Title: Lesson 5 First Row d-Block

Learning Objectives:

• Describe the properties of transition metals

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By reference to the structure and bonding in NaCl and SiCl4:

a) State and explain the differences in electrical conductivity in the liquid state.

b) Predict an approximate pH value for the solutions formed by adding each compound separately to water. Explain your answer.

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The TraditionalBased on Mendeleev’s work. Easiest to use and display.

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First row d-block elements A transition elements can be defined as different from

‘d block elements’: ‘A transition element is an element that forms at least

one stable ion with a partially filled d subshell’

For example, Cu2+: 1s2 2s2 2p6 3s2 3p6 (4s0) 3d9

Small range in atomic radii similar properties

Why is the range small? We need to look at the electron configuration…

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Decreasing atomic radii due to small increase in effective nuclear charge.

Increase in nuclear charge from added proton is offset by addition to the electron in the 3d inner shell – SHIELDING! (Without the shielding the decrease in atomic radii would be even more!)

Small range of ionisation energies because electrons are removed from the 4s shell.

Notice Cr and Cu – more stable to have half filled and filled 3d subshells respectively…

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Solution

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Properties of Transition Metals

Physical Properties Chemical Properties

High electrical and thermal conductivity

Form compounds with more than one oxidation number

High melting point Form a variety of complex ions

Malleable Form coloured compounds

High tensile strength Act as catalysts

Ductile

Iron, cobalt and nickel are ferromagnetic

Other things to note:• Physical properties can be explained in terms of the strong metallic bonding• 3d and 4s electrons are close in energy, so are all involved in the bonding – they form part of the delocalised sea of electrons which holds the lattice together.

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Scandium and Zinc

Although in the first row of the d-block, these are not transition metals.

To understand why, write the full electron configuration for: Sc and Sc3+

Zn and Zn2+

However, please note that for our syllabus we count Scandium as a transition metal!

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Zinc is not a transition metal...

‘A transition element is an element that forms at least one stable ion with a partially filled d subshell’

WHY??? Only ion it forms is Zn2+ ion This means that 2 electrons are removed from the 4s shell A fully filled 3d shell will be left

ALSO...Does not form coloured compounds/solutions.

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Variable Oxidation Number of Transition ElementsComparison between group 2 and Transition Metals: Ca only has oxidation state of 2+ Titanium shows +4, +3 and + 2 states

Can you relate the oxidation states formed to the pattern in successive ionisation energy? (HINT: Look at the jumps in energy...)

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Variable Oxidation Number of Transition ElementsComparison between group 2 and Transition Metals: Ca only has oxidation state of 2+ Titanium shows +4, +3 and + 2 states

Can you relate the oxidation states formed to the pattern in successive ionisation energy?

Ca3+ would be unstable as 3rd electron would be removed from the 3p orbital...

Successive IE for Ti is more gradual as 3d and 4s are close in energy hence can make

+2, +3 and +4 oxidation states. No +5 as this would mean removing from 3p as well.

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Oxidation States of d Block Elements

States highlighted blue are the more common ones

All transition metals show both 2+ and 3+ oxidation states Note: Unlike other groups, electrons are not removed to in

order to generate the nearest stable noble gas configuration...

M3+ Ions more common from Sc to Cr

M2+ Ions more common from Mn to Zn

States respond to 3d and 4s electrons in bonding...

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High Oxidation States

Oxidation of above 3+ generally show covalent character

High charge means large charge density

They polarise negative ions Increases covalent character of

the compound

Think about electro-negativities to explain why large charge density will polarise ions…

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Variable oxidation numbers (ions) Transition metals have large numbers of electrons in d-

orbitals, This means the amount of energy required to remove the

second electron is not much different to that required to remove the first and so on.

Some common oxidation states we need to know: All of them in the +2 oxidation state Cr(III), Cr(VI) Mn(IV), Mn(VII) Fe(III) Cu(I)

Task: select 4 of these and write the electron configuration

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Solutions