Electrovalent bonding Questions in English

(A) The element $A$ has $3$ valence electrons,so it will lose $3$ electrons to form an ion with a charge of $+3$ $(A^{3+})$.
The element $B$ has $6$ valence electrons,so it will gain $2$ electrons to complete its octet,forming an ion with a charge of $-2$ $(B^{2-})$.
To form a neutral compound,the total positive charge must equal the total negative charge.
Using the criss-cross method: $A^{3+}$ and $B^{2-}$ combine to form $A_2B_3$.

(A) Among the given compounds,$LiCl$ is an ionic compound with a high lattice energy,resulting in a high melting point.
$BeCl_2$ is covalent but exists as a polymeric chain in the solid state,which gives it a higher melting point compared to the molecular covalent compounds $BCl_3$ and $CCl_4$.
However,comparing the ionic character,$LiCl$ has the highest ionic character among the choices,leading to the highest melting point.

(D) The solubility of an ionic compound in water depends on the balance between its lattice energy and hydration energy.
For a compound to be soluble,the hydration energy must be greater than the lattice energy.
In the case of $Na_2SO_4$,the hydration energy is greater than the lattice energy,making it soluble.
In the case of $BaSO_4$,the lattice energy is significantly higher than the hydration energy due to the high charge density and strong ionic interaction between $Ba^{2+}$ and $SO_4^{2-}$ ions,making it insoluble.

(A) The correct order is $LiF > LiCl > LiBr > LiI$.
Lattice energy is inversely proportional to the inter-ionic distance between the cation and the anion.
As the size of the halide ion increases from $F^-$ to $I^-$,the inter-ionic distance increases,which leads to a decrease in the lattice energy.

(D) Ionic crystals are formed by the electrostatic attraction between oppositely charged ions.
This electrostatic force is non-directional in nature,meaning it acts equally in all directions.
Therefore,the statement that ionic crystals exhibit directional properties of the bond is incorrect.
Thus,the correct option is $D$.

(C) The distance between the centres of cations and anions in an ionic crystal is given by the sum of their ionic radii,$d = r_{+} + r_{-}$.
To maximize this distance,we need to select the cation and anion with the largest ionic radii.
Among the given options,$Cs^{+}$ has the largest ionic radius among the cations $(Li^{+}, Cs^{+})$ and $I^{-}$ has the largest ionic radius among the anions $(F^{-}, I^{-})$.
Therefore,the compound with the maximum interionic distance is $CsI$.

(C) $CsBr_3$ is an ionic compound that crystallizes in a structure containing $Cs^{+}$ cations and $Br_3^{-}$ anions. The $Br_3^{-}$ ion is a linear tri-bromide ion.

(C) Elements of group $14$ exhibit $+4$ and $+2$ oxidation states.
Compounds in the $+4$ oxidation state are generally covalent.
As we move down the group,the inert pair effect becomes more prominent,making the $+2$ oxidation state more stable.
$PbCl_2$ contains $Pb^{2+}$ ions,which exhibits more ionic character compared to the covalent chlorides of lighter elements like $C$,$Si$,or $Sb$ in their higher oxidation states.

(D) The lattice energy $(U)$ of an ionic crystal is directly proportional to the product of the charges of the ions $(q_1, q_2)$ and inversely proportional to the inter-ionic distance $(r_0)$. The formula is given by $U \propto \frac{|q_1 q_2|}{r_0}$. Thus,it depends on both the magnitude of the ionic charges and the size of the ions.

(A) The bond strength in ionic compounds depends on the lattice energy,which is inversely proportional to the interionic distance $(r_+ + r_-)$.
$CsF$ has a smaller interionic distance compared to other alkali metal halides due to the high electronegativity of $F$ and the small size of the $F^-$ ion,resulting in a very strong ionic bond.

(B) In the molten state,$NaCl$ dissociates into its constituent ions,$Na^+$ and $Cl^-$.
These free ions are responsible for the electrical conductivity of molten $NaCl$.

(B) The melting point of ionic compounds depends on the lattice energy. $NaF$ has the highest lattice energy among the given halides due to the small size of the $F^-$ ion,which results in a stronger electrostatic attraction with the $Na^+$ ion. While $Na_2S$ is also ionic,$NaF$ is generally considered to have a very high melting point due to its compact crystal lattice structure.

(D) The ratio of the size of the anion to the cation is given by $\frac{r_{anion}}{r_{cation}}$.
To minimize this ratio,we need the largest possible cation and the smallest possible anion.
Among the given options,$Cs^{+}$ is the largest cation and $F^{-}$ is the smallest anion.
Therefore,the compound $CsF$ has the smallest ratio of the size of the anion to the cation.

(A) Lattice energy is inversely proportional to the interionic distance $(r_+ + r_-)$.
As the size of the alkali metal cation increases down the group $(Na^+ < K^+ < Rb^+ < Cs^+)$,the interionic distance increases.
Therefore,the lattice energy decreases as we move down the group.
The correct decreasing order is $NaF > KF > RbF > CsF$.

(C) $CsBr_3$ is a polyhalide compound. In polyhalide compounds,the alkali metal exists in a $+1$ oxidation state. Therefore,$CsBr_3$ is composed of $Cs^+$ and $Br_3^-$ ions.

(A) The ionic character of a bond is directly proportional to the difference in electronegativity between the bonded atoms.
$Cs$ (Cesium) is an alkali metal with the lowest electronegativity,while $Cl$ (Chlorine) is a non-metal with high electronegativity.
Therefore,the electronegativity difference $(\Delta EN)$ is maximum for the $Cs-Cl$ bond,making it the most ionic.

(D) An electrovalent or ionic bond is formed by the complete transfer of electrons from one atom to another,typically between a metal and a non-metal.
In $NaBr$,$Na$ (a metal) transfers one electron to $Br$ (a non-metal) to form $Na^+Br^-$,which is an ionic compound.
$O_2$,$CCl_4$,and $CHCl_3$ are covalent compounds formed by the sharing of electrons.

(A) Element $X$ has $2$ valence electrons,so it loses $2$ electrons to form $X^{2+}$ ion $(X \to X^{2+} + 2e^-)$.
Element $Y$ has $6$ valence electrons,so it gains $2$ electrons to form $Y^{2-}$ ion $(Y + 2e^- \to Y^{2-})$.
Combining these ions in a $1:1$ ratio to maintain electrical neutrality,the formula of the compound is $XY$.

(D) The lattice energy of an ionic compound is directly proportional to the product of the charges on the ions and inversely proportional to the distance between them (ionic size).
Therefore,it depends on both the charge on the ions and the size of the ions.

(B) The ratio of cation to anion size is given by $\frac{r_{+}}{r_{-}}$.
To maximize this ratio,we need the largest cation $(r_{+})$ and the smallest anion $(r_{-})$.
Comparing the cations: $Li^{+} < Na^{+} < Cs^{+}$. Thus,$Cs^{+}$ is the largest cation.
Comparing the anions: $F^{-} < I^{-}$. Thus,$F^{-}$ is the smallest anion.
Therefore,the ratio is highest for $CsF$.

(C) The melting point of ionic compounds is directly related to their lattice energy.
As the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the lattice energy decreases due to weaker electrostatic attraction between the cation and the larger,more polarizable anion.
Therefore,the melting point decreases in the order: $CaF_2 > CaCl_2 > CaBr_2 > CaI_2$.
Thus,$CaI_2$ has the lowest melting point.

(B) $CsI_3$ is an ionic compound that dissociates in the solid state as $CsI_3 \rightarrow Cs^{+} + I_3^{-}$.
Here,$Cs^{+}$ is the cesium cation and $I_3^{-}$ is the triiodide anion.

(C) First,determine the valency of each element based on its valence shell configuration:
$L$ has $6$ valence electrons $(2s^2 2p^4)$,so it gains $2$ electrons to complete its octet,valency = $-2$.
$Q$ has $7$ valence electrons $(3s^2 3p^5)$,so it gains $1$ electron to complete its octet,valency = $-1$.
$P$ has $1$ valence electron $(3s^1)$,so it loses $1$ electron,valency = $+1$.
$R$ has $2$ valence electrons $(3s^2)$,so it loses $2$ electrons,valency = $+2$.
Now,combine them to form neutral ionic compounds:
Between $P(+1)$ and $L(-2)$: $P_2L$
Between $R(+2)$ and $L(-2)$: $RL$
Between $P(+1)$ and $Q(-1)$: $PQ$
Between $R(+2)$ and $Q(-1)$: $RQ_2$
Thus,the correct formulae are $P_2L, RL, PQ$ and $RQ_2$.

