LECTURE 19: Inorganic Chemistry II Important Notes 1. Transitional metal elements have their electrons taken away from the ‘s’ subshells before the ‘d’ subshells (Co is originally [Ar] 3d7 4s2 ! Co2+ becomes [Ar] 3d7. 2. Electrons are taken away from 4s before 3d in the transition metals. 3. Also, note that 3d7 is written before 4s2 in Co configuration, as even though 3d is a higher energy level than 4s, you write it down in numerical order. Electronic Configurations For neutral transition metal atoms: x • [Noble gas] ns2 (n-1)d . • The ns orbitals generally fill before the 3d orbitals, however, this is not always the case - Cr and Cu fill up 4s1 then d orbital to 5 or 10. • Memorise copper and chromium. • The electrons in d orbital do not pair up, because pairing up costs more energy - according to Hund's rule, each orbital fills up with one electron and then once all the orbitals have one electron in them, they then fill up with the second electron. • 4p orbitals are considerably higher in energy than the 4s and 3d therefore they fill up after the 4s and 3d orbital energy levels. • In chromium 2+, the 4s1 electron is taken away first, and then the remaining electron is taken from the d orbital ! becomes [Ar] 3d4. Trends of the Periodic Table • Atomic radius: As you go across the rows of the periodic table (from left to right), the z- effective/core effective nuclear charge from the protons in the nucleus increases and thus electrons are pulled more strongly towards the nuclear core of the atom. Therefore, atomic radius decreases. • D orbitals shield very well, proton and electrons can cancel each other out. • As you go down the groups of the periodic table, the atomic radius increases. • Electronegativity: increases as you go from left to right. • First ionisation energy: increases from left to right of the periodic table, as the electrons are held more strongly by the protons due to the higher z-effective ! more energy is required for the element to ionise by removing one of its valence electrons. Decreases as you go down the group, as you go down the group, the principle quantum number increases by one, meaning that one extra shell occurs. As the number of shells increases, the electrons are able to counter-oppose the force of the protons as they are further from the positively charged nucleus, therefore, the energy required for the first valence electron to be removed decreases. • Electron affinity: Trend across the periodic table (as you go from left to right) is as a result of z effective (effective nuclear charge). Trend down the periodic table (down the same group) occurs as a result of the growing principal quantum number, n, due to the number of electron shells, meaning that valence electrons are further from the nucleus and thus can more effectively resist the electrostatic forces of attraction from the nucleus. Oxidation State • The charge of the metal ion. • Can accept and donate electrons - able to participate in redox reactions. • +2 and +3 most common oxidation states. • Up to Mn, highest oxidation state equals the group number, e.g. V5+, Cr6+ etc. • After Mn, high oxidation states are less common, due to very high ionisation energy along the rows of the periodic table. • When you go to ions, ns electrons go first and then remove d electrons. • E.g. Cu = 4s1 3d10, Cu+ = 4s0 3d10 (take electrons from s orbital before the d orbital). LECTURE 20: Oxidation of Transition Metals (Workshop Lecture) Similarities between the Transitional Metals • Can form paramagnetic compounds – have some sort of magnetism from unpaired electrons. • Can form coloured complexes. • Have variable stable oxidation states, can go from one oxidation state to the other, by still being stable. Transition Metals, Oxidation States, Colour Changes • Even if it is the same metal element, different oxidation states will impact the colour and/or the appearance of the metal. • To work out the highest oxidation states of metals ! look at the group number. • If take a metal in its highest oxidation state (very oxidised), and put it with a strong reducing agent (reductant), the oxidation state will be reduced, therefore the metal will be less oxidised. • E.g.2 V5+ (aq) + Zn (s) ! 2V4+ (aq) + Zn2+ (aq). • Only common oxidation state of zinc (Zn) = 2+. • Each colour change will represent a different oxidation state. • V5+ ! V2+: Yellow ! green ! blue ! green again ! violet (5 different colours). • But if started off with V5+ and finished with V2+, why were there 5 different colours? Because the green colour represents the mixture between the two oxidation states. The transition between oxidation states. • To take V5+ to V2+, use a strong oxidising agents, Zn (in excess), to keep on reducing the V5+ until it reaches V2+. Important Note: the colours that are observed from the ion solutions occur as a result of the electron excitation, where the electrons are excited and move to higher energy levels by absorbing energy. The colour observed is not the wavelength of energy absorbed, but the complementary wavelength being reflected from the aqueous ion solution. .