Using SPECTROSCOPIC notation write the complete electron configuration for the iron(III) ion. Using NOBLE GAS notation write the electron

Using SPECTROSCOPIC notation write the complete electron configuration for the iron(III) ion. Using NOBLE GAS notation write the electron configuration for the copper(I) ion.

Answers

1) H 1s^1

2) [Ar] 3d⁸ 4s²

3) [He] 2s2 2p3

4) [Kr] 4d10 5s2 5p5

5) [Ar] 4s² and [Ar] 4s²

6) [He] 2s2 2p2

7)  [He] 2s² 2p⁴

8) [Ar] 3d7 4s2

9) [Kr] 4d¹⁰ 5s¹  

10) [Ne] 3s² 3p⁶

Explanation:

Answers and Explanation:

b) Complete electron configuration of Se:

1 s² 2 s²p⁶ 3 s²p⁶ 4 s²p⁴

Se has 4 energy levels, with an incomplete 4th level (they are necessary 2 electrons to complete p orbital in the 4th level).

c) The first ionization energy of Se

i) is less that of bromine (Br) because the first ionization energy increases from left to right in a period in the Periodic Table. Se and Br are in the same period, and Se is located to the left of Br so Se has lesser ionization energy than Br.

ii) is greater than that of tellurium (Te) because the first ionization energy decreases from top to bottom in a group in the Periodic Table. Se and Te are in the same group, and Se is at the top of Te. Hence, Se has a greater first ionization energy than Te.

Explanation:

Answer 1:

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 element has such electonic configuration.

D. Bromine but the correct electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5

i) 1s²2s²2p⁶3s²3p³

ii) [Ne]3s²3p³

iii) ↑↓       ↑↓    ↑↓  ↑↓  ↑↓       ↑↓   ↑ ↑ ↑

    1s²      2s²         2p⁶             3s²    3p³    

Explanation:

i) The phosphorus, with atomic number (Z) 15, has the following  electron configuration:

Z=15 → Number of electrons = 15

1s²2s²2p⁶3s²3p³  

ii) The noble gas configuration of the phosphorus is the next:

[Ne]3s²3p³  

Neon is the noble gas that precedes phosphorus, its atomic number is 10. Therefore, the first ten electrons in the electronic configuration of P correspond to the noble gas Ne.  

iii) The orbital configuration of the P is given by the spins of the electrons in each orbital:

   ↑↓            

1   s²

   ↑↓      ↑↓  ↑↓  ↑↓

2  s²             p⁶  

   ↑↓        ↑ ↑ ↑

3  s²           p³  

Where ↑ and ↓ represents electron spins.

We can consider every orbital as a box that contains the electrons, and using the Pauli exclusion principle we can fill the boxes with the electrons.

So, following that principle, we have that every box will have a maximum of two electrons, which are represented as ↑↓, when they are paired up, and ↑ when they are not paired up.          

In phosphorous case, every s orbital contain 2 electrons represented as ↑↓, and every p orbital has three orbitals: p_{x}, p_{y} and p_{z}, for a total of 6 electrons (two electrons for each orbital), except for the 3p orbital, which has only 3 electrons, represented as ↑ ↑ ↑.                                  

I hope it helps you!

Iodine has an atomic number of 53

Electronic configuration - 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2  4d^10 5p^5

Noble gas configuration - [kr]4d^10 5s^2 5p^5

Iodine have 7 valence shell electron and 46 inner shell electron in the ground state.

S=  [Ne] 3s²3p⁴

S2- = [Ne] 3s²3p6 = [Ar]

Si 1s^2 2s^2 2p^6 3s^2 3p^2

Co (actual) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1

Co (expected) 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9

i'd be careful with Copper. i listed the actual e- configuration for an element and lost points cause they were looking for the expected e- configuration

Si: 1s²2s²2p⁶3s²3p2

Explanation:

Other electron configuration summaries include ...

Si:[Ne]3s²3p2; [Ne] = electron configuration of noble gas Neon (1s²2s²2p⁶)

Si:[Ne]3s²2p₋₁¹p₀¹p₊₁⁰ <=> e⁻ configuration with orbital orientations

Si:[Ne]3s(↑↓)3p₋₁(↑)p₀(↑) <=> Orbital Diagram

We are to write the orbital notation and electronic configuration of sulfur and iron.

The orbital notation shows the filling of electrons into orbitals or sublevels.

Electron configuration  shows the distribution of electrons into shells;

       Number of electrons     Electron configuration     Orbital notation

S                  16                                   2 8 6                      1s² 2s² 2p⁶ 3s² 3p⁴

Fe                26                                2, 8, 14 2           1s² 2s² 2p⁶ 3s² 3p⁶3d⁶4s²        

Explanation:

YOLO

Explanation:

You only live once, might as well give people the wrong answers.

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