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3 years ago

The structures of TeF4 and TeCl4 in the gas phase have been studied by electron diffraction.

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

5 0

Answer:

From VSPER theory, the compound TeF₄ and TeCl₄ AX₄E type of models. Hence, the molecular geometry of this molecule is trigonal bipyramidal. The plausible structure of this molecule are shown on the first uploaded image

Looking at the image we see that structure 1 has three lone pairs repulsion with 90° angle. From VSPER theory, the repulsive force between the electron pair is given as

lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

The above shows that bond pair -bond pair has less repulsion than lone pair- bond pair, and next is lone pair - lone pair

This means that structure 2 is more stable structure for TeF₄ or TeCl₄ .,because the lone pair is present at equatorial position.

This stable structure is shown on the second uploaded image

A lone pair of electron occupy more space around the central atom than a bond pair electron that are present at axial positions, the two atoms that are present at axial position are tilt away from the lone pair

We can see the structure on the third uploaded image

Due to the repulsive interaction , the bond distance between Te - X(axial) increases, which means that the Te - X(axial) distance is longer than Te - X(equitorial) distance.

It is a general concept that TeX₄ ha a bond angle of 90° for axial position and 120° in between equitorial position. This is shown on the fourth uploaded image

Considering TeF₄ and TeCl₄ , F is more electro-negative element Than Cl.This mean that the fluorine would pull strongly the shared electron away from the Te which reduces the electron density near the central Te atom.This would result in decrease bond angle be TeCl₄

So , F(axial) - Te - F(axial) has smaller bond angle than Cl(axial) - Te - Cl(axial)

Considering the equitorial position we see that because of the highly electro - negative fluorine atom, the bond angle decrease much in TeF₄ than in TeCl₄

This means that F(equitorial) - Te - F(equitorial) has smaller bond angle than Cl(equitorial) - Te - Cl(equitorial)

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A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

3 years ago

Someone please help me figure this out!!

zavuch27 [327]

Answer: from what i can tell it looks like a heterogeneous mixture

Explanation:

heterogeneous is multiple different components, homogenous includes only the same component

7 0

3 years ago

What is the boiling point of water at an elevation of 1.55×104 ft ?

horsena [70]

The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)

4 0

3 years ago

What do chemists use percent yield calculations for in the real world?

STALIN [3.7K]

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation. false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need. false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are. true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain the amount of product you want.

8 0

2 years ago

Some glucose produced by gluconeogenesis is stored in the body as glycogen. Order the steps of glycogen synthesis.

Harlamova29_29 [7]

Answer:

e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi

a. Pyrophosphatase converts PPi and water into two Pi

b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP

Explanation:

Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.

The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.

In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.

Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.

Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.

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3 years ago