Answer
(a) An excited state of neon.
(b) A ground state of titanium.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have $ 1s^2 \; 2s^2 \; 2p^5 \; 3s^1 $ where
$ 1s^2 $: 2 electrons
$ 2s^2 $: 2 electrons
$ 2p^5 $: 5 electrons
$ 3s^1 $: 1 electron
So, the total Electrons: $ 2 + 2 + 5 + 1 =\bf 10 \;\rm e^-$
The element with 10 electrons is $\underline{\color{red}{\text{ neon (Ne)}}}$, which has an atomic number of $Z=10$.
The ground-state configuration for neon should be $ 1s^2 \; 2s^2 \; 2p^6 $, with a completely filled 2p subshell. In this configuration, an electron has been excited from the 2p to the 3s orbital. Therefore, this represents $\underline{\color{red}{\text{ an excited state configuration for neon}}}$.
$$\color{blue}{\bf [a]}$$
We have $ 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 4s^2 \; 3d^2 $ where
$ 1s^2 $: 2 electrons
$ 2s^2 $: 2 electrons
$ 2p^6 $: 6 electrons
$ 3s^2 $: 2 electrons
$ 3p^6 $: 6 electrons
$ 4s^2 $: 2 electrons
$ 3d^2 $: 2 electrons
So, the total Electrons:$ 2 + 2 + 6 + 2 + 6 + 2 + 2 =\bf 22\;\rm e^- $
The element with 22 electrons is $\underline{\color{red}{\text{
titanium (Ti)}}}$, which has an atomic number of $Z=22$.
The ground-state electron configuration for titanium is $ 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 4s^2 \; 3d^2 $, which matches the given configuration. Therefore, this represents $\underline{\color{red}{\text{ the ground state for titanium}}}$.
