Create the Lewis structure of CF4. NOTE: do only the step asked for in each part and then click Check–don't work ahead to solve
the final structure.
Step 1. Count valence electrons in the molecule or ion. Do this by adding the periodic group numbers for each atom in the structure and adjusting for charge. Enter and Check.
Step 1. Count valence electrons in the molecule or ion. Do this by adding the periodic group numbers for each atom in the structure and adjusting for charge. Enter and Check.
1 answer:
To create the Lewis structure we need to take into account the octet rule: atoms tend to gain, lose or share electrons to complete their valence shell with 8 electrons.
C belongs to Group 4A in the periodic table so it has 4 valence electrons. It needs to share 4 pairs of electrons to complete the octet.
F belongs to Group 7A in the periodic table so it has 7 valence electrons. Each F needs to share 1 pair of electrons to complete the octet.
As a consequence, in CF₄, C will form a single bond with each F and all the octets will be complete.
You might be interested in
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
