JEE 2018 :: Chemistry (2018) :: Solved questions (2018)

• Chemistry (2018) - Solved questions (2018)

When metal 'M' is treated with NaOH , a while gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X'  when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

Answer:

Explanation:

Among the given metals Al forms white gelatinous ppt. with NaOH . Hence , the probable metal can be Al. This ppt. is dissolved in excess of NaOH due to the formation of sodium metal Aluminate . Both the reactions are shown below,

$Al^{3+}$          $\underrightarrow{NaOH}$      $Al(OH)_{3+}\underrightarrow{ excess -of-NaOH}$         $NaAlO_{2}$

Aluminum hydroxide on strong heating gives alumina (Al₂O₃) which is used as an adsorbent in chromatography. This reaction can be seen as :

 $2Al(OH)_{3}$        $\underrightarrow{\triangle}$          $Al_{2}O_{3}+3H_{2}O $

Thus, metal M is Al

Ca  being below sodium in electrochemical reactivity series, cannot displace Na from its aqueous solution.

Zn reacts with NaOH to form sodium zincate which is a soluble compound.

Fe  reacts with sodium hydroxide to form tetrahydroferrate (II) sodium which is again a soluble complex.

The oxidation states of Cr, in [Cr(H₂O)₆Cl₃ , [Cr(C₆H₆)₂] and K₂[Cr (CN)₂(O)₂ (O₂ )(NH₃)] respectively are 

Answer:

Explanation:

Let the oxidation state of Cr in all cases is 'x'

(i)  Oxidation state of Cr in  [Cr(H₂O)₆ ]Cl₃

x + (0 × 6) + (-1 × 3)=0

or       x+0-3=0

or        x= +3

 (ii)  Oxidation state of Cr in [Cr(C₆H₆)₂]

   x+(2× 0)=0

or x=0

(iii) Oxidation state of Cr in

     K₂[Cr (CN)₂(O)₂ (O₂ )(NH₃)] 

    1×2+x+ (-1× 2) + (-2× 2) + (-2)+0=0

  or    2 +x-2-4-2=0

 or     x-6=0

 hence x=+6

 Thus, +3,0 and +6 is the answer

The increasing order of basicity  of the following compound is 

Answer:

Explanation:

Key Idea Among the given compounds the basic nature depends upon their tendency to donate electron pair

Among the given compounds in 

This marginally increases the electronegativity  of nitrogen which in turn decreases the electron donation tendency of nitrogen. Thus making compound least basic.

differnent from others as in this  compound lone pair of one nitrogen are in conjugation with π bond 

This equivalent resonance is cation

basic among all.

Categorisation is very simple between rest two as

Hence, the correct order is

( II)<( I)<( IV)<( III) i.e option ( c) is correct.

How long ( approximate ) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diborane?  (  Atomic weight of B = 10.8μ )

Answer:

Explanation:

Given that , i= 100 amp. also 27.66 g of diborane ( B₂H₆)

 Molecular mass of B₂H₆ = $10.8\times 2+6 =27.6$

  Number of moles of B₂H₆  in 27.66 g=  $\frac{Given mass}{Molecular mass}=\frac{27.66}{27.6}=1$

Now consider the equation

  $B_{2}H_{6}+3O_{2}\rightarrow B_{2}O_{3}+3H_{2}O$

From the equation  we can interpret that 3 moles of oxygen is required to burn 1 mole ( i.e 27.6 g) B₂H₆ completely.

Also consider the electrolysis reaction of water i.e

    $H_{2}O\rightleftharpoons 2H^{+}+O$

From the above  equation it can be easily interpreted that in electrolysis of water for the production of 1 mole of oxygen from 1 mole of H₂O at anode 4 moles electrons are required.

Likewise  for the production of 3 moles of O₂ 12 ( 3× 4)  moles of electrons will be needed.

 So, the total amount of charge required to produce 3 moles of oxygen will be 12 × F or 12 × 96500

   We know Q= it

So,       12 × 96500 = 100 × t in seconds

or           $\frac{12\times 96500}{100\times 3600}=t$ in hours = 3.2 hours

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing pointt?

Answer:

Explanation:

Key Idea " Addition of solute particles to a pure solvent results to depression in its freeing point."

   All the compounds given in question are ionic in nature  so, consider their van't Hoff factor ( i) to reach at final conclusion.

     The solution with maximum freezing point must have minimum number of solute particles. This generalisation can be done with the help of van't Hoff factor (i) i.e.

Number of solute particles α van't Hoff factor (i)

   THus , we can stay directly

 Solution with maximum freezing point will be the one in which solute with minimum van't Hoff factor is present 

  Now, for  $[Co( H_{2}O)_{6}]Cl_{3}\rightleftharpoons [Co(H_{2}O)_{6}]^{3+}+3Cl^{-}$

van't Hoff factor (i) is 4. Similarly for,

$[Co( H_{2}O)_{5}Cl_{}]Cl_{2}.H_{2}O\rightleftharpoons [Co(H_{2}O)_{5}Cl]^{2+}+2Cl^{-}$   'i' is 3

$[Co( H_{2}O)_{4}Cl_{2}]Cl_{}.2H_{2}O\rightleftharpoons [Co(H_{2}O)_{4}Cl_{2}]^{+}+Cl^{-}$   'i' is 2

   and for  $[Co( H_{2}O)_{3}Cl_{3}].3H_{2}O$. 'i' is 1 as it does not show ionisation. Hence,

       $[Co( H_{2}O)_{3}Cl_{3}].3H_{2}O$ have minimum number of particles in the solution. So freezing point of its solution will be maximum.

• Chemistry (2018) - Solved questions (2018)