Subjective Type

What characteristic is given by the angular momentum quantum number?

Solution

The angular momentum quantum number $$l$$ describes the shape of an orbital. It corresponds to each orbital type i.e. $$(0, 1, 2, 3, ...., n-1)=(s, p, d, f, g, h, .....)$$
Note that:
$$l_{max}=n-1$$
$$l=0, 1, 2, 3, ....,n-1$$
$$2l+1$$ is the number of orbitals in a given $$l$$ subshell
$$l$$ is the number of angular nodes in an orbitla.
So two examples:
$$1s$$ orbital:
$$n=1$$ for the principal quantum number
Thus, $$l_{max}=1=0$$. This also means there is no angular node in the $$1s$$ orbital.
$$2l+1=2(0)+1=1$$. Thus, there is only one $$1\ s$$ orbital.
That is also determined by realizing that for the magnetic quantum number, $$m_l=\left\{ 0 \right\}$$ due to $$l=0$$ i.e. the only $$1\ s$$ orbital there is has $$l=0$$.
$$2s, 2p$$ orbital(s):
$$n=2$$ for the principal quantum number
Thus, $$1=0, 1$$ and $$l_{max}=1$$. This also means there is no angular node in the $$2\ s$$ orbital, but $$1$$ angular node in each $$2p$$ orbital.
For the $$2\ s$$ orbital(s), $$2l+1=2(0)+1=1$$. Thus, there is only one $$2\ s$$ orbital.
That is also determined by realizing that for the magnetic quantum number,$$m_l =\left\{ 0\right\}$$ due to $$l=0$$ i.e. the only $$1\ s$$ orbital there is has $$l=0$$
For the $$2p$$ orbital(s), $$2l+1=2(1)+1=3$$. Thus, there are three $$2p$$ orbitals
That is also determined by realizing that for the magnetic quantum number, $$m_l \left\{ -14, 0+1\right\}$$ due to $$l=1$$ i.e.each $$2p$$ orbital uniquely corresponds to either $$l=-1, 0$$ or $$+1$$. As there are three $$m_l$$ values, there are three $$2p$$ orbitals.

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