Chemistry - Atomic Structure Question with Solution | TestHub

$l$ = orbital (0=s, 1=p, 2=d)

$m$ = $-l$ to $+l$

Cr configuration:

Cr = $[Ar]3d^{5}4s^1$

Condition: $l+m\ne 2$

In d-subshell ($l=2$) → $m=-2,-1,0,+1,+2$

Only case where $l+m=2$:

→ $l=2,m=0$ exclude

So, 1 orbital excluded, remaining 4 orbitals valid

→ max 4 electrons from d

1 electron from 4s (always allowed)

Total = 5 electrons