Why is the s-orbital always spherical in shape?

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Why is the s-orbital always spherical in shape?

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Answer 1 · 2017-12-21T22:06:47

Because its wave function has no angular dependence.

By definition, an $s$ $s$ orbital has zero angular momentum, and $l = 0$ $l = 0$ . Any nonzero angular momentum leads to atomic orbitals having non-spherical shapes.

Some explicit wave functions for the hydrogen atomic orbitals are:

$ψ_{1 s} (r, θ, ϕ) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} e_{0}^{- r / a_{0}}$ $psi_(1s)(r,theta,phi) = 1/(sqrtpi) (1/a_0)^(3//2) e^(-r//a_0)$

$ψ_{2 s} (r, θ, ϕ) = \frac{1}{4 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{3 / 2} (2 - \frac{r}{a_{0}}) e_{0}^{- r / 2 a_{0}}$ $psi_(2s)(r,theta,phi) = 1/(4sqrt(2pi)) (1/a_0)^(3//2) (2 - r/a_0)e^(-r//2a_0)$

$ψ_{3 s} (r, θ, ϕ) = \frac{1}{81 \sqrt{3 π}} {(\frac{1}{a_{0}})}^{3 / 2} [27 - 18 (\frac{r}{a_{0}}) + 2 {(\frac{r}{a_{0}})}^{2}] e_{0}^{- r / 3 a_{0}}$ $psi_(3s)(r,theta,phi) = 1/(81sqrt(3pi)) (1/a_0)^(3//2) [27 - 18(r/a_0) + 2(r/a_0)^2]e^(-r//3a_0)$

The main thing you should notice is that all of these $s$ $s$ orbital wave functions have no $θ$ $theta$ or $ϕ$ $phi$ in them, which are angles in spherical coordinates.

That means there is no way the angles could deviate from a straight integration in spherical coordinates at a constant radius (giving a spherical integration path).

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Why is the s-orbital always spherical in shape?

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