Predict the ground-state electron configuration of the following ions. write the answer in abbreviated form beginning with noble gas brackets for ru2+ and w3+

Ru ²⁺ :[Kr] 4d⁶

W ³⁺ : [Xe] 4f¹⁴5d³

The Atomic Number (Z) indicates the number of protons in an atom of an element. If the atom is neutral then the number of protons will be equal to the number of electrons. So the atomic number can also indicate the number of electrons.

This energy level is expressed in the form of electron configurations.

Charging electrons in the sub shell use the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Writing the electron configuration can be abbreviated using the electron configuration of the noble gas in brackets because it has the same configuration

It is in group 8B and period 5 with atomic number 44

So the electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s¹, 4d⁷

This configuration includes an exception to the Aufbau rule which is that Electrons occupy orbitals of the lowest energy level, which should be 5s² 4d⁶ (exceptions occur mainly in the transition group), so that with the configuration of the noble gases from Kr [1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶], the Ru element can be written:

Ruthenium-Ru: [Cr] 5s¹4d⁷

Because it releases 2 electrons to form Ru²⁺ ions, the configuration becomes

Ru ²⁺ :[Kr] 4d⁶

It is in group 6B and period 6 with atomic number 74

So the electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s²4f¹⁴5d⁴

Because according to Xe configuration [1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶] it can be written:

W: [Xe] 6s²4f¹⁴5d⁴

Because it releases 3 electrons to form a W³⁺ ion, the configuration becomes (electrons with the largest n , 6s and 5d, the electrons are removed first):

W ³⁺ : [Xe] 4f¹⁴5d³

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