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3 years ago

Which element has atoms that can bond with each other to form ring, chain, and network structures?

2 answers:

6 0

This is basics of organic chemistry - the element that has atoms that can bond with each other to form ring, chain, and network structures is (3) carbon.
The other elements available above cannot form such structures, whereas carbon, which is abundant in nature, can.

Ipatiy [6.2K]3 years ago

3 0

Answer: The correct answer is Option 3.

Explanation:

The property in which atoms of the same element can form ring, chain or network structures is known as catenation.

From the elements given in the options, only carbon shows this property.

Carbon can bond with other carbon atom to form ring. Example: cyclopentane, cyclohexane etc..

Carbon can bond with other carbon atom to form long chains. Example: fatty acids.

The atoms of carbon bond with each other to form network structures. Example: Buckminster fullerene

Hence, the correct answer is Option 3.

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Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$ *100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$ *100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$ *100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$ *100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$ *100%

%IC = 55.5%

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