JEE 2016 :: Advanced 2016 :: Chemistry 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

The compound(s)  with two lone pairs of electrons on the central atom is (are

Answer:

Explanation:

 A plot of the number of neutons (n) against the number of protons (p) of stable nuclei exhibits upward deviation from linearity  for atomic number, Z >20, For an unstable nucleas having n/p ratio less than 1, the possible mode (s) of decay is (are)

Answer:

Explanation:

For the elements with atomic number (Z) larger than 20,

 Neutrons  (n)  > Protons (p);  Thus, n/p>1.   Thus , there is upward  deviation from linearity

    If n<p, thus n/p <1, then

  (a) By  $\beta^{-}$   - decay ,  $_{0}^{1}n\rightarrow ^{1}_{1}p+$ $_{-1}^{0}e$

 neutron changes to proton. Thus (n/p) ratio further decreases below 1.

 Thus, this decay is not allowed

   (b)By orbital or K - electron capture.

 $^{1}_{1}p+$ $_{-1}^{0}e\rightarrow _0^1n^{}$ proton changes to neutron, hence , (n/p) ratio increases.

  Thus stability increases. Thus correct.

 (c) Neutron  emission further decreases n/p ratio.

   (d)  By   $\beta^{+}$  - emission . $^{1}_{1}p\rightarrow _0^1n^{}+$ $_{+1}^{0}e$  proton changes to neutron.

    Hence,n/p ratio increases, Thus correct

Plot of the number of neutrons against the number of protons in stable nuclei (shown by dots)

The reagent(s)  that can selectively precipitate $S^{2-}$  from a mixture of  $S^{2-}$ and  $SO_{4}^{2-}$ in aqueus  solution is (are)

Answer:

Explanation:

$S^{2-}+CuCl_{2}\rightarrow CuS\downarrow$  (black ppt.)

$SO_{4}^{2-}+CuCl_{2}\rightarrow Soluble,  $ thus

   (a)  $CuCl_{2}$ selectively precipitates  $S^{2-}$

   (b)    $S^{2-}  (soluble)+BaCl_{2}\rightarrow BaS\downarrow$

          $SO_{4}^{2-}(white  ppt.)+BaCl_{2}\rightarrow BaSO_{4}\downarrow$

(b)   precipitates    $SO_{4}^{2-}$  and not  $S^{2-}$

 (c) $S^{2-} (black .ppt)+Pb^{2+}\rightarrow PbS\downarrow$

     $SO_{4}^{2-} (white.ppt)+Pb^{2+} \rightarrow PbSO_{4}\downarrow$

  $S^{2-} $ and    $SO_{4}^{2-}$, both are precipitated.

  (d)     $  S^{2-}+Na_{2}[Fe(CN)]_{5}NO\rightarrow Na _{4}[Fe(CN)]_{5}NOS]$

                             Sodium nitroprusside                          (purple colour)

 But no colour with  $SO_{4}^{2-}$

THe crystalline form of borax has 

Answer:

Explanation:

$Na_{2}B_{4}O_{7}.10H_{2}O$  (borax) is actually made of two tetrahedral and two triangular units and is actually written as 

 $Na_{2}[B_{4}O_{5}(OH)_{4}].5H_{2}O$ 

(a)  Thus, correct

(b) Boron atoms are in different planes thus, incorrect

(c) Two  $sp^{2}$ and two  $sp^{3}$ - hybridised B atoms thus, correct

 (d) Each boron has one- OH group thus, correct

The correct statement (s) about the following reaction sequence is (are)  

 $ Cumene (C_{6}H_{12})$ $\xrightarrow[{(i)O_{2}}]{(ii)H_{2}O}$ P   $\underrightarrow {CHCl_{3}/NaOH}$Q(major) +R (minor),

Q$\xrightarrow[\text{NaOH}]{{PhCH_{2}Br}}$S

Answer:

Explanation:

(a) R is not steam volatile , but Q is steam volatile thus, incorrect.

(b) Q has enolic group thus, gives violet colour with 1% aqueous  $FeCl_{3}$ solution thus, correct.

 (c) S has carbonyl group hence, gives yellow precipitate with 2, 4-DNP  thus, correct.

(d) S does not give colour with $FeCl_{3}$ thus, incorrect.

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016