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Answer:
(a) Cr²⁺: 1s² 2s²2p⁶ 3s²3p⁶ 3d⁴; (b) eight
Explanation:
(a) Electron configuration of Cr
(i). Locate chromium in the Periodic Table.
It's Element 24, so in Period 4, Group 3.
(ii). Write its electron configuration
Remember that Cr is one of those exceptions to the Aufbau principle.
Instead of the predicted electron configuration of [Ar] 4s²3d⁴, we get the configuration [Ar] 4s3d⁵.
The reason is that this this configuration minimizes electron repulsion.
(b) Write the electron configuration for Cr²⁺.
Now, we must remove two electrons from a chromium atom.
The electron in the 4s orbital is removed first, because this orbital is further from the nucleus, making electrons easier to remove.
So, we remove the 4s electron, and then a 3d electron.
The electron configuration of Cr²⁺ is
[Ar]3d⁴ or 1s² 2s²2p⁶ 3s²3p⁶ 3d⁴
(c). Electrons with n + l = 3
The orbitals with n + l = 3 are those covered by the third arrow down in the order of orbital energies:
2p (n = 2, l = 1) and 3s (n = 3, l = 0)
The 2p orbitals can hold six electrons, and the 3s orbital can hold two.
Both subshells are filled in a Cr²⁺ ion, so there are eight electrons
with n+ l = 3.
Answer:
1.
Explanation:
Hello,
In this case, for the given reaction we first assign the oxidation state for each species:
Whereas the half reactions are:
Next, we exchange the transferred electrons:
Afterwards, we add them to obtain:
By adding and subtracting common terms we obtain:
Finally, by removing the oxidation states we have:
Therefore, the smallest whole-number coefficient for Sn is 1.
Regards.