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Answer:
Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷
Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰
S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶
Explanation:
Cobalt (Co): atomic number 27
The electronic configuration of Co in ground state:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
The electronic configuration of Co in +2 oxidation state (Co²⁺) :
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷
Tin (Sn): atomic number 50
The electronic configuration of Sn in ground state:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²
The electronic configuration of Sn in +2 oxidation state (Sn²⁺) :
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
Zirconium (Zr): atomic number 40
The electronic configuration of Zr in ground state:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²
The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
Silver (Ag): atomic number 47
The electronic configuration of Ag in ground state:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰
The electronic configuration of Ag in +1 oxidation state (Ag⁺) :
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰
Sulphur (S): atomic number 16
The electronic configuration of S in ground state:
1s² 2s² 2p⁶ 3s² 3p⁴
The electronic configuration of S in -2 oxidation state (S²⁻) :
1s² 2s² 2p⁶ 3s² 3p⁶
The electronic configurations of the various elements in their oxidation or reduction states have been written below.
We want to write the electronic configurations of;
Co²⁺, Sn²⁺, Zr⁴⁺, Ag⁺, S²⁻.
1) Co²⁺; This Cobalt with atomic number of 27
Its' electronic configuration in its' ground state is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
However, due to the fact that it is in its' in +2 oxidation state, then it will lose two electrons in the 4s orbital to form:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷
2) Tin (Sn): It has an atomic number of 50
Its' electronic configuration of in its's ground state is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²
However, due to the fact that it is in +2 oxidation state, then it will lose 2 electrons in the 5p orbital to form;
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
3) Zirconium (Zr): It has an atomic atomic number of 40.
Its' electronic configuration in its' ground state is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²
However, due to the fact that it is in +4 oxidation state, then it will lose 4 electrons with 2 each from the 4d orbital and 5s orbital to form;
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
4) Silver (Ag): It has an atomic number of 47.
Its' electronic configuration in its' ground state is:1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰
However, due to the fact that it is in +1 oxidation state, then it will lose 1 electron from the 5s orbital to form:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰
5) Sulphur (S): It has an atomic number of 16
Its' electronic configuration in its' ground state is:
1s² 2s² 2p⁶ 3s² 3p⁴
However, due to the fact that it is in -2 reduction state, it will gain 2 additional electrons to the 3p orbital to form:
1s² 2s² 2p⁶ 3s² 3p⁶
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