Chemistry, 05.07.2019 11:00, josephicarusmarrujo

For elements 1–36, there are two exceptions to the filling order as predicted from the periodic table. draw the atomic orbital diagrams for the two exceptions and indicate how many unpaired electrons are present.

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Here we have to draw the atomic orbital diagram of the elements present in the range of ¹H to ³⁶Kr having exception and its number of unpaired electrons present in the molecule.

The two exceptional elements are chromium (Cr) and Copper (Cu) the atomic orbital diagrams are-

²⁴Cr = 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.

²⁹Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹.

The number of unpaired electrons in Cr and Cu are 6 and 1 respectively.

The atomic orbital diagram of the elements in periodic table is governed by the three rules or principles. Those are (i) The Aufbau principle, (ii) The Pauli exclusion principle and (iii) Hund''s rule of maximum multiplicity.

Among it the Aufbau principle explain the energy level of the atomic orbitals 1s < 2s 2p < 3s < 3P < 3d < 4s etc.

The Pauli's principle tells number of maximum electrons can be present in a particular orbital. As per the principle the s, p and d orbital can posses maximum 2, 6 and 10 electrons.

the Hund's rules explains the extra stability of the half filled and full filled orbitals.

In ²⁴Cr (after ²³V: 1s²2s²2p⁶3s²3p⁶3d³4s²) the expected electronic configuration is 1s²2s²2p⁶3s²3p⁶3d⁴4s² having four unpaired electron but as the half filled d-orbital (5 electrons) is more stable thus it posses 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ electronic configuration having 6 unpaired electrons.

In ²⁹Cu (after ²⁸Ni: 1s²2s²2p⁶3s²3p⁶3d⁸4s²) the expected electronic configuration is 1s²2s²2p⁶3s²3p⁶3d⁹4s² having 1 unpaired electron in d orbital, but due to high stability of the fulfilled d orbital (10 electrons) the atomic orbital diagram become 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ having same number of unpaired electron is s orbital.

The two exceptional atomic orbital diagram of the element in between ¹H to ³⁶Kr is drawn and number of unpaired electrons are calculated.

answered: Guest

The answer is letter c

answered: Guest

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.

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For elements 1–36, there are two exceptions to the filling order as predicted from the periodic tabl...