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3 years ago

Explain halogens reactivity in terms of outer electron’s distance from nucleus and electronic shielding

Chemistry

1 answer:

yan [13]3 years ago

3 0

Answer:

Fluorine is the most reactive halogen followed by chlorine, bromine and iodine

Explanation:

The reactivity of halogens decreases down the group, that is, fluorine is the most reactive halogen followed by chlorine and bromine. The least reactive halogen is iodine.

There are several reasons and viewpoints explaining this concept:

Fluorine is the smallest atom of all halogens, since: (a) it has the lowest number of electron shells and (b) it has the greatest electronegativity, so that the electrons are attracted with the greatest force towards the nucleus. Therefore, it has the lowest radius. This means when the protons in the nucleus attract the electrons from another atom, the distance between the nucleus of fluorine and the electrons of another atom will be the smallest of all halogens. The lower the distance, the greater the attraction force, so that we obtain the strongest bond with fluorine and the weakest with iodine, as iodine has the greatest radius of all. From this standpoint, fluorine is the most reactive.
Similarly, fluorine is least shielded atom, as it only has two valence shells. Going down the group, the number of shells increases, so iodine is the most shielded atom. Shielding increases the atomic radius, so that it's easier for fluorine to approach the electrons from the valence shell of another atom compared to more shielded chlorine, bromine and iodine.

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Help! Which of these is a compound?: Carbon, chlorine, uranium, ammonia.

MrRa [10]

Ammonia is the compound.

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3 years ago

I need an example of a chemical change involving water.Pleaseeee.

slava [35]

Chemical change occur when two substances are combined and produces a new substance or decomposes into two or more substances which are entirely different from the original two substances.

There are three types of chemical changes. These are 1) Inorganic Changes, 2) Organic Changes, and 3) Biochemical Changes

Here are some examples of chemicsal changes.

If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.

Sodium + Water ==> Sodium Hydroxide and Hydrogen

Na + H2O ====> NaOH and H

Another example of chemical change is:

Carbon Dioxide and Water will decompose into Sugar and Oxygen

Carbon Dioxide + Water ==> Sugar and Oxygen

CO2 + H2O ==> CnH2nOn (where n is between 3 and 7) and O

7 0

3 years ago

How do atoms in molecules bond? plz answer I will give brainliest

spin [16.1K]

Answer:

Via covalent bonds

Explanation:

Atoms in molecules are bonded together via covalent bonds. Covalent bonds are bonds that are formed by sharing of the valence electrons between two atoms.

The atoms can be of the same kind or of different kinds.

In most molecules, the two atoms are connected by sharing of their valence electrons.
This way, each atom can attain stability by becoming isoelectric with the nearest noble gas.
Some molecules are monoatomic, some are polyatomic.

5 0

3 years ago

The label on a bottle of medicine reads "Each 5 mL teaspoonful contains glucose, 1.87 g; levulose, 1.87 g; and phosphoric a

Alexandra [31]

(a) 43.6 mg; (b) 520 mg

(a) Mass of phosphoric acid (PA) in a dose

Mass of PA = 2 tsp × (21.8 mg PA/1 tsp) = 43.6 mg PA

(b) Mass of PA in the bottle

Step 1. Convert ounces to millilitres

Volume = 4 oz × (30 mL/1 oz) = 120 mL

Step 2. Calculate the mass of PA

Mass of PA = 120 mL × (21.8 mg PA/5 mL) ≈ 520 mg PA

4 0

4 years ago

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

5 0

3 years ago