How do you find unpaired electrons in a metal ion in CuSO4·5H20?
Oh, finding unpaired electrons in a metal ion is like playing hide-and-seek with electrons! In the case of CuSO4·5H20, let's focus on the metal ion, which is copper (Cu). Now, copper usually loves to party with its electrons, so it tends to have a +2 charge to keep things balanced.
To figure out whether there are any unpaired electrons, we need to look at the electron configuration of copper. In its ground state, copper has the electron configuration [Ar] 3d^10 4s^1. But when it becomes a Cu(II) ion, it loses 2 electrons from the 4s orbital.
Since the 3d^10 subshell is fully occupied with electrons, there are no unpaired electrons in the Cu(II) ion. It's like a party with no plus-one! However, don't worry, there's still chemistry fun to be had with copper and its buddies in that CuSO4·5H20 party mix.
To find the number of unpaired electrons in a metal ion, you need to determine the oxidation state of the metal ion and its electronic configuration. In the compound CuSO4·5H2O, the metal ion is copper (Cu).
1. Determine the oxidation state of copper (Cu):
To find the oxidation state of copper in the CuSO4·5H2O compound, we need to know the oxidation states of the other elements involved. In sulfate (SO4), oxygen has an oxidation state of -2, and the overall charge of the sulfate ion is -2. Since there are four oxygen atoms, they contribute to a total charge of -8. The charge of the entire compound is neutral, so copper must have an oxidation state that balances the charge. The total charge due to sulfate is -2, and there is one copper ion. Therefore, the oxidation state of copper (Cu) in this compound is +2.
2. Determine the electronic configuration of Cu2+:
Copper (Cu) has an atomic number of 29. In the Cu2+ ion, copper has lost two electrons, giving it an electronic configuration of [Ar] 3d^9.
3. Calculate the number of unpaired electrons:
In the 3d orbital, there are five suborbitals, each with a capacity to hold two electrons. Since the electronic configuration of Cu2+ is 3d^9, it means that there are nine electrons present. Out of these, one electron will occupy each of the five suborbitals, resulting in five unpaired electrons.
Therefore, in the Cu2+ ion in CuSO4·5H2O, there are five unpaired electrons.
To find the number of unpaired electrons in a metal ion, we need to determine the electron configuration of the metal ion. In the case of Cu2+ in CuSO4·5H2O, we start by looking at the electron configuration of the neutral copper atom (Cu).
The atomic number of copper is 29, which means it has 29 electrons in its neutral state. The electron configuration of copper is 1s2 2s2 2p6 3s2 3p6 4s2 3d9. When copper is oxidized to form Cu2+, it loses two electrons from its 4s orbital, leaving us with an electron configuration of 1s2 2s2 2p6 3s2 3p6 3d9.
To determine the number of unpaired electrons, we need to examine the last partially filled subshell, which is the 3d orbital. In the case of Cu2+, the 3d orbital is half-filled with 9 electrons, meaning there are one unpaired electron.
Therefore, the Cu2+ ion in CuSO4·5H2O has one unpaired electron.
In summary, to find the number of unpaired electrons in a metal ion, we need to determine the electron configuration and examine the last partially filled subshell.
You write the electron structure like this.
1s2 2s2 2p6 3s2 3p6 3d10 4s1 is the neutral Cu atom.
Remove the two outside electrons to make the Cu^+2 ion. That leaves us with
3d9 as the outside and since that is an odd number, one electron must be unpaired. And you will note that many Cu(II) compounds are colored.