JEE 2014 :: Advanced 2014 :: Chemistry 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

Hydrogen  peroxide in its reaction with KIO₄  and NH₂OH respectively is acting as a

Answer:

Explanation:

Plan This problem can be solved by using concept of oxidant and reductant.

 Oxidant increases the oxidation number of the species with which it is reacted.

Reductant decreases the oxidation number of the species with which it is reacted.

 H₂O₂   reacts with KIO₄  in the following manner

                $Kl^{+7}O_{4}+H_{2}O_{2\rightarrow} Kl^{+5}O_{3}+H_{2}O+O_{2}$

On reaction of KlO₄ with H₂ O₂, Oxidation state of I varies from +7 to +5 i.e,

Thus, KlO₄ Thus gets reduced hence H₂O₂ is a reducing agent here.

  -1                                                 +3

$NH_{2}OH+H_{2}O_{2}\rightarrow  N_{2}O_{2}+H_{2}O$

In the above reaction, oxidation state of N  varies from -1 to +3 Here, oxidation number increases hence H₂O₂ is acting as an oxidising agent here. 

Hence, (a)  is the correct choice.

The product formed in the reaction of SOCl₂ with white phosphorus is 

Answer:

Explanation:

This  problem is based on chemical properties of phosphorus

 White  phosphorus on reaction  with  thionyl  chloride (SOCl₂) produces  phosphorus trichloride

           $P_{4}(s)+8SOCl_{2}(l)\rightarrow 4PCl_{3}(l)+4SO_{2}(g)+2S_{2}Cl_{2}(g)$

  But if amount  of thionyl chloride (SOCl2) is in excess then it produces phosphrous pentachloride.

    $P_{4}(s)+10SOCl_{2}(l)\rightarrow 4PCl_{5}+10SO_{2}$

 Hence, (a)  is the correct choice.

Under ambient conditions  , the total number of gases released as products in the final step of the reaction scheme shown below is

Answer:

Explanation:

 Plan  This problem can be solved by using concept involved in chemical properties  of xenon oxide  and xenon fluoride

 XeF₄   on complete hydrolysis produce XeO₃

 XeO₃ on reaction with OH^- produces HXeO₄^- which on further treatment with OH^- undergo slow disproportionation  reaction and produces XeO₆^4-  along with Xe(g) , H₂O (l) and O₂ (g)  as a by-product

 OXidation half cell in basic aqueous solution

 $HXeO_4^-+5OH^{-}\rightarrow XeO_{6}^{4-}+3H_{2}O+2e$

 Reduction half-cell in basic aqueous solution

      $HXeO_4^-+3H_{2}O+6e^{-}\rightarrow Xe+7OH^{-}$

  Balanced  overall disproportionation reaction is

  $4HXeO_4^-+8OH^{-}\rightarrow 3XeO_6^{4-}+Xe+6H_{2}O$

                                                       2 products

 Complete sequence of reaction can be shown as

$XeF_{6}+3H_{2}O\rightarrow XeO_{3}+3H_{2}F_{2}$  $\underrightarrow{OH^{-}}$  $HXeO_4^-$

   $\underrightarrow{OH^{-}/H_{2}O (disproportionation)}XeO^{4-}_{6}(s)+Xe(g)+H_{2}O(l)+O_{2}(g)$

Isomers of hexane, based on their branching, can be divided in to  three distinct  classes as shown  in the figure.

  The correct order of their boiling point is 

Answer:

Explanation:

 Plan   This problem  is based on the boiling point of isomeric alkanes

 As we know more the branching in an alkane, lesser will be its surface area and lesser will be the boiling point

    On moving left to right (III to I)

•  branching increases
•  surface area decreases
•  boiling point decreases

 Hence, correct choice is (b)

For the identification of   $\beta$- naphthol using dye test, it is necessary to use

Answer:

Explanation:

 Plan    This problem can be solved by using the concept of synthesis of dye using electrophilic aromatic substitution reaction.

  In basic (alkaline) solution naphthol exists as naphthoxide ion which is a strong o, p -directing group.

Thus, formation  of dye can be shown as

  Thus (d) is the correct choice

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014