Give the oxidation state,$d$-orbital occupation,and coordination number of the central metal ion in the following complexes:
$(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$

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(N/A) $(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$: Central metal is $Co$. Coordination number is $6$. Oxidation state: $x + 3(-2) = -3 \implies x = +3$. $Co^{3+}$ $(d^{6})$ is $t_{2g}^{6} e_{g}^{0}$.
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$: Central metal is $Cr$. Coordination number is $6$. Oxidation state: $x + 2(0) + 2(-1) = +1 \implies x = +3$. $Cr^{3+}$ $(d^{3})$ is $t_{2g}^{3} e_{g}^{0}$.
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$: Central metal is $Co$. Coordination number is $4$. Oxidation state: $x + 4(-1) = -2 \implies x = +2$. $Co^{2+}$ $(d^{7})$ is $t_{2g}^{5} e_{g}^{2}$.
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$: Central metal is $Mn$. Coordination number is $6$. Oxidation state: $x + 6(0) = +2 \implies x = +2$. $Mn^{2+}$ $(d^{5})$ is $t_{2g}^{3} e_{g}^{2}$.

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Similar Questions

Given below are two statements:
Statement $I$ : Dimethyl glyoxime forms a six-membered covalent chelate when treated with $NiCl_2$ solution in presence of $NH_4OH$.
Statement $II$ : Prussian blue precipitate contains iron both in $(+2)$ and $(+3)$ oxidation states.
In the light of the above statements,choose the most appropriate answer from the options given below:

$A$ blue colouration is not obtained when:

Match List-$I$ with List-$II$ and select the correct code:
List-$I$ List-$II$
$A$. Ziegler-Natta $i$. $Fe_4[Fe(CN)_6]_3$
$B$. Brown ring complex $ii$. $[Fe(H_2O)_5NO]SO_4$
$C$. Prussian Blue $iii$. $Al(C_2H_5)_3 + TiCl_4$
$D$. Turnbull Blue $iv$. $Fe_3[Fe(CN)_6]_2$

Correct code is:

Give the correct order of initials $T$ or $F$ for following statements. Use $T$ if statement is true and $F$ if it is false.
$(I)$ Sulphide ions react with $Na_2[Fe(CN)_5(NO)]$ to form a purple coloured compound $Na_4[Fe(CN)_5(NOS)]$. In the reaction,the oxidation state of iron changes.
$(II)$ $Pt(IV)$ compounds are relatively more stable than $Ni(IV)$ compounds.
$(III)$ The welding of magnesium can be done in the atmosphere of Helium.
$(IV)$ $LiAlH_4$ on hydrolysis will give $H_2$.

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Match the following:
Column-$1$ (Reaction)Column-$2$ (Main product)
$A$. $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow$$I$. $Fe_4[Fe(CN)_6]_3$
$B$. $ZnCl_2 + 4NaOH \rightarrow$$V$. $Na_2ZnO_2$
$C$. $2FeCl_3 + H_2S \rightarrow$$II$. $FeCl_2$

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