Question

Define the atomic number of elements which is present just below the Ru.
A) 76
B) 44
C) 74
D) 77

Answer 1

First determine the atomic number of ruthenium. Then add 32 to the atomic number of ruthenium to obtain the atomic number of the element which is present just below ruthenium.

Complete answer:
The number of electrons in s,p,d and f subshells is 2, 6, 10 and 14 respectively.
Calculate the total number of electrons present in a principle energy level containing s,p,d and f orbitals.
${\text{2 + 6 + 10 + 14 = 32}}$
When you want to find out the atomic number of an element that is placed just below ruthenium in the group, then you need to add 32 to the atomic number of ruthenium.
The element which is present just below the ${\text{Ru}}$ is ${\text{Os}}$ .
The chemical symbols ${\text{Ru}}$ and ${\text{Os}}$represent the elements ruthenium and osmium respectively.
The atomic number of ${\text{Ru}}$ is 44.
Add 32 to 44.
${\text{44 + 32 = 76}}$
The atomic number of elements which is present just below the ${\text{Ru}}$ is 76.

Hence, the correct option is the option (A) 76.

Note: When only s,p and d orbitals are present, we should add 18 to the atomic number of another element to obtain the atomic number of the element placed just below another element.
For chlorine and bromine, the atomic numbers are 17 and 35 respectively. ${\text{17 + 18 = 35}}$
When only s, and p orbitals are present, you should add 8 to the atomic number of another element to obtain the atomic number of the element placed just below another element.
For fluorine and chlorine, the atomic numbers are 9 and 17 respectively. ${\text{9 + 8 = 17}}$