Question

Consider the following Born-Haber's cycle for formation of MX3(s).

$M(s) + \frac{3}{2}X_2(g) \xrightarrow{∆H_f = -740 \text{ kJ mol}} MX_3(s)$

Then calculate value $q_1$, here $q_1$ is electron affinity of X(g) in kJ/mol. Given B.E. of $X_2 = 200$ kJ/mole The group number is (a) and period number is (b) of the element with atomic number Z = 100. Report value of (10a + b).

Answer 1

37

Answer 2

The problem consists of two independent parts.

Part 1: Calculate the electron affinity ($q_1$) using the Born-Haber cycle.

The Born-Haber cycle is an application of Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction.

The overall reaction for the formation of $MX_3(s)$ is: $M(s) + \frac{3}{2}X_2(g) \xrightarrow{∆H_f = -740 \text{ kJ mol}} MX_3(s)$

Let's list the enthalpy changes for each step in the cycle as provided in the diagram:

1. Sublimation of M(s): $M(s) \rightarrow M(g)$ $∆H_{sub} = 150 \text{ kJ/mol}$

2. Ionization of M(g) to M³⁺(g): $M(g) \rightarrow M^{3+}(g) + 3e^-$ $∆H_{ion} = 350 \text{ kJ/mol}$ (This represents the sum of the first three ionization energies of M).

3. Dissociation of X₂(g): $\frac{3}{2}X_2(g) \rightarrow 3X(g)$ Given Bond Energy (B.E.) of $X_2 = 200 \text{ kJ/mol}$. This is the energy to break 1 mole of $X_2$ into 2 moles of $X$. So, for $\frac{3}{2}$ moles of $X_2$, the energy required is: $∆H_{diss} = \frac{3}{2} \times \text{B.E.}(X_2) = \frac{3}{2} \times 200 \text{ kJ/mol} = 300 \text{ kJ/mol}$. This value is denoted as 'p' in the diagram, so $p = 300 \text{ kJ/mol}$.

4. Electron Affinity of X(g): $3X(g) + 3e^- \rightarrow 3X^{-}(g)$ The energy change for this step is 'q'. The question states that $q_1$ is the electron affinity of X(g) in kJ/mol, which means $X(g) + e^- \rightarrow X^{-}(g)$. Therefore, $q = 3 \times q_1$.

5. Lattice Energy of MX₃(s): $M^{3+}(g) + 3X^{-}(g) \rightarrow MX_3(s)$ The lattice energy ($U$) is given as $-1000 \text{ kJ/mol}$.

According to Hess's Law: $∆H_f = ∆H_{sub} + ∆H_{ion} + ∆H_{diss} + q + U$

Substitute the known values into the equation: $-740 \text{ kJ/mol} = 150 \text{ kJ/mol} + 350 \text{ kJ/mol} + 300 \text{ kJ/mol} + q + (-1000 \text{ kJ/mol})$ $-740 = 150 + 350 + 300 + q - 1000$ $-740 = 800 + q - 1000$ $-740 = -200 + q$ $q = -740 + 200$ $q = -540 \text{ kJ/mol}$

Since $q = 3 \times q_1$: $-540 = 3 \times q_1$ $q_1 = \frac{-540}{3}$ $q_1 = -180 \text{ kJ/mol}$

Part 2: Determine the group number (a) and period number (b) of the element with atomic number Z = 100.

The element with atomic number Z = 100 is Fermium (Fm). To determine its position in the periodic table, we write its electron configuration. The noble gas preceding Z=100 is Radon (Rn), with Z=86. The electron configuration of Fermium (Fm) is $[Rn] 5f^{12} 7s^2$.

• Period Number (b): The period number is determined by the highest principal quantum number (n) of the valence shell. For Fermium, the highest principal quantum number is 7 (from $7s^2$). So, b = 7.
• Group Number (a): Fermium is an actinide, which is an f-block element. All f-block elements (lanthanides and actinides) are conventionally placed in Group 3 of the periodic table. So, a = 3.

Finally, we need to report the value of $(10a + b)$: $10a + b = 10(3) + 7$ $10a + b = 30 + 7$ $10a + b = 37$

The question asks to report the value of (10a + b).

The final answer is $\boxed{37}$.

Explanation:

1. Born-Haber Cycle Calculation: Apply Hess's Law: $\Delta H_f = \Delta H_{sub} + \sum IE + \Delta H_{diss} + \sum EA + U$. Substitute the given values: $-740 = 150 + 350 + (\frac{3}{2} \times 200) + 3q_1 + (-1000)$. Solve for $q_1$: $-740 = 150 + 350 + 300 + 3q_1 - 1000 \implies -740 = 800 + 3q_1 - 1000 \implies -740 = -200 + 3q_1 \implies 3q_1 = -540 \implies q_1 = -180 \text{ kJ/mol}$.
2. Periodic Table Position: For Z=100 (Fermium, Fm), the electron configuration is $[Rn] 5f^{12} 7s^2$. The highest principal quantum number is 7, so period (b) = 7. Fermium is an actinide, an f-block element, which belongs to Group 3. So, group (a) = 3.
3. Final Value: Calculate $10a + b = 10(3) + 7 = 30 + 7 = 37$.