Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $MC{l}_2.H_{2}O\left(x\right)andN{H}_{4}Cl$ 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 coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.
Among the following options, which statemet(s) is (are) correct?

• A. The hybridization of the central metal ion in Y is $d^{2}s{p}^3$
• B. Addition of silver nitrate to Y gives only two equivalents of silver chloride
• C. When X and Z are in equilibrium at $0^{\circ },$ the colour of the solution is pink
• D. Z is a tetrahedral complex
  

Answer 1

Answer: (A), (C), (D)

The correct options are
A

The hybridization of the central metal ion in Y is $d^{2}s{p}^3$

C

When X and Z are in equilibrium at $0^{\circ },$ the colour of the solution is pink

D

Z is a tetrahedral complex