Give the oxidation state,$d$-orbital occupation,and coordination number of the central metal ion in the following complexes:
$(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$
$(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$
(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 $t_{2g}^{3} e_{g}^{2}$.
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$: 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 $t_{2g}^{3} e_{g}^{2}$.
Explore More
Similar Questions
The correct option(s) about entropy $(S)$ is(are)
$[R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, if $\frac{dE_{cell}}{dT} = \frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes in entropy and internal energy are temperature independent).
$(B)$ The cell reaction, $Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 0.01 \ M) \parallel H^{+}(aq, 0.1 \ M) \mid H_2(g, 1 \ bar) \mid Pt_{(s)}$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S > 0$.
$(D)$ $\Delta S > 0$, for $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ (where $en =$ ethylenediamine).
$[R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, if $\frac{dE_{cell}}{dT} = \frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes in entropy and internal energy are temperature independent).
$(B)$ The cell reaction, $Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 0.01 \ M) \parallel H^{+}(aq, 0.1 \ M) \mid H_2(g, 1 \ bar) \mid Pt_{(s)}$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S > 0$.
$(D)$ $\Delta S > 0$, for $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ (where $en =$ ethylenediamine).
$[Ni(CO)_4]$ is diamagnetic. Which of the following statements is correct for this complex?
Medium
View SolutionExcess of $aq. NH_3$ can dissolve
Medium
View SolutionSelect the incorrect statement.
Medium
View SolutionConsider the following reaction and identify the correct statement:
$[M(gly)_2(NH_3)_2]^+ \xrightarrow[-2NH_3]{+gly} [M(gly)_3]$
$[M(gly)_2(NH_3)_2]^+ \xrightarrow[-2NH_3]{+gly} [M(gly)_3]$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo