Polarisation and Fajan’s rule Questions in English

(C) $LiCl$ is a covalent compound due to the high polarizing power of the small $Li^+$ cation,which causes polarization of the electron cloud of the large $Cl^-$ anion (Fajans' rule).
The electronegativity difference between $Li$ $(1.0)$ and $Cl$ $(3.0)$ is $2.0$,which is significant,not small.
Therefore,the Assertion is correct,but the Reason is incorrect.

(N/A) $LiF$ is almost insoluble in water due to its very high lattice energy,which arises from the small size of both $Li^+$ and $F^-$ ions.
In contrast,$LiCl$ has a lower lattice energy and possesses significant covalent character due to the polarization of the $Cl^-$ ion by the small $Li^+$ ion (Fajans' rule).
This covalent character makes $LiCl$ soluble in organic solvents like acetone,while its lower lattice energy compared to $LiF$ allows it to dissolve more readily in water.

(N/A) The smaller the size of the cation and the larger the size of the anion,the greater the covalent character of an ionic bond.
The greater the charge on the cation,the greater the covalent character of the ionic bond.
For cations of the same size and charge,the one with electronic configuration $(n-1)d^n s^0$,typical of transition metals,is more polarising than the one with a noble gas configuration,$ns^2 np^6$,typical of alkali and alkaline earth metal cations.
The cation polarises the anion,pulling the electronic charge toward itself and thereby increasing the electronic charge density between the two nuclei. This is precisely what happens in a covalent bond,i.e.,the build-up of electron charge density between the nuclei.
The polarising power of the cation,the polarizability of the anion,and the extent of distortion (polarisation) of the anion are the factors that determine the percent covalent character of the ionic bond.

(N/A) $(i)$ $BeO$ is almost insoluble in water because $Be^{2+}$ is a small cation with high charge density,leading to very high lattice energy that hydration energy cannot overcome. In $BeSO_4$,the large $SO_4^{2-}$ anion reduces the lattice energy,making it soluble.
$(ii)$ $BaO$ is soluble because $Ba^{2+}$ and $O^{2-}$ have a size mismatch,resulting in lower lattice energy that is easily overcome by hydration energy. In $BaSO_4$,both $Ba^{2+}$ and $SO_4^{2-}$ are large,leading to high lattice energy,making it insoluble.
$(iii)$ $LiI$ is more soluble than $KI$ in ethanol due to the higher polarising power of the small $Li^+$ ion,which imparts greater covalent character to $LiI$ compared to $KI$,facilitating its dissolution in organic solvents like ethanol.

(N/A) The distortion of the electron cloud of an anion by the influence of a nearby cation is known as polarization. This phenomenon is described by Fajan's rules,which explain the partial covalent character in ionic bonds.

(N/A) Fajan's rule is used to predict the covalent character in an ionic bond. The rules are as follows:
$1$. Smaller size of the cation: $A$ smaller cation has a higher charge density,which increases its polarizing power.
$2$. Larger size of the anion: $A$ larger anion is more easily polarized by the cation due to its loosely held valence electrons.
$3$. Higher charge on the cation or anion: An increase in the charge on either the cation or the anion increases the extent of polarization.
$4$. Electronic configuration of the cation: Cations with a pseudo-noble gas configuration (e.g.,$Cu^+$,$Ag^+$,$Zn^{2+}$) have a greater polarizing power than those with a noble gas configuration (e.g.,$Na^+$,$K^+$) due to the poor shielding effect of $d$-electrons.

(D) The solubility of ionic compounds in water depends on their lattice energy and hydration energy.
According to Fajan's rule,compounds with higher covalent character exhibit lower solubility in polar solvents like water.
$MgS$ has a high degree of covalent character due to the small size and high charge density of the $Mg^{2+}$ ion,which polarizes the $S^{2-}$ ion significantly.
Therefore,$MgS$ is the least soluble among the given options.

(D) According to Fajan's rule,the covalent character in an ionic bond increases with the polarizing power of the cation.
Lithium $(Li^+)$ has a very small ionic radius,which results in a high charge density and high polarizing power.
This high polarizing power allows the $Li^+$ ion to distort the electron cloud of the halide anion,thereby introducing significant covalent character into the lithium halides.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(A) $(A)$ $LiCl$ and $MgCl_2$ are covalent in nature due to polarization (Fajans' rule), hence they are soluble in ethanol.
$(B)$ $Li$ and $Mg$ do not form superoxides; they form oxides and peroxides respectively.
$(C)$ $LiF$ has a very high lattice energy due to the small size of both $Li^+$ and $F^-$ ions, making it less soluble in water compared to other alkali metal fluorides.
$(D)$ $Li_2O$ is less soluble in water compared to other alkali metal oxides due to high lattice energy.
Therefore, statements $(A)$ and $(C)$ are correct.

(A) According to $Fajan's$ rule,the covalent character of an ionic bond increases with the decrease in the size of the cation.
The order of the size of the alkali metal cations is: $Li^+ < Na^+ < K^+ < Cs^+$.
Since the size of the cation increases from $Li^+$ to $Cs^+$,the polarizing power decreases.
Therefore,the decreasing order of covalent character is: $LiCl > NaCl > KCl > CsCl$,which corresponds to $(A) > (B) > (C) > (D)$.

(A) According to Fajan's rule,the covalent character of an ionic bond increases as the size of the anion increases.
The order of the size of the anions is $F^- < Cl^- < Br^- < I^-$.
Therefore,the increasing order of covalent character is $CaF_{2} < CaCl_{2} < CaBr_{2} < CaI_{2}$.

(A) According to Fajan's rule,the covalent character of an ionic compound increases with a decrease in the size of the cation.
As we move down the group $2$ $(Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+})$,the size of the cation increases.
Consequently,the polarising power of the cation decreases,leading to a decrease in the covalent character.
Therefore,the correct order of covalent character is $BaCl_2 < SrCl_2 < CaCl_2 < MgCl_2$.

(B) According to $Fajans'$ rule,the polarisability of an ion depends on its size and charge.
For ions with the same charge,the larger the size of the anion,the more easily its electron cloud can be distorted,leading to higher polarisability.
Comparing the given anions $(F^{-}, Cl^{-}, I^{-})$,the size increases down the group as $F^{-} < Cl^{-} < I^{-}$.
Therefore,the $I^{-}$ ion has the largest size and is the most polarisable.

(C) According to Fajan's Rule,covalent character increases with higher polarization of the anion by the cation.
$1.$ For $KF$ vs $KI$: $I^-$ is larger than $F^-$,so $KI$ has more covalent character. Thus,$KF < KI$ is true.
$2.$ For $LiF$ vs $KF$: $Li^+$ is smaller than $K^+$,so $LiF$ has more covalent character. Thus,$LiF > KF$ is true.
$3.$ For $SnCl_4$ vs $SnCl_2$: $Sn^{4+}$ has higher charge than $Sn^{2+}$,so $SnCl_4$ is more covalent. Thus,$SnCl_4 > SnCl_2$ is true.
$4.$ For $CuCl$ vs $NaCl$: $Cu^+$ (pseudo-noble gas configuration) has higher polarizing power than $Na^+$ (noble gas configuration). Thus,$CuCl > NaCl$ is true.
Evaluating the options:
$A.$ $KF > KI$ (False),$LiF > KF$ (True)
$B.$ $KF < KI$ (True),$LiF > KF$ (True)
$C.$ $SnCl_4 > SnCl_2$ (True),$CuCl > NaCl$ (True)
$D.$ $LiF > KF$ (True),$CuCl < NaCl$ (False)
$E.$ $KF < KI$ (True),$CuCl > NaCl$ (True)
Therefore,statements $B, C,$ and $E$ are correct.

(B) According to Fajan's rule,the covalent character in an ionic bond is determined by the following factors:
$1$. Polarising power of the cation $(a)$.
$2$. Polarisability of the anion $(c)$.
$3$. The extent of distortion of the anion $(b)$,which is a direct consequence of the first two factors.
Factor $(d)$,the polarising power of the anion,is generally negligible as anions are typically large and have low charge density compared to cations.
Therefore,there are $3$ factors ($a$,$b$,and $c$) that affect the covalent character.

(B) According to Fajan's rule,the ionic character depends on the polarizability of the cation.
$Ag^+$ has a pseudo-inert gas configuration ($18$ electrons in the outermost shell),which makes it highly polarizing compared to $Ba^{2+}$,$K^+$,and $Co^{2+}$.
Greater polarization leads to more covalent character and less ionic character.
The order of ionic character is $KCl > BaCl_2 > CoCl_2 > AgCl$.
Therefore,$AgCl$ is the least ionic.

(A) According to Fajan's rule,small cations with high charge density have a high polarizing power,which leads to the formation of covalent bonds.
Boron has a very small atomic size and a high ionization energy,making it impossible to form $B^{3+}$ ions in a typical ionic lattice.
Instead,boron shares its valence electrons to form covalent bonds.
Therefore,both statements are true,and $Statement-2$ is the correct explanation for $Statement-1$.

(C) According to Fajans' Rules,ionic character increases as the polarizing power of the cation decreases.
Polarizing power depends on the charge and size of the cation.
Cations with a noble gas configuration ($ns^2 np^6$,$8e^-$) have lower polarizing power and thus form more ionic bonds compared to those with pseudo-noble gas configurations $(18e^-)$.
In option $C$,$Ca^{2+}$ has a noble gas configuration,while $Zn^{2+}$ and $Sn^{2+}$ have pseudo-noble gas configurations.
Therefore,the correct order of ionic character is $CaCl_2 > SnCl_2 > ZnCl_2$.
Thus,the order $ZnCl_2 > SnCl_2 > CaCl_2$ is incorrect.

(B) According to Fajan's rule,the covalent character of a compound increases with the increase in the oxidation state of the central metal atom.
In $UF_6$,the oxidation state of Uranium is $+6$,whereas in $UF_4$,it is $+4$.
Since the oxidation state of $U$ in $UF_6$ is higher,it has higher polarizing power,making $UF_6$ more covalent than $UF_4$.
The Reason states that Fluorine is smaller in size,which is a true statement,but it does not explain why $UF_6$ is more covalent than $UF_4$. The correct explanation relates to the oxidation state of the central metal atom.

(B) According to Fajan's rule,covalent character increases with:
$1$. Smaller size of the cation.
$2$. Larger size of the anion.
Comparing the given compounds:
- The cations are $Li^+$ and $Na^+$. Since $Li^+$ is smaller than $Na^+$,$Li^+$ compounds have more covalent character.
- The anions are $Cl^-$ and $I^-$. Since $I^-$ is larger than $Cl^-$,$I^-$ compounds have more covalent character.
Therefore,$LiI$ has the smallest cation and the largest anion,resulting in the maximum covalent character.

(A) According to $Fajan's$ rule,the covalent character of an ionic bond increases with an increase in the polarising power of the cation.
Polarising power is directly proportional to the charge density of the cation.
The charge on the cations is $Al^{+3} > Mg^{+2} > Na^{+}$.
Therefore,the polarising power follows the order $Al^{+3} > Mg^{+2} > Na^{+}$.
Consequently,the decreasing order of covalent character is $AlCl_3 > MgCl_2 > NaCl$.

(A) According to Fajan's rule,as the size of the anion increases,its polarizability increases,which leads to an increase in covalent character and a decrease in ionic character.
Since the size of the fluoride ion $(F^-)$ is the smallest among the given halide ions $(F^- < Cl^- < Br^- < I^-)$,it has the least polarizability.
Therefore,$MF$ exhibits the highest ionic character.
The order of ionic character is: $MF > MCl > MBr > MI$.

(C) According to Fajan's rule,the covalent character of a compound increases with the increase in the oxidation state of the metal cation.
Higher oxidation state leads to higher polarising power of the cation,which results in greater covalent character.
Comparing the oxidation states of the metals in the given compounds:
$SbCl_3$: $Sb$ is in $+3$ oxidation state.
$PbCl_2$: $Pb$ is in $+2$ oxidation state.
$SnCl_4$: $Sn$ is in $+4$ oxidation state.
$SnCl_2$: $Sn$ is in $+2$ oxidation state.
Since $Sn$ in $SnCl_4$ has the highest oxidation state $(+4)$,it has the highest polarising power and thus $SnCl_4$ is the most covalent compound among the given options.

