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Total number of angular nodes of orbitals associated with third shell $(n=3)$ of an atom is
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The correct answer is:
$3$
Total number of angular nodes $=l$, for a given orbital. For $n=3$, there will be three subshells having $3 \mathrm{~s}$, $3 \mathrm{p}$ and $3 \mathrm{~d}$ orbitals.
For $3 \mathrm{~s}, l=0$, so zero angular nodes.
For $3 \mathrm{p}, l=1$, so one angular nodes.
For $3 \mathrm{~d}, l=2$, so two angular nodes.
$\Rightarrow$ Total angular nodes $=0+1+2=3$
For $3 \mathrm{~s}, l=0$, so zero angular nodes.
For $3 \mathrm{p}, l=1$, so one angular nodes.
For $3 \mathrm{~d}, l=2$, so two angular nodes.
$\Rightarrow$ Total angular nodes $=0+1+2=3$
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