Question

State and explain the Aufbau principle, write electronic configuration of Cr, Fe, Cu.

Answer 1

Hint: The filling of electrons in the orbital relative to its energy level, to attain stable configuration of element. The shielding effect of d-electrons is also taken into account with the increase of the atomic number.

Complete step by step answer:
The Aufbau principle describes the model-building method in atoms by progressively adding electrons, stating that “in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.”
This way it forms the most stable electron configuration possible. Thereafter, the electronic configuration can be represented using the Aufbau diagram which uses the Madelung's Rule. That is,
- the electrons are filled in order of increasing value of (n + l) in the orbitals.
- the subshells with the same value of (n + l), then, the lower n value subshell is filled first. Here, n is the principal quantum number or and l is the azimuthal quantum number.
Therefore, the order of occupancy of electrons is as follows:
\[1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p\]

Then, according to the Aufbau principle, the electronic configuration of C with atomic number 6, is $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}\,or\,\left[ He \right]\,2{{s}^{2}}2{{p}^{2}}$.

In case of, Cr (atomic number = 24), in accordance to Aufbau principle, the 4s-orbital (n + l = 4 + 0 = 4) is filled before the 3d-orbital (n + l = 3 + 2 = 5) giving the electronic configuration as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{4}}4{{s}^{2}}\,or\,\left[ Ar \right]\,3{{d}^{4}}4{{s}^{2}}$, where $\left[ Ar \right]$ is the configuration of the preceding noble gas. However, to obtain a lower energy state from the half-filled 3d-orbital, one electron from 3d to 4s orbital. Then, the configuration of Cr is $\left[ Ar \right]\,3{{d}^{5}}4{{s}^{1}}$.

In case of, Fe (atomic number = 26), the electronic configuration will be $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}\,or\,\left[ Ar \right]\,3{{d}^{6}}4{{s}^{2}}$.

In Cu (atomic number = 29), the electron configuration is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{9}}4{{s}^{2}}\,or\,\,\left[ Ar \right]\,3{{d}^{9}}4{{s}^{2}}\], with respect to the Aufbau rule. However, to obtain a lower energy state by the fully-filled 3d-orbital, one electron from 3d moves to 4s orbital. Then, the configuration is $\left[ Ar \right]\,3{{d}^{10}}4{{s}^{1}}$.

 The occupancy in 7s, 5f, 6d, 7p subshells is done only after the lower energy orbitals 1s to 6p are fully occupied.
Also, as the atomic number increases, the energy of the atomic orbitals vary. From hydrogen (Z = 1) to calcium (Z = 20), the unoccupied 3d-orbitals are shielded effectively from the increasing nuclear charge. Whereas, the energies of the 4s and 4p orbitals decrease as it penetrates the $\left[ Ar \right]$ core more.
However, the 3d orbitals penetrate the 4s and 4p orbitals, as after calcium addition of electrons in 3d are not fully shielded by the 4s electrons. Thus, its energy falls between the 4s and 4p orbital. The next electron therefore enters the 3d-orbital. Since the d-electrons are poorly shield from the nuclear charge.

Note: The energy of the 3d-orbital is lower than the 4s-orbital, still the electrons occupy the 4s-orbital first before 3d-orbital, because the 3d electrons are found closer to the nucleus; hence, repelling each other more strongly.