Why is the s-orbital always spherical in shape?

Answer 1

due to the lack of angular dependence in its wave function.

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

The following are some clear wave functions for the atomic orbitals of hydrogen:

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

#psi_(2s)(r,theta,phi) = 1/(4sqrt(2pi)) (1/a_0)^(3//2) (2 - r/a_0)e^(-r//2a_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# orbital wave functions have no #theta# or #phi# in them, which are angles in spherical coordinates.

This implies that the angles cannot diverge from a spherical integration path that is a straight integration in spherical coordinates at a constant radius.

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Answer 2

Luna Chen

Given a probability distribution function that is symmetrical around the nucleus (i.e., there is an equal chance of finding an electron at any point in space around the nucleus), the s-orbital is always spherical in shape. This spherical symmetry results from the mathematical solutions to the Schrödinger equation for the hydrogen atom, which regulates the behavior of electrons in atoms.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.