(A) The reaction is: $\underline{Te}F_6 + 6H_2O \longrightarrow H_6TeO_6 + 6HF$.Here,$H_6TeO_6$ is orthotelluric acid,which can be written as $Te(OH)_

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

If the product is an oxyacid with an $-ic$ suffix.

(A) The reaction is: $\underline{Te}F_6 + 6H_2O \longrightarrow H_6TeO_6 + 6HF$.Here,$H_6TeO_6$ is orthotelluric acid,which can be written as $Te(OH)_6$ or $H_2TeO_4 \cdot 2H_2O$.The oxidation state of $Te$ in $H_6TeO_6$ is $+6$.Since $Te$ is in its highest oxidation state $(+6)$,the corresponding oxyacid is named telluric acid,which ends with the $-ic$ suffix.Therefore,the product is an oxyacid with an $-ic$ suffix.

Class 12 Chemistry p-Block Elements (Class 12)Oxygen family

Difficult

In the following reaction,identify the nature of the product formed by the complete hydrolysis of the underlined atom at $R.T.$:
$\underline{Te}F_6 + H_2O \longrightarrow H_6TeO_6 + HF$

A
If the product is an oxyacid with an $-ic$ suffix.
B
If the product is an oxyacid with an $-ous$ suffix.
C
If the products are two oxyacids,one with an $-ic$ suffix and the other with an $-ous$ suffix.
D
If the product is not an oxyacid,neither with an $-ic$ suffix nor with an $-ous$ suffix.