Question

3. a) Explain the reason for the variation of the first ionization energies of the third period  elements. (4) b) Calculate the lattice energy (in Units kJ mol1 ) for ZnO crystal using Eq. 3.4 based on  electrostatic model and using a Born-Haber cycle. Compare the two answers and comment  on any difference. Useful data: Medelung constant (A) = 1.6411 Born Constant (n) = 8 Internuclear distance (a) = 199 pm Zn(s) + ½ O2(g)  ZnO(s) Hf =  350.5 kJ mol1 Zn(s)  Zn(g) HS = 130.4 kJ mol1 Zn(g)  Zn+ (g) I(Zn) = 906.3 kJ mol1 Zn+ (s)  Zn2+(g) I(Zn+ ) = 1733 kJ mol1 ½ O2(g)  O(g) ½Hd = 248.5 kJ mol1 O(g)  O  (g) EA(O) = 141 kJ mol1 O - (g)  O 2 (g) EA(O ) = 780 kJ mol1

Answer 1

Question 1

3.a) Explain the reason for the variation of the first ionization energies of the third period elements.

Solution

First ionisation energies of the period 3 elements

On this diagram we can see that the first ionization energy increases but there two drops: between Mg and Al and P and S. The explanation lies in the electronic structures of atoms.

Na [Ne]3S

Mg [Ne]3S²

Al [Ne]3S²3p¹

We can see in atom of Al one more electron is added but it is added to the p orbital. P orbital is higher in energy than s orbital and electron on this orbital is further from the nucleus. The first ionization energy decreases 1) due to the screening effect of $3S^2$ electrons that reduces the pull from the nucleus and 2) due to the increased distance between 3p electron and nuclei that reduces attraction between them.

Si [Ne]3S²3p²

P Al [Ne]3S²3p³

S [Ne]3S²3p⁴

We can see that three p orbitals in P atom are half- filled $(\mathsf{p_x}^1\mathsf{p_y}^1\mathsf{p_z}^1)$ whereas in S atom one p orbital is fully filled $(\mathsf{p_x}^2\mathsf{p_y}^1\mathsf{p_z}^1)$ . The screening effect is the same for all p electrons, we remove an electron from the same p orbital $(\mathsf{p_x})$ but for the pair of electrons repulsion exists therefore it easier to remove electron from the pair. For the next atoms (Cl and Ar) addition of one more proton has more influence on p electrons than repulsion between electrons of a pair.

Question 2

(4) b) Calculate the lattice energy (in Units kJ mol 1) for ZnO crystal using Eq. 3.4 based on electrostatic model and using a Born-Haber cycle. Compare the two answers and comment on any difference. Useful data: Medelung constant (A) = 1.6411 Born Constant (n) = 8 Internuclear distance (a) = 199 pm

$$\begin{array}{l} Z n (s) + \frac {1}{2} O 2 (g) \rightarrow Z n O (s) \Delta H _ {f} = - 3 5 0. 5 k J / m o l \\ Z n (s) \rightarrow Z n (g) \Delta H _ {s} = 1 3 0. 4 k J / m o l \\ Z n (g) \rightarrow Z n ^ {+} I (Z n) = 9 0 6. 3 k J / m o l \\ Z n ^ {+} \rightarrow Z n ^ {+ 2} I (Z n ^ {+}) = 1 7 3 3 k J / m o l \\ 1 / 2 O _ {2} (g) \rightarrow O (g) \frac {1}{2} \Delta H _ {d} = 2 4 8. 5 k J / m o l \\ O (g) \rightarrow O ^ {-} (g) E A (O) = 1 4 1 k J / m o l \\ O ^ {-} (g) \rightarrow O ^ {- 2} E A (O ^ {-}) = 7 8 0 k J / m o l. \\ \end{array}$$

Solution

Find lattice energy value using electrostatic model.

$$\Delta U = N _ {A} \cdot A \frac {Z _ {1} \cdot Z _ {2} \cdot e _ {0} ^ {2}}{4 \pi \varepsilon_ {0} a} (1 - 1 / n)$$

Medelung constant (A) = 1.6411

Born Constant (n) = 8

Internuclear distance (a) = 199 pm = 199·10 $^{-12}$ m

$$Z _ {1} = + 2$$ $$Z _ {2} = - 2$$ $$e _ {0} = 1. 6 0 2 2 \cdot 1 0 ^ {- 1 9} C$$ $$\varepsilon_ {0} = 8. 8 5 4 \cdot 1 0 ^ {- 1 2} C ^ {2} / m$$ $$\Delta U = 6. 0 2 2 \cdot 1 0 ^ {2 3} \cdot 1. 6 4 1 1 \frac {(1 . 6 0 2 2 \cdot 1 0 ^ {- 1 9}) ^ {2} \cdot 2 \cdot (- 2)}{4 \cdot 3 . 1 4 \cdot 8 . 8 5 4 \cdot 1 0 ^ {- 1 2} 1 9 9 \cdot 1 0 ^ {- 1 2}} (1 - 1 / 8) = 4 0 1 2 3 1 5 \mathrm {J / m o l} = 4 0 1 2 \mathrm {k J / m o l}.$$

Find lattice energy value using a Born-Haber cycle.

According to Hess Law $\Sigma \Delta H i = 0$

$\Delta H (ZnO) = -(-1733 - 906.3 - 130.4 - 248.5 + 141 + 780 - 350.5) = 2447.7 \, \text{kJ/mol}$ .

We can see that lattice energy calculated using electrostatic model is $4012\mathrm{kJ / mol}$ and when using a Born-Haber cycle lattice energy is $2447.7\mathrm{kJ / mol}$ . The accuracy of measuring of lattice energy using a Born-Haber cycle is limited by the accuracy of measuring of EA and I. Moreover good results were obtained only for singly charged ions, metal halides, metal hydride and noble gases. For ionic compounds good results are obtained when electrostatic model is used and we can see that $4012\mathrm{kJ / mol}$ is more close to the table meaning of lattice energy (4061 $\mathrm{kJ / mol}$ ), than $2447.7\mathrm{kJ / mol}$ .

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