Question

Assume set of four quantum no. for last electron of an element X in ground state is n = 4, l =1, m = 1 & s = –1/2. and spin multiplicity of element X in its ground state is 4. Identify the element X.

Answer 1

Answer:

n=4,l=0,m=0 and x=+

2

1

set represents that the last electron enters in a 4s orbital.

Among the given options, only Cr is the element, the last electron of which enters in a 4s orbital.

Cr=[Ar]3d

5

4s

1

.