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3 years ago

(a) (1)

Chemistry

1 answer:

Elis [28]3 years ago

8 0

Explanation:

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g) ® M+(g) + e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge. The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g) ® M2+(g) + e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g) ® M3+(g) + e-

and so on. Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms. The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus. Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same. The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons. The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

The following charts illustrate the general trends in the first ionization energy:

Dunno kung tama beng pero trysorry kung mali

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Answer:

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$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

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Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
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First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?

The equation for this process is $\text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}$ .

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