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weqwewe [10]
2 years ago
7

How might atomic radius have any influence on electronegativity trends or no influence on ionizatation

Chemistry
1 answer:
VMariaS [17]2 years ago
7 0
The radius is proportional to the ionization energy.

This is because electrons are drawn closer to protons, which have opposite charges and hence cling to them, in a small-radius atom.
If the radius is bigger, the electrons on the outside edge of the atom are not as tightly bound and are therefore more easily lost, requiring less energy to ionize.
Factors are more shielding (from core electrons) in the lowest elements of a family, allowing electrons to escape more easily. For those who are currently in a period, the effective nuclear charge grows as the period progresses (more protons, but no more energy levels, so the electrons are the same distance from the nucleus). This causes the electrons to be held closer together (smaller radius), requiring more energy to ionize them.
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3 years ago
The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12
Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu

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