# Electronic configuration of the first transition series > From Sc to Zn: the series where the 3d sublevel is filled successively ## Objectives - Write the full and the [Ar] short electron configuration of Sc (Z = 21) to Zn (Z = 30). - Draw the 4s and 3d boxes using Hund's rule and count the unpaired electrons. - Know that Cr and Cu are different from the general pattern, and write their configurations as the book does. ## 1. Electronic configurations and oxidation states ## 2. Where does the extra electron go? Predict: where does electron number 21 of Sc go? - $4p$ - $4s$ - $3d$ ✓ Electron 21 goes into 3d. In the filling order 4s comes before 3d, and 4s is already full at Ca, so Sc is [Ar] $4s^2\ 3d^1$. [ar-EG] الـ electron رقم 21 بيروح الـ 3d. في ترتيب الملء الـ 4s قبل الـ 3d، والـ 4s اتملى خلاص عند الـ Ca، فالـ Sc بيبقى [Ar] $4s^2\ 3d^1$. ## 3. The general pattern $[\text{Ar}]$ is short for the first 18 electrons. After it comes $4s^2$, which was filled at Ca, and then 3d. Full configuration The same configuration, shortened, in the order of the book's table: 4s, then 3d $\ce{Sc}$: one electron in 3d General pattern: number of 3d electrons = Z − 20 (except Cr and Cu) ## 4. Worked example: Fe (Z = 26) Step 1: Take away the [Ar] $26 - 18 = 8$ electrons are left after $[\text{Ar}]$, because $[\text{Ar}]$ = the first 18 electrons. Step 2: Share them between 4s and 3d The first 2 go to $4s$ (just like Ca), and the other 6 go to $3d$. Step 3: Draw the boxes Hund's rule: the first five go in singly, and the sixth pairs with the first one. $3d^6$: 4 unpaired electrons Fe can be written as $[\text{Ar}]\ 3d^6$, without $4s^2$. ## 5. Your turn: Co (Z = 27) Same steps. Step 1 is done for you: $27 - 18 = 9$ electrons after $[\text{Ar}]$. The first ___ electrons go to 4s, and the rest (___) go to 3d. So 3d has ___ unpaired electrons. Show the boxes $\ce{_{27}Co}$: $[\text{Ar}]\ 4s^{2}\ 3d^{7}$ ## 6. The whole series in your hands Move the slider from Sc to Zn. Watch the 3d boxes and the number of unpaired electrons. Z (atomic number) $\ce{_{21}Sc}$: $[\text{Ar}]\ 4s^{2}\ 3d^{1}$ $\ce{_{22}Ti}$: $[\text{Ar}]\ 4s^{2}\ 3d^{2}$ $\ce{_{23}V}$: $[\text{Ar}]\ 4s^{2}\ 3d^{3}$ $\ce{_{24}Cr}$: $[\text{Ar}]\ 4s^{1}\ 3d^{5}$ (one of the book's two exceptions; it has its own lesson) $\ce{_{25}Mn}$: $[\text{Ar}]\ 4s^{2}\ 3d^{5}$ $\ce{_{26}Fe}$: $[\text{Ar}]\ 4s^{2}\ 3d^{6}$ $\ce{_{27}Co}$: $[\text{Ar}]\ 4s^{2}\ 3d^{7}$ $\ce{_{28}Ni}$: $[\text{Ar}]\ 4s^{2}\ 3d^{8}$ $\ce{_{29}Cu}$: $[\text{Ar}]\ 4s^{1}\ 3d^{10}$ (one of the book's two exceptions; it has its own lesson) $\ce{_{30}Zn}$: $[\text{Ar}]\ 4s^{2}\ 3d^{10}$ Tried all ten? Tap to see the pattern The five 3d orbitals are filled singly up to Mn ($3d^5$). After Mn, pairing takes place up to Zn ($3d^{10}$). Cr and Cu are the exceptions: $4s^1$, as the book writes them. So 4s can change too. Compare Cr and Mn: both are $3d^5$, but 4s goes from 1 to 2. ## 7. Why no early pairing? In Fe the 3d electrons can pair up while orbitals are still empty: $[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\ \ ][\ \ ]$. Mn (Z = 25) $3d^5$: every orbital has one electron Fe (Z = 26) $3d^6$: the first pair in the series The book says it clearly: the five 3d orbitals are filled singly up to Mn ($3d^5$). After Mn, pairing takes place up to Zn ($3d^{10}$). ## 8. Draw Ni $\ce{_{28}Ni}$: which diagram of 4s and 3d is correct? - $4s\,[\uparrow\downarrow]\quad 3d\,[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow][\uparrow]$ ✓ - $4s\,[\ \ ]\quad 3d\,[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow]$ - $4s\,[\uparrow]\quad 3d\,[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow]$ - $4s\,[\uparrow\downarrow]\quad 3d\,[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\ \ ]$ $\ce{_{28}Ni}$: $[\text{Ar}]\ 4s^2\ 3d^8$, so 2 unpaired electrons in 3d. [ar-EG] $\ce{_{28}Ni}$: $[\text{Ar}]\ 4s^2\ 3d^8$، يعني فيه 2 unpaired electrons في الـ 3d. ## 9. Write the configuration Which configuration is correct for $\ce{_{23}V}$? - $[\text{Ar}]\ 3d^5$ - $[\text{Ar}]\ 3d^3$ - $[\text{Ar}]\ 4s^2\ 3d^3$ ✓ - $[\text{Ar}]\ 4s^1\ 3d^4$ General pattern $[\text{Ar}]\ 4s^2\ 3d^{\,Z-20}$: $23 - 20 = 3$. [ar-EG] القاعدة العامة هي $[\text{Ar}]\ 4s^2\ 3d^{\,Z-20}$: يعني $23 - 20 = 3$. ## 10. Summary From Sc to Zn the 3d sublevel is filled successively. General pattern: $[\text{Ar}]\ 4s^2\ 3d^{\,Z-20}$. 3d follows Hund's rule: singly up to Mn ($3d^5$), then pairing up to Zn ($3d^{10}$). In an atom, never leave out 4s: the electrons must add up to Z. Two exceptions: $\ce{_{24}Cr}$: $[\text{Ar}]\ 4s^1\ 3d^5$ and $\ce{_{29}Cu}$: $[\text{Ar}]\ 4s^1\ 3d^{10}$ (the reason is in its own lesson). The five **3d orbitals** are filled **singly** up to Mn ($3d^5$), then **pairing** takes place up to Zn ($3d^{10}$): **Hund's rule**. Can you explain why Fe has 4 unpaired electrons in 3d, not just 2? $\ce{_{26}Fe}$: $[\text{Ar}]\ 4s^2\ 3d^6$. By Hund's rule the first five 3d electrons go in singly, and only the sixth pairs with the first one, so 4 electrons stay unpaired. $3d^6$: 4 unpaired electrons Chapter: Transition Elements (https://yaani.ai/en/learn/chemistry/chem-s3-en.ch1) Source: https://yaani.ai/en/lesson/chem-s3-en.lesson-te.electron-configuration