Write the electronic configuration of cr^2+ (z=24) and predict the number of electrons having n+1 value equal to 3

(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.