Question

The outer electronic configuration of two members of lanthanide series are as follows $4{f^1}5{d^1}6{s^2}$ and $4{f^7}5{d^0}6{s^2}$.What are their atomic numbers? Predict their oxidation states.

Answer 1

We should first know what lanthanides are and where they lay on the periodic table. From the given atomic configuration we can find out the atomic number easily and we can know about the name of the element. We can know about the stable oxidation state of a compound.

Complete step by step solution:
Step1: Lanthanides are the rare earth metals. They are presented separately in the periodic table. They start from 57 and end at 71. First lanthanide us lanthanum and last is lutetium. In general $Ln$ symbol is used to represent the lanthanides. They are f-block elements.
Step2. Atom with configuration $4{f^1}5{d^1}6{s^2}$ is the second element in the lanthanide series. It has one electron in the f shell.
$[Xe]4{f^1}5{d^1}6{s^2} = 58$, the atomic number is 58. 58 atomic numbers represent the Cerium whose symbol is $Ce$ .
Step3: Atom with atomic configuration $4{f^7}5{d^0}6{s^2}$ is the 7th element in the series.
$[Xe]4{f^7}5{d^0}6{s^2} = 63$, The atomic number is 63. It represents Europium. Its symbol is $Cu$.
Step4: The stable oxidation state of lanthanides is $+ 3$ in general but it varies often. Oxidation state of Cerium is $+ 3$ and $+ 4$. Stable oxidation state of Europium is $+ 2$.It also depicts $+ 3$ oxidation state.

Note: All the lanthanide shows almost similar properties, They are used in hybrid cars, to make permanent magnets and superconductors. Lanthanides were first discovered in 1787. They usually adopt a higher coordination number than 6 in compounds. Lanthanoids oxidize rapidly in the moist air. They dissolve quickly in the acids. They can also react with halogens upon heating.