Answer:
Precent yield
Explanation:
This is takes into account how much of a substance should have been created (theoretical yield) and compares it to what was actually created (the actual yield).
Answer:
They test it using the scientific method.
The molecule have TETRAHEDRAL HYBRID ORBITAL.
SP3 hybridization involves the mixing of one orbital of S sub level and three orbitals of P sub level of the valence shell. All the orbitals possess equivalent energies and shapes. The SP3 orbital has 25% S character and 75% P character. S and P refers to the s and p sub shells.<span />
Here's the numbers
H = +1
Cl = -1
First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.
The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.
Sum of the oxidation numbers of each element = Charge of the compound
a + 6 x (-1) = -4
a -6 = -4
a = +2
Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2.
Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26.
Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²
When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.
Hence, the electron configuration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
= [Ar] 3d⁶
Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
