Answer:
The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>
Explanation:
The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.
<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.
Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.
Hence , the answer is true .
The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
Learn more about equilibrium of forces here:
brainly.com/question/12980489
We need the reading for this I think
I think the answer is [Xe] 6s2
