# Why do we say that #s# orbitals have zero angular momentum?

##### 1 Answer

The angular momentum of any **no angular dependence**.

In other words, recall that angular momentum gives rise to irregular shapes of a given atomic orbital. Well, all **spherically symmetric**, so angular momentum has no influence on the shape.

Consider some of the

#psi_(1s)(r,theta,phi) = 1/sqrtpi (Z/a_0)^(3//2)e^(-Zr//a_0)#

#psi_(2s)(r,theta,phi) = 1/(4sqrt(2pi)) (Z/a_0)^(3//2)(2 - (Zr)/a_0)e^(-r//2a_0)#

#psi_(3s)(r,theta,phi) = 1/(81sqrt(3pi)) (Z/a_0)^(3//2) (27 - 18 (Zr)/a_0 + 2((Zr)/(a_0))^2)e^(-Zr//3a_0)#

If you look closely, there is no **zero angular momentum**.