(B) $CsF$ (Cesium fluoride) is an ionic (electrovalent) compound.
Ionic compounds consist of strong electrostatic forces of attraction between oppositely charged ions,which require a significant amount of energy to overcome.
In contrast,$He$ is a noble gas with weak London dispersion forces,$NH_3$ exhibits hydrogen bonding,and $CHCl_3$ exhibits dipole-dipole interactions.
Since ionic bonds are much stronger than intermolecular forces,$CsF$ has the highest boiling point.

(D) The melting point of an ionic compound depends on the lattice energy.
Since the ionic radius of $Na^{+}$ $(0.95 \ \mathring{A})$ is smaller than $K^{+}$ $(1.33 \ \mathring{A})$,the lattice energy of $NaCl$ is higher than that of $KCl$,resulting in a higher melting point for $NaCl$.
Regarding solubility,$KCl$ is more soluble in water than $NaCl$ because the hydration energy of $K^{+}$ and $Cl^{-}$ ions is more favorable relative to the lattice energy compared to $NaCl$.
Therefore,$NaCl$ has a higher melting point but lower solubility than $KCl$.

(A) Ionic compounds are characterized by strong electrostatic forces of attraction between oppositely charged ions.
$1$. They possess high melting and high boiling points due to these strong forces.
$2$. The ionic bond is non-directional in nature.
$3$. They are generally soluble in polar solvents (like water) and insoluble in non-polar solvents.
$4$. They form three-dimensional crystal lattice structures and conduct electricity in the molten or aqueous state.
Therefore,the statement that they have 'high melting points and low boiling points' is incorrect.

(C) $CsBr_3$ is an ionic compound that dissociates into $Cs^{+}$ and $Br_3^{-}$ ions. The $Br_3^{-}$ ion is a linear polyhalide ion.

(C) The stability of an ionic bond is determined by the lattice energy of the resulting ionic compound.
Lattice energy is directly proportional to the product of the charges of the ions and inversely proportional to the interionic distance $(U \propto \frac{q_1 q_2}{r_0})$.
For $MgF_2$,the charges are $Mg^{2+}$ and $F^-$,giving a product of charges equal to $2 \times 1 = 2$.
For $NaF$,$LiF$,and $NaCl$,the product of charges is $1 \times 1 = 1$.
Since $MgF_2$ has a higher charge product and relatively small ionic radii,it possesses the highest lattice energy,making it the most stable ionic bond among the given options.

(B) The ionization energy $(IE)$ is the energy required to remove an electron from a neutral gaseous atom to form a cation.
Lower $IE$ facilitates the formation of $P^{+}$.
Electron affinity is the energy released when an electron is added to a neutral gaseous atom to form an anion.
Higher electron affinity facilitates the formation of $Q^{-}$.
Therefore,an ionic bond $P^{+} Q^{-}$ is most likely to be formed when the ionization energy of $P$ is low and the electron affinity of $Q$ is high.

(D) The solubility of ionic compounds is determined by the balance between lattice energy and hydration energy.
For compounds with small,highly charged anions (like $F^-$),the lattice energy is very high due to strong electrostatic attraction,making them less soluble compared to compounds with larger,less charged anions (like $Cl^-$).
In the case of $Ag_2O$ and $Ag_2S$,$Ag_2O$ is more soluble than $Ag_2S$ because the lattice energy of $Ag_2S$ is significantly higher due to the greater polarizability of the $S^{2-}$ ion compared to $O^{2-}$,which increases the covalent character and decreases solubility.
Therefore,the correct order is $Ag_2O > Ag_2S$.

(A, B, C, D) The question asks for the match that is not incorrect,meaning all provided options are actually correct representations of the ionic nature or structure of these compounds.
$1$. $CsI_3$ is an ionic compound consisting of $Cs^{+}$ and $I_3^{-}$ ions.
$2$. $TiCl_3$ is an ionic compound consisting of $Ti^{3+}$ and $3Cl^{-}$ ions.
$3$. $GaCl_3$ is an ionic compound consisting of $Ga^{3+}$ and $3Cl^{-}$ ions.
$4$. $InCl_3$ is an ionic compound consisting of $In^{3+}$ and $3Cl^{-}$ ions.
Since all options correctly represent the dissociation of these compounds into their respective ions,all are correct.

(C) The melting point of ionic compounds is directly related to their lattice energy.
Lattice energy is inversely proportional to the interionic distance between the ions.
Since the size of the halide ion increases in the order $F^{-} < Cl^{-} < Br^{-} < I^{-}$,the interionic distance in $NaF$ is the smallest.
Consequently,$NaF$ possesses the maximum lattice energy among the given sodium halides,leading to the highest melting point.

