Question

Which of the following primary and secondary valencies are not correctly marked against compound.
A. $\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3},{\text{ p}} = 3,{\text{ s}} = 6$
B. ${K_2}\left[ {PtC{l_4}} \right],{\text{ p}} = 2,{\text{ s}} = 4$
C. $\left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right],{\text{ p}} = 0,{\text{ s}} = 4$
D. $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4},{\text{ p}} = 4,{\text{ s}} = 4$

Answer 1

Primary valency is normally ionisable and is satisfied by anions only. Secondary valency is non – ionisable and is satisfied by ions or neutral electron pair donor molecules. Primary valency is determined by the oxidation number of the molecule and secondary valency is determined by the coordination number. Also, primary valency is non-directional while secondary valency is directional.

Complete step by step answer:
According to Werner’s theory in coordination compounds, the central metal atoms exhibit two types of valencies, the primary and the secondary valencies. The number of primary valencies depend upon oxidation state of central metal atom or ion and the number of secondary valencies depend upon size and charge of central metal atom / ion. Every metal atom has a fixed number of secondary valencies. It has a fixed coordination number. The metal atom tends to satisfy its both primary and secondary valencies.
In the given complexes:
A. $\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}\xrightarrow{{}}{\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{ + 3}} + 3C{l^ - }$ (Ionisable complex)
Primary valency $= 3$ and secondary valency $= 6$ ( as given in the question)
Since six ammonia molecules are coordinated to $C{r^{3 + }}$, the secondary valency is equal to six.
Now, calculating the oxidation number of chromium.
$x + 0\, \times \,6 = + 3 x = + 3$
Hence, primary valency $= 3$
Which matches with that given in the question.
B. ${K_2}\left[ {PtC{l_4}} \right] \to 2{K^ + } + {\left[ {PtC{l_4}} \right]^{2 - }}$ (Ionisable complex)
Primary valency $= 2$, secondary valency $= 4$( as given in the question)
Since four chloride ions are coordinated to the metal, the secondary valency is equal to four.
Now, calculating the oxidation number of platinum,
$x - 4 = - 2 x = - 2 + 4 = 2$
Hence, primary valency $= 2$
Which matches with that given in the question.
C. $\left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]\xrightarrow{{}}$non – ionisable.
$\therefore$Primary valency $= 0$, secondary valency $= 4$ ( as given in the question)
Since four ligands are coordinated to the metal, the secondary valency is equal to four.
Now, calculating the oxidation number of platinum,
$x - 2 = 0 x = + 2$
Which does not match with that given in the question.
D. $\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}\xrightarrow{{}}{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{ + 2}} + SO_4^{2 - }$ (Ionisable complex)
Primary valency $= 4$, secondary valency $= 4$ ( as given in the question).
Since, four ligands are coordinated to the metal, the secondary valency is equal to four.
Now, calculating the oxidation number of copper
$x - 4 = 2 x = + 2$
Hence, primary valency $= 2$
Which does not match with that given in the question.

Hence, the correct option is (C) and (D).

Note: Primary valencies are non – directional while secondary valencies are directional in nature. Primary valency is ionic valency and secondary valency is non– ionic valency. The secondary valencies are always directed towards a fixed position in space and this causes definite geometry of the coordinate compound. For example: If a metal ion has six secondary valencies, these are arranged octahedrally around the central metal ion. If the metal ion has four secondary valencies, these are arranged in either tetrahedral or square planar arrangement around the central metal ion. The secondary valency determines the stereochemistry of complex ions.