JEE 2014 :: Advanced 2014 :: Chemistry 2014

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

As an aqueous solution of metal ion M₁ reacts separately with reagents  Q and R  in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M₂  always forms tetrahedral complexes with these reagents. An aqueous solution of M₂ on  reaction with  reagent S gives white precipitate  which dissolves in  excess of S. The  reactions are summarized in the scheme given below

$ M_{1}$, Q, R  respectively are

Answer:

Explanation:

 Plan This problem can be solved by using the concept of chemical reaction of transition metal ions (.)  colour and structure of transition metal compounds.

   Here, among given four option Ni²⁺ and Zn²⁺ has  ability to form tetrahedral as well as square planar complex depending upon types of reagent used.

 Ni²⁺ on reaction with KCN forms square planar complex [Ni(CN)₄]^2- due to strong field strength of CN.

Ni²⁺ + KCN $\rightarrow$ [Ni(CN)₄]^2- (Square Planar)

while on reaction with HCl.Ni²⁺  forms a stable tetrahedral complex [Ni(Cl)₄]^2-. 

Zn²⁺, on the other hand on reaction with KCN as well as HCl, produces tetrahedral complex because of its d¹⁰ electronic configuration.

Complete reaction sequence can be shown as

Schemes 1 and 2 describe sequential transformation of alkynes M and  N. Consider only the major products formed in each step for both schemes.

The correct statement  with respect to  product Y is

Answer:

Explanation:

Plan  This problem can be solved by using the concept of iodoform test and functional isomerism. The compound containing  -COCH₃  or -CH(OH) group will undergo Iodoform test.

Thus, X  and Y  are functional isomers of each other and Y gives iodoform test due to the presence of CH₃CHO group as indicated

 Hence, correct choice is (c)

X and Y  are two volatile liquids with molar weights of 10 gmol^-1 and 40 g mol^-1 respectively, two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24 cm, as shown in the figure.

 The tube is filled with an inert gas at 1 atm pressure and a temperature of 300K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take  X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This due to

Answer:

Explanation:

Collision frequency  is the measurement of number of molecular collision taking place per second  per unit volume of a gas

 As the collision frequency increases, molecular size increases and hence covered distance decreases than the expected value.

 Since speed is higher for x so it produces more collisions per second. i.e, more collision frequency and hence

 covers lesser distance than that predicted by Graham's law  

 Hence (d) is the correct choice

X and Y  are two volatile liquids with molar weights of 10 gmol^-1 and 40 g mol^-1 respectively, two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24 cm, as shown in the figure.

 The tube is filled with an inert gas at 1 atm pressure and a temperature of 300K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take  X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

  The value of d in cm ( shown in the figure) as estimated from Graham's law, is

Answer:

Explanation:

Plan  This problem can be solved by using the  concept  of Graham's law  of diffusion according to  which rate of diffusion of non-reactive gases under similar conditions of temperature and pressure  are inversely proportional to square  root of their density

Rate of diffusion $\propto$  $\frac{1}{\sqrt{molar weight of gas}}$

 Let distance covered  by X is d, then  distance covered by Y is 24-d 

If r_x and r_y are the rate of diffusion of gases X and Y.

$\frac{r_{x}}{r_{y}}=\frac{d}{24-d}=\sqrt{\frac{40}{10}}=2$

            d=48-2d

                        3d=48= d=16 cm

Hence,(c) is the correct choice

The acidic hydrolysis of ether (X)  shown below is fastest when

Answer:

Explanation:

Plan  This problem can be solved by using the concept of stability of carbocation and S_N1  reaction when two phenyl groups are replaced by two para methoxy group, carbocation formed will be more stable. As the stability of carbocation formed increases, rate of acidic hydrolysis increases.

Hence, (c)  is the correct, choice

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