Based on the provided periodic table segment, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The element with atomic number $40$ is Zirconium $(Zr)$,which belongs to the $5^{th}$ period and group $4$. Element $B$ is directly below $Zr$ in

Vedclass · Accepted Answer

$71, 72$

Vedclass · Answer

(A) The element with atomic number $40$ is Zirconium $(Zr)$,which belongs to the $5^{th}$ period and group $4$. Element $B$ is directly below $Zr$ in the same group,so it belongs to the $6^{th}$ period. Elements in the $6^{th}$ period,group $4$ have an atomic number of $40 + 18 + 14 = 72$ (Hafnium,$Hf$). Element $A$ is to the left of $B$ in the same period ($6^{th}$ period),so its atomic number is $72 - 1 = 71$ (Lutetium,$Lu$). Therefore,the atomic numbers of $A$ and $B$ are $71$ and $72$ respectively.

Based on the provided periodic table segment,calculate the atomic numbers of $A$ and $B$ respectively.

Similar Questions

Which elements are considered exceptions in the classification of the periodic table?

Based on the provided periodic table segment,identify the correct statement.

Where is helium placed in the periodic table and why?

Explain the electronic configuration of elements according to groups in the extended periodic table.

What is the position of copper in the modern periodic table?

Vedclass Test Series

Exam Paper Generator

Online Exam Module