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Montano1993 [528]
3 years ago
8

Why are the trends and exceptions to the trends in ionization energy observed? Check all that apply. Ionization energy tends to

increase down a group because the electrons get farther away from the nucleus. Ionization energy tends to increase across a period because the nuclear charge increases. Ionization energy tends to increase across a period because electrons are added to the same main energy level. The ionization energies of the elements in Group 16 tend to be slightly smaller than the elements in Group 15 because the fourth electron is added to an unfilled p orbital. The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.
Chemistry
2 answers:
NNADVOKAT [17]3 years ago
5 0
The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus. 
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
Georgia [21]3 years ago
4 0

answers are

2) Ionization energy tends to increase across a period because the nuclear charge increases.

3)Ionization energy tends to increase across a period because electrons are added to the same main energy level.

5) The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.


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The complex [Ni(CN)4]2- is diamagnetic and the complex [NiCl4]2- is paramagnetic. What can you conclude about their molecular ge
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[Ni(CN)4]2- square planar

[NiCl4]2- tetrahedral

Explanation:

For a four coordinate complex such as [Ni(CN)4]2- and [NiCl4]2-, we can decide its geometry by closely considering its magnetic properties. Both of the complexes are d8 complexes which could be found either in the tetrahedral or square planar crystal field depending on the nature of the ligand.

CN^- being a strong field ligand leads to the formation of a square planar diamagnetic d8 complex of Ni^2+. Similarly, Cl^- being a weak field ligand leads to the formation a a tetrahedral paramagnetic d8 complex of Ni^+ hence the answer given above.

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Which statement is true for the given reaction? Na2SO4(aq) + Mg(s) →
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Does the pH increase or decrease, and does it do so to a large or small extent, with each of the following additions?(c) 5 drops
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The pH decreases to a large or small extent with each of the given additions.

<h3>What is common name of NaOH?</h3>

The common name of NaOH is sodium hydroxide. Lye and caustic soda are other names for sodium hydroxide, an inorganic substance having the formula NaOH. It is a white, solid ionic substance made up of the cations sodium (Na+) and the anions hydroxide (OH). Sodium hydroxide is a chemical that manufacturers utilize to make things like soap, rayon, paper, explosives, colors, and petroleum products. Processing cotton fabrics, metal cleaning and processing, oxide coating, electroplating, and electrolytic extraction are further uses for sodium hydroxide. A caustic metallic base is sodium hydroxide (NaOH), sometimes referred to as lye or caustic soda. Caustic soda, an alkali, is commonly employed in a variety of sectors, primarily as a potent chemical base in the production of pulp and paper, textiles, drinking water, and detergents.

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For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re
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Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

Ba2+(aq) + SO42-(aq) → BaSO4(s)

Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

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Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

this will help, I used this for my work x

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