# Which orbital do these transition metals are filling in? Sc, Ti, V, Cr?

Aug 11, 2018

Well, you are filling the $d - \text{subshell...}$

#### Explanation:

For scandium, $Z = 21$, we got...$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{1}$

For vanadium, $Z = 23$, we got...$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{3}$

For titanium, $Z = 22$, we got...$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{2}$

For chromium, $Z = C r$, $Z = 24$, we got...$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{4}$

The valence electrons with RESPECT to the atom are the $\text{d-electrons}$... Upon ionization, we conceive that the $\text{s-electrons}$ are removed first...