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julia-pushkina [17]
3 years ago
9

A single bond is almost always a sigma bond, and a double bond is almost always made up of a sigma bond and a pi bond. There are

very few exceptions to this rule. Which of the following species have violated this generalization?
A. N2
B. O2
C. F2
D. C2
E. B2.
Chemistry
1 answer:
katrin [286]3 years ago
8 0

Answer:

The exception here is

C2 and B2

both the bonds are pi bond in the above cases

Explanation:

The fault here is the unforeseen energy order of molecular orbitals. Since if filled, sigma 2px will be firmly pushed away by the electron density of orbitals already filled with sigma 2s (both electron densities lie on the inter-nuclear axis). Because of this, sigma 2px's energy is found shockingly greater than pi 2py and pi 2pz.

The electrons therefore occupy pi 2py and pi 2pz orbitals as opposed to normal sigma 2px (which remains vacant) while filling.

The existence of these 4 electrons in orbitals pi 2py and pi 2pz results in two pi bonds being formed.

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How many grams of sodium sulfate Na2SO4 (mw = 142.04 g/mol) is needed to prepare 350.0 ml of a solution having a concentration o
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According to the definition of molarity, 350 mL of a solution with a sodium ion concentration of 0.125 M requires 6.2125 g of Na2SO4 to manufacture.

<h3><u>How to find Molarity ?</u></h3>

The number of moles of a solute that are dissolved in a given volume is what is meant by the definition of molarity, which is a measurement of a solute's concentration.

By dividing the moles of the solute by the volume of the solution, molarity, also known as the molar concentration of a solution, is obtained.

Molarity = No. of moles of solute / Volume

Molarity is expressed in units mole/litre

In this case, you know that:

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Replacing in the definition of molarity:

0.125M = No. of moles of solute / 0.350l

Solving:

number of moles of solute= 0.125 M× 0.350 L

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Being 142 g/mole the molar mass of Na₂SO₄, that is, the mass of one mole of the compound, the amount of mass that contains 0.04375 moles is calculated as:

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In summary, 6.2125 g of Na₂SO₄ is needed to prepare 350 mL of a solution having a sodium ion concentration of  0.125 M.

To view more about concentrations, refer to;

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