Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock = 
molarity of stock = 4M
volume = 
= 0.0625L or 62.5mL
Answer:
a) nitrogen
b) nitrogen =5
Oxygen = 6
Fluorine =7
Explanation:
Usually, if we have two or more elements in a compound, the central atom in the compound is the atom having the least value of electro negativity.
If we consider fluorine, oxygen and nitrogen; nitrogen is the least electronegative of the trio hence it should be the central atom of the triatomic molecule.
The number of valence electrons on the valence shell of each atom is shown below;
nitrogen =5
Oxygen = 6
Fluorine =7
