CsCl , CsI , TlCl and TlI all adopt a caesium chloride structure. The inter-ionic distances are: Cs - Cl ⇒ 3 . 06 A ° , Cs - I ⇒ 3 . 41 A ° , Tl - Cl ⇒ 2 . 55 A ° and Tl - I ⇒ 2 . 90 A ∘ . Assuming that the ions behave as hard spheres and that the radius ratio in TlI has the limiting value, calculate the ionic radii for Cs + , Tl + , Cl - , I - in eight-coordination.

Radius ratio r + r - for compounds having co-ordination number 8 and BCC structures like CsCl , CsI , TlCl , TlI is 0 . 732 < r + r - < 1 . 0 Here, r + is radius of cation and r - is radius of anion. Here, 0 . 732 and 1 . 0 are smallest and largest value for the stability of these types of lattice structure. Beyond this range these lattice become unstable. The smallest value of radius radio for which lattice is stable is called limiting radius ratio. As given TlI has radius ratio of limiting value, that means Tl + I - = 0 . 732 . . . ( i ) For BCC structure of TlI given that Tl - I ⇒ 2 . 90 A ° it means Tl + + I - = 2 . 90 A ° . . . ( ii ) But by equation ( i ) Tl + I - = 0 . 732 or Tl + = 0 . 732 × I - putting this value in equation ( ii ) we get 0 . 732 × I - + I - = 2 . 90 A ° I - = 1 . 67 A ° putting this value in equation ( ii ) Tl + + I - = 2 . 90 A ° Tl + + 1 . 67 A ° = 2 . 90 A ° Tl + = 1 . 23 A ° Now, given Tl - Cl ⇒ 2 . 55 A ° It means Tl + + Cl - = 2 . 55 A ° . . . ( iii ) but we calculated above Tl + = 1 . 23 A ° putting this value in equation ( iii ) 1 . 23 A ° + Cl - = 2 . 55 A ° Cl - = 1 . 32 A ° Now given Cs - Cl ⇒ 3 . 06 A ° or Cs + + Cl - ⇒ 3 . 06 A ° . . . ( iv ) but we calculated above Cl - = 1 . 32 A ° putting this value in equation ( iv ) Cs + + 1 . 32 A ° ⇒ 3 . 06 A ° Cs + = 1 . 74 A ° So ionic radii for Cs + , Tl + , Cl - , I - is 1 . 74 A ° , 1 . 23 A ° , 1 . 32 A ° , 1 . 67 A ° respectively.

Explain the term lattice energy as applied to an ionic solid. Calculate the lattice energy of caesium chloride using the following data. Cs ( s ) → Cs ( g ) ΔH = + 79 . 2 kJ mol - 1 Cs ( g ) → Cs + ( g ) ΔH = + 374 . 05 kJ mol - 1 Cl 2 ( g ) → 2 Cl ( g ) ΔH = + 241 . 84 kJ mol - 1 Cl ( g ) + e → Cl - ( g ) ΔH = - 397 . 90 kJ mol - 1 Cs ( s ) + 1 2 Cl 2 → CsCl ( s ) ΔH = - 623 . 00 kJ mol - 1

Given, Cs ( s ) → Cs ( g ) ΔH = + 79 . 2 kJ mol - 1 Cs ( g ) → Cs + ( g ) ΔH = + 374 . 05 kJ mol - 1 Cl 2 ( g ) → 2 Cl ( g ) ΔH = + 241 . 84 kJ mol - 1 Cl ( g ) + e → Cl - ( g ) ΔH = - 397 . 90 kJ mol - 1 Cs ( s ) + 1 2 Cl 2 → CsCl ( s ) ΔH = - 623 . 00 kJ mol - 1 Sublimation energy = 79 . 2 kJ mol - 1 Ionization energy = 374 . 05 kJ mol - 1 Bond energy = 241 . 84 kJ mol - 1 Electron Affinity = - 397 . 90 kJ mol - 1 Formation energy = - 623 . 00 kJ mol - 1 Lattice energy is defined as the energy released, when one mole solid is formed from component gaseous ions. Here Cs ( g ) + + Cl ( g ) - → CsCl ( s ) ∆ H = Lattice energy ∆ H cycle = 0 = Formation energy + Ionization energy + Bond Energy + Sublimation energy - Lattice energy - Electron Affinity ⇒ ∆ H lattice = - 623 + 374 . 05 + 241 . 84 + 79 . 2 + 397 . 90 ⇒ ∆ H lattice = 467 kJ mol - 1

The standard enthalpy changes ΔH ° at 298 K for the reaction: M ( s ) + 3 2 Cl 2 ( g ) → MCl 3 ( s ) are given for the first row transition metals: Sc Ti V Cr Mn Fe Co Ni Cu ΔH ° / kJ mol - 1 - 339 - 209 - 138 - 160 + 22 - 59 - 131 - 280 + 357 Use a Born-Haber cycle to account for the change in ΔH ° as the atomic number of the metal increases. Comment on the relative stabilities of the + II and + III oxidation states of the 3 d metals.

