$TiI_{4}$ on heating gives

  • A
    $TiI_{2} + I_{2}$
  • B
    $Ti + 2I_{2}$
  • C
    $TiI_{3} + \frac{1}{2}I_{2}$
  • D
    None of these

Explore More

Similar Questions

Silver can be separated from lead by

Match the following:
List-$I$ (Process)List-$II$ (Metal)
$A$. Hall-Heroult process$I$. $Ti$
$B$. Mond process$II$. $In$
$C$. van-Arkel process$III$. $Al$
$D$. Zone refining process$IV$. $Ni$

The correct answer is:

Given below are two statements.
Statement $I:$ During electrolytic refining,blister copper deposits precious metals.
Statement $II:$ In the process of obtaining pure copper by electrolysis method,blister copper is used to make the anode.
In the light of the above statements,choose the most appropriate answer from the options given below.

Explain the purification of Zirconium and Titanium metals.

Match Column $-I$ with Column $-II$ and select the correct answer using the codes given below.
Column $-I$ Column $-II$
$(i)$ Iron and copper $(P)$ Poling
$(ii)$ Zirconium and Titanium $(Q)$ Bessemerisation
$(iii)$ Lead and Tin $(R)$ Van Arkel
$(iv)$ Copper and Tin $(S)$ Liquation

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo