Coordination Compounds 2 Question 50

####52. Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl26H2O(X) and NH4Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue colured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whersas it is zero for complex Y.

Among the following options, which statement(s) is (are) correct?

(2017 Adv.)

(a) The hybridisation of the central metal ion in Y is d2sp3

(b) Addition of silver nitrate to Y given only two equivalents of silver chloride

(c) When X and Y are in equilibrium at 0C, the colour of the solution is pink

(d) Z is a tetrahedral complex

Assertion and Reason

Read the following questions and answer as per the direction given below:

(a) Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I.

(b) Statement I is true; Statement II is true; Statement II is not the correct explanation of Statement I.

(c) Statement I is true; Statement II is false.

(d) Statement I is false; Statement II is true.

Show Answer

Answer:

Correct Answer: 52. (c)

Solution:

  1. [Co(H2O)6 Pink (X)]Cl2O2 (Air)  Excess NH4OH/NH4ClCo(NH3)6Y]Cl3

[Co(H2O)6]2+X+4Cl (Excess) [CoCl4]2 blue Z 

(a) Since NH3 is moderately strong ligand, hybridisation of cobalt in Y is d2sp3.

(b) Cobalt is sp3-hybridised in [CoCl4]2.

(c) [Co(NH3)6]Cl3+3AgNO3( aq )3AgCl

(d) [CoCl4]2+6H2O[Co(H2O)6 Blue ]2++4Cl;ΔH<0



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