HF present as impurity in gaseous $F_2$ can be removed by passing over:

(1) $P_2{O}_5$

(2) NaF

(3) $H_{2}S{O}_4$

(4)_ $CaC{l}_2$

गैसीय $F_2$ में अशुद्धता के रूप में उपस्थित HF को किसे पारित करके निष्कासित किया जा सकता है:

(1) $P_2{O}_5$

(2) NaF

(3) $H_{2}S{O}_4$

(4) $CaC{l}_2$

Concentrated nitric acid reacts with iodine to give :

(1) $\mathrm{HOI}$                             

(2) $\mathrm{HI}$

(3) ${\mathrm{HIO}}_3$                         

(4) ${\mathrm{HIO}}_4$

सांद्र नाइट्रिक अम्ल आयोडीन के साथ अभिक्रिया कर देता है:

(a) $\mathrm{HOI}$         (b) $\mathrm{HI}$

(c) ${\mathrm{HIO}}_3$        (d) ${\mathrm{HIO}}_4$

${\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{Ba}{\left(\mathrm{OH}\right)}_2<mspace\; width="1em"></mspace>\stackrel{}{\to }<mspace\; width="1em"></mspace>\mathrm{X}(\mathrm{aq}.)+{\mathrm{BaCl}}_2+{\mathrm{H}}_2\mathrm{O}$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{X}+{\mathrm{H}}_2{\mathrm{SO}}_4<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace>}{\to }<mspace\; width="1em"></mspace>\mathrm{Y}+{\mathrm{BaSO}}_4$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{Y}<mspace\; width="1em"></mspace>\underset{∆>365\mathrm{K}}{\overset{∆}{\to }}<mspace\; width="1em"></mspace>\mathrm{Z}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

Y and Z are respectively :

$\left(\mathrm{a}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_4,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_2<mspace\; width="1em"></mspace>$

$\left(\mathrm{b}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_3,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_2$

$\left(\mathrm{c}\right)<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>{\mathrm{HCIO}}_3,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_6$

$\left(\mathrm{d}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_4,<mspace\; width="1em"></mspace>{\mathrm{CI}}_2{\mathrm{O}}_7$

${\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{Ba}{\left(\mathrm{OH}\right)}_2<mspace\; width="1em"></mspace>\stackrel{}{\to }<mspace\; width="1em"></mspace>\mathrm{X}(\mathrm{aq}.)+{\mathrm{BaCl}}_2+{\mathrm{H}}_2\mathrm{O}$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{X}+{\mathrm{H}}_2{\mathrm{SO}}_4<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace>}{\to }<mspace\; width="1em"></mspace>\mathrm{Y}+{\mathrm{BaSO}}_4$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{Y}<mspace\; width="1em"></mspace>\underset{∆>365\mathrm{K}}{\overset{∆}{\to }}<mspace\; width="1em"></mspace>\mathrm{Z}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

Y और Z क्रमशः हैं:

$\left(\mathrm{a}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_4,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_2<mspace\; width="1em"></mspace>$

$\left(\mathrm{b}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_3,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_2$

$\left(\mathrm{c}\right)<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>{\mathrm{HCIO}}_3,<mspace\; width="1em"></mspace>{\mathrm{CIO}}_6$

$\left(\mathrm{d}\right)<mspace\; width="1em"></mspace>{\mathrm{HCIO}}_4,<mspace\; width="1em"></mspace>{\mathrm{CI}}_2{\mathrm{O}}_7$

S-S bond is present in

(1)H₂S₂O₇

(2)H₂SO₅

(3)H₂S₂O₈

(4)H₂S₂O₆

S-S बंध किसमें उपस्थित है?

(1) H₂S₂O₇

(2) H₂SO₅

(3) H₂S₂O₈

(4) H₂S₂O₆

Spirit of salt is:

(1) HBr

(2) HCl

(3) $H_{2}S{O}_4$

(4) HI

लवण का अम्ल है:

(1) HBr

(2) HCl

(3) $H_{2}S{O}_4$

(4) HI

The molecule having the smallest bond angle is

(1) H₂O

(2) H₂S

(3) H₂Te

(4) H₂Se

सबसे छोटा बंधन कोण वाला अणु है

(A) H₂O

(B) H₂S

(C) H₂Te

(D) H₂Se

Which hydride has the lowest boiling point?

1. $H_{2}O$

2. $H_{2}S$

3. $H_{2}Se$

4. $H_{2}Te$

किस हाइड्राइड में सबसे कम क्वथनांक होता है?

(1) $H_{2}O$

(2) $H_{2}S$

(3) $H_{2}Se$

(4) $H_{2}Te$

Correct order of boiling point of group 16 hydrides 

1.H₂O<H₂S<H₂Se<H₂Te

2.H₂Te<H₂Se<H₂S<H₂O

3.H₂S<H₂Se<H₂Te<H₂O

4.H₂O<H₂Te<H₂Se<H₂S

वर्ग 16 हाइड्राइड के क्वथनांक का सही क्रम है 

(1) H₂O<H₂S<H₂Se<H₂Te

(2) H₂Te<H₂Se<H₂S<H₂O

(3) H₂S<H₂Se<H₂Te<H₂O

(4) H₂O<H₂Te<H₂Se<H₂S

Decreasing order of reducing power hydrogen halides is:

(1) HI > HBr > HCl > HF

(2) HF > HI > HBr > HCl

(3) HI > HF > HBr > HCl

(4) None of these

हाइड्रोजन हैलाइडों की अपचयन क्षमता का घटता हुआ क्रम है:

(1) HI > HBr > HCl > HF

(2) HF > HI > HBr > HCl

(3) HI > HF > HBr > HCl

(4) इनमें से कोई नहीं

HI reacts with $HN{O}_3$ to form:

(1) $O_2$

(2) $N_{2}O$

(3) $HI{O}_3$

(4) $N{O}_2+I_{\mathit{2}}$

HI क्या बनाने के लिए $HN{O}_3$के साथ अभिक्रिया करता है?

(1) $O_2$

(2) $N_{2}O$

(3) $HI{O}_3$

(4) $N{O}_2+I_{\mathit{2}}$