Abnormal electronic configurations are observed in which pair?
A.Cu, Cr
B.Pd, Pt
C.Cr, Ni
D.Both A and B

Hint:To answer this question, you must recall the electronic configuration of an element. The electronic configuration of an element depends on three rules namely, the Hund’s rule, Pauli’s exclusion principle and Auf- bau principle. We shall write the electronic configuration of each element to check.

Complete answer:
We know that according to the Auf- Bau principle, the electrons in an atom fill the orbitals in the order $1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d$.
Chromium element has an exceptional electronic configuration. Considering the rules and general trend of electronic configurations, it is expected to have an electronic configuration as,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^2}{\text{3}}{{\text{d}}^4}$
But, one electron from the $4s$ orbital is shifted to the one empty $3d$ orbital to make the $3d - $orbital half filled. This happens on account of the extra stability of a half- filled orbital due to symmetry in the electronic arrangement and greater exchange energy in the orbitals. Thus, the electronic configuration of chromium is given as,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$.
It is necessary that the adjacent orbitals have comparable energies, so that the electron is easily transferred without the acceptance of much energy.
Similarly in copper, palladium and platinum, the electronic configuration is abnormal due to the greater stability of half- filled and fully- filled orbitals. Their electronic configurations are given as:
${\text{Cr}}:\left[ {{\text{Ar}}} \right]{\text{4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$
${\text{Cu:}}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{1}}}$
${\text{Pd:}}\left[ {{\text{Kr}}} \right]{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{0}}}$
$Pt:\left[ {4{f^{14}}5{d^{10}}6{s^0}} \right]$

Hence, the correct answer is D.

Note:
The laws that determine the electronic configuration of an element are:
The Hund’s rule of maximum multiplicity: It states that all orbitals in a sub- shell are first singly- filled before pairing of electrons starts.
The Pauli’s exclusion principle: It states that each electron has a different set of quantum numbers associated to it.
The Auf- bau principle: It states that atomic orbitals in an atom are filled in the increasing order of the energy level of the orbitals.