Question

The group number of the element in the periodic table with the electronic configuration $(n - 1)d^2 ns^2$. For $n = 4$ is:

• A. 4
• B. 13
• C. 3
• D. 5

Hint:

The electronic configuration of transition metals in the form $(n - 1) d^2 ns^2$ corresponds to elements in the \( d \)-block, and the group number is found by summing the electrons in the \( s \)- and \( d \)-orbitals.

Answer 1

The Correct Option is A

The given electronic configuration is $(n - 1)d^2 ns^2$. For \( n = 4 \), the electronic configuration becomes: \[ % Option (3)d^2 4s^2 \] This indicates that the element has 2 electrons in the \( 4s \)-orbital and 2 electrons in the \( 3d \)-orbital. The number of electrons in the \( s \)-orbital (2 electrons) and \( d \)-orbital (2 electrons) determines the group number. - The \( d \)-block elements (transition metals) follow the \( (n - 1) d \) and \( ns \) filling rule. The sum of the electrons in the \( s \)-orbital and \( d \)-orbital gives the group number. - The number of electrons in the \( 4s \) and \( 3d \) orbitals is 4, and the element will belong to Group 4. Thus, the group number of the element is 4.