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Delvig [45]
3 years ago
13

Determine the number of bonding electrons and the number of nonbonding electrons in the structure of OF2. Enter the number of bo

nding electrons followed by the number of nonbonding electrons in the dot structure of this molecule separated by a comma (e.g., 1,2).

Chemistry
2 answers:
lesya [120]3 years ago
5 0

OF2-

O has 6 electrons in its outer shell and F has 7 electrons in its outer shell

<h2>Further Explanation </h2>

Therefore, you want to account for a complete of 20 electrons in

structure (7 + 7 + 6 = 20)

therefore draw linear first. F ---- O ----- F

Two bonds make sure of 4 electrons now you have got to feature 16 others.

Therefore 3 free pairs on each F and a pair of free pairs on O.

If you check the formal charges, all atoms are neutral

F will have 3 electron pairs + 1 bond = 7 electrons (bond = 1/2 electron for the formal charge distribution) therefore both F's are neutral

Now look O: it must have 6. it's two negatron pairs and a pair of bonds = 4 electrons and a pair of bonds = 1 electron each = 2 electrons from the bond = 6 total electrons for the precise formal charge # should have. there's no need for double bonds during this case because there aren't any costs to separate.

Now if you observe # domains around O you'll see if you enter a free pair that has sp3 hybridization (4 domains) therefore a tetrahedron that has 2 electron pairs and a pair of bonds .. because there are two electron pairs, repulse single pairs/pairs the bond is so high that it expels both fluorine and forms a bent structure, noticeably like H2O.

Learn More

Electrons Bond  brainly.com/question/11486787

Outer Shell  brainly.com/question/11486787

Details

Grade: College

Subject: Chemistry

Keyword: electron, bond, shell

kvasek [131]3 years ago
3 0

Answer:

(4,20)

Explanation:

Oxygen di-fluoride or OF2 is a polar molecule, formed through the covalent bonding between one Oxygen and two Fluorine atoms.

Oxygen has atomic no. 8 and its electronic configuration is as below:

  • 1s^2, 2s^2, 2p^4

Oxygen has 4 electrons in the valence shell. It has the ability to share two electrons with other atoms through covalent bond and get stable.

The atomic number of Fluorine is 7, and its electronic configuration is as:

  • 1s^2, 2s^2, 2p^5

So, Fluorine atom  has the capacity to accept 1 electron to form 1 covalent bond or in the formation of the F- ion.  

The structure of OF2 is a bent structure with O atom sandwiched by two fluorine atoms like F-O-F. So there are two electrons on oxygen atom that are taking part in the formation of covalent bond with each Fluorine atom (Please dot structure in figure).

Now we can easily calculate no of bonding and non-bonding electrons.

  • No. of bonding electrons = 4 (two from oxygen and two from fluorine atoms each)
  • No. of non-bonding electrons= 20 (6+6 from two Fluorine atoms and 6 from Oxygen atom)

Hope it helps! :)


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If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?
vfiekz [6]

Answer:  0.745 g of H_2SO_4 will be produced from  1.08 g of sodium sulfate

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles  

3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4

Na_2SO_4 is the limiting reagent as it limits the formation of product and H_3PO_4 is the excess reagent.

According to stoichiometry :

3 moles of Na_2SO_4 produce = 3 moles of H_2SO_4

Thus 0.0076 moles of Na_2SO_4 will require=\frac{3}{3}\times 0.0076=0.0076moles  of H_2SO_4

Mass of H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g

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3 0
2 years ago
Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

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