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gizmo_the_mogwai [7]
2 years ago
6

What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?

Chemistry
1 answer:
zhuklara [117]2 years ago
5 0

This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time,  define the molecular geometry.

<h3>Molecular geometry:</h3>

In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.

Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.

In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)

Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

Learn more about molecular geometry: brainly.com/question/7558603

Learn more about the VSEPR theory:  brainly.com/question/14225705

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Which of the following statements is true about both nuclear fusion and nuclear fission?
creativ13 [48]

The correct answer is A.

B is incorrect because that only applies to nuclear fission.

C is incorrect because it only applies to nuclear fusion.

D is incorrect because energy can be neither created nor destroyed meaning that this statement is physically impossible,

8 0
3 years ago
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Es un cambio natural o provocado, donde se produce energía para agua y dióxido de carbono: a)Precipitación. b)Fermentación c) Ef
Andre45 [30]

Answer:

d) Combustión.

Explanation:

¡Hola!

En este caso, dado que estamos enfocados en el concepto de cambio químico, el cual se caracteriza por exhibir un cambio en la composición e identidad de las sustancias iniciales (reactivos) a otras finales (productos).

Ahora bien, como se nos dice que los productos de este cambio químico son energía, agua y dióxido de carbono, inferimos que el nombre de este proceso es d) Combustión, por ejemplo la combustion del gas natural para calentar nuestras comidas en la cocina.

¡Saludos!

6 0
3 years ago
Write the formulas for the following compounds: (a) rubidium nitrite, (b) potassium sulfide, (c) sodium hydrogen sulfide, (d) ma
choli [55]

Answer:

All are having different valent cation and anion like mono,di and trivalent polyatomic ions .

A. RbNO3

B. K2S

C. NaHS

D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-

E. CaHPO4

F. PbCO3 , lead is in Pb+2 form

G. SnF2

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6 0
3 years ago
Why is the atmosphere of Earth the outermost layer of the planet?
nasty-shy [4]
Option C but i am not sure
3 0
2 years ago
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____ 26. what is the weight percentage of nitrogen in urea, cn2h4o
Paraphin [41]
Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams

therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams

Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%
7 0
3 years ago
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