M ( s ) + 3 2 Cl 2 ( g ) → MCl 3 ( s ) . . . 298 K M= Transition metals Born-Haber cycle for above reaction is given as . Enthalpy of sublimation of metal increase with increase in atomic number of metal. Hence ∆ H ° increase as atomic number of the metal increases. From given data Scandium Sc has more negative value of standard enthalpy change - 339 KJ mol - 1 , hence greater the stability of Sc + 3 metal ion. Copper has more positive value of standard enthalpy change + 357 kJ mol - 1 , hence copper does not form stable ion with + 3 oxidation state. Manganese also has positive value of standard enthalpy change, it will also not form stable ion with + 3 oxidation state. The relative stabilities of the + 2 to + 3 oxidation states of the 3 d metals is are order: Sc > Ni > Ti > Cr > V > Co > Fe > Mn > Cu .

EASY

JEE Advanced

IMPORTANT

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Why are ionic compounds have usually high melting, while most simple covalent compounds have low melting points? Explain the high melting point of diamond.

Important Questions on The Ionic Bond

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 4

How are the minimum values of radius ratio arrived at for various coordination numbers and what are these limits? Give examples of the types of crystal structures associated with each coordination number.

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 5

Show by means of a diagram, and a simple calculation, the minimum value of the radius ratio

r^{+} / r^{-}

, which permits a salt to adopt a cesium chloride type of structure.

EASY

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 6

Give the co-ordination number of the ions and describe the crystal structures of zinc blende, wurtzite and sodium chloride in terms of close packing and the occupancy of the tetrahedral and the octahedral holes.

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 7

CsCl, CsI, TlCl

and

TlI

all adopt a caesium chloride structure. The inter-ionic distances are:

Cs - Cl \Rightarrow 3.06 A^{°}, Cs - I \Rightarrow 3.41 A^{°}, Tl - Cl \Rightarrow 2.55 A^{°}

and

Tl - I \Rightarrow 2.90 A^{\circ}

. Assuming that the ions behave as hard spheres and that the radius ratio in

TlI

has the limiting value, calculate the ionic radii for

{Cs}^{+}, {Tl}^{+}, {Cl}^{-}, I^{-}

in eight-coordination.

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 8

Write down the Born-Lande equation and define the terms in it. Use the equation to show why some crystals, which according to the radius ratio concept should adopt a coordination number of

8

, in fact have a coordination number of

6

MEDIUM

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 10

Explain the term lattice energy as applied to an ionic solid. Calculate the lattice energy of caesium chloride using the following data.

$Cs (s) \to Cs (g)$ $ΔH = + 79.2 kJ {mol}^{- 1}$

$Cs (g) \to {Cs}^{+} (g)$ $ΔH = + 374.05 kJ {mol}^{- 1}$

${Cl}_{2} (g) \to 2 Cl (g)$ $ΔH = + 241.84 kJ {mol}^{- 1}$

$Cl (g) + e \to {Cl}^{-} (g)$ $ΔH = - 397.90 kJ {mol}^{- 1}$

$Cs (s) + \frac{1}{2} {Cl}_{2} \to CsCl (s)$ $ΔH = - 623.00 kJ {mol}^{- 1}$

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 11

Draw the structures of $CsCl$ and ${TiO}_{2}$ , showing clearly the coordination of the cations and anions. Show how the Born-Haber cycle may be used to estimate the enthalpy of the hypothetical reaction:

$Ca (s) + \frac{1}{2} {Cl}_{2} (g) \to CaCl (s)$

Explain why $CaCl (s)$ has never been made even though the enthalpy for this reaction is negative.

HARD

JEE Advanced

IMPORTANT

Concise Inorganic Chemistry>Chapter 2 - The Ionic Bond>Exercise 1>Q 12

The standard enthalpy changes $ΔH °$ at $298 K$ for the reaction: $M (s) + \frac{3}{2} {Cl}_{2} (g) \to {MCl}_{3} (s)$ are given for the first row transition metals:

$\begin{matrix} Sc & Ti & V & Cr & Mn & Fe & Co & Ni & Cu \\ ΔH ° / kJ {mol}^{- 1} & - 339 & - 209 & - 138 & - 160 & + 22 & - 59 & - 131 & - 280 & + 357 \end{matrix}$

Use a Born-Haber cycle to account for the change in $ΔH °$ as the atomic number of the metal increases. Comment on the relative stabilities of the $+ II$ and $+ III$ oxidation states of the $3 d$ metals.

Why are ionic compounds have usually high melting, while most simple covalent compounds have low melting points? Explain the high melting point of diamond.

Important Questions on The Ionic Bond

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