The equation which is balanced and represents the correct product$(s)$ is:

  • A
    $Li_2O + 2KCl \rightarrow 2LiCl + K_2O$
  • B
    $[CoCl(NH_3)_5]^{2+} + 5H^{+} \rightarrow Co^{2+} + 5NH_4^+ + Cl^{-}$
  • C
    $[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \rightarrow [Mg(EDTA)]^{2-} + 6H_2O$
  • D
    $CuSO_4 + 4KCN \rightarrow K_2[Cu(CN)_4] + K_2SO_4$

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Match List-$I$ with List-$II$ and select the correct option:
List-$I$List-$II$
$(A). [Ag(CN)_2]^-$$1. \text{Square planar, } 1.73 \, B.M.$
$(B). [Cu(CN)_4]^{3-}$$2. \text{Linear, } 0 \, B.M.$
$(C). [Cu(CN)_6]^{4-}$$3. \text{Octahedral, } 0 \, B.M.$
$(D). [Cu(NH_3)_4]^{2+}$$4. \text{Tetrahedral, } 0 \, B.M.$
$(E). [Fe(CN)_6]^{4-}$$5. \text{Octahedral, } 1.73 \, B.M.$

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$A$ blue colouration is not obtained when:

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

Excess of $aq. NH_3$ can dissolve

Match the following complexes in List-$I$ with their colors in List-$II$:
List-$I$ (Complex)List-$II$ (Color)
$A. [Ni(en)_3]^{2+}$$I. \text{Green}$
$B. [Ni(H_2O)_4(en)]^{2+}$$II. \text{Blue}$
$C. [Ni(H_2O)_6]^{2+}$$III. \text{Pale blue}$
$D. [Ni(H_2O)_2(en)_2]^{2+}$$IV. \text{Violet}$

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