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Coordination Compounds Test - 54

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Coordination Compounds Test - 54
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  • Question 1
    1 / -0
    Green house effect is caused by
    Solution

  • Question 2
    1 / -0
    The correct name of $$[Pt{({NH}_{3})}_{4}{Cl}_{2}][Pt{Cl}_{4}]$$ is:
    Solution
    Tetraaminedichloro platinum (IV) tetrachloroplatinate(II)
    First, we have to write the cationic part, of the coordinate compound
    Here, the ligands are Amine and Chloride. Since there are $$4$$ amine groups we’ll name it as tetramine and $$2$$ chloride groups will be named as dichloro.  Also, we have platinum metal as the central metal atom in a cation, and it’s oxidation state is $$+4$$.

    Now for the other (anionic part):

    Here the ligand is chloride. Since there are $$4$$ chlorides so we’ll call it tetrachloride. We have the central metal atom as platinum again, but since it is in the anionic coordination sphere.

  • Question 3
    1 / -0
    Which of the following complex is useful in the dehydrogenation of alkanes?
    Solution
    Rhodium (I) complex are used for dehydrogenation of alkanes as $$[(Ph_3P)_3RhCl]$$
    Hence, option $$C$$ is correct.
  • Question 4
    1 / -0
    The sum of coordination number and oxidation number of the metal $$M$$ in the complex $$[M{(en)}_{2}({C}_{2}{O}_{4})]Cl$$ (where $$en$$ is ethylenediamine) is :
    Solution
    The given complex compound is [M$$(e{ n) }_{ 2 }({ C }_{ 2 }{ O }_{ 4 })$$]Cl

    Let oxidation number of M is X

    $$\therefore X-2-2= - 1$$

    $$X=+3$$

    Now, as the coordination number is defined as the total no of binding sites attached to the metal. Hence, in the given complex coordination number is 6.
    Sum of Coordination no. and oxidation state $$=3+6=9$$

    Hence, the correct option is C
  • Question 5
    1 / -0
    The crystal field stabilisation energy [CFSE] and the spin only magnetic moment in Bohr magneton [BM] for the complex $$K_3[Fe(CN)_6]$$ respectively are:
    Solution
    Electronic Configuration of $$_{ 26 }^{  }{ Fe }$$: $$_{ 26 }^{  }{ Fe }\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3{ p }^{ 6 }3{ d }^{ 6 }4{ s }^{ 2 }$$
    Electronic Configuration of $$_{ 26 }^{  }{ Fe }{  }^{ 3+ }$$ : $$_{ 26 }^{  }{ Fe }{  }^{ 3+ }\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3p^{ 6 }3{ d }^{ 5 }4{ s }^{ 0 }$$

    In $$[fe(CN)_{ 6 }]{  }^{ 3- }$$ ,hexacyanoferrate(iii) ion.We have 5-d electrons.

    Here strong cyanide ligand causes greater d-orbital splitting.The octahedral splitting energy,becomes high.

    During filling up process,the first 3 electrons go into $${ t }_{ 2g }$$ orbitals.Now however, the splitting energy is much greater so it is lesser energitically  costly for electrons to pair up  in the $${ t }_{ 2g }$$  orbitals than to go into the $${ e }_{ g }$$  orbitals and low spin complex is formed.

    Here the crystal field splitting energy:

    = -no. of electrons in $${ e }_{ g }$$ X0.6$${ \Delta  }_{ 0 }$$+no. of electrons in $${ t }_{ 2g }$$ X0.4$${ \Delta  }_{ 0 }$$

    = -0 X0.6$${ \Delta  }_{ 0 }$$+5X0.4$${ \Delta  }_{ 0 }$$=2$${ \Delta  }_{ 0 }$$ 

    The spin only moment $${ \mu  }_{ s }=\sqrt { n(n+2) } $$

    So..,

    $${ \mu  }_{ s }=\sqrt { n(n+2) } $$=$${ \mu  }_{ s }=\sqrt { 1X(1+2) } $$=$$\sqrt { 3 } $$=1.7BM
  • Question 6
    1 / -0
    In the separation of $$Cu^{2+}$$ and $$Cd^{2+}$$ ions in the second group of qualitative analysis of cations, tetraamine copper (ll) sulphate and tetraamine cadmium (ll) sulphate react with $$KCN$$ to form the corresponding cyano complexes. 

    Which of the following pairs of complexes and their relative stability enables the separation of $$Cu^{2+}$$ and $$Cd^{2+}$$?
    Solution

  • Question 7
    1 / -0
    Highest conductivity is seen in:
    Solution
    The complex $$K_4[Fe(CN)_6]$$ will show the higher conductivity because among the given option it will dissociate into maximum 5 ions as $$4 K^+$$ and a $$[Fe(CN)_6]^{4-}$$. Hence the correct option will be B.
  • Question 8
    1 / -0
    Consider the following three halides:
    (i) $${CH}_{3}-{CH}_{2}Cl$$
    (ii) $${CH}_{2}=CH-Cl$$
    (iii) $${C}_{6}{H}_{5}-Cl$$
    Arrange $$C-Cl$$ bond lenght of these compounds in decreasing order:
    Solution

  • Question 9
    1 / -0
    Choose the incorrect order of stability:
  • Question 10
    1 / -0
    Select the incorrect order of stability:
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