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Coordination Compounds Test - 11

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Coordination Compounds Test - 11
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  • Question 1
    1 / -0
    On treatment of $$100\, mL$$ of $$0.1\, M$$ solution of $$CoCl_{3}.6H_{2}O$$ with excess $$AgNO_{3};1.2\times 10^{22}$$ ions are precipitated. The complex is :
    Solution
     Avogadro's number is  $$\displaystyle 6.023\times 10^{23}$$

    $$\displaystyle 1.2\times 10^{22}$$ ions corresponds to $$\displaystyle \dfrac {1.2\times 10^{22}}{6.023\times 10^{23}}=0.02$$ moles.

    $$100$$ mL of 0.1M solution of $$\displaystyle CoCl_{3}.6H_{2}O$$ corresponds to $$\displaystyle \dfrac {100 \: mL}{1000 \: mL/L} \times 0.1 \: M = 0.01 \: moles$$.

    $$0.01$$ moles of $$\displaystyle CoCl_{3}.6H_{2}O$$ reacts with excess $$\displaystyle AgNO_3$$ to form $$0.02$$ moles of $$\displaystyle AgCl$$ precipitate, which means $$0.01$$ moles of $$Ag^+$$ and $$0.01$$ moles of $$Cl^-$$ ions.

    Thus, from one molecule of $$\displaystyle CoCl_{3}.6H_{2}O$$, one chlorine atom is displaced. Hence, the complex is 
    $$\displaystyle \left [ Co\left ( H_{2}O \right )_{5}Cl\right ]Cl_2.H_{2}O$$ as number of chloride ion outside co-ordination sphere should be one.

    Hence, the option (C) is the correct answer.
  • Question 2
    1 / -0
    Which one of the following has largest number of isomers? ( $$\mathrm{R}=$$ alkyl group, en $$=$$ ethylenediamine) 
    Solution
    $$[\mathrm{C}\mathrm{o}(\mathrm{e}\mathrm{n})_{2}\mathrm{C}1_{2}]^{+}$$  has largest number of isomers which includes geometrical isomers (cis and trans isomers) as well as stereoisomers (d form, l form and optically inactive form). $$en$$ is a bidentate ligand.
  • Question 3
    1 / -0
    Co-ordination compounds have great importance in biological systems. In this context which of the
    following statements is incorrect? 
    Solution
    Chlorophyll is the green pigment in plants that helps in photosynthesis. The metal inside the ring structure of chlorophyll is magnesium $$(Mg^{2+})$$. The four nitrogen atoms in the ring structure of chlorophyll are bonded to the magnesium ion.
  • Question 4
    1 / -0
    In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $$(Ca^{2+})$$ and fluoride ion $$(F^-)$$ are:
    Solution

    Calcium fluoride crystallizes in a Face-Centered Cubic unit cell (FCC) having an edge length of 5.463 Angstroms.

    The cell is displayed here with calcium cations (in blue) defining FCC lattice sites, and fluoride anions (in green) occupying all tetrahedral sites.

    Fluoride anions have 4 neighbors of opposite charge arranged at vertices of an tetrahedron. Calcium cations have EIGHT neighbors of opposite charge arrange at corners that outline a smaller cube. So the Ca: F coordination ratio is 8 : 4 or 2 : 1.

    In $$4CaF_2$$ the the co ordination number of

    $$Ca^{+2}$$ is 8

    $$F^{-}$$ is 4
    Hence option A is correct
  • Question 5
    1 / -0
    Iron carbonyl, $$Fe(CO)_{5}$$ is __________.
    Solution
    Iron carbonyl, $$Fe(CO)_{5}$$ is mononuclear. In $$Fe(CO)_{5}$$, one Fe atom is surrounded by 5 CO ligands. Mononuclear complexes are those complexes in which one metal atom/ion is surrounded by ligands.

  • Question 6
    1 / -0
    The correct increasing order of trans-effect of the following species is:
    Solution

    Trans effect can be defined as the effect of a ligand towards substitution on other ligand which is trans to it.

    (weak) $$F^- < HO^- < H_2O < NH_3 < py < Cl^- < Br^- < I^-< SCN^-< NO_2^{-}< SC(NH_3)_2< Ph^-$$$$ < SO_3^{2-}< PR_3 < AsR_3< SR_2< H_3C^- < H^-< NO< CO< NC^-< C_2H_4 $$(strong).

    Trans effect is nothing but those ligands which are trans directing. That means, which prefer a trans structure. Mostly in Square planar complexes.

    Hence option D is correct.

  • Question 7
    1 / -0
    Which of the following complexes is used to be as an anticancer agent?
    Solution
    $$Cis-[PtCl_2NH_3)_2]$$ known as cis platin is used as an anticancer agent.
  • Question 8
    1 / -0
    The hybridizations of atomic orbitals of nitrogen in $$NO_2^+ ,\ N{ O }_3^-$$ and $$NH_4^+$$ respectively are :
    Solution
    $$NO_2^{+}$$ contains 2 sigma bonds so hybridization is $$sp$$ and shape is linear.

    $$N{ O }_3^{-}$$ contains 3 sigma bonds so hybridization is $$sp^{2}$$ and shape is trigonal.

    $$NH_4^{+}$$ contains 4 sigma bonds so hybridization is $$sp^{3}$$ and shape is tetrahedral.

    Hence, option A is correct.
  • Question 9
    1 / -0
    Correct increasing order for the wavelength of absorption in the visible region for the complexes of $$Co^{3+}$$ is?
    Solution
    Strength of ligands attached with $${Co}^{3+}$$ ion is in the order of $${CN}^{-} > N{H}_{3} > {H}_{2}O > {Cl}^{-}$$. So order of spliting (value of $$\Delta_{0}$$) will be in the same order.

    Therefore,
    Wavelength $$\propto \cfrac{1}{{\Delta}_{0}}$$

    Thus the correct increasing order of wavelength is-
    $$[Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+}< [Co(NH_3)_5Cl]^{+2}$$

    Hence, the correct option is $$A$$
  • Question 10
    1 / -0
    The hypothetical complex chloro-diaquatriamminecobalt(III) chloride can be represented as:
    Solution
    The hypothetical complex chloro-diaquatriamminecobalt(III) chloride can be represented as $$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 } \right] { Cl }_{ 2 }$$.

    The complex $$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) Cl }_{ 3 } \right] $$ is tetra $$chloro-aquatriamminecobalt(III)$$.

    The complex $$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 }Cl \right] $$ is di $$chloro-diaquatriamminecobalt(III)$$.

    The complex $$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 3 }{ Cl }_{ 3 } \right] $$ is tetra $$chloro-triaquatriamminecobalt(III)$$.
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