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4 years ago

Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?

Chemistry

2 answers:

DochEvi [55]4 years ago

8 0

Answer:

option b is correct

2 p electron makes aromatic

Explanation:

An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)

Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.

where n is the number of pi electrons

where n = 0

(4n +2) pi electrons = 2pi electrons

attached is an example of aromatic which is cyclic, planar and a complete conjugate ring

den301095 [7]4 years ago

8 0

Answer:

2 p electrons.

Explanation:

For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.

n can represent an integer from 2, 6,10, 14,........

The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.

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4 years ago

Which of the following weathering processes involves the constant freezing and thawing of water?

Artemon [7]

Frost wedging

Explanation:

Weathering is the physical disintegration and chemical decomposition of rocks to form sediments and soils.

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3 years ago

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will r

dusya [7]

The question is incomplete, the complete question is:

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will result in a reaction.

$Li(s)\rightarrow Li^+(aq)+e^-$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

A) Li(s) with Al(s)

B) Li(s) with $Al^{3+}$ (aq)

C) $Li^+$ (aq) with Al(s)

D) $Li^+$ (aq) with $Al^{3+}$ (aq)

Answer: The correct option is B): Li(s) with $Al^{3+}$ (aq)

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction.

The chemical species will undergo a reduction reaction if the value of standard reduction potential is more positive or less negative.

For the given half-reactions:

$Li(s)\rightarrow Li^+(aq)+e^-;E^o_{Li^+/Li}=-3.04V$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-;E^o_{Al^{3+}/Al}=-1.662V$

As the value of standard reduction potential of aluminium is less negative. Thus, it undergoes reduction reaction and lithium will undergo oxidation reaction.

The half-reaction follows:

Oxidation half-reaction: $Li(s)\rightarrow Li^+(aq)+e^-$ ( × 3)

Reduction half-reaction: $Al^{3+}(aq)+3e^-\rightarrow Al(s)$