(D) According to Fajan's rules,the covalent character of an ionic bond increases with an increase in the charge on the cation.
$SbCl_5$ contains the $Sb^{5+}$ cation,which has the highest positive charge among the given options ($Sn^{2+}$,$Pb^{2+}$,$Sb^{3+}$,$Sb^{5+}$).
$A$ higher charge on the cation results in greater polarising power,which distorts the electron cloud of the anion more effectively,leading to increased covalent character.
Therefore,$SbCl_5$ is the most covalent compound among the choices.

(D) According to Fajan's rule,the covalent character of an ionic bond increases with the increase in the size of the anion due to higher polarisability.
The size of the halide ions follows the order: $I^{-} > Br^{-} > Cl^{-} > F^{-}$.
As the size of the anion increases,the covalent character increases,which means the ionic character decreases.
Therefore,the order of ionic character is $MgF_2 > MgCl_2 > MgBr_2 > MgI_2$.
Thus,magnesium fluoride $(MgF_2)$ has the highest ionic character.

(B) $AlCl_3$ is a covalent compound due to the high polarizing power of the $Al^{3+}$ ion (Fajans' rule).
Because it is covalent,it is soluble in organic solvents like ether.
$LiCl$ and $NaCl$ are ionic compounds and are insoluble in ether.
Therefore,only $AlCl_3$ is extracted into ether.

(B) According to Fajan's rule,the larger the size of the anion,the greater the covalent character.
Since $Cl^-$ is larger than $F^-$,$AlCl_3$ exhibits significant covalent character and exists as a molecular dimer $(Al_2Cl_6)$,making it electron-deficient.
In contrast,$AlF_3$ is primarily ionic due to the high electronegativity and small size of $F^-$,forming a giant ionic lattice structure.
Therefore,$AlCl_3$ is electron-deficient,while $AlF_3$ is not.

(C) According to Fajan's rule,covalent character increases when the cation is smaller and has a higher positive charge,and the anion is larger and has a higher negative charge.
$(i)$ $I^-$ has a larger size than $F^-$,so $KI$ has more covalent character than $KF$. Thus,$KF < KI$ is correct.
$(ii)$ $Li^+$ is smaller than $K^+$,so $LiF$ has more covalent character than $KF$. Thus,$LiF > KF$,making $LiF < KF$ incorrect.
$(iii)$ $Sn^{4+}$ has a higher positive charge than $Sn^{2+}$,so $SnCl_4$ has more covalent character than $SnCl_2$. Thus,$SnCl_2 < SnCl_4$ is correct.
$(iv)$ $Cu^+$ has a pseudo-noble gas configuration $(d^{10})$,which causes higher polarization than the noble gas configuration of $Na^+$. Thus,$CuCl > NaCl$,making $NaCl < CuCl$ correct.
Therefore,the correct orders are $(i), (iii),$ and $(iv)$.

(A) According to Fajan's rule,the covalent character of an ionic bond increases with:
$1$. Smaller size of the cation.
$2$. Higher charge on the cation.
Comparing the cations $Li^{+}$,$Be^{2+}$,$B^{3+}$,and $C^{4+}$:
- The charge increases in the order: $Li^{+} < Be^{2+} < B^{3+} < C^{4+}$.
- The ionic radius decreases in the order: $Li^{+} > Be^{2+} > B^{3+} > C^{4+}$.
Both factors favor increased polarization of the anion by the cation as we move from $LiCl$ to $CCl_4$.
Therefore,the covalent character increases in the order: $LiCl < BeCl_2 < BCl_3 < CCl_4$.

(C) According to Fajan's rules,the covalent character of an ionic bond increases with the polarization of the anion.
Polarization is favored by a smaller cation size and a larger anion size.
Comparing $LiF$ and $KF$: $Li^+$ is smaller than $K^+$,so $Li^+$ has a higher polarizing power,making $LiF$ more covalent than $KF$.
Comparing $KF$ and $KI$: $I^-$ is larger than $F^-$,so $KI$ is more covalent than $KF$.
Comparing $SnCl_4$ and $SnCl_2$: $Sn^{4+}$ has a higher charge density than $Sn^{2+}$,making $SnCl_4$ more covalent.
Comparing $ZnCl_2$ and $NaCl$: $Zn^{2+}$ has a pseudo-inert gas configuration ($18$ electrons in the valence shell),which is more polarizing than the inert gas configuration of $Na^+$,making $ZnCl_2$ more covalent than $NaCl$.
Therefore,the correct statement is that $LiF$ is more covalent than $KF$.

