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s2008m [1.1K]
4 years ago
7

When atoms lose more than one electron, the ionization energy to remove the second electron is always more than the ionization e

nergy to remove the first. Similarly, the ionization energy to remove the third electron is more than the second and so on. However, the increases in ionization energy upon the removal of subsequent electrons is not necessarily uniform. For example, consider the first three ionization energies of magnesium: First ionization energy 738kJ/mol Second ionization energy 1450kJ/mol Third ionization energy 7730kJ/mol The second ionization energy is roughly twice the first ionization energy, but then the third ionization energy is over five times the second. Use the electronic configuration of magnesium to explain why this is so? Would you expect the same behavior in sodium? Why or why not?
Chemistry
2 answers:
Ugo [173]4 years ago
4 0

Answer:

The behaviors of magnesium and sodium are different.

Explanation:

It can be said that in the periodic table the elements are organized according to their atomic number. In the periodic table the electronic configuration of each element is identified, in addition to its physical and chemical properties. On the other hand, ionization energy is defined as the energy needed to remove an electron from an atom. In the case of magnesium, its electronic configuration is as follows:

When the first ionization occurs, its electronic configuration is: [Ne] 3s²

A second ionization, its electronic configuration is: [Ne] 3s¹

In sodium, its electronic configuration is [Ne]

When the first ionization occurs, sodium has an electron configuration similar to that of a noble gas ([Ne] 3s¹). A second ionization energy increases dramatically. In conclusion, the behaviors of magnesium and sodium are very different.

elena-s [515]4 years ago
3 0

The electronic configuration of magnesium is:

1s² 2s² 2p⁶ 3s² = [Ne] 3s²

This means that for a magnesium atom, in order to have its outermost orbital full, the easiest way would be to lose the two outermost electrons. This is seen in its relatively low first two ionization energies. <u>The third ionization energy is several times higher because the ion would move from a stable form to an highly unstable form.</u> (Mg⁺² → Mg⁺³ + e⁻).

  • Sodium only has one electron in its outermost orbital, so its first ionization energy would be several times lower than the second.
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At 1000 K the Kc value for the following reaction is 1.5 × 10⁶ (1) N₂O₄ (g) ⇌ 2 NO₂ (g) At 1000 K the Kc value for the following
Ahat [919]

Answer:

The value of Kc is 9.33 * 10^-17

Explanation:

Step 1: Data given

Temperature = 1000 K

N₂O₄ (g) ⇌ 2 NO₂ (g) Kc = 1.5 * 10^6

N₂ (g) + 2 O₂(g) ⇌ 2 NO₂ (g)  Kc = 1.4 * 10^-10

Step 2: Calculate Kc for  N₂ (g) + 2 O₂ ⇌ N₂O₄

N₂O₄ (g) ⇌ 2 NO₂ (g) Kc = 1.5 * 10^6

2 NO₂ (g) ⇌  N₂O₄ (g)    = 1/Kc = 1 / 1.5 * 10^6   = 6.67 *10^-7

N₂ (g) + 2 O₂(g) ⇌ 2 NO₂ (g)  Kc = 1.4 * 10^-10

Kc = Kc1 * Kc2

Kc = (6.67 *10^-7) * (1.4 * 10^-10)

Kc = 9.33 * 10 ^-17

The value of Kc is 9.33 * 10^-17

4 0
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Explanation:

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V = Volume of gas in L = ?

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