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Daniel [21]
1 year ago
15

What is the axmen classification for benzene (C6H6)?

Chemistry
2 answers:
Tatiana [17]1 year ago
8 0

Answer is: Benzene is trigonal (or triangular) planar.

VSEPR theory (The Valence Shell Electron Pair Repulsion Theory) uses the AXE notation (m and n are integers, m + n = number of regions of electron density).

For benzene molecule (C₆H₆):

m = 3; the number of atoms bonded to the central atom.

n = 0; the number of lone pairs on the central atom.

kirza4 [7]1 year ago
3 0

The{\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}classification for each carbon in benzene is {\text{A}}{{\text{X}}_{\text{3}}}.

Further explanation:

VSEPR Theory:

The VSEPR theory is also known as valence shell electron pair repulsion theory. This theory is used to determine the geometry of a central atom or ion. It is based on the repulsion between bond pairs and lone pair in the valence shell of the central atom.

The {\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}classification is used to expresses the geometry of the central atom in VSEPR theory. In {\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}classification, A is a central atom, X is covalent bond, E is the lone pair, m is the total number of covalent bond present on central atom, and n is the total number of lone pair present on central atom.

Since {\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}classification is only calculated for the central atoms and benzene contains 6 central carbon atoms thus we can calculate {\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}classification of each central carbon atom.

In benzene, each carbon atom has 4 valence electrons. These 4 valence electron is are used in forming 3 covalent bonds, one C-H single bond, one C-C single bond, and one C=C double bond. Therefore, the value of m is 3. There is no lone pair on any central carbon atom thus the value of n is zero.

Therefore the configuration for each carbon in benzene is {\text{A}}{{\text{X}}_{\text{3}}}.

Learn more:

1.The neutral element represented by the excited state electronic configuration:  

2. Number of covalent bond formed by carbon:brainly.com/question/5974553

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: VSEPR theory

Keywords: axmen, axmen classification, VSEPR theory central atom, benzene, six carbon, no lone pair, AX3 classification.

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Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water  + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g  = 4*(719.42)g

    473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

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