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

• Chemistry (2018) - Solved questions (2018)

The total number of compounds having at least one bridging oxo group amoung the molecule given below...

N₂O₃ ,N₂O₅ ,P₄ O₆ , P₄O₇, H₄P₂O_5, H₅P₃O₁₀, H₂S₂O₃, H₂S₂O₅

Answer:

Explanation:

The structures of various molecules given in problem are discussed below  -

 1. N₂O₃ It is the tautomeric mixture of following two structures-

Conclusion  1 bridging oxo group is present in the compound.

2. N₂O₃     It has following structure.

Conclusion 1  bridging oxo group is present in the compound.

             where, p₁  = initial pressure, p₁= total pressure

                    y= number of gaseous products per mole of reactant

3.   P₄ O₆

Conclusion   6 bridging oxo groups are present in the compound.

4. P₄ O7

Conclusion 6 bridging oxo groups are present in the compound.

5.  H₄P₂O₅

Conclusion 1 bridging oxo groups is present in the compound.

6.  H₅P₃O₁₀

Conclusion   2 bridging oxo groups are present in the compound.

7.   H₂S₂O3

Conclusion   This compound doesnot contain any bridging oxo group.

8.  H₂S₂O₅

Conclusion This compound also doesnot contain any bridging oxo group.

For a reaction , $A\rightleftharpoons P,$  the plots of [A]  and [P] with time at temperature T₁  and T₂ are given below.

 If T₂ > T₁ , the correct statement (s) is /are 

     [ Assume $\triangle H^{\ominus}$ and $\triangle S^{\ominus}$ are independent of temperature and ratio of ln K at T₁ to ln K at T₂ is greater than T2/T1. Here  H, S, G and K are enthalpy, entropy, Gibbs energy  and equilibrium constant, respectively.)

Answer:

Explanation:

For the reaction, $A\rightleftharpoons P,$

Then  T₁ <T₂

     $\frac{ln K_{1}}{ln K_{2}}>\frac{T_{2}}{T_{1}}$

It shows,On increasing the temperature,  K decreases so reaction is exothermic i.e.,

 Δ H^{0 <  0}

Besides, graph shows K >1

     So , Δ G⁰ <0

   Now from equation (i)

    T₁ ln K₁  > T₂ ln K₂

   -Δ G⁰ ₁  >  -Δ G⁰₂  

 Like wise

   (- Δ H⁰ +T₁Δ S⁰ )> (- Δ H⁰ +T₂Δ S⁰ )

or simply

                 T₁Δ S ⁰  >T₂ Δ S⁰

So,     (T₂-T₁) Δ S⁰ < 0

           Δ S⁰ <0

In other words, the increase of Δ G with an increase in temperature is possible only when Δ S⁰ <0.

   Hence, options (a) and (c) are correct.

The correct options (s) to distinguish nitrate salts to Mn²⁺ and Cu²⁺ taken separately is (are)

Answer:

Explanation:

Statement wise explanation is

Statement   (a)    Mn²⁺    produces yellow-green colour in the flame test while Cu²⁺ produce bluish-green colour in a flame test. Thus, due to the presence of green colour in both the cases, flame test is not a suitable method to distinguish between nitrate salts of Cu²⁺ and Mn^2+. Hence this statement is wrong.

Statement  (b)  Cu²⁺ belong to group II of cationic or basic radicals. It gives black ppt. of CuS if H₂S  is passed through it in the presence of acid (eg HCl). Mn²⁺ does not show this property hence this can be considered as a suitable method to distinguish between Mn²⁺ and Cu2+. Hence, this statement is correct.

Statement (c)     In faintly basic medium when H₂S is passed both Cu²⁺ and Mn²⁺ forms precipitates. Thus, it is not a suitable method to distinguish between them. Hence, this statement is incorrect.

Statement (d)   The standard reduction potential of Cu²⁺ /Cu is +0.34 V while that of Mn²⁺/Mn is -1.18V. This can be used to distinguish between Cu²⁺  and Mn2+. In general, less electropostive metals have higher SRP.

               Hence, this statement is correct.

