The secondary valences (coordination numbers) | vedclass

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(D) The secondary valence of a metal complex is equal to its coordination number ($C$.$N$.),which is the total number of ligand donor atoms directly b

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$6 \ 6 \ 6$

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(D) The secondary valence of a metal complex is equal to its coordination number ($C$.$N$.),which is the total number of ligand donor atoms directly bonded to the central metal ion.$1$. For $(I) \ CoCl_3 \cdot 6 H_2O$:Since $3$ moles of $AgCl$ are precipitated,all $3$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Co(H_2O)_6]Cl_3$. The coordination number is $6$ (six $H_2O$ ligands).$2$. For $(II) \ NiCl_3 \cdot 6 H_2O$:Since $2$ moles of $AgCl$ are precipitated,$2$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Ni(H_2O)_5Cl]Cl_2 \cdot H_2O$. The coordination number is $6$ (five $H_2O$ and one $Cl^-$ ligand).$3$. For $(III) \ Co(SO_4)Br \cdot 5 NH_3$:Since $1$ mole of $AgBr$ is precipitated,the $Br^-$ ion is outside the coordination sphere. The formula is $[Co(NH_3)_5(SO_4)]Br$. The coordination number is $6$ (five $NH_3$ and one $SO_4^{2-}$ ligand).Thus,the secondary valences for $(I), (II),$ and $(III)$ are $6, 6, 6$ respectively.

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Formula of the complexes	Moles of $AgCl$ precipitated per mole of complex
$(I) \ CoCl_3 \cdot 6 H_2O$	$3$
$(II) \ NiCl_3 \cdot 6 H_2O$	$2$
$(III) \ Co(SO_4)Br \cdot 5 NH_3$	$1$