Answer:
See explanation
Explanation:
Two structures are pictured in the image attached, structure 2 is a more stable structure for TeF₄ or TeCl₄ since the lone pair in the structure is present at an equatorial position. Note that the structure of the compound is based on a trigonal bipyramid but of the structure type AX4E, where the lone pair could be positioned at axial or equatorial positions as shown in the images attached.
As a result of repulsion between electron pairs, the bond distance between Te - X(axial) is greater than Te-X (equatorial). This implies that the Te - X(axial) bonds are longer than Te - X(equatorial) bonds.
In answering the question, we must take cognisance of the fact fluorine is more electronegative than chlorine. This implies that fluorine will draw more electron density towards itself than chlorine and this shortens the bond length.
The structures of the both compounds are based on a trigonal bipyramid with a bond angle of 90° for axial position and 120° for the equatorial position.
Owing to the greater electo negativity of fluorine, both the fluorine F-axial and F-equatorial bonds of TeF4 are shorter than the Cl-axial and Cl-equatorial bonds of TeCl4. Hence the axial and equatorial angles in TeF4 are smaller than the axial and equatorial bond angles in TeCl4 due to the greater electro negativity of the fluorine atom.
The magnitude of the orbital angular momentum of the electron is doubled when the radius of the electron orbit increases by a factor of four.
<h3>
What is orbital angular momentum?</h3>
The orbital angular momentum is the sum of angular momenta of all electrons, it is the rotational analog of the linear momentum.
The formula of angular momentum

Thus, the magnitude of the orbital angular momentum of the electron is doubled when the radius of the electron orbit increases by a factor of four.
Learn more about orbital angular momentum
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Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
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1.00mol is the correct answer
Answer:
The answer for number one is .A
The answer for number two is heat rises
Explanation: