbCl$ has the maximum ionic character and $BeCl_2$ has the minimum ionic character.

(A) According to Fajan's rule,the ionic character increases as the polarizing power of the cation decreases.
Polarizing power depends on the charge and size of the cation.
$Pb^{4+}$ has a higher charge and smaller size than $Pb^{2+}$,so $PbCl_4$ is more covalent than $PbCl_2$.
$CaCl_2$ and $NaCl$ are ionic compounds,where $NaCl$ is more ionic than $CaCl_2$ due to the lower charge on the $Na^+$ ion compared to $Ca^{2+}$.
Thus,the increasing order of ionic character is $PbCl_4 < PbCl_2 < CaCl_2 < NaCl$.

(A) The solubility of ionic compounds is governed by the lattice energy and hydration energy.
Compounds with high lattice energy are generally less soluble in water.
According to Fajan's rule,the covalent character increases with the increase in the size of the anion and the charge density of the cation.
$MgS$ has a high degree of covalent character due to the presence of a small $Mg^{2+}$ cation and a large $S^{2-}$ anion,which leads to high lattice energy and low solubility in water.
Therefore,the order of solubility is $NaCl > MgCl_2 > Na_2S > MgS$.
Thus,$MgS$ is the least soluble.

(B) $HgCl_2$ (mercury$(II)$ chloride) is a covalent compound with a low melting point $(276 \ ^\circ C)$ and a boiling point of $302 \ ^\circ C$.
Due to its covalent nature and weak intermolecular forces,it easily volatilises upon heating.
In contrast,$MgCl_2$,$CaCl_2$,and $FeCl_3$ are primarily ionic in nature and have significantly higher melting and boiling points.

(C) According to Fajan's rule,a smaller anion is polarized to a lesser extent than a larger anion by the same cation $(Ag^+)$.
Since the fluoride ion $(F^-)$ is the smallest among the halide ions $(F^-, Cl^-, Br^-, I^-)$,$AgF$ exhibits the most ionic character.
Greater ionic character leads to stronger electrostatic forces of attraction,resulting in a higher melting point for $AgF$ compared to the other silver halides.

(D) $HgCl_2$,$HgBr_2$,and $HgI_2$ are covalent in nature due to Fajan's rule.
Since ethanol is a polar covalent solvent,these covalent mercury halides are soluble in ethanol.
$HgF_2$ is ionic in nature and is less soluble in organic solvents like ethanol compared to other mercury halides.
However,in the context of general solubility of mercury halides in organic solvents,$HgCl_2$,$HgBr_2$,and $HgI_2$ show significant solubility.

(A) According to Fajan's rule,the solubility of an ionic compound in water is inversely proportional to its covalent character.
Covalent character increases with higher charge density and smaller size of the cation (polarizing power) and larger size of the anion (polarizability).
Comparing the given compounds,$MgS$ consists of $Mg^{2+}$ and $S^{2-}$ ions. Both ions have high charges,and $S^{2-}$ is a large anion,which leads to significant covalent character.
Compounds with higher covalent character are generally less soluble in polar solvents like water.
Therefore,$MgS$ is the least soluble among the given options.

(A) Lewis acid is defined as an electron-pair acceptor.
According to Fajan's rule and the concept of charge density,the Lewis acidity of a cation increases with an increase in positive charge and a decrease in ionic size.
Comparing the given ions:
$B^{3+}$ has a high positive charge $(+3)$ and a very small ionic radius.
$Li^{+}$,$Na^{+}$,and $Cs^{+}$ have a $+1$ charge and larger ionic radii compared to $B^{3+}$.
Due to the highest charge-to-size ratio,$B^{3+}$ has the strongest tendency to accept electron pairs,making it the strongest Lewis acid among the options.

(B) According to $Fajans'$ rule,the ionic character of a compound depends on the polarizing power of the cation.
Smaller cations with higher charge have higher polarizing power,leading to greater covalent character and lower ionic character.
Larger cations with lower charge have lower polarizing power,leading to greater ionic character.
Comparing the cations: $Rb^+$ (largest,$+1$ charge) has the least polarizing power,making $RbCl$ the most ionic.
$Be^{2+}$ (smallest,$+2$ charge) has the highest polarizing power,making $BeCl_2$ the least ionic.
Therefore,$RbCl$ has the highest ionic character and $BeCl_2$ has the lowest ionic character.

