Advertisements
Advertisements
Question
What are the consequences of lanthanoid contraction?
Solution
(1) Resemblance of second and third transition series - This significantly affects the relative properties of the elements coming before and after the lanthanides in the periodic table. It is clear from the following table that there is a regular increase in size from Sc to Y and Y to La.
Table Atomic radii of d-block elements (in pm)
Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
164 | 147 | 135 | 129 | 137 | 126 | 125 | 125 | 128 | 137 |
Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd |
180 | 160 | 146 | 139 | 136 | 134 | 134 | 137 | 144 | 154 |
La* | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg |
187 | 158 | 146 | 139 | 137 | 135 | 136 | 138 | 144 | 157 |
Similarly, we can expect a general increase in size in other groups, though after the lanthanoids, the increase in radii from the second to the third transition series is almost negligible.
\[\ce{Ti -> Zr -> Hf}\]
\[\ce{V -> Nb -> Ta}\]
Pairs of primary elements, such as Zr – Hf, Nb – Ta, Mo – W, etc., have similar (almost) sizes, and the properties of these elements are also similar. Hence, as a result of lanthanoid contraction, the elements of the second and third transition series have greater similarity with each other than the elements of the first and second transition series.
(2) Similarity among lanthanides - Due to very small variations in the radii of lanthanoids, their chemical properties are almost similar. Hence, it is very difficult to separate the elements in a pure state. Modern methods based on repeated fractional crystallization or ion exchange techniques separate them based on very small differences in the sizes of their trivalent ions. These methods separate them based on very small differences in the properties of the elements, such as solubility, complex ion formation, hydration etc.
(3) Basicity differences - Due to lanthanoid contraction, the size of lanthanoid ions decreases regularly with an increase in atomic number. As a result of a decrease in size, the covalent character between lanthanoid ions and OH ions increases from La3+ to Lu3+. Hence, the basic strength of hydroxides decreases with an increase in atomic number. Thus, La(OH)3 is the most basic, while Lu(OH)3 is the least basic.
APPEARS IN
RELATED QUESTIONS
Define lanthanoid contraction.
Account for the following :
Zr and Hf have almost similar atomic radii.
What are the different oxidation states exhibited by the lanthanoids?
Compare the chemistry of actinoids with that of lanthanoids with special reference to electronic configuration.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to oxidation state.
The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements.
Explain the cause of lanthanoid contraction?
Explain the cause of Lanthanoids contraction.
Name an element of lanthanoid series which is well knwon to shown +4 oxidation state. Is it a strong oxidising agent or reducing agent?
What is the action of the following on lanthanoids?
a. water
b. Sulphur, heat
c. nitrogen, heat
The f-block elements are known as ____________.
General electronic configuration of actinoids is `(n-2)f^(1-14)(n - 1)d^(0-2)ns^2`.Which of the following actinoids have one electron in 6d orbital?
(i) U (Atomic no. 92)
(ii) Np (Atomic no.93)
(iii) Pu (Atomic no. 94)
(iv) Am (Atomic no. 95)
Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because:
(i) it has variable ionisation enthalpy
(ii) it has a tendency to attain noble gas configuration
(iii) it has a tendency to attain f 0 configuration
(iv) it resembles Pb4+
Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?
Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
Match the compounds/elements given in Column I with uses given in Column II.
Column I (Compound/element) | Column II (Use) | |
(i) | Lanthanoid oxide | (a) Production of iron alloy |
(ii) | Lanthanoid | (b) Television screen |
(iii) | Misch metal | (c) Petroleum cracking |
(iv) | Magnesium based alloy is constituent of | (d) Lanthanoid metal + iron |
(v) | Mixed oxides of lanthanoids are employed | (e) Bullets |
(f) In X-ray screen |
Match the property given in Column I with the element given in Column II.
Column I (Property) | Column II (Element) | |
(i) | Lanthanoid which shows +4 oxidation state |
(a) Pm |
(ii) | Lanthanoid which can show +2 oxidation state |
(b) Ce |
(iii) | Radioactive lanthanoid | (c) Lu |
(iv) | Lanthanoid which has 4f7 electronic configuration in +3 oxidation state |
(d) Eu |
(v) | Lanthanoid which has 4f14 electronic configuration in +3 oxidation state |
(e) Gd |
(f) Dy |
On the basis of Lanthanoid contraction, explain the following:
Nature of bonding in \[\ce{La2O3}\] and \[\ce{Lu2O3}\] .
On the basis of Lanthanoid contraction, explain the following:
Radii of 4d and 5d block elements.
Which of the following pairs has the same ionic size?
The titanium (Z = 22) compound that does not exist is:-
In lanthanoid the last electron enters (n - 2)f subshell where n is equal to ______.
Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of _______.
Write any two consequences of Lanthanoid Contraction.
The lathanide ion that would show colour is ______.
Cerium (Z = S8) is an important member of lanthanoids. Which of the following statements about cerium is incorrect?