(A) Lattice energy is inversely proportional to the inter-ionic distance $(r_+ + r_-)$ between the cation and the anion.
As the size of the cation increases down the group $(Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+})$,the inter-ionic distance increases.
Consequently,the lattice energy decreases as we move from $MgO$ to $BaO$.
Therefore,$MgO$ has the smallest cation size,resulting in the shortest inter-ionic distance and the maximum lattice energy.

(D) The ionic character of a bond is directly proportional to the difference in electronegativity between the bonded atoms.
The electronegativity values are: $Cl = 3.0$,$Si = 1.8$,$P = 2.1$,$B = 2.0$,and $C = 2.5$.
The electronegativity differences are:
$Si-Cl$: $|3.0 - 1.8| = 1.2$
$P-Cl$: $|3.0 - 2.1| = 0.9$
$B-Cl$: $|3.0 - 2.0| = 1.0$
$C-Cl$: $|3.0 - 2.5| = 0.5$
Since the $Si-Cl$ bond has the largest electronegativity difference,it possesses the highest ionic character.

(B) The ionization energy $(IE)$ is the energy required to remove an electron from a neutral gaseous atom to form a cation. Lower $IE$ facilitates the formation of $A^{+}$.
Electron affinity $(EA)$ is the energy released when an electron is added to a neutral gaseous atom to form an anion. Higher $EA$ facilitates the formation of $B^{-}$.
Therefore,for the formation of an ionic compound $A^{+} B^{-}$,the ionization energy of $A$ should be low and the electron affinity of $B$ should be high.

(D) The strength of an ionic bond is determined by the lattice energy,which is directly proportional to the product of the charges and inversely proportional to the interionic distance $(U \propto \frac{q_1 q_2}{r_0})$.
$Al$ has a charge of $+3$ in all these compounds.
Comparing the anions: $O^{2-}$,$N^{3-}$,$Se^{2-}$,and $As^{3-}$.
While $N^{3-}$ has a higher charge than $O^{2-}$,the bond between $Al$ and $O$ is considered to have high ionic character due to the high electronegativity difference and the small size of the $O^{2-}$ ion,leading to a very stable lattice.
However,based on the options provided and standard chemical trends,$Al$ and $O$ form a very strong ionic bond in $Al_2O_3$.

(A) Lattice energy is inversely proportional to the inter-ionic distance between the cation and the anion.
As the size of the halide ion increases from $F^-$ to $I^-$,the inter-ionic distance increases,leading to a decrease in the lattice energy.
The order of ionic radii is $F^- < Cl^- < Br^- < I^-$.
Therefore,the order of lattice energies is $KF > KCl > KBr > KI$.

(C) The lattice energy of an ionic compound is directly proportional to the product of the charges of the ions and inversely proportional to the inter-ionic distance (sum of ionic radii).
$Lattice \, Energy \propto \frac{|q_+ \times q_-|}{r_+ + r_-}$
Comparing the charges:
$NaF$: $Na^+ (+1), F^- (-1) \rightarrow \text{Product of charges} = 1$
$MgF_2$: $Mg^{2+} (+2), F^- (-1) \rightarrow \text{Product of charges} = 2$
$AlF_3$: $Al^{3+} (+3), F^- (-1) \rightarrow \text{Product of charges} = 3$
$AlN$: $Al^{3+} (+3), N^{3-} (-3) \rightarrow \text{Product of charges} = 9$
Since $AlN$ has the highest product of charges and the ions ($Al^{3+}$ and $N^{3-}$) are relatively small,it possesses the largest lattice energy.

(C) The formation of an ionic bond $X^{+}Y^{-}$ is favored by the following factors:
$1$. Low ionization energy of the metal $X$ to facilitate the formation of $X^{+}$ ion.
$2$. High electron affinity of the non-metal $Y$ to facilitate the formation of $Y^{-}$ ion.
$3$. High lattice energy of the resulting ionic compound $XY$ to provide stability to the crystal lattice.
Therefore,statements $I, II,$ and $III$ are correct.