(A) The ionic character of halides of a given alkaline earth metal decreases in the order: $MF_2 > MCl_2 > MBr_2 > MI_2$.
This is due to the increasing polarizability of the halide ion as the size of the anion increases (Fajans' rule).
Greater ionic character corresponds to higher lattice energy and higher melting points.
Therefore,the decreasing order of melting points is: $CaF_2 > CaCl_2 > CaBr_2 > CaI_2$.

(B) According to Fajan's rule,$Be^{2+}$ has a small size and high charge density,which gives $BeF_2$ significant covalent character.
Due to this covalent nature,$BeF_2$ has the lowest melting point compared to the other ionic fluorides $(MgF_2, CaF_2, SrF_2)$.

(B) According to Fajan's rule,the ionic character of a compound increases as the size of the cation increases.
In Group $2$ elements,the size of the cation increases as we move down the group: $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,the ionic character of their chlorides increases in the order: $BeCl_2 < MgCl_2 < CaCl_2 < BaCl_2$.

(B) According to $Fajans'$ rule,the covalent character increases with the increase in polarising power of the cation.
$Zn^{2+}$ has a pseudo-noble gas configuration $(n-1)d^{10}ns^0$,which makes it more polarising than $Mg^{2+}$ (noble gas configuration).
Among $FeCl_2$,$ZnCl_2$,$CdCl_2$,and $MgCl_2$,$ZnCl_2$ exhibits the highest covalent character due to the high polarising power of the $Zn^{2+}$ ion.
Therefore,$ZnCl_2$ has the least ionic character.

(C) According to $Fajan's$ rule,the covalent character increases with a decrease in the size of the cation and an increase in its charge.
In $BeCl_2$,the $Be^{2+}$ ion has the smallest size and the highest charge density among the given options ($K^+$,$Ca^{2+}$,$Be^{2+}$,$Ba^{2+}$).
Therefore,$BeCl_2$ has the highest polarizing power,leading to the highest covalent character.

(D) According to $Fajan's$ rule,the polarizability of an anion increases as its size increases.
Since the ionic size increases down the group in the periodic table,the order of size is $F^- < Cl^- < Br^- < I^-$.
Therefore,the correct decreasing order of polarizability is $I^- > Br^- > Cl^- > F^-$.

(D) According to Fajan's rule,the covalent character in an ionic bond is favored by:
$1$. Small size of the cation.
$2$. Large size of the anion.
$3$. High charge on the cation or anion.
Therefore,a small cation and a large anion lead to greater polarization of the electron cloud,which favors the formation of a covalent bond.

(C) According to Fajan's rule,the covalent character increases as the charge density of the cation increases (i.e.,smaller size and higher positive charge).
As we move from $LiCl$ to $CCl_4$,the charge on the central metal/non-metal atom increases $(Li^+ < Be^{2+} < B^{3+} < C^{4+})$ and the size decreases.
Therefore,the covalent character increases in the order: $LiCl < BeCl_2 < BCl_3 < CCl_4$.

(D) The polarizing power of a cation is directly proportional to its charge density,which is defined by the ratio of charge to size (ionic radius).
$1$. Higher charge and smaller size lead to higher polarizing power.
$2$. Comparing the given cations:
- $K^+$: Charge $+1$,largest size.
- $Ca^{2+}$: Charge $+2$,smaller than $K^+$.
- $Mg^{2+}$: Charge $+2$,smaller than $Ca^{2+}$.
- $Be^{2+}$: Charge $+2$,smallest size.
$3$. The increasing order of polarizing power is $K^+ < Ca^{2+} < Mg^{2+} < Be^{2+}$.

(B) Ionic compounds are generally more soluble in water or aqueous media.
According to Fajans' rule,the covalent character increases with the polarising power of the cation,which is inversely proportional to the ionic character.
Ionic character is higher for cations with larger size and lower charge.
The order of size of the cations is $Na^{+} > Zn^{2+} > Cu^{2+}$.
Since $Na^+$ has the largest size and lowest charge,$Na_2S$ is the most ionic and thus the most soluble.
$CuS$ has the highest covalent character due to the smaller size and higher charge of $Cu^{2+}$,making it the least soluble.
Therefore,the correct order of solubility in water is $Na_2S > ZnS > CuS$.

