Given that 1 moles of $N i C l_{2} .6 H_{2} O$ with excess $A g N O_{3}$ precipitate 2 moles of $A g C l$ , what is the coordination number of $N i$ ?

Open in App

Solution

$N i C l_{2} .6 H_{2} O$ on reacting with excess $A g N O_{3}$ , it gives 2 moles of $A g C l$ . This indicates that the $2 C l^{-}$ ions in the complex is not bounded by the coordination sphere and hence are present as counter ions of the complex.
Thus, the coordination complex is $[N i (H_{2} O)_{6}]^{2 +} 2 C l^{-}$ .
Thus, 6 water molecules are bounded to the $N i$ and hence, the coordination number is 6.

Suggest Corrections

Similar questions

P t

P t C l_{4} .2 H C l

A g C l

A g N O_{3}

N i C l_{2} . 6 H_{2} O

A g N O_{3}, 2

C o C l_{3} .4 N H_{3}

A g N O_{3}

C o C l_{3} .3 N H_{3}

{A g N O}_{3}

C o C l_{3} .5 N H_{3}

A g C l

{A g N O}_{3}

Formula	Moles of $A g C l$ precipitated per mole of the compounds with excess $A g N O_{3}$
(i) $P d C l_{2} .4 N H_{3}$	$2$
(ii) $N i C l_{2} .6 H_{2} O$	$2$
(iii) $P t C l_{4} .2 H C l$	$0$
(iv) $C o C l_{3} .4 N H_{3}$	$1$
(v) $P t C l_{2} .2 N H_{3}$	$0$

Join BYJU'S Learning Program