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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.

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