Question

Write the electron configuration for ru and ru3+ ion. How many d electrons are there in the outermost energy level of ru3+ ion?

Answer 1

Explanation:

Electronic configuration of Ru = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s0

or Ru = (Kr) 4d8 5s0

electronic configuration of [Ru]3+ = [Kr] 4d5 5s0

There are 5d electrons in the outermost energy level of Ru3+ ion.

Answer 2

Answer:

The electron configuration for the neutral atom of ruthenium (Ru) is [Kr] $4d^7 5s^1.$

Explanation:

Ruthenium is the 44th element in the periodic table and its symbol is ‘Ru’. The total number of electrons in ruthenium is forty-four. These electrons are arranged according to specific rules of different orbits.

The ground state electron configuration of ruthenium is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^7 5s^1.$(By Aafbau principle)

When ruthenium forms a cation (Ru3+), it loses three electrons. Therefore, the electron configuration for the Ru3+ ion is [Kr] $4d^4.$

In the outermost energy level of Ru3+ ion, there are 4d electrons. In general, the electron configuration of a cation is obtained by subtracting electrons from the outermost energy level (valence electrons) of the neutral atom.

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