Chapter 6, Problem 16CO

• rank various orbitals in terms of size and energy.

Interpretation Introduction

To determine:

Rank various orbitals in terms of size and energy.

Explanation of Solution

Size of atomic orbitals mainly depends upon principal quantum number $\left(n\right).$

Size of atomic orbitals $\gamma n\left(\text{principle quantum number}\right)$

So,

Size of atomic orbital: $\left(\text{incresing order}\right)$

$1s<2s=2p<3s=3p=3d<4s=4p=4d=4f<5s=5p=5d=5f<6s=6p=6d<7s=7p.$

Energy of atomic orbital $\gamma \left(n+l\right)$

$n$ is principal quantum number

$l$ is secondary quantum number

If $\left(n+l\right)$ are equals for two orbitals, then higher

$1s:n=1\text{ and }l=0$

$\left(n+l\right)=1$

$2s:n=2\text{ and }l=0$

$\left(n+l\right)=2$

$2p:n=2\text{ and }l=1$

$\left(n+l\right)=3$

$3s:n=3\text{ and }l=0$

$\left(n+l\right)=4$

$3d:n=3\text{ and }l=2$

$\begin{array}{c}\left(n+l\right)=\left(3+2\right)\\ =5\end{array}$

$4s:n=4\text{ and }l=0$

$\begin{array}{c}\left(n+l\right)=4+0\\ =4\end{array}$

$4p:n=4\text{ and }l=1$

$\begin{array}{c}\left(n+l\right)=4+1\\ =5\end{array}$

$4d:n=4\text{ and }l=2$

$\begin{array}{c}\left(n+l\right)=4+2\\ =6\end{array}$

$4f:n=4\text{ and }l=3$

$\begin{array}{c}\left(n+l\right)=4+3\\ =7\end{array}$

$5s:n=5\text{ and }l=0$

$\begin{array}{c}\left(n+l\right)=5+0\\ =5\end{array}$

$5p:n=5\text{ and }l=1$

$\left(n+l\right)=6$

$5d:n=5\text{ and }l=2$

$\left(n+l\right)=7$

$5f:n=5\text{ and }l=3$

$\begin{array}{c}\left(n+l\right)=7+3\\ =8\end{array}$

$6s:n=6\text{ and }l=0$

$\begin{array}{c}\left(n+l\right)=6+0\\ =6\end{array}$

$6p:n=6\text{ and }l=1$

$\begin{array}{c}\left(n+l\right)=6+1\\ =7\end{array}$

$6d:n=6\text{ and }l=2$

$\begin{array}{c}\left(n+l\right)=6+2\\ =8\end{array}$

$7s:n=7\text{ and }l=0$

$\begin{array}{c}\left(n+l\right)=7+0\\ =7\end{array}$

$7p:n=7\text{ and }l=1$

$\begin{array}{c}\left(n+l\right)=7+1\\ =8\end{array}$

Order of orbital with in creasing $\left(n+l\right)$

$1s<2s<2p=3s<3p=4s<4p=5s=4d<5p=6s=4f<5d=6p=7s<5f=6d=7p.$

If $\left(n+l\right)$ are same, then higher value of $n$ containing orbitals has higher energy.

So,

Energy order

$1s<2s<2p<3s<3p<4s<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p.$

There are no element in which: electrons in ground state found is higher in energy than $7p.$

Conclusion

Size atomic orbital $:\left(\text{in creasing order}\right)$

$1s<2s\approx 2p<3s=3p=3d<4s=4p=4d=4f<5s=5p=5p\approx 5f<6s=6p=6d<7s=7p.$

Energy of atomic orbitals $\left(\text{increasing order}\right)$

$1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p.$