Home
Class 12
CHEMISTRY
Assertion (A) Actinoids form relatively ...

Assertion (A) Actinoids form relatively less stable complexes as compared to lanthanoids.
Reason (R ) Actinoids can utilise their 5f orbitals alongwith 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.

A

Both assertion and reason are true, and reason is the correct explanation of the assertion.

B

Both assertion and reason are true but reason is not the correct explanation of assertion.

C

Assertion is not true but reason is true.

D

Both assertion and reason are false.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we will analyze both the assertion and the reason provided. ### Step 1: Understand the Assertion The assertion states that "Actinoids form relatively less stable complexes as compared to lanthanoids." We need to evaluate whether this statement is true or false. ### Step 2: Understand the Reason The reason states that "Actinoids can utilize their 5f orbitals along with 6d orbitals in bonding, but lanthanoids do not use their 4f orbital for bonding." We need to assess the validity of this statement. ### Step 3: Analyze the Stability of Complexes - **Actinoids**: The actinoids (elements 89-103) have 5f electrons that can participate in bonding. They also have the ability to utilize their 6d orbitals, which allows for a greater variety of oxidation states and complex formation. This flexibility generally leads to the formation of more stable complexes. - **Lanthanoids**: The lanthanoids (elements 57-71) primarily utilize their 4f orbitals for bonding. However, the 4f orbitals are more shielded and less available for bonding compared to the 5f orbitals in actinoids. This results in lanthanoids forming less stable complexes compared to actinoids. ### Step 4: Conclusion on Assertion and Reason - The assertion that actinoids form less stable complexes than lanthanoids is **incorrect**. - The reason provided is **correct** as actinoids can utilize their 5f and 6d orbitals for bonding, while lanthanoids do not effectively use their 4f orbitals. ### Final Answer The assertion is **false**, and the reason is **true**. Therefore, the correct conclusion is that the assertion is not true, but the reason is true.
To solve the question, we will analyze both the assertion and the reason provided. ### Step 1: Understand the Assertion The assertion states that "Actinoids form relatively less stable complexes as compared to lanthanoids." We need to evaluate whether this statement is true or false. ### Step 2: Understand the Reason The reason states that "Actinoids can utilize their 5f orbitals along with 6d orbitals in bonding, but lanthanoids do not use their 4f orbital for bonding." We need to assess the validity of this statement. ...
Promotional Banner

Topper's Solved these Questions

  • D AND F-BLOCK ELEMENTS

    NCERT EXEMPLAR ENGLISH|Exercise Short Answer Type Question|20 Videos

  • COORDINATION COMPOUNDS

    NCERT EXEMPLAR ENGLISH|Exercise Long Answer Type Question|5 Videos

  • ELECTROCHEMISTRY

    NCERT EXEMPLAR ENGLISH|Exercise All Questions|67 Videos

Similar Questions

Explore conceptually related problems

Assertion (A) : Ionisation potential of Be (atomic number 4 ) is more than B (atomic number 5) Reason (R ) : The first electron released fromm Be is of p orbitals but that from B is of s orbitals.

Assertion : In formaldehyde all the four atoms lie in one plane. Reason : Carbonyl carbon forms a pi -bond with oxygen by overlapping of p -orbitals.

Assertion: Yb^(+2) is more stable in compare to Gd^(+2) Reason: The electronic configuration of Gd is [Xe] 4f^(7)5d^(2)6s^(2)

Assertion : Inner orbital complexes are low spin complexes . Reason : In low spin complexes , inner d-orbital (3d) is used in hybridisation .

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important for oxygen compounds. Oxygen is never more than divalent beacasue when it hs formed two covalent bonds, there are no low energy orbitals which can be used for form. further bonds. However, the elements S,Se,Te and Po have empty d- orbitals which may be used for bonding, and they can form four or six bonds by unpairing electrons. The higher oxidation states become less stable on decending the group. The bond between S and O, or Se and O, are much shorter than might be expected for a single bond owing to rho pi - dpi interaction between the p- orbital of oxygen and d- orbital of S or Se. Which of the following statemet is incorrect ?

