How many outer-shell d electrons are there in each of the following transition-metal ions?
Re(III)
Sc(III)
Ru(IV)
Hg(II)
To determine the number of outer-shell d electrons in each transition-metal ion, we need to determine the electron configurations of the corresponding neutral atoms first. Then, we can determine the configurations of the ions by adding or removing electrons based on the oxidation state.
Let's find the electron configurations for each neutral atom:
1. Re (Rhenium)
The atomic number of Re is 75. The electron configuration of the neutral atom can be written as: [Xe] 4f^14 5d^5 6s^2.
2. Sc (Scandium)
The atomic number of Sc is 21. The electron configuration of the neutral atom can be written as: [Ar] 3d^1 4s^2.
3. Ru (Ruthenium)
The atomic number of Ru is 44. The electron configuration of the neutral atom can be written as: [Kr] 4d^7 5s^1.
4. Hg (Mercury)
The atomic number of Hg is 80. The electron configuration of the neutral atom can be written as: [Xe] 4f^14 5d^10 6s^2.
Now, let's determine the configurations of the ions by adding or removing electrons based on the oxidation states:
1. Re(III) (Re with a +3 charge)
Re(III) loses three electrons, so the electron configuration becomes: [Xe] 4f^14 5d^4 6s^2.
Therefore, there are 4 outer-shell d electrons in Re(III).
2. Sc(III) (Sc with a +3 charge)
Sc(III) loses three electrons, so the electron configuration becomes: [Ar] 3d^0 4s^0.
Therefore, there are 0 outer-shell d electrons in Sc(III).
3. Ru(IV) (Ru with a +4 charge)
Ru(IV) loses four electrons, so the electron configuration becomes: [Kr] 4d^3 5s^0.
Therefore, there are 3 outer-shell d electrons in Ru(IV).
4. Hg(II) (Hg with a +2 charge)
Hg(II) loses two electrons, so the electron configuration becomes: [Xe] 4f^14 5d^10 6s^0.
Therefore, there are 10 outer-shell d electrons in Hg(II).
In summary, the number of outer-shell d electrons in each of the transition-metal ions is as follows:
Re(III): 4
Sc(III): 0
Ru(IV): 3
Hg(II): 10
To determine the number of outer-shell d electrons in transition-metal ions, we need to know the electron configurations of the elements involved. The electron configuration indicates how the electrons are distributed among different energy levels and orbitals.
Let's break down the electron configurations for each transition-metal ion:
1. Re(III):
The atomic number of rhenium (Re) is 75. To determine the electron configuration of the neutral atom, we can look at the periodic table. Rhenium is in the 6th period, corresponding to n=6.
The electron configuration for Re is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^5
When Re forms a +3 ion, it loses three electrons. Therefore, the electron configuration of Re(III) becomes:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^2
In this case, the number of outer-shell d electrons is 2.
2. Sc(III):
The atomic number of scandium (Sc) is 21. Scandium is in the 4th period, corresponding to n=4.
The electron configuration for Sc is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1
When Sc forms a +3 ion, it loses three electrons. Therefore, the electron configuration of Sc(III) becomes:
1s^2 2s^2 2p^6 3s^2 3p^6 3d^0
In this case, there are no outer-shell d electrons.
3. Ru(IV):
The atomic number of ruthenium (Ru) is 44. Ruthenium is in the 5th period, corresponding to n=5.
The electron configuration for Ru is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^7
When Ru forms a +4 ion, it loses four electrons. Therefore, the electron configuration of Ru(IV) becomes:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^3
In this case, the number of outer-shell d electrons is 3.
4. Hg(II):
The atomic number of mercury (Hg) is 80. Mercury is in the 6th period, corresponding to n=6.
The electron configuration for Hg is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 6p^2
When Hg forms a +2 ion, it loses two electrons. Therefore, the electron configuration of Hg(II) becomes:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10
In this case, there are no outer-shell d electrons.
Thus, the number of outer-shell d electrons for each transition-metal ion is as follows:
Re(III): 2
Sc(III): 0
Ru(IV): 3
Hg(II): 0
Not sure if you'll ever see this, Bob, but I really like how you force students to learn instead of just outright giving them the answers. Every single one of your posts have been really helpful. thank you very much!
Here is a link. Click on the element you want, scroll down the left side of the page and click on electron properties. You will find the electron configurations there. For example, click on Sc. You will see it has a 3d1 4s2. If it's the 3+ ion, it will have lost the 2 4s and the 1 3d electrons and will have zero in the 3d orbital.
http://www.webelements.com/