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9966 [12]
3 years ago
9

Using principles of atomic structure, explain why the atomic radius of Te is less than the ionic radius of Te2− . Photoelectron

spectroscopy data for the 1s sublevel of Te and the 1s sublevel of O are represented below. In terms of Coulomb's law and atomic structure, explain why the peak for O is positioned so far to the right of the peak for Te .
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

See explanation

Explanation:

Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.

In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.

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6 0
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Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
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[C₂H₃O₂⁻] = 0,25 M - x.
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