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3 years ago

PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!

2 answers:

7 0

The result of introducing a solution of alkali metal cations to a solution of halogen anions will form <u>halide compounds </u>which is <u>soluble compound</u>

<h3>Further explanation</h3>

The noble gas element is a stable of elements. The noble gas element is monatomic. Stability of noble gases is caused by an electron configuration that has a stable configuration of 8 (has 8 valence electrons) except He with a duplet configuration (has 2 valence electrons)

Other elements that do not yet have electron configurations such as noble gases will try to achieve their stability by forming bonds between elements.

This is generally called the octet rule

Group 1A elements from top to bottom, the easier the elements release electrons, and the more reactive

For this reason, IA elements (also called alkali metal elements) are highly reactive metals.

Alkaline element in the positive ion state (cation) is 1 (+)

The halogen elements have a high electronegativity so that it is easy to attract electrons to fill the electrons so that they are like noble gases

The halogen element in the negative ion state (anion) is 1 (-)

In modern periodic systems, the metal element is generally located on the left while the non-metal is on the right.

The formation of electron configurations such as noble gases can be done by sharing electrons with covalent bonds or hand over electrons with ionic bonds.

Alkali metal reacts with the halogen element to form halide compounds. General equation

Examples of the reaction of Sodium alkali metals with halogens:

2Na (s) + Cl₂ (g) -> 2NaCl (s) (Sodium chloride)

For the solution, for example, the neutralization reaction of NaOH containing alkali Na⁺ and HCl which has the element Halogen Cl⁻

NaOH + HCl ---> NaCl + H2O

Bonding that occurs due to electron handover. Can occur in metal and non-metal atoms.

Like the alkali metal Na will release electrons to form Na + so that the electron configuration is stable as the noble gas element Ne

electron configuration Ne: [He] 2s2 2p6

Na electron configuration: [Ne] 3s¹

electron configuration Na + = [Ne] = [He] 2s2 2p6

While non-metallic Halogen elements like Cl will attract 1 electron so that the electron configuration is like Ar

electron configuration Ar: [Ne] 3s² 3p⁶

electron configuration Cl: [Ne] 3s² 3p⁵

electron configuration Cl- = [Ar] = [Ne] 3s² 3p⁶

The bond between alkali and halogen will form ionic bonding

<h3>Learn more</h3>

the octet rule

brainly.com/question/6979933

calcium obey the octet rule when reacting to form compounds

brainly.com/question/1503241

an exception to the octet rule

brainly.com/question/1412629

fomenos3 years ago

4 0

Explanation:

Alkali metal cations reacts vigorously with halogens to form ionic/electrovalent compounds.

The alkali metals are about the most reactive metals and they belong to group I .
In their outermost shell, they have one valence shell electron.
Alkali metals are largely electropositive willing to release their valence electron to attain stability.
The halogens have seven electrons in their outermost shell. To complete their configuration, they need just one electron.
All the halogens are strongly electronegative and they are readily reactive.
In the vicinity of group 1 elements, they react vigorously.
This is due to the electrostatic attraction between the metal and non-metal ion formed.
Therefore, they form strong ionic bonds between their compounds.

Learn more:

sodium alkali and halogens brainly.com/question/6324347

#learnwithBrainly

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Explanation:

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3 years ago

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(d) $1s^22s^22p^63s^23p^63d^4s^1$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^63d^4s^1$ → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: $(n-1)d^{1-10}ns^{0-2}$ where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^63d^4s^1$

The element having this electronic configuration belongs to the transition family.

4 0

3 years ago

When the metal was placed in the calorimeter its

guajiro [1.7K]

Answer:

Exam 3 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Explanation:

5 0

3 years ago

Read 2 more answers

The chemical equation, 2 Cr + 3 Fe(NO3)2 - 3 Fe + 2 Cr(NO3)3, is an

lana66690 [7]

Answer:

Single replacement reaction (aka single displacement reaction)

Explanation:

In a single replacement reaction, one element is substituted for another in a compound to create a new compound and a new element in the products. The general form is:

A + BC --> B + AC

In the case of this question, Cr and Fe "trade places."

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3 years ago