Period 4 transition element that forms 2+ ion with a half‐filled d sub level is
Manganese (Mn)
What is the half-filled d sub-level?
Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.
Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.
Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.
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Answer;
294.13 amu
Solution;
-293nv is 293.10 amu and that of 295nv is 295.45 amu
293.05+295.2=588.25
588.25/2= 294.13 amu
an amu of 294.13 using significant figures
It becomes ionized and attains its stable electronic configuration.
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Answer:
24.4 amu or g/mole
Explanation:
24 x 0.790 = 19.0 amu
25 x 0.100 = 2.50 amu
26 x 0.110 = 2.86 amu
(Because of the 19.0, the sig figs go only to the 1/10 decimal place)
19.0 + 2.5 + 2.9 = 24.4 amu or g/mole
