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Olenka [21]
3 years ago
12

Rank the following atoms in order of decreasing first ionization energies (i.e., highest to lowest): Li, Be, Ba, F.

Chemistry
1 answer:
Minchanka [31]3 years ago
5 0

Answer:

The order is:

F >Be >Li >Ba

Explanation:

Electrons are held in atoms by their  attraction to the nucleus which means that to remove an electron from the atom energy is needed.

The ionization energy is the minimum energy necessary to remove an electron from an atom in the gas phase and ground state, the electron removed being the outermost, that is, the furthest from the nucleus. The further away the electron is from the nucleus, the easier it is to remove it, that is, the less energy is needed.

By increasing the atomic number of the elements of the same group, the nuclear attraction on the outermost electron decreases, since the atomic radius increases. Then the ionization energy decreases. In other words, in a group it decreases from top to bottom because the size of the atom increases and it is easier to remove an external electron.

By increasing the atomic number of the elements of the same period, the nuclear attraction on the outermost electron increases, since the atomic radius decreases. Therefore, in a period, as the atomic number increases, the ionization energy increases. In summary, in a period it increases from left to right as the effective nuclear charge increases and it increases thanks to the decrease in the size of the atom.

Taking these considerations into account, the order is:

<u><em>F >Be >Li >Ba</em></u>

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\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

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\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

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