Write the electron configuration for nickel, whose atomic number is \(28 .\) Remember that the 4\(s\) orbital has lower energy than the 3\(d\) orbital does and that the \(d\) sublevel can hold a maximum of 10 electrons.

Short Answer

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The electron configuration for Nickel (Ni), whose atomic number is 28, is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8\).

Step by step solution

01

Identify atomic number

The atomic number of Nickel (Ni) is 28. This means that in a neutral atom, there are 28 electrons to account for in the electron configuration.
02

Start to fill orbitals by increasing energy levels

Start to fill orbitals from lower to higher energy levels. The filling order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p and so on. However, for Nickel, we need to go up to 3d only, as it contains 28 electrons.
03

Fill the 1s orbital

The 1s orbital can hold up to 2 electrons. So, it will be filled first. This gives us: 1s^2.
04

Fill the 2s and 2p orbitals

The 2s orbital can hold 2 electrons and the 2p orbital can hold 6. Filling these gives us: 1s^2 2s^2 2p^6 (
05

Fill the 3s and 3p orbitals

Like the 2s, the 3s orbital can also hold 2 electrons and the 3p, like the 2p, can hold 6. So, after filling these, our configuration becomes: 1s^2 2s^2 2p^6 3s^2 3p^6.
06

Fill the 4s orbital

According to the energy ordering, the 4s orbital is filled before the 3d. Again, the 4s can hold 2 electrons, so after filling it, we obtain: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2.
07

Fill the 3d orbital

Finally, fill the 3d orbital. It can hold up to 10 electrons, but in this case, we already have filled up 20. So, we only need put the remaining 8 electrons into the 3d orbital, resulting in the final electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8.

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