What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?
1 answer:
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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Answer:
the energy of the third excited rotational state
Explanation:
Given that :
hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm
Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.
Thus; the reduced mass μ =
μ =
μ =
∵ 1 μ = 1.66 × 10⁻²⁷ kg
μ =
μ = 1.6139 × 10⁻²⁷ kg
The rotational level Energy can be expressed by the equation:
where ;
J = 3 ( i.e third excited state) &
We know that :
1 J =
Answer:
Explanation:
Hello,
In this case, for latent heat (phase change) we need to consider the enthalpy associated with the involved process, here, melting or fusion; thus, the enthalpy of fusion of copper is 13.2 kJ/mol, therefore, the heat is computed as:
Nevertheless, since the given enthalpy is per mole of copper, we need to use its atomic mass to perform the correct calculation as follows:
Which is positive as it needs to be supplied to the system.
Best regards.