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Alecsey [184]
3 years ago
15

The structures of TeF4 and TeCl4 in the gas phase have been studied by electron diffraction (S. A. Shlykov, N. I. Giricheva, A.

V. Titov, M. Szwak, D. Lentz, G. V. Girichev, Dalton Trans., 2010, 39, 3245).
a. Would you expect the Te i X (axial) distances in these molecules to be longer or shorter than than Te i X (equatorial) distances? Explain briefly.
b. Which compound would you predict to have the smaller X(axial) i Te i X(axial) angles? The smaller X(equatorial) i Te i X(equatorial) angles? Explain briefly.

Chemistry
1 answer:
Studentka2010 [4]3 years ago
7 0

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.

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The volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid is 0.12 L

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Molarity of stock solution (M₁) = 15.7 M
  • Volume of diluted solution (V₂) = 12 L
  • Molarity of diluted solution (M₂) = 0.156 M
  • Volume of stock solution needed (V₁) = ?

<h3>How to determine the volume of the stock solution needed</h3>

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M₁V₁ = M₂V₂

15.7 × V₁ = 0.156 × 12

15.7 × V₁ = 1.872

Divide both side by 15.7

V₁ = 1.872 / 15.7

V₁ = 0.12 L

Thus, the volume of the stock solution needed to prepare the solution is 0.12 L

Learn more about dilution:

brainly.com/question/15022582

#SPJ1

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