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During the formation of $MgC{l_2}$ by Born-Haber cycle, which of the following occurs?
A. Ionisation of Mg to $M{g^{2 + }}$.
B. Bond dissociation of $C{l_2}$.
C. Sublimation of Mg.
D. Electron gained by Cl.
E. All of these.

Last updated date: 25th Jun 2024
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Hint: In the Born Haber cycle, the ionic compound is formed by a combination of elements. It Born Haber cycle summation of enthalpy of all the processes taking place determines the net enthalpy formation of ionic compounds from its respective element in its standard condition.

Complete answer:Born-Haber cycle of $MgC{l_2}$
The formation of magnesium chloride from magnesium metal and chlorine gas is shown below.
$Mg + 2Cl \to MgC{l_2}$
The formation of ionic magnesium chloride from solid magnesium metal and chlorine gas takes place in a single step but it goes through several steps.
Solid magnesium sublimes to form gaseous atoms by absorbing heat energy $(\Delta {H_{sub}})$.
$Mg(s) \to Mg(g)$
Gaseous magnesium atoms absorb ionization energy to release one electron to further generate gaseous magnesium ions.
$Mg(g) \to M{g^{2 + }}$
The diatomic chlorine gas absorbs bond dissociation energy to form gaseous chlorine atoms.
$\frac{1}{2}C{l_2} \to Cl(g)$
STEP 4: The chlorine atom accepts two electrons to form chloride ions.
$Cl(g) + 2e \to C{l^{2 - }}(g)$
Thus, during the formation of $MgC{l_2}$ by Born-Haber cycle, the processes ionisation of Mg to $M{g^{2 + }}$, Bond dissociation of $C{l_2}$, sublimation of Mg and electron gained by Cl takes place.
Therefore, the correct option is E.

 In the last step the gaseous magnesium and gaseous chloride ion combine to form magnesium chloride by releasing energy same as the lattice energy which cannot be calculated experimentally but all the other processes can be calculated experimentally.