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important for oxygen compounds. Oxygen is never more than divalent beacasue when it hs formed two covalent bonds, there are no low energy orbitals which can be used for form. further bonds. However, the elements S,Se,Te and Po have empty d- orbitals which may be used for bonding, and they can form four or six bonds by unpairing electrons. The higher oxidation states become less stable on decending the group. The bond between S and O, or Se and O, are much shorter than might be expected for a single bond owing to rho pi - dpi interaction between the p- orbital of oxygen and d- orbital of S or Se. Which one of the following orders represents the correct order for the properties indicated against them?

Assertion : According to crystal field theory , during complex formation , the d-orbitals split and form two sets of orbitals t_(2g) and e_(g) . Reason : Splitting of d-orbitals occurs only in case of strong field ligands .

Assertion: Ionisation potential of Be (atomic no.4) is less than B (atomic no.5). Reason: The first electron released from Be is of p-orbital but that from B is of s-orbital.

The concept of hybridisation has been introduced to explain the shapes of molecules. It involves the intermixing of two or more atomic orbitals belonging to same atom but in or more atomic orbitals belonging to same atom but in different sub-shells so as to intermix and redistribute energies to from equivalent orbitals called hybrid orbitals. Depending upon total number and nature of the orbitals involved, the hybridisation may be divided into sp (linear), sp^(2) (trigonal), sp^(3) (tetrahedral), sp^(3)d (trigonal bipyramidal), sp^(3)d^(3) (octahedral) and sp^(3)d^(3) (pentagonal bipyramidal) types. it may be noted that the orbitals of isolated atoms never hybridize and they do so at the time of bond formation. A hybrid orbital from s-and p-orbitals can contribute to

Two important physical evidences supporting the synergic bonding in non-classical complexes are bond lengths and vibrational spectra. Vibrational spectra is based on the fact that the compression and extension of a bond may be analogous to the behaviour of a spring and obeys Hook's law. overline(v)=(1)/(2pic)sqrt((k)/(mu))cm^(-1) where, k=force force constant of the bond which is directly proportional to bonnd strength of CO mu= reduced mass of ligand overline(v) =stretching frequency of the CO bond c=velocity of light Q. In Mn_(2)(CO)_(10) carbonyl complex, the d-orbital of Mn-atom which can not be involved in synergic bonding betwee Mn and CO ligands:

NCERT EXEMPLAR ENGLISH-D AND F-BLOCK ELEMENTS-Short Answer Type Question
  1. Assertion (A) Actinoids form relatively less stable complexes as compa...

    Text Solution

    |

  2. Why does copper not replace hydrogen from acids?

    Text Solution

    |

  3. Why E^(-) values for Mn, Ni and Zn are more negative than expected?

    Text Solution

    |

  4. Why first ionisation enthalpy of Cr is lower than that of Zn?

    Text Solution

    |

  5. Transition elements show high melting points. Why?

    Text Solution

    |

  6. When Cu^(2+) ion is treated with KI, a white precipitate is formed. Ex...

    Text Solution

    |

  7. Out of Cu(2)Cl(2) and CuCl(2), which is more stable and why?

    Text Solution

    |

  8. When a brown compound of manganese (A) is treated with HCl it gives a ...

    Text Solution

    |

  9. Although fluorine is more electronegative than oxygen, but the ability...

    Text Solution

    |

  10. Although Cr^(3+) and Co^(2+) ions have same number of unpaired electro...

    Text Solution

    |

  11. Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. W...

    Text Solution

    |

  12. Although Zr belongs to 4d and Hf belongs to 5d transition series but i...

    Text Solution

    |

  13. Cerium shows oxidation state of +4 because

    Text Solution

    |

  14. Explain why does colour of KMnO(4) disappear when oxalic acid is added...

    Text Solution

    |

  15. When orange solution containing Cr(2)O(7)^(2-) ion is treated with an ...

    Text Solution

    |

  16. A solution of KMnO(4) on reduction yields either a colourless solution...

    Text Solution

    |

  17. The second and third rows of transition elements resemble each other m...

    Text Solution

    |

  18. E^(Theta) of Cu is +0.34V while that of Zn is -0.76 V. Explain.

    Text Solution

    |

  19. The halides of transition elements become more covalent with increasin...

    Text Solution

    |

  20. While filling up of electrons in the atomic orbitals, the 4s orbital i...

    Text Solution

    |

  21. Reactivity of transition elements decreases almost regularly from Sc t...

    Text Solution

    |