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In the second row of the transition metals there are five elements that do not follow...

In the second row of the transition metals there are five elements that do not follow the normal orbital filling. Five - Nb, Mo, Ru, Rh, and Ag-result in a [Kr] 5s14dx configuration and Pd results in a [Kr] 4d10 configuration. Write the ground state electron configuration for the following species.

Ag

Ag+

Ru

Ru3+

Rh

Rh2+

Pd

Pd+

Pd2+

Solutions

Expert Solution

Solution :-

While writing the electronic configuration the electrons are filled in the atomic orbitals with increasing energy orbital.

Following is the electronic configuration for the given species

Ag = 47 electrons

Electron configuration. = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d10

Ag+ here +1 charge is present on Ag that means Ag looses 1 electrons so the total electrons present in the Ag+ = 47 -1 = 46 ( electron is lost from the 5s orbital)

Electronic configuration. = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10

Ru total electron = 44 electrons.

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d7

Ru3+ = 41 electrons because it loses 3 electrons to form Ru3+ ion (1electron from 5s and 2 electron from 4d are loosed)

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d5

Rh = 45 electrons

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d8

Rh2+ = 43 electrons because it loses 2 electrons to form Rh2+ from 5s1 and 4d orbital

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d7

Pd = 46 electrons

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10

Pd+ = 45 electrons because it looses 1 electron from the 4d shell

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d9

Pd2+ = 44 electrons because it loses 2 electrons from the 4d shell

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d8

queen_honey_blossom answered 1 month ago
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