Question

Electronic configuration $\left[Kr\right]4d^{10}$ belongs to:

• A. Period(4) Group(10) Block($d$)
• B. Period(5) Group(12) Block($d$)
• C. Period(5) Group(10) Block($d$)
• D. Period(4) Group(12) Block($d$)

Answer 1

The correct option is A Period(4) Group(10) Block($d$)

1. Here the outermost electron enters into the $N-shell$ i.e $n=4$ hence the it belongs to $4th$ period.
   
2. The outermost electron enters into the $d$ orbital hence it belongs to $d$ block.
3. For d-block elements no. of group= no. of $(n-1)d$electrons + no. of electrons in $nth$ shell=10+0=10