(D) The Born-Haber cycle relates the standard enthalpy of formation $(\Delta H_f^\circ)$ of an ionic compound to its constituent elements' properties.
The equation is: $\Delta H_f^\circ = \Delta H_{sub} + IE + \Delta H_{diss} + EA + U$
Using the values from the cycle:
$\Delta H_{sub} = +108 \ kJ/mol$
$IE = +496 \ kJ/mol$
$\Delta H_{diss} = +122 \ kJ/mol$
$EA = -349 \ kJ/mol$
$U = -788 \ kJ/mol$
Comparing the absolute magnitudes of these energy terms,the lattice energy $(U = -788 \ kJ/mol)$ is the largest in magnitude. This large negative value is the primary driving force for the formation and stability of the solid ionic lattice.

(B) Ionic bonds are non-directional in nature.
$RbCl$ is an ionic compound because $Rb$ is a highly electropositive alkali metal and $Cl$ is an electronegative non-metal.
In contrast,$NCl_3$,$BeCl_2$,and $BCl_3$ are covalent compounds,which possess directional bonds due to the overlap of atomic orbitals.

(C) $MgO$ has the highest melting point due to its significantly higher lattice energy.
All these compounds possess a similar $FCC$ (rock salt) crystal structure.
Lattice energy is directly proportional to the product of the charges of the ions $(Z^{+} \cdot Z^{-})$. For $MgO$,the charges are $+2$ and $-2$,whereas for $NaCl$ and $KCl$,the charges are $+1$ and $-1$. Thus,the lattice energy of $MgO$ is much higher.
Comparing $MgO$ and $BaO$,both have $+2$ and $-2$ charges,but $Mg^{2+}$ is smaller than $Ba^{2+}$. Since lattice energy is inversely proportional to the interionic distance $(r)$,the smaller size of $Mg^{2+}$ results in a shorter bond length and higher lattice energy for $MgO$ compared to $BaO$.
Therefore,$MgO$ has the highest melting point.

(D) The melting point of ionic compounds is inversely proportional to the internuclear distance $(r_c + r_a)$ between the ions.
As the size of the ions increases,the electrostatic force of attraction decreases,leading to a lower melting point.
Since the ionic radii of $Rb^+$ and $Br^-$ are larger than those of $Na^+$ and $F^-$,the internuclear distance $(r_c + r_a)$ is greater for $RbBr$ than for $NaF$.
Therefore,$NaF$ has a higher melting point than $RbBr$.

(B) Anhydrous aluminium chloride $(Al_2Cl_6)$ is a covalent compound. When it dissolves in water,it undergoes hydration and forms an octahedral complex.
Specifically,it forms the hexaaquaaluminium$(III)$ ion,$[Al(H_2O)_6]^{3+}$,and chloride ions,$Cl^-$.
The reaction is: $AlCl_3(s) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq) + 3Cl^-(aq)$.

(B) When $AlCl_3$ dissolves in water,it dissociates into $Al^{3+}$ and $Cl^-$ ions.
The high charge density of the $Al^{3+}$ ion results in a very large hydration energy,which compensates for the lattice energy required to break the covalent structure of $AlCl_3$.
Thus,the formation of hydrated ionic species is driven by the large hydration energy of $Al^{3+}$.

(B) The covalent character of a $P-H$ bond in various compounds can be determined by the distribution of formal charge per bond. The higher the formal charge distribution per bond,the lower the covalent character.
Formal charge distribution per bond = $\frac{\text{Total Charge}}{\text{Number of } P-H \text{ bonds}}$
For $PH_3$: $\frac{0}{3} = 0$
For $P_2H_6^{2+}$: $\frac{2}{8} = 0.25$ (Note: $P_2H_6^{2+}$ has $8$ $P-H$ bonds)
For $P_2H_5^+$: $\frac{1}{7} \approx 0.14$
For $PH_4^+$: $\frac{1}{4} = 0.25$
Comparing the values,$P_2H_6^{2+}$ and $PH_4^+$ show higher charge distribution,but based on standard chemical analysis of these species,$P_2H_6^{2+}$ exhibits the least covalent character due to the delocalization of charge across the $P-P$ framework.

(D) The trihalides of Group $15$ elements are generally covalent in nature.
However,$BiF_3$ is an exception because the electronegativity difference between $Bi$ and $F$ is significantly large,which imparts it with ionic character.

(C) $CsBr_3$ is an ionic compound that dissociates into $Cs^+$ and $Br_3^-$ ions in the solid state.
The $Br_3^-$ ion is a polyhalide ion formed by the interaction of $Br^-$ with $Br_2$ molecule.
Therefore,the correct statement is that it contains $Cs^+$ and $Br_3^-$ ions.