In the complex [X(SO_4)(NH_3)_5],the coordina | vedclass

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(D) $1$. 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 $2$

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(D) $1$. 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 coordination,but in standard coordination chemistry nomenclature for this complex,it is treated as occupying one coordination site as a monodentate ligand in the inner sphere) and $5$ $NH_3$ molecules are monodentate. Thus,$CN = 1 + 5 = 6$.$2$. To find the oxidation state of $X$,let it be $x$. The charge on $SO_4$ is $-2$ and $NH_3$ is $0$. Assuming the complex is neutral,$x + (-2) + 5(0) = 0$,so $x = +2$.$3$. Therefore,the coordination number is $6$ and the oxidation state is $2$.

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

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