(A) According to Fajan's rule,the covalent character in an ionic compound increases with a smaller size of the cation and a larger size of the anion.
Comparing the cations: $Mg^{2+}$ is smaller than $Na^{+}$ due to higher charge density.
Comparing the anions: $O^{2-}$ is smaller than $Cl^{-}$ and $Br^{-}$. However,the polarizing power of $Mg^{2+}$ is significantly higher than $Na^{+}$.
Among the given options,$Mg^{2+}$ and $Cl^{-}$ provides a combination where the cation has high charge and the anion is relatively large and polarizable,leading to significant covalent character. Specifically,$MgCl_2$ is more covalent than $NaCl$ or $MgO$ due to the balance of cation size and anion polarizability.

(B) According to Fajan's rule,the covalent character of an ionic bond depends on the polarising power of the cation and the polarisability of the anion.
Smaller cations have higher polarising power,leading to greater covalent character.
In the given group $15$ trihalides,the size of the central metal cation increases down the group $(N < P < Sb < Bi)$.
Since $Bi^{3+}$ is the largest cation among the given options,it has the lowest polarising power.
Therefore,$BiF_3$ exhibits the least covalent character and is the most ionic in nature.

(D) According to Fajan's rules,the covalent character of an ionic bond increases as the size of the cation decreases and the charge on the cation increases.
The correct order of covalent character is $NaCl < LiCl < MgCl_2 < BeCl_2$.
In the given assertion,the order $NaCl < MgCl_2 < BeCl_2 < LiCl$ is incorrect because $LiCl$ should be between $NaCl$ and $MgCl_2$.
Therefore,$A$ is false and $R$ is true.

(B) According to Fajan's rule,the covalent character of an ionic bond increases with an increase in the size of the anion and a decrease in the size of the cation.
In the given compounds,$Al^{3+}$ is the smallest cation,which has the highest polarizing power.
Among the anions ($Cl^{-}$ and $I^{-}$),the iodide ion $(I^{-})$ is larger than the chloride ion $(Cl^{-})$.
Therefore,$AlI_3$ has the maximum polarization of the anion electron cloud by the cation,resulting in the highest covalent character.

(D) According to Fajan's rule,covalent character increases as the size of the anion increases for a fixed cation.
The polarizability of the anion increases with its size,leading to greater distortion of the electron cloud by the cation.
The size of the halide ions follows the order: $F^- < Cl^- < Br^- < I^-$.
Since $I^-$ is the largest anion among the given options,it is the most polarizable.
Therefore,$CaI_2$ exhibits the highest covalent character.

(B) According to Fajan's rule,the covalent character of an ionic bond increases with the increase in the charge on the cation.
In the given compounds,the oxidation states of the metal ions are:
$FeF_3$: $Fe^{3+}$
$VF_5$: $V^{5+}$
$VF_2$: $V^{2+}$
$TiF_2$: $Ti^{2+}$
Since the vanadium ion in $VF_5$ has the highest positive charge $(+5)$,it exerts the strongest polarizing power on the fluoride anion.
Therefore,$VF_5$ exhibits the maximum covalent character among the given options.

(B) According to Fajan's rules,covalent character increases with:
$1$. Smaller size of the cation.
$2$. Larger size of the anion.
$3$. Higher charge on the cation or anion.
Evaluating the options:
- $KF < KI$: $I^-$ is larger than $F^-$,so $KI$ is more covalent. (Correct)
- $LiF < KF$: $Li^+$ is smaller than $K^+$,so $LiF$ should be more covalent than $KF$. Thus,$LiF > KF$. (Incorrect)
- $SnCl_2 < SnCl_4$: $Sn^{4+}$ has a higher charge than $Sn^{2+}$,so $SnCl_4$ is more covalent. (Correct)
- $NaCl < CuCl$: $Cu^+$ has a pseudo-noble gas configuration ($18$ electrons in the valence shell) and is more polarizing than $Na^+$. (Correct)
Therefore,the statement $LiF < KF$ is incorrect.

