Question

${\text{Pd}}$ has exceptional valence shell electronic configuration of ${\text{4}}{{\text{d}}^{10}}5{{\text{s}}^0}$. It is a member of:
A) 5th period, group 10
B) 4th period, group 12
C) 6th period, group 10
D) 5th period, group 14

Answer 1

Hint:We know that ${\text{Pd}}$ is palladium. Palladium belongs to d-block of the periodic table. The atomic number of palladium is 46. We can write the electronic configuration of palladium using Aufbau’s principle.

Complete answer:
We know that the atomic number 46. Thus, palladium has 46 electrons. The electrons fill the orbitals according to Aufbau's principle. The Aufbau’s principle states that in the ground state of the atoms, the orbitals are filled with electrons in order of the increasing energies.

The order of energy of different orbitals in an atom is as follows:
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$ and so on.
Thus, the electronic configuration of palladium is,
$1{s^2}\,2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^8}$
But we are given that palladium has exceptional valence shell electronic configuration of ${\text{4}}{{\text{d}}^{10}}5{{\text{s}}^0}$.
The period of the element is the principal quantum number of the valence shell.
From the electronic configuration, we can see that the valence shell of palladium is 5s. The principal quantum number of the valence shell is 5. Thus, palladium belongs to the 5th period.
The group of the d-block element is equal to the number of electrons in a $\left( {n - 1} \right)d$ subshell + the number of electrons in valence shell.
From the electronic configuration, we can see that the number of electrons in a 4d subshell are 8 and the number of electrons in the 5s valence shell are 2. Thus,
Group $ = 8 + 2$
Group $ = 10$
Thus, palladium belongs to group 10. Thus, ${\text{Pd}}$ is a member of 5th period, group 10.

Thus, the correct answer is option (A) 5th period, group 10.

Note:Remember that the period of the element is the principal quantum number of the valence shell. And the group of the d-block element is equal to the number of electrons in a $\left( {n - 1} \right)d$ subshell + the number of electrons in valence shell.