is the type of orbital hybridization of a central atom that has one lone pair and bonds to four 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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Answer:
there are 9 Chlorine atoms
Explanation:
Answer:
The first 50 elements along with their valences are given below :
1. Hydrogen = 1
2. Helium = 0
3. Lithium = 1
4. Beryllium = 2
5. Boron = 3
6. Carbon = 4
7. Nitrogen = 3
8. Oxygen = 2
9. Fluorine = 1
10. Neon = 0
11. Sodium = 1
12. Magnesium = 2
13. Aluminium = 3
14. Silicon = 4
15. Phosphorus = 3
16. Sulphur = 2
17. Chlorine = 1
18. Argon = 0
19. Potassium = 1
20. Calcium = 2
21. Scandiun = 3
22. Titanium = 3
23. Vanadium = 4
24. Chromium = 3
25. Manganese = 4
26. Iron = 2
27. Cobalt = 2
28. Nickel = 2
29. Copper = 2
30. Zinc = 2
31. Gallium = 3
32. Germanium = 4
33. Arsenic = 3
34. Selenium = 2
35. Bromine = 1
36. Krypton = 0
37. Rubidium = 1
38. Strontium = 2
39. Yttrium = 3
40. Zirconium = 4
41. Niobium = 3
42. Molybdenum = 3
43. Technetium = 7
44. Ruthenium = 4
45. Rhodium = 3
46. Palladium = 4
47. Sliver = 1
48. Cadmium = 2
49. Indium = 3
50. Tin = 4
<u>Note</u> :
An element like Iron, copper can have more than one valencies.
Answer:
2.01 moles of P → 1.21×10²⁴ atoms
2.01 moles of N → 1.21×10²⁴ atoms
4.02 moles of Br → 2.42×10²⁴ atoms
Explanation:
We begin from this relation:
1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br
Then 2.01 moles of PNBr₂ will have:
2.01 moles of P
2.01 moles of N
4.02 moles of Br
To determine the number of atoms, we use the relation:
1 mol has NA (6.02×10²³) atoms
Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms
2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms
4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms