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kramer
3 years ago
12

Which of the following atoms has the largest atomic radius and explain why: potassium, gallium, or germanium.

Chemistry
2 answers:
kotegsom [21]3 years ago
8 0

Answer:

<em>Potassium</em>

Explanation:

<em>An atom's size is defined by the distance from the nucleus to the valence (outer) shell of electrons. Although difficult to to actually measure, due to electrons existing in a probability field with no defined boundaries, it is common knowledge that the more valence shells of an atom, the larger the atom.</em>

<em>Knowing this, lets look at the electron configuration of the elements mentioned:Oxygen = 2,6Sodium = 2,8,1Magnesium = 2,8,2Potassium = 2,8,8,2As we can see, only Potassium has four electron shells, therefore Potassium has the largest atomic size.Alternatively, one can look at a Van der Waal radii table to find that the radii of the four elements are (in picometres):Oxygen = 152 pmSodium = 227 pmMagnesium = 173 pm</em>

<em>Showing once again how Potassium has the largest atomic radius, and therefore the largest atomic size. Interestingly enough, Sodium has a larger atomic radius then magnesium, despite having less electrons and protons.</em>

Leni [432]3 years ago
3 0

Answer:

Potassium

Explanation:

The atom with the largest atomic radius is potassium from the given list.

The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between nuclei in the solid metal state.

Across a period in the periodic table, atomic radii decreases progressively from left to right.

This is due to the progressive increase in the nuclear charge without an attendant increase in the number of electronic shells.

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Answer:

See explanation

Explanation:

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3 years ago
The atomic radii of a divalent cation and a monovalent anion are 0.62 nm and 0.121 nm, respectively. (a) Calculate the force of
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Answer:

\boxed{8.4\times 10^{-10} \text{ N}}

Explanation:

The formula for the force exerted between two ions is  

F = \dfrac{kq_{1}q_{2}}{r^{2}} = \dfrac{kz_{1}z_{2}e^{2}}{r^{2}}

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z₁ = +2; r₁ = 0.62    nm

z₂ =  -1;  r₂ = 0.121 nm

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F =\dfrac{8.988 \times 10^{9} \text{ N $\cdot$ m$^{2} \cdot$ C$^{-2}$} \times 2 \times (-1) \times ( 1.602 \times 10^{-19} \text{ C})^{2}}{(0.741 \times 10^{-9} \text{ m})^{2}}\\\\ = -\mathbf{8.4\times 10^{-10}} \textbf{ N}\\\text{The attractive force between the ions is } \boxed{\mathbf{8.4\times 10^{-10}} \textbf{ N}}

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