JEE 2016 :: Advanced 2016 :: Chemistry 2016

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

Reagent(s)  which  can be  used to bring about the following transformation is (are)

Answer:

Explanation:

Only -CHO group is to be reduced to -CH₂OH

   It can be done using $NaBH_{4}$  in  $C_{2}H_{5}OH$

(a)$LiAlH_{4} / (C_{2}H_{5})_{2}O$ reduces I,II and III  into - $CH_{2}OH$ , and IV  into diol.

(b) $BH_{3}$ / THF show same properties as (a)

(c) $NaBH_{4}$ /  $C_{2}H_{5}OH$ reduces  III into -$CH_{2}OH$

(d) Raney nickel, same as (a) and (b) thus (c) is correct reagent

The correct statement (s) for cubic close packed (ccp) three dimensional structure is (are)

Answer:

Explanation:

(a)  Nearest neighbour in the topmost  layer of ccp structure is 9 thus, incorrect

(b) Packing efficiency is 74% thus, correct

(c) Tetrahedral voids=2

 Octahedral  voids= 1 per atom thus, correct.

 (d) Edge length,

                   $a=\frac{4}{\sqrt{2}}r=2\sqrt{2}r$

thus, correct explanation

Edge length=a, Radius=r

 $AC^{2}=AB^{2}+BC^{2}$

$(4r)^{2}=a^{2}+a^{2}=2a^{2}$

$4r=\sqrt{2}a\Rightarrow r=\frac{\sqrt{2}}{4}a=\frac{a}{2\sqrt{2}}$

$\therefore  a=2\sqrt{2}r$

In ccp structure, the number of sphere is 4, 

Hence , volume of 4 spheres=  $4(\frac{4}{3}\pi r^{3})$

Total volume of unit cell= $a^{3}=(2\sqrt{2}r)^{3}$

% of packing efficiency= $\frac{volume. of 4 spheres}{ volume .of unit cell}$

=  $\frac{4(\frac{4}{3}\pi r^{3})}{(2(\sqrt{2}r))^{3}}\times100=74.05$%= 74%

According to molecular orbital theory, which of the following statement(s) is (are) correct?

Answer:

Explanation:

Thus , (a) is correct.

(b) Bond order  $O^{2+}_{2}>O_{2}$  thus,

Bond length of  $O^{2+}_{2}<O_{2}$ thus, incorrect

(c)$N^{+}_{2}$ and $N^{-}_{2}$ have  same bond order thus correct.

(d) $He^{+}_{2}$ with bond order =0.5 is more stable thus, less energy than isolated He atoms . Thus, (d)  is incorrect.

For the following electrochemical cell at 298 K,

$Pt(s) | H_{2}(g,1bar) | H^{+}(aq,1M) $

                                    $|| M^{4+}(aq), M^{2+}(aq)| Pt (s)$

$E_{cell}$=0.092 V

$when \frac{[M^{2+}(aq)]}{[M^{4+}(aq)]}=10^{x}$

Given:  $E^{0}_{M^{4+/}M^{2^{+}}}=0.15V:$

$2.303 \frac{RT}{F}=0.059 V$

 The value of x is

Answer:

Explanation:

Oxidation at anode

$H_{2}(g) \rightarrow 2H^{+}(aq) +2e^{-}; E^{0}_{SHE}=0.00V$

Reduction at cathode

$M^{4+}(aq) +2e^{-}\rightarrow M^{2+}(aq):$

$E^{0}_{M^{4+}/M^{2+}}=0.151V$

$Net: M^{4+}(aq)+H_{2}(g)\rightarrow M^{2+}(aq)+2H^{+}(aq):$

$K= \frac{[M^{2+}][H^{+}]^{2}}{[M^{4+}]PH_{2}}(E^{0}_{cell}=0.151 V)$

$K= \frac{[M^{2+}]}{[M^{4+}]}$

$E_{cell}=E^{0}_{cell}-\frac{0.059}{2}\log K$

$0.092= 0.151_{}-\frac{0.059}{2}log \frac{[M^{2+}]}{[M^{4+}]}$

$0.059= \frac{0.059}{2}log 10^{x}$

$\therefore \log 10^{x}=2$

$\therefore x=2$

The geometries of the ammonia  complexes of Ni²⁺, Pt²⁺ and Zn²⁺ respectively are

Answer:

Explanation:

$[Ni(NH_{3)_{6}}]^{2+} sp^{3}d^{2}octahedral$

$[Pt(NH_{3)_{4}}]^{2+} dsp^{2}squareplanar$

$[Zn(NH_{3)_{4}}]^{2+} sp^{3}tetrahedral$

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