The correct option (s) regarding the complex [Co(en)(NH₃)₃ (H₂O)]³⁺   (en=H₂NCH₂CH₂NH₂) is (are)

Answer:

Explanation:

Statement wise explanation is 

Statement (a)

  [Co(en)(NH₃)₃H₂O]³⁺ has the following 2 geometrical isomers.

Hence, this is the correct statement.

Statement (b)

   If bidentate ligand 'en' is replaced by two cyanide ligands then

          [Co (NH₃)₃(H₂O) (CN)₂]⁺ is formed.

  It is  [Ma₃b₂c] type complex which has the following 3 geometrical isomers.

Hence, this statement is also correct.

Statement (c)

Co metal has   [Ar] 3d⁷4s² configuration while in [Co(en) (NH₃)₃ (H₂O)]³⁺   , It is in +3 oxidation state . Thus, Co³⁺ has [Ar]3 d⁶ configuration, and d² sp³ in complex.

As en is a strong ligand, so pairing will occur

Due to the presence of unpaired electrons it shows diamagnetic behavior rather than paramagnetic.

          Hence, the statement is incorrect.

Statement (d)

  According to CFT, absorption of light by coordination complexes depends upon CFSE  i.e, crystal field splitting energy (Δ₀) as

        $\triangle_{0}\alpha\frac{1}{\lambda}$

  Among the complexes given [Co(en) (NH₃)4]³⁺ has more Δ₀ value as compared to complex  [Co(en) (NH₃)₃ (H₂O)]³⁺   . Thus , [Co (en) (NH₃)₃(H₂O)]³⁺ absorbs the light at longer wavelength for d-d  transition. Hence , this statement is also correct.

   Note. For any complex , the value of Δ₀ can be calculated via the difference or gap between e_g and t_2g values.

The plot given below shows p-T curves (where p is the pressure and T is the temperature ) for two solvents X and Y and isolmolal solution of NaCl  these solvents. NaCl completely dissociates in both the solvents

On the addition of the equal number of moles of a non -volatile S in equal amount (in kg) of these solvents, the elevation of boiling point X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is........... 

Answer:

Explanation:

From the graph we can note $\triangle T_{b}$ for solution X i.e.,

 $\triangle T_{b(X)}=362-360=2$

 Like wiise, $\triangle T_{b}$ for solution Y i.e.,

 $\triangle T_{b(Y)}=368-367=1$

Now by using the formula

    $\triangle T_{b}$ =i × molality of solution × K_b

For solution X

       2= i × m_NaCl × K_b(X)          ...............(i)

Similarly for solution Y 

 1= i × m_Nacl × Kb(Y)            ..................(ii)

from Eq .(i) and (ii) above

          $\frac{K_{b(X)}}{K_{b(Y)}}=\frac{2}{1}=2$

or    K_b(X) = Kb(Y)

  For solute S 

           $2S\rightarrow S_{2}$    (Given due to dimerisation)

initial         α                 0

final         (i - α )           α/2

So, here     $i=[1-\frac{\alpha}{2}]$

      $\triangle T_{b[X](s)}= (1-\frac{\alpha_{1}}{2})K_{b(X)}$

     $\triangle T_{b[Y](s)}= (1-\frac{\alpha_{2}}{2})K_{b(Y)}$

 Given    $\triangle T_{b(X)(s)}=3\triangle T_{b(Y)(s)}$

    $(1-\frac{\alpha_{1}}{2})K_{b(X)}= 3 \times (1-\frac{\alpha_{2}}{2}) \times K_{b(Y)}$

 or    $2(1-\frac{\alpha_{1}}{2})= 3  (1-\frac{\alpha_{2}}{2})$   [ $\because  K_{b(X)}=2K_{b(Y)}$ ]

  or           $2(1-\frac{\alpha_{1}}{2})= 3  (1-\frac{0.7_{}}{2})$   [ as given,α₂ =0.7 ]

  or         4- α₁ =6 -2.1          or  2 α₁ =0.1

            so,      α  =0.05

• Chemistry (2018) - Solved questions (2018)