Most acidic among the given compounds is:

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In the lanthanoid series,the basic character of the hydroxides $Ln(OH)_3$ decreases as the atomic number increases due to the lanthanoid contracti

Vedclass · Accepted Answer

$Er(OH)_3$

Vedclass · Answer

(D) In the lanthanoid series,the basic character of the hydroxides $Ln(OH)_3$ decreases as the atomic number increases due to the lanthanoid contraction. As the atomic number increases,the ionic radius of $Ln^{3+}$ ions decreases,which increases the covalent character of the $Ln-OH$ bond. According to Fajan's rule,higher covalent character leads to lower basicity and higher acidity. Among the given elements,$Pr$ $(Z=59)$,$Sm$ $(Z=62)$,$Gd$ $(Z=64)$,and $Er$ $(Z=68)$. Since $Er$ has the highest atomic number,$Er(OH)_3$ has the smallest ionic radius and the highest covalent character,making it the most acidic among the given options.

Similar Questions

The relation between the coefficient of real expansion $(\gamma_r)$ and coefficient of apparent expansion $(\gamma_a)$ of a liquid and the coefficient of linear expansion $(\alpha_g)$ of the material of the container is

The condition for diffraction is

Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them,the initial force of $12 \times 10^3 ~N$ will change to

If one of the lines $2x^2 - xy + by^2 = 0$ passes through the point $(-4, -2)$,then $b^2 =$

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$ where $k \neq 0$,then the distance from the origin to the line $l$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module