For the coordination compound [Co(H_2O)_4SO_3 | vedclass

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(A) $1$. The coordination number is determined by the number of donor atoms directly bonded to the central metal ion. Here,$4$ $H_2O$ molecules (monod

Vedclass · Accepted Answer

$6, 3, 6, 0$

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(A) $1$. The coordination number is determined by the number of donor atoms directly bonded to the central metal ion. Here,$4$ $H_2O$ molecules (monodentate) and $1$ $SO_3^{2-}$ ion (monodentate) are bonded,so the coordination number is $4 + 1 = 6$.$2$. To find the oxidation state of $Co$,let it be $x$. The charge on $H_2O$ is $0$,$SO_3$ is $-2$,and $Cl$ is $-1$. Thus,$x + 4(0) + 1(-2) + 1(-1) = 0$,which gives $x = +3$.$3$. The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,it is $[Ar] 3d^6$.$4$. In the presence of strong field ligands like $SO_3^{2-}$,the $d$-electrons in $Co^{3+}$ $(3d^6)$ pair up in the $t_{2g}$ orbitals,resulting in $0$ unpaired electrons.

For the coordination compound $[Co(H_2O)_4SO_3]Cl$,the coordination number,oxidation state,number of $d$-electrons,and number of unpaired $d$-electrons of the metal ion are respectively:

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