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3 years ago

What is a consequence of electron-pair repulsion around an atom?

Chemistry

2 answers:

BARSIC [14]3 years ago

7 0

Answer:

c. The atoms sharing the electron pairs will spread out around the

central atom

Explanation:

If an atom is bonded to two or more kinds of atoms for example many covalent molecule or ions, the shape of the system is determined by the geometry of the bonds around the central atom. The variations in the bond angle of such molecules and ions can be explained by the Valence shell electron pair repulsion theory.

The Valence shell electron pair repulsion theory (VSEPR) model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs consists of lone pair and bond pairs. All the electron pairs will orient themselves in such a way as to minimize the electrostatic repulsion between them.

wolverine [178]3 years ago

6 0

Answer:

Option C is correct.

The atoms sharing the electron pairs will spread out around the central atom.

Explanation:

In covalent bonding, the atoms involved form a bond by sharing a pair of electrons. These pairs of electrons are repulsive towards one another, when they occur together around a central atom.

Infact, the Valence shell electron pair repulsion (VSEPR) theory explains further that 'electron pairs around a central atom arrange themselves so that they can be as far apart as possible from each other'.

So, it becomes obvious that the atoms that share these electron pairs around a central atom are pushed to spread out as far as possible around that central atom.

Hope this Helps!!!

You might be interested in

How abundant is Polonium in the earth's crust?

artcher [175]

Even in pitchblende, polonium is exceedingly rare: 1,000 tons of ore must be treated to yield 40 milligrams of polonium. It has a one-part-in-ten-thousandth-thousandth-thousandth-thousandth-thousandth-thousandth It is found in nature as a byproduct of the radioactive decay of uranium, thorium, and actinium.

Atomic Number: 210

3 0

3 years ago

A mixture of NaBrO 3 , NaBrO3, NaHCO 3 , NaHCO3, Na 2 CO 3 , Na2CO3, and NaBr NaBr was heated, producing H 2 O , H2O, CO 2 , CO2

nata0808 [166]

Answer:

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

Explanation:

For the reactions:

2NaBrO₃(s) ⟶ 2NaBr(s) + 3O₂(g)

2NaHCO₃(s) ⟶ Na₂O(s) + H₂O(g) + 2CO₂(g)

Na₂CO₃(s) ⟶ Na₂O(s)+CO₂(g)

All H₂O(g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

1.83g H₂O ₓ (1 mol H₂O / 18.02g) ₓ (2 mol NaHCO₃ / 1 mol H₂O) = 0.203moles NaHCO₃ ₓ (84g / 1mol NaHCO₃) =

17.06g of NaHCO₃

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

15.51g CO₂ ₓ (1mol / 44.01g) = 0.352moles of CO₂

As 2 moles of NaHCO₃ produce 2 moles of CO₂, moles of CO₂ that comes from NaHCO₃ are 0.203moles NaHCO₃. Moles of CO₂ that comes from Na₂CO₃ are:

0.352mol CO₂ - 0.203mol CO₂ = 0.149mol CO₂

These moles of CO₂ are produced from:

0.149mol CO₂ ₓ (1 mol Na₂CO₃ / 1 mol CO₂) ₓ (106g / 1mol Na₂CO₃) =

15.84g of Na₂CO₃

And all O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

2.87g O₂ ₓ (1 mol O₂ / 32g) ₓ (2 mol NaBrO₃ / 3 mol O₂) ₓ (150.9g / 1mol NaBrO₃) =

9.02g of NaBrO₃

If initial mass of the mixture was 124.5g, mass of NaBr was:

124.5g - 9.02g of NaBrO₃ - 15.84g of Na₂CO₃ - 17.06g of NaHCO₃ =

82.58g NaBr

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

5 0

4 years ago

In a balanced equation, coefficients always represent

steposvetlana [31]

Answer:

D. The number of atoms present

Explanation:

In a balanced equation, coefficients always represent the number of atoms present.

8 0

3 years ago

A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in

mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is $5.90\times 10^{-4}M$

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock magnesium fluoride solution

$M_2\text{ and }V_2$ are the molarity and volume of chemist's magnesium fluoride solution

We are given:

$M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL$

Putting values in above equation, we get:

$0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M$

Hence, the concentration of chemist's working solution is $5.90\times 10^{-4}M$

8 0

3 years ago

The temperature of a gas is 100 K and its volume is 500.0 ml. If the volume increases to 1,000.0 ml,

olga55 [171]

1000000000000000 that’s the answer

5 0

3 years ago