(D) The polarizing power of a cation is directly proportional to its charge density,which is defined by the ratio of charge to size.
As the charge on the cation increases and its ionic radius decreases,the polarizing power increases.
For the given ions,the ionic radii follow the order: $K^{+} > Ca^{2+} > Mg^{2+} > Be^{2+}$.
Since $Be^{2+}$ has the highest charge-to-size ratio and $K^{+}$ has the lowest,the increasing order of polarizing power is $K^{+} < Ca^{2+} < Mg^{2+} < Be^{2+}$.

(C) According to Fajan's rule,the covalent character of an ionic bond increases with an increase in the polarizing power of the cation.
Polarizing power is directly proportional to the charge density of the cation,which depends on the charge and size of the cation.
Comparing the charges on the cations: $Fe^{2+}$,$Sn^{2+}$,$Al^{3+}$,and $Mg^{2+}$.
$Al^{3+}$ has the highest positive charge $(+3)$ among the given cations.
Therefore,$AlCl_3$ exhibits the maximum covalent character.

(A) According to Fajan's rule,the covalent character of a compound increases with an increase in the oxidation state of the metal cation.
In the $+4$ oxidation state,the metal ion has a smaller size and a higher charge density,which leads to greater polarizing power.
Consequently,$MCl_4$ exhibits more covalent character,whereas $MCl_2$ exhibits more ionic character.
Therefore,$MCl_2$ is more ionic than $MCl_4$.

(D) According to Fajan's rule,the polarizing power of a cation increases with an increase in its charge.
$Fe^{3+}$ has a higher charge and smaller size than $Fe^{2+}$,making it more polarizing and more covalent in character.
Consequently,$Fe^{3+}$ compounds are more easily hydrolysed than $Fe^{2+}$ compounds.
Therefore,the statement that ferrous compounds are more easily hydrolysed than ferric compounds is incorrect.

(A) According to $Fajan's \ rule$,the polarising power of a cation is directly proportional to its charge density (charge/size ratio).
For the given ions $K^{+}$,$Ca^{2+}$,$Mg^{2+}$,and $Be^{2+}$,the charge increases from $+1$ to $+2$ and the ionic size decreases in the order $K^{+} > Ca^{2+} > Mg^{2+} > Be^{2+}$.
Since polarising power increases as the charge increases and the size decreases,the correct increasing order is $K^{+} < Ca^{2+} < Mg^{2+} < Be^{2+}$.

(B) According to Fajan's rule,the covalent character increases as the size of the cation decreases and its charge density increases. Conversely,the ionic character increases as the size of the cation increases.
Comparing the cations $Rb^+$,$Mg^{2+}$,$Li^+$,and $Be^{2+}$:
$1$. The size order is $Rb^+ > Mg^{2+} > Li^+ > Be^{2+}$.
$2$. Since $Rb^+$ has the largest size,$RbCl$ has the highest ionic character.
$3$. Since $Be^{2+}$ has the smallest size and highest charge density,$BeCl_2$ has the lowest ionic character (highest covalent character).
Therefore,the compounds with the highest and lowest ionic characters are $RbCl$ and $BeCl_2$ respectively.

(D) According to Fajan's rule,the polarizability of anions increases with their size.
$Br^-$ and $I^-$ ions are large and highly polarizable.
This leads to a significant amount of covalent character in $MgBr_2$ and $MgI_2$.
Since acetone is a polar aprotic solvent with a lower dielectric constant than water,it favors the dissolution of compounds with covalent character.
Therefore,$MgBr_2$ and $MgI_2$ are soluble in acetone due to their covalent nature.

(D) According to $Fajan's$ rule,the covalent character in an ionic compound increases as the size of the anion increases.
For the given halides of $Cu^{2+}$,the size of the halide ions follows the order: $F^- < Cl^- < Br^-$.
$CuF_2$ is primarily ionic due to the small size and high electronegativity of the $F^-$ ion.
$CuCl_2$ and $CuBr_2$ exhibit significant covalent character due to the larger size and higher polarizability of $Cl^-$ and $Br^-$ ions.
Therefore,both $(a)$ and $(b)$ are correct statements.