<span>After 1 half-life, 1/2 the sample has decayed. After 2 half-lives 3/4 of the matter has decayed. So after, 3 half-lives about 11/12 would have decayed.</span>
The electron group arrangement of CH₂Cl₂ is penta-atomic. The molecular shape is tetrahedral, and the bond angle is 109.5°.
<h3>What is the bond angle?</h3>
The bond angle is the angle between the atoms of the compound. The bond angle is defined in degree. The bond length is also there. It is the distance between the nuclei of the two atoms.
The bond angle of CH₂Cl₂ is 109.5°. According to the VSEPR theory, as the structure will be tetrahedral.
Thus, the penta-atomic configuration of CH2Cl2's electrons. Tetrahedral is the molecular shape, the bond angle is 109.5°.
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Answer:
Chemistry is a branch of science that involves the study of the composition, structure and properties of matter. Often known as the central science, it is a creative discipline chiefly concerned with atomic and molecular structure and its change, for instance through chemical reactions.Explanation:Chemistry is a branch of science that involves the study of the composition, structure and properties of matter. Often known as the central science, it is a creative discipline chiefly concerned with atomic and molecular structure and its change, for instance through chemical reactions.
The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A