Thee question is incomplete; the complete question is;
Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?
Select one or more:
A: Chlorine
B: Carbon
C: Hydrogen
D: Sulfur
E: Fluorine
F: Oxygen
Answer:
chlorine
sulphur
Explanation:
The octet rule states that, for atoms to be stable, they must have eight electrons on their outermost shells.
This rule is not strictly followed by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.
These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.
Answer:
a. the maximum number of σ bonds that the atom can form is 4
b. the maximum number of p-p bonds that the atom can form is 2
Explanation:
Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.
The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.
This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.
And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.
Answer:D
Explanation:because The farther an object is from a magnet are apart from each other, the weaker the repulsion force will be.
Answer: the HCO3- to act as a base and remove excess H by the formation of H2CO3
Explanation:
H2CO3 in an aqueous solution is a buffer. This means the reaction is the following:
H2CO3 ------ HCO3- + H+
Then, the HCO3- that was formed acts as a base (absorbing a proton) like this
HCO3- + H+ ------- H2CO3
If there was an increase in H+, there would be an increase in the second reaction in an effort to neutralize that acid, thus making the H2CO3 more concentrated