Question

For $ \left[ {CrC{l_3}.xN{H_3}} \right] $ , elevation in boiling point of one molal solution is double of one molal solution of glucose; hence $ x $ is, if complex is $ 100\% $ ionised:
A. $ 4 $
B. $ 5 $
C. $ 6 $
D. $ 3 $

Answer 1

Hint: The boiling point of the solution changes due to the dissociation of the ions in different molal concentration in the solution. Formation of the complex involves different residues and the ionization depends on the different residues.

Complete step by step solution:
The boiling point changes or becomes elevated in the case of solution of glucose. This means that there is a $ \Delta {T_b} $ value for glucose and that is denoted by $ \Delta {T_{b(glu)}} $ . Therefore, for the glucose: $ \Delta {T_{b(glu)}} = i \times {K_b} \times m $
 $ \Rightarrow 1 \times {K_b} \times 1 = {K_b} $
For the complex $ \left[ {CrC{l_3}.xN{H_3}} \right] $ , the solution the changes in the boiling point is denoted by $ \Delta {T_{b(complex)}} $ . This means that the $ \Delta {T_{b(complex)}} = 2 \times \Delta {T_{b(glu)}} $
 $ \Rightarrow {i_{complex}} \times {K_b} \times 1 = 2 \times {K_b} $
 $ \Rightarrow {i_{complex}} = 2 $
If the complex is completely ionised then there is a chance of getting the $ \alpha $ value for the given complex. The ionization of the complex takes place in the presence of specific molality of the given solution of glucose molecules. Therefore, the alpha value can be calculated as the ionization level, which is $ \alpha = \dfrac{{100}}{{100}} $ for the given complex, which means $ \alpha = 1 $ .
 $ \alpha = \dfrac{{i - 1}}{{n - 1}} $
 $ \Rightarrow 1 = \dfrac{{i - 1}}{{n - 1}} $
 $ \Rightarrow i = n $
Therefore, according to the given condition of the ionization process, $ {n_{complex}} = 2 $ . Here the complex that is involved is $ \left[ {CrC{l_3}.xN{H_3}} \right]Cl $ which is subjected to ionization. It is known that for $ C{r^{2 + }} $ the coordination number is $ 6 $ . The complex according to the given condition of ionization can yield a maximum of $ 2 $ ions. Therefore, the value of $ x $ can be calculated as $ x = 6 - 2 = 4 $ . Therefore, the formation of the complex structure involves formation of a bonded structure with $ 4 $ of $ N{H_3} $ residues associated with the central metal ion.

Note
The ionization depends on the number of residues which are associated with the central metal ions forming the main structure of the complex. The ligands that are involved in the formation of a complex are dissociated in the specific concentration.