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Pani-rosa [81]
2 years ago
8

In general, the higher the charge on the ions in an ionic compound, the more favorable is the lattice energy. Why do some stable

ionic compounds have +1 charged ions even though +4, +5, and +6 charged ions would have a more favorable lattice energy?
Chemistry
1 answer:
Sever21 [200]2 years ago
7 0

Answer:

The main reason why some stable ionic compounds have +1 charged ions, even though other higher would have a more favorable lattice energy is because;

1. Energy is required to remove each of the electrons in an atom, and the more more the electrons are removed, the smaller the atom radius becomes, with increasing nuclear attraction between the opposite charges.

Note, it is a bit easy to remove a single electron, when compared to 4, 5 or 6 electrons. hence, the reason why Ionic compounds are more stable with + charged ions than others.

2. The stability of an ionic compound is dependent on his lattice energy. And the ions with large charge will have a greater attraction between the cations and anions, hence have higher lattice energy. However, the distance between the ions and the nucleus of the atom plays a major role in the stability of the ion.

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Which electron transition represents a gain of energy
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The transition from lower energy level to higher energy level require a gain of energy.

Explanation:

When transition occur from lower energy level to higher energy level require a gain of energy. Electron could not jump unto higher energy level without gaining thew energy.

When electron jump into lower energy level from high energy level it loses the energy.

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The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

3 0
3 years ago
Which of these particles are lost in the oxidation process ?
elena-14-01-66 [18.8K]
The answer is electrons !. Hope it helps !! :)
8 0
2 years ago
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a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula
kolezko [41]

The chemical compound's empirical formula is NS.

The chemical compound's molecular formula is N4S4.

<h3>What does a chemical empirical formula look like?</h3>
  • The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
  • This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
  • The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
  • Glucose has the empirical formula CH2O.
  • Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.

learn more about empirical formula here

brainly.com/question/1603500

#SPJ4

the question you are looking for is

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?

3 0
1 year ago
In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0
2 years ago
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