Chapter 6: Problem 19

Give reason for the following : (a) \(\mathrm{Agl}_{2}^{-}\) is stable but \(\mathrm{AgF}_{2}^{-}\) is unstable. (b) \(\mathrm{CF}_{4}\) is more stable than \(\mathrm{CH}_{3} \mathrm{~F}\).

Short Answer

Expert verified

\(\mathrm{Agl}_{2}^{-}\) is stable due to the stable electronic configuration of the silver ion, while \(\mathrm{AgF}_{2}^{-}\) is unstable due to the higher effective nuclear charge of fluorine. \(\mathrm{CF}_{4}\) is more stable than \(\mathrm{CH}_{3} \mathrm{~F}\) due to its non-polar nature and the cancellation of bond dipoles in its symmetrical tetrahedral structure, which makes it more able to resist molecular motion.

Step by step solution

Stability of \(\mathrm{Agl}_{2}^{-}\) vs \(\mathrm{AgF}_{2}^{-}\)

The stability of these entities can be attributed primarily to the effective nuclear charge of the central atom. In \(\mathrm{Agl}_{2}^{-}\), the silver ion Ag+ gets a stable configuration after accepting an electron, making \(\mathrm{Agl}_{2}^{-}\) stable. However, in \(\mathrm{AgF}_{2}^{-}\), the fluorine's higher effective nuclear charge destabilizes the ion, making it unstable.

Stability of \(\mathrm{CF}_{4}\) vs \(\mathrm{CH}_{3} \mathrm{~F}\)

The stability of these compounds can be related to the polarity and strength of the bonds within the molecule. \(\mathrm{CF}_{4}\) is a symmetrical tetrahedral molecule, so the bond dipoles of the C-F bonds cancel each other out, resulting in a non-polar molecule, hence it is more stable. On the other hand, \(\mathrm{CH}_{3} \mathrm{~F}\) is a polar molecule. Even though C-F bond is stronger due to electronegativity difference, the dipole-dipole intermolecular forces present within \(\mathrm{CH}_{3} \mathrm{~F}\) are much less effective at preventing molecular motion compared to the symmetrical \(\mathrm{CF}_{4}\), which makes \(\mathrm{CF}_{4}\) more stable than \(\mathrm{CH}_{3} \mathrm{~F}\).

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Give reason for the following : (a) \(\mathrm{Agl}_{2}^{-}\) is stable but \(\mathrm{AgF}_{2}^{-}\) is unstable. (b) \(\mathrm{CF}_{4}\) is more stable than \(\mathrm{CH}_{3} \mathrm{~F}\).

Short Answer

Step by step solution

Stability of \(\mathrm{Agl}_{2}^{-}\) vs \(\mathrm{AgF}_{2}^{-}\)

Stability of \(\mathrm{CF}_{4}\) vs \(\mathrm{CH}_{3} \mathrm{~F}\)

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