Assertion: UF_6 is more covalent than UF_4.Re | vedclass

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

Question

(B) According to Fajan's rule,the covalent character of a compound increases with the increase in the oxidation state of the central metal atom. In $U

Vedclass · Accepted Answer

Both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.

Vedclass · Answer

(B) According to Fajan's rule,the covalent character of a compound increases with the increase in the oxidation state of the central metal atom. In $UF_6$,the oxidation state of Uranium is $+6$,whereas in $UF_4$,it is $+4$. Since the oxidation state of $U$ in $UF_6$ is higher,it has higher polarizing power,making $UF_6$ more covalent than $UF_4$. The Reason states that Fluorine is smaller in size,which is a true statement,but it does not explain why $UF_6$ is more covalent than $UF_4$. The correct explanation relates to the oxidation state of the central metal atom.

Assertion: $UF_6$ is more covalent than $UF_4$.
Reason: Fluorine is smaller in size.

Similar Questions

$AlCl_3$ is an electron-deficient compound,but $AlF_3$ is not. This is because

The charge/size ratio of a cation determines its polarizing power. Which of the following sequences represents the increasing order of the polarizing power of the cations $K^+$,$Ca^{2+}$,$Mg^{2+}$,and $Be^{2+}$?

According to Fajan's rule,electrovalent bond formation is favoured by

How does the covalent character vary in $LiCl, BeCl_2, BCl_3$ and $CCl_4$?

Assertion $(A)$: Covalent character follows $NaCl < MgCl_2 < BeCl_2 < LiCl$.
Reason $(R)$: According to Fajan's rules,covalent character of ionic compounds increases with decrease in size of cation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion: $UF_6$ is more covalent than $UF_4$.Reason: Fluorine is smaller in size.