What is the electron configuration for the transition metal ion(s) in each of the following compounds?
Na2[TaF7]
Cant seem to figure this one out...
Ta is [Xe] 4f14 5d3 6s2. It's fairly obvious that to make the Na salt the two 6s electrons are removed to make the 2Na^+ + TaF7^-2. I suspect, but I have not been able to confirm it, that what is left is [Xe] 4f14 5d3 6s0 for Ta with the F atoms doing their thing.
To determine the electron configuration for the transition metal ion in the compound Na2[TaF7], we first need to identify the transition metal ion present in the compound. In this case, the transition metal ion is tantalum (Ta).
The electron configuration of an atom or ion is a way to represent the distribution of electrons in the atomic orbitals. Tantalum (Ta) belongs to the transition metal group, specifically Group 5B in the periodic table. The electron configuration for a neutral tantalum atom is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6.
In order to determine the electron configuration for the transition metal ion in the compound, we need to consider the charge on the ion. In this case, the compound is Na2[TaF7], which means the compound has a -2 charge overall due to the presence of two sodium ions (Na+) each with a +1 charge.
Since there are seven fluoride ions (F-) in the compound, and each F- ion has a -1 charge, the total negative charge from the fluorine ions is -7. To balance the overall charge of -2, the tantalum ion (Ta) in this compound must have a +5 charge (7-2 = +5).
So, the electron configuration for the Ta ion in Na2[TaF7] is obtained by removing five electrons from the neutral electron configuration of Ta. Therefore, the electron configuration for the Ta ion in Na2[TaF7] can be written as 1s2 2s2 2p6 3s2 3p6 4s2 3d10.
To summarize, the electron configuration for the transition metal ion (Ta) in the compound Na2[TaF7] is 1s2 2s2 2p6 3s2 3p6 4s2 3d10.
To determine the electron configuration of the transition metal ion in Na2[TaF7], we need to look at the electron configurations of sodium (Na), fluorine (F), and tantalum (Ta).
1. Start by determining the electron configuration of sodium (Na), which has an atomic number of 11. The electron configuration for sodium is: 1s2 2s2 2p6 3s1.
2. Next, consider the electron configuration of fluorine (F), which has an atomic number of 9. The electron configuration for fluorine is: 1s2 2s2 2p5.
3. Finally, examine the electron configuration of tantalum (Ta), which has an atomic number of 73. The electron configuration for tantalum is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d3.
4. In the compound Na2[TaF7], the number 2 in Na2 indicates that there are two sodium ions present. Since sodium (Na) is a Group 1 element, it loses one electron to form a positive charge. Therefore, each sodium ion has an electron configuration of 1s2 2s2 2p6 3s0.
5. Similarly, the number 7 in TaF7 indicates that there are seven fluorine (F) ions present. Since fluorine is a Group 17 element, it gains one electron to form a negative charge. Therefore, each fluorine ion has an electron configuration of 1s2 2s2 2p6 3s2 3p6.
6. The compound Na2[TaF7] contains one tantalum (Ta) ion, which forms a positive charge to balance the negative charges of the fluorine ions. Therefore, the electron configuration for the transition metal ion (tantalum) in Na2[TaF7] is the same as the electron configuration of tantalum itself: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d3.
So, the electron configuration for the transition metal ion (tantalum) in Na2[TaF7] is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d3.