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Sphinxa [80]
2 years ago
9

What is the orbital hybridization of a central atom that has one lone pair and bonds to:

Chemistry
1 answer:
nirvana33 [79]2 years ago
5 0

sp^3  is the orbital hybridization of a central atom that has one lone pair and bonds to three other atoms.

<h3>What is orbital hybridization?</h3>

In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.

For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.

Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.

Learn more about Hybridization

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Also: "Neon is rare on Earth, found in the Earth's atmosphere at 1 part in 55,000, or 18.2 ppm by volume (this is about the same as the molecule or mole fraction), or 1 part in 79,000 of air by mass."

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5 0
3 years ago
Write a<br> paragraph or two on electromagnetism.
patriot [66]

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How many moles are present in
olasank [31]

Answer: (a) There are 0.428 moles present in 12 g of N_{2} molecule.

(b) There are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of N_{2} is 28 g/mol so its number of moles are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol

So, there are 0.428 moles present in 12 g of N_{2} molecule.

(b). According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

Therefore, moles present in 12.044 \times 10^{23} particles are calculated as follows.

Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol

So, there are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

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3 years ago
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ohaa [14]

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It should b KNO3

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