In [Ni(NH_3)_4]SO_4,the oxidation state and c | vedclass

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(B) The complex is $[Ni(NH_3)_4]SO_4$. Let the oxidation state of $Ni$ be $x$. The oxidation state of $NH_3$ is $0$ and $SO_4$ is $-2$. $x + 4(0) + (-

Vedclass · Accepted Answer

$2$ and $4$

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(B) The complex is $[Ni(NH_3)_4]SO_4$. Let the oxidation state of $Ni$ be $x$. The oxidation state of $NH_3$ is $0$ and $SO_4$ is $-2$. $x + 4(0) + (-2) = 0$ $x - 2 = 0 \Rightarrow x = +2$. The coordination number is determined by the number of ligands directly bonded to the central metal ion. Here,$4$ $NH_3$ molecules are bonded to $Ni$,so the coordination number is $4$. Thus,the oxidation state is $2$ and the coordination number is $4$.

In $[Ni(NH_3)_4]SO_4$,the oxidation state and coordination number of $Ni$ will be respectively:

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