1
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { Match the following. } $$

$$ \begin{array}{llll} \hline & \begin{array}{c} \text { List-I } \\ \text { (Refining method) } \end{array} & & \begin{array}{c} \text { List-II } \\ \text { (Metals to be refined) } \end{array} \\ \hline \text { (A) } & \text { Zone refining } & \text { (I) } & \text { Titanium } \\ \hline \text { (B) } & \text { Poling } & \text { (II) } & \text { Tin } \\ \hline \text { (C) } & \text { Liquation } & \text { (III) } & \text { Gallium } \\ \hline \text { (D) } & \text { Vapour phase refining } & \text { (IV) } & \text { Copper } \\ \hline \end{array} $$

The correct answer is

A
A-IV, B-II, C-I, D-III
B
A-III, B-I, C-IV, D-II
C
A-III, B-IV, C-II, D-I
D
A-II, B-IV, C-I, D-III
2
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Assertion (A) : In group 15 elements nitrogen does not form pentahalides.

Reason (R) : Nitrogen can exhibit +5 oxidation state.

The correct option among the following is

A
(A) and (R) are true and (R) is the correct explanation of (A).
B
(A) and (R) are true but (R) is not correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.
3
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In which of the following options, molecules are correctly arranged with respect to their bond angles.

A
$\mathrm{S}_6<\mathrm{O}_3<\mathrm{S}_8<\mathrm{P}_4$
B
$\mathrm{P}_4<\mathrm{O}_3<\mathrm{S}_6<\mathrm{S}_8$
C
$\mathrm{P}_4<\mathrm{S}_6<\mathrm{S}_8<\mathrm{O}_3$
D
$\mathrm{S}_6<\mathrm{P}_4<\mathrm{S}_8<\mathrm{O}_3$
4
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Which of the following reaction represents Deacon's method?

A
$2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Cl}_2 \xrightarrow{\text { Sunlight }} 4 \mathrm{HCl}+\mathrm{O}_2$
B
$4 \mathrm{HCl}+\mathrm{O}_2 \xrightarrow[723 \mathrm{~K}]{\mathrm{CaCl}_2} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$
C
$2 \mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Electrolysis }} 2 \mathrm{NaOH}+\mathrm{H}_2+\mathrm{Cl}_2$
D
$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{Cl}_2 \longrightarrow \mathrm{CaOCl}_2 \cdot \mathrm{H}_2 \mathrm{O}$
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