Solution:
Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons.
This assertion is true. 5f electrons, being more spatially extended, have greater interaction with ligand orbitals and thus participate in bonding to a greater extent compared to the more buried 4f electrons.
Reason R: 5f orbitals are not as buried as 4f orbitals.
This reason is also true. 5f orbitals extend further into space than 4f orbitals, meaning they are less shielded by inner electrons and are more available for bonding.
Furthermore, the Reason R directly explains Assertion A. Because 5f orbitals are less buried, they are more available for bonding, which is why 5f electrons participate in bonding to a far greater extent than 4f electrons.
Therefore, both A and R are true, and R is the correct explanation of A.
Correct Answer: (2) Both A and R are true and R is the correct explanation of A.
Given below are the quantum numbers for 4 electrons.
A. n=3, l=2, ml=1,ms=+\(\frac{1}{2}\)
B. n=4, l=1, ml=0,ms=+\(\frac{1}{2}\)
C. n=4, l=2, ml=–2,ms=–\(\frac{1}{2}\)
D. n=3, l=1, ml=–1,ms=+\(\frac{1}{2}\)
The correct order of increasing energy is
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: