Question

1

a. One of the properties of ionic compounds is hard but brittle. Briefly explain why? b. The data is known as follows: atomization energy Cu and Cl = 338 and 121 kJ / mol, respectively; first and second ionization energies Cu 747 and 1958 kJ / mol; electron affinity (energy produced to release electrons) Cl = 349 kJ / mol, lattice energy of CuCl and CuCl2 973 and 2772 kJ/ mol, respectively. (If needed, atomic number Cu = 29 CI= 17) i. Explain why the lattice energy of CuCl2 is greater than CuCl ii. With the Bohr-Harber cycle and from the data above, calculate the energy formation of CuCl and CuCl2 salts. ii. Which salt is more stable in nature. Explain

2. for molecules CS2 and CH3F

a.describe its lewis structure

b.the kind of orbitals used by the central atoms to bond

c.polarity

d.approximate geometry

3.explain the formation of sigma and phi bonds in the XeO3.molecule illustrate its structure and explain its polarity

4.explain why

a. melting point NaF> NaCl but CF4 <CCl4

b. Boiling point HF> HCl but CF4 <CCl4

Answer 1

Q1 (a) in ionic compounds , the ions are placed at their fixed positions. They do not slide over each other and hence these are not ductile.

When a high pressure is applied , ionic compounds would break unevenly.

(b)

(I) lattice energy of CuCl₂ is more than CuCl because lattice energy is directly proportional to the product of charges on cation and anion.

(ii) Cu(s) + 1/2 Cl₂(g) $\rightarrow$CuCl(s). ∆H = ?

Applying Born-harber cycle:

∆H = enthalpy of atomisation of Cu + enthalpy of atomization of Cl + I.E.₁ of Cu + enthalpy of electron gain of Cl + lattice energy

= 338 + 121 + 747 + (-349) +( - 973)

= -116kJ/mol

Cu(s) + Cl₂ (g) $\rightarrow$ CuCl₂(s). ∆H =?

∆H = enthalpy of atomisation of Cu + 2 × (enthalpy of atomization of Cl) + (I.E.₁ + I.E.₂) of Cu + 2×(enthalpy of electron gain of Cl) + lattice energy of CuCl₂

= 338 + 2×121 + (747 +1958) + 2×(-349) + (-2772)

= -185kJ/mol

(iii) CuCl₂ salt is more stable.

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Q4:

(a) melting point of NaF is more than that of NaCl because of higher lattice energy of NaF due to smaller size of F^- ion. But CCl₄ has higher melting point than CF₄ becuase of the higher molecular mass of CCl₄.

(b)

Boiling point of HF is more than that of HCl due to Hydrogen bonding in HF.