In [Ni(NH_3)_4]SO_4,the E.A.N. of Ni is | vedclass

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Question

(A) The complex is $[Ni(NH_3)_4]SO_4$.
First,determine the oxidation state of $Ni$: Let the oxidation state be $x$.
$x + 4(0) + (-2) = 0$,so $x = +2

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) The complex is $[Ni(NH_3)_4]SO_4$.
First,determine the oxidation state of $Ni$: Let the oxidation state be $x$.
$x + 4(0) + (-2) = 0$,so $x = +2$.
The atomic number of $Ni$ is $28$.
The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = (	ext{Atomic number}) - (	ext{Oxidation state}) + (2 	imes 	ext{Coordination number})$.
$E.A.N. = 28 - 2 + (2 	imes 4) = 26 + 8 = 34$.

In $[Ni(NH_3)_4]SO_4$,the $E.A.N.$ of $Ni$ is

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