In [X(SO_4)(NH_3)_5]Cl,the coordination numbe | vedclass

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(C) $1$. The coordination number is the total number of ligand donor atoms bonded to the central metal atom. Here,$SO_4^{2-}$ acts as a bidentate liga

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$6$ and $+3$

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(C) $1$. The coordination number is the total number of ligand donor atoms bonded to the central metal atom. Here,$SO_4^{2-}$ acts as a bidentate ligand (or monodentate depending on bonding,but typically in this complex,it occupies one coordination site) and $5$ $NH_3$ molecules are monodentate. However,in $[X(SO_4)(NH_3)_5]Cl$,$SO_4$ is a bidentate ligand in some contexts,but standard coordination chemistry for this formula assumes $SO_4$ acts as a monodentate ligand. Thus,$1 (SO_4) + 5 (NH_3) = 6$ coordination sites.$2$. To find the oxidation state of $X$,let it be $x$. The charge on $SO_4$ is $-2$,$NH_3$ is $0$,and $Cl$ is $-1$. The total charge of the complex is $0$.$3$. $x + (-2) + 5(0) + (-1) = 0$$4$. $x - 3 = 0$,so $x = +3$.$5$. Therefore,the coordination number is $6$ and the oxidation state is $+3$.

In $[X(SO_4)(NH_3)_5]Cl$,the coordination number and oxidation state of $X$ are respectively ...............

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