Answer:
1. For the central boron atom:
A. The number of lone pairs = 0.
B. The number of single bonds = 3.
C. The number of double bonds = 0.
2. The central boron atom B. Has an incomplete octet.
Explanation:
Hello,
In this case, as shown on the attached picture containing the required Lewis structure of boron chloride, we can see that:
1. For the central boron atom:
A. The number of lone pairs = 0 since all the three valance electrons are bonded with the chlorine atoms.
B. The number of single bonds = 3 as each chlorine atom needs only one bond to obey the octet.
C. The number of double bonds = 0 considering the previous explanation.
2. The central boron atom B. Has an incomplete octet since it reaches six electrons only from its initial three and other three provided by the three chlorine atoms.
Regards.
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
The answer is Human behavior.
Answer:
A
Explanation:
In this question, we are to calculate the enthalpy of change of the reaction. ΔH
To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.
Thus, mathematically, the enthalpy of change of the reaction would be;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]
We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.
Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.
Thus, we can have the modified equation as follows;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]
Substituting the values we have according to the question, we have;
-95.98 + 4(-92.3) -(-17.9)
= -95.98 - 369.2 + 17.9
= -447.28 KJ/mol