(D) $LiCl$ exhibits both ionic and covalent character.
$LiCl$ is ionic because $Li$ is a metal and $Cl$ is a non-metal,resulting in an electrostatic force of attraction between the positive $(Li^{+})$ and negative $(Cl^{-})$ ions.
$LiCl$ also exhibits covalent character because the small size of the $Li^{+}$ cation causes significant polarization of the larger $Cl^{-}$ anion,as described by Fajan's rule.
Therefore,the bonding in $LiCl$ is both ionic and covalent.

(D) According to Fajan's rule,the covalent character of an ionic bond increases with the increase in the size of the anion.
Since the cation $K^{+}$ is common in all the given compounds,the covalent character depends on the size of the halide ions $(F^{-}, Cl^{-}, I^{-})$.
The size order of the anions is $F^{-} < Cl^{-} < I^{-}$.
Therefore,the order of covalent character is $KF < KCl < KI$.

(C) Many simple compounds of elements such as $AlCl_3, GaCl_3$ and $InCl_3$ are covalent while anhydrous,but in aqueous solution,these are ionic in nature.
In anhydrous condition,the (charge/radius) ratio,i.e.,polarisability of $Al^{3+}$ is high and hence,according to Fajans' rule,$Al^{3+}$ polarises $Cl^{-}$ ions to a large extent,thereby introducing covalent character in the compound,i.e.,$AlCl_3$ behaves as a covalent compound in anhydrous conditions.
In aqueous medium,the ions get hydrated because the amount of hydration enthalpy released exceeds the sum total of ionization enthalpy required.
Since the (charge/radius) ratio of the hydrated aluminium ion is much smaller compared to that of $Al^{3+}$,the tendency of $[Al(H_2O)_6]^{3+}$ to polarise the hydrated $Cl^{-}$ ion decreases,and the resulting hydrated compound is ionic in nature.

(B) Assertion $(A)$ is true because $LiCl$ and $MgCl_2$ exhibit significant covalent character due to the high polarizing power of small $Li^+$ and $Mg^{2+}$ ions (Fajans' rule),making them soluble in organic solvents like ethanol.
Reason $(R)$ is true because lithium and magnesium are indeed harder than other elements in their respective groups due to their small atomic size and strong metallic bonding.
However,the hardness of the metals is not the reason for the solubility of their chlorides in ethanol; the solubility is due to the covalent nature of the compounds.
Therefore,both statements are true,but the reason is not the correct explanation for the assertion.

(A) According to $Fajan's$ rule,the ionic character is inversely proportional to the polarising power of the cation.
Polarising power depends on the charge density of the cation.
As the size of the cation increases $(B^{3+} < Al^{3+} < Ga^{3+})$,the polarising power decreases.
Therefore,the ionic character increases as the size of the cation increases.
The correct order of ionic character is $BCl_{3} < AlCl_{3} < GaCl_{3}$.

(B) Key Point: $A$ compound with more ionic character has a higher melting point.
$Be$ and $Ca$ belong to the same group $(2)$,and ionic character increases down the group due to the decrease in polarizing power of the cation.
Thus,$BeCl_2$ is more covalent (less ionic) than $CaCl_2$,so $CaCl_2$ has a higher melting point than $BeCl_2$.
$HgCl_2$ is highly covalent due to the pseudo-noble gas configuration of $Hg^{2+}$ ($18$ electrons in the outer shell),which exerts a high polarizing power on the chloride ion.
Therefore,the order of melting points is $CaCl_2 > BeCl_2 > HgCl_2$,which corresponds to $(ii) > (i) > (iii)$.
This is equivalent to $iii < i < ii$.

(A) Statement $I$: The bond dissociation enthalpy of $F_{2}$ is lower than $Cl_{2}$ and $Br_{2}$ due to high inter-electronic repulsion between the lone pairs of small $F$ atoms. Thus,the correct order is $Cl_{2} > Br_{2} > F_{2} > I_{2}$. Statement $I$ is true.
Statement $II$: According to Fajan's rule,higher oxidation state of the metal cation leads to greater covalent character. Therefore,$SnCl_{4} > SnCl_{2}$ and $PbCl_{4} > PbCl_{2}$ are correct. However,for uranium halides,$UF_{6}$ has a higher oxidation state $(+6)$ than $UF_{4}$ $(+4)$,so $UF_{6} > UF_{4}$ is the correct order. The statement claims $UF_{4} > UF_{6}$,which is false. Thus,Statement $II$ is false.