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topjm [15]
3 years ago
9

What is the predominant intermolecular force in the liquid state of each of these compounds: water (H2O), carbon tetrabromide (C

Br4), and dichloromethane (CH2Cl2)? Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp Dipole-dipole forces Hydrogen bonding Dispersion forces
Chemistry
1 answer:
Valentin [98]3 years ago
8 0

Explanation:

  • Water -

It shows , Hydrogen bonding , Since ,

The molecule of water have a strong electronegative atom of oxygen and a hydrogen atom ,

So , the electronegative oxygen atom can interact with the other hydrogen atom of the other water molecule .

  • Carbon tetrabromide -

The molecule of Carbon tetrabromide have polar bonds of carbon and bromine atom , but overall the molecule is non - polar in nature , due to its tetraderal structure , hence ,

the molecule shows only dispersion forces .

  • dichloromethane -

The molecule of dichloromethane has Dipole Dipole forces due to difference in electronegativity between atoms .

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In order: LOWER, HIGHER, NEGATIVE, POSITIVE, INCREASE

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First you must consider the definition of electronegativity.

All in all, electronegativity is the ability of an atom to attract electrons to itself. For this explanation, let's just consider the groups 1, 2, 13-17, and the noble gases of the periodic table. Furthermore, let's also consider the octet rule. The octet rule says that the last orbit of an atom tends to have 8 electrons in order to be stable.

If you take an atom like Sodium (Na) and search its distribution of electrons by levels, you'll see these number in the last two orbits: 8.1

The penultimate orbit has 8 electrons and the last one has only 1. Therefore, in order to futfill the octet rule what do you think it's easier? Losing the last electron and leaving the 8 electrones of the penultimate orbit as last orbit? Or maybe taking 7 electrons for the last orbit and, hence, futfilling the octet? Well, the answer is the first case.

The more you move from group to group (i.e, rightwards in the periodic table), more electrons will have in their last orbit.

So, in order to respond your question: Yes you should display it like this:

A --> B

Consider A and B two random atoms, for which B's electronegativity is higher than A's. This arrow shows the tendency of the electron flow.

As electrons go to the more electronegative atom, the whole electronic ''cloud'' will tend to be closer to B, because this atom is more electronegative. Hence, there will be a partial (not fixed) negative charge on B rather than on A. This is due to B's high electronegativity. This means that the negative charge will spend more time on B rather than on A, this is what partial negative charge means.

The higher the electronegativity is, the higher the magnitudes of the partial charges --> Therefore, these partial charges are increased and sometimes it might lead to the chemical bond breakage.

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