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ladessa [460]
3 years ago
6

What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo- nal planar arra

ngement of electron pairs? A linear ar- rangement of electron pairs? How many unhybridized p atomic orbitals are present when a central atom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? What are the unhybridized p atomic orbitals used for?
Chemistry
1 answer:
Reika [66]3 years ago
5 0

Answer:

sp³;

sp²;

sp;

None;

One;

Two;

They're used to pi bonds.

Explanation:

The central atom in a molecule is generally the one that can make a greater number of bonds. The covalent bonds are made by the sharing of electrons, and, for that, the electron must be alone in the orbital.

To explain this, the hybridization theory was created, which states that, the orbitals are joined to form hybrids ones, and so, by the Hund's law, the electrons are alone in them.

The sigma bonds are done in the hybrids orbitals, and at the pure orbitals, the pi bonds are done. The lone pair of electrons are at a pure orbital. So, to know the hybridization of the central atom, we must know how many sigma bonds it does, and it will be the number of hybrids orbitals (each orbital may have two electrons, thus each bond are done in one orbital).

Double bonds and triple bonds have always only one sigma bond, so the number of sigma bonds is equal to the number of bonds, it's not necessary to know if they are simple, double or triple.

When the arrangement is tetrahedral, the central atom does 4 bonds, so it has 4 sigma bonds, and 4 hybrids orbitals (one of s and three for p), does its hybridization is sp³. Because exists only 3 p orbitals, there are no unhybridized p orbitals in this case.

When the arrangement is trigonal, the central atom does 3 bonds, so it has 3 hybrids orbitals (one of s and two of p), thus the hybridization is sp². So there are one unhybridized p orbitals.

When the arrangement is linear, the central atom does 2 bonds, so it has 2 hybrids orbitals (one of s and one of p), thus the hybridization is sp. So, there are two unhybridized p atoms.

As stated before, the unhybridized p orbitals are used to pi bonds.

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In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

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