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Q3.31(e):
The first ($\Delta_{i}H_{1}$) and the second ($\Delta_{i}H_{2}$) ionization enthalpies (in kJ $mol^{-1}$) and the ($\Delta_{eg}H$) electron gain enthalpy (in kJ $mol^{-1}$) of a few elements are given below: Elements $\Delta H_{1}$ $\Delta H_{2}$ $\Delta_{eg}H$ I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 Which of the above elements is likely to be : (e) the metal which can form a stable binary halide of the formula $MX_{2}$($X$=halogen).

Solution :

Step 1: Theoretical Analysis of Binary Halide Formation

A stable binary halide of the formula $MX_2$ implies that the metal $M$ typically exhibits an oxidation state of $+2$. For an element to form a stable $MX_2$ compound, the energy required to remove two electrons (the sum of the first and second ionization enthalpies, $\Delta_i H_1 + \Delta_i H_2$) must be low enough to be compensated by the lattice energy of the resulting ionic crystal or the bond dissociation energy in a covalent network [Per Born-Haber Cycle principles].

Step 2: Evaluating the Energetics of Potential Candidates

We analyze the sum of the first two ionization enthalpies ($\Sigma\Delta_i H = \Delta_i H_1 + \Delta_i H_2$) for the provided elements to identify which are most likely to form divalent cations ($M^{2+}$):

Element $\Delta_i H_1$ $\Delta_i H_2$ $\Sigma\Delta_i H$ (kJ $mol^{-1}$)
I 520 7300 7820
II 419 3051 3470
III 1681 3374 5055
IV 1008 1846 2854
V 2372 5251 7623
VI 738 1451 2189

Step 3: Justification for Element VI

  • Elements I and V: The jump between $\Delta_i H_1$ and $\Delta_i H_2$ is massive (I: $7300 - 520 = 6780$; V: $5251 - 2372 = 2879$), indicating that these are likely Alkali Metals (Group 1), which form $M^+$ ions, not $M^{2+}$.
  • Element VI: The $\Sigma\Delta_i H$ value is the lowest in the table ($2189$ kJ $mol^{-1}$). A relatively low $\Delta_i H_2$ suggests that the second electron is removed from the same shell or subshell without a massive energy penalty, which is characteristic of Alkaline Earth Metals (Group 2, e.g., Magnesium). Group 2 elements characteristically form stable $MX_2$ halides (e.g., $MgCl_2$).
  • Comparison: While Element IV also has a relatively low $\Sigma\Delta_i H$, Element VI fits the periodic trend of an Alkaline Earth Metal more precisely, where the configuration allows for the stable loss of two valence $s$-electrons.

Step 4: Verification of Periodic Properties

The ionization energy profile of Element VI ($738$ kJ $mol^{-1}$ for $\Delta_i H_1$ and $1451$ kJ $mol^{-1}$ for $\Delta_i H_2$) matches the expected trend for a metal located in Group 2 of the Periodic Table. By the Law of Conservation of Energy and Lattice Energy stability, the formation of $M^{2+}$ is exothermic or energetically favorable for such metals when forming halides [Per Born-Haber Cycle].

Elements Comparison VI I V

Final Solution: Element VI is the metal which can form a stable binary halide of the formula $MX_2$.


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