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प्रश्न
What are the consequences of lanthanoid contraction?
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उत्तर
(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.
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संबंधित प्रश्न
Define lanthanoid contraction.
Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
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.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to atomic and ionic sizes.
Write the electronic configurations of the element with the atomic number 61.
Explain, why lanthanum (Z = 57) forms La3+ ion, while cerium (Z = 58) forms Ce4+ ion?
Account for the following:
Zn, Cd and Hg are soft metals.
Explain the cause of Lanthanoids contraction.
Which trivalent ion has the maximum size in the Lanthanoid series, i.e., Lanthanum ion (La3+) to Luteium ion (Lu3+)?
(at. no. of Lanthanum = 57 and Lutetium = 71)
Name an element of lanthanoid series which is well knwon to shown +4 oxidation state. Is it a strong oxidising agent or reducing agent?
The f-block elements are known as ____________.
Gadolinium belongs to 4f series. It’s atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
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 statements given in Column I with the oxidation states given in Column II.
| Column I | Column II | |
| (i) | Oxidation state of Mn in MnO2 is | (a) + 2 |
| (ii) | Most stable oxidation state of Mn is | (b) + 3 |
| (iii) | Most stable oxidation state of | (c) + 4 |
| Mn in oxides is | (d) + 5 | |
| (iv) | Characteristic oxidation state of lanthanoids is | (e) + 7 |
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:
Trends in the stability of oxo salts of lanthanoids from \[\ce{La}\] to \[\ce{Lu}\].
On the basis of Lanthanoid contraction, explain the following:
Stability of the complexes of lanthanoids.
On the basis of Lanthanoid contraction, explain the following:
Radii of 4d and 5d block elements.
On the basis of Lanthanoid contraction, explain the following:
Trends in acidic character of lanthanoid oxides.
How would you account for the following:
There is a greater range of oxidation states among the actinoids than among the lanthanides.
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?
Write a note on lanthanoids.
State a reason for the following:
La(OH)3 is more basic than Lu(OH)3.
Trivalent Lanthanoid ions such as La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in their solution. Give a reason.
Assertion: There is a continuous increase in size among Lanthanoids with an increase in atomic number.
Reason: Lanthanoids do not show Lanthanoid contraction.
Mention uses of alloys.
The pair of lanthanoid ions which are diamagnetic is:
