Using SPECTROSCOPIC notation create the complete electron construction for the iron(III) ion. Using NOBLE GAS notation compose the electron construction for the copper(I) ion.
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Answers and also Explanation:
b) complete electron configuration of Se:
1 s² 2 s²p⁶ 3 s²p⁶ 4 s²p⁴
Se has actually 4 energy levels, through an incomplete 4th level (they are essential 2 electron to complete p orbital in the fourth level).
c) The an initial ionization power of Se
i) is much less that the bromine (Br) due to the fact that the very first ionization energy increases from left to appropriate in a period in the routine Table. Se and also Br space in the same period, and Se is located to the left that Br therefore Se has lesser ionization energy than Br.
ii) is higher than the of tellurium (Te) due to the fact that the very first ionization power decreases from height to bottom in a group in the periodic Table. Se and Te room in the same group, and Se is at the top of Te. Hence, Se has a greater an initial ionization energy than Te.
Lithium : 1s2 2s1 Fluorine: 1s2 2s2 2p5 Carbon: 1s2 2s2 2p2
Argon : 1s2 2s2 2p6 3s2 3p6 Sulphur: 1s2 2s2 2p6 3s2 3p4
Nickel: 1s2 2s2 2p6 3s2 3p6 3d8 4s2 Rubidium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 Xenon: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6
Answer 2: A. Fluorine B. Calcium
C. It is Tellurium if this was the exact electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4 you intend to write, if not, no facet has such electonic configuration.
D. Bromine but the correct digital configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
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iii) ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑
1s² 2s² 2p⁶ 3s² 3p³
i) The phosphorus, with atomic number (Z) 15, has actually the following electron configuration:
Z=15 → variety of electrons = 15
ii) The noble gas configuration of the phosphorus is the next:
Neon is the noble gas that comes before phosphorus, its atomic number is 10. Therefore, the first ten electron in the electronic configuration of ns correspond come the noble gas Ne.
iii) The orbital construction of the ns is offered by the spins the the electrons in each orbital:
↑↓ ↑↓ ↑↓ ↑↓
2 s² p⁶
↑↓ ↑ ↑ ↑
3 s² p³
Where ↑ and ↓ to represent electron spins.
We can think about every orbital together a box that includes the electrons, and using the Pauli exclusion principle we have the right to fill the boxes through the electrons.
So, following that principle, we have actually that every box will have a best of 2 electrons, which are represented as ↑↓, once they space paired up, and also ↑ as soon as they space not combine up.
In phosphorous case, every s orbit contain 2 electrons represented as ↑↓, and also every p orbital has actually three orbitals: