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fenix001 [56]
3 years ago
15

Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.

Chemistry
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

Yes. Example: <u>Sulfur hexafluoride (SF₆) molecule</u>

Explanation:

According to the octet rule, elements tend to form chemical bonds in order to have <u>8 electrons in their valence shell</u> and gain the stable s²p⁶ electronic configuration.

However, this rule is generally followed by main group elements only.

Exception: <u>SF₆ molecule</u>

In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.

<u>Each fluorine atom has 8 electrons in their valence shells</u>. Thus, it <u>follows the octet rule.</u>

Whereas, there are <u>12 electrons around the central sulfur atom</u> in the SF₆ molecule. Therefore, <u>sulfur does not follow the octet rule.</u>

<u>Therefore, the SF₆ molecule is known as a </u><u>hypervalent molecule</u><u> or expanded-valence molecule.</u>

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3 0
2 years ago
The density of a material is a/an ___________.
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The density of a material is an intensive property.

<h3>What is intensive property?</h3>

An intensive property of matter is one that does not change with the amount of matter. It is a bulk property, which means that it is a physical property that is independent of sample size or mass. An extensive property, on the other hand, is one that is affected by sample size.

<h3>What factors influence an intensive property?</h3>

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5 0
1 year ago
What are precautions to be taken when doing titrations? A-Make sure to always vigorously shake that flask after every drop if th
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Answer:

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6 0
2 years ago
Can someone answer these two questions for me ill give brainliest
gladu [14]
Answer:
1. No
2.a. Nothing will happen to figure 1 as both the sides have 30 N.
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6 0
2 years ago
45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution
Stells [14]
As we know that Molarity is given as,

                                       M = moles / V 
Solving for V,
                                        V = moles / M ------------------(1)
Also, moles is equal to,
                                       moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
                                       V = (mass / M.mass) / M
Putting values,
                                       V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³ 
6 0
3 years ago
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