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3 years ago

Among the elements of the main group, the first ionization energy increases

2 answers:

7 0

Among the elements of the main group, the first ionization energy increases $\boxed{{\text{a}}{\text{. from left to right across a period}}}$ .

Further Explanation:

Ionization energy:

It is the minimum amount of energy that is required to remove the most loosely bound electron of a neutral isolated gaseous atom. It is usually represented as IE. If less amount of energy is needed for electron removal, it is said to have less ionization energy and vice-versa.

When the first electron is to be removed from the neutral atom, the required energy is called the first ionization energy, denoted by . Similarly, when the second electron is removed, the ionization energy is termed as the second ionization energy and is represented by .

The representation of first ionization energy is as follows:

${\text{A}} - {e^ - }\xrightarrow{{{\text{ I}}{{\text{E}}_1}}}{{\text{A}}^ + }$

The representation of second ionization energy is as follows:

${{\text{A}}^ + } - {e^ - }\xrightarrow{{{\text{ I}}{{\text{E}}_2}}}{{\text{A}}^{2 + }}$

While moving from left to right in a period, the atomic number increases but the electrons are to be added in the same valence shell. This results in an increase in the effective nuclear charge across a period due to which the outermost electrons experience stronger attractive forces with the atomic nucleus and atomic size decreases. Since atomic size decreases in a period, the first ionization energy increases from left to right while moving across a period. The main group elements show regular trends in the first ionization energy throughout the period.

When we move from top to bottom, the atomic number increases as well as the number of shells increases which results in an increase in the atomic size down the group. Since size increases down the group, the first ionization energy decreases in a group. The main group elements show regular trends in the first ionization energy down the group.

Learn more:

Rank the elements according to first ionization energy: brainly.com/question/1550767
Write the chemical equation for the first ionization energy of lithium: brainly.com/question/5880605

Answer details:

Grade: Senior School

Chapter: Periodic classification of elements

Subject: Chemistry

Keywords: ionization energy, first ionization energy, group, period, atomic number, IE, IE1, IE2, energy, main group elements.

Contact [7]3 years ago

4 0

Answer is: a. from left to right across a period.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

Alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons (one electron, earth alkaline metals (right next to alkaline metals) have higher ionization energy than alkaline metals, because they have two valence electrons.

Nonmetals are far right in the main group and they have highest ionization energy, because they have many valence electrons.

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Answer:

Explanation:

The cell reaction properly written is shown below:

Cu|Cu²⁺ $_{aq}$ || Ag⁺ $_{aq}$ | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

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Reduction half:

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At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

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we multiply the second reaction by 2 to balance up:

2Ag⁺ $_{aq}$ + 2e⁻ ⇄ 2Ag $_{s}$

The net reaction equation:

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We then cancel out the electrons from both sides since they appear on both the reactant and product side:

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