MCQ
Which of these statements about $[Co(CN)_6]^{3-}$ is true?
  • A
    $[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a high-spin configuration.
  • B
    $[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
  • $[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
  • D
    $[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a low-spin configuration.

Answer

Correct option: C.
$[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
c
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{-3}$

$\mathrm{Co}^{+3}=3 \mathrm{d}^{6} 4 \mathrm{s}^{0} 4 \mathrm{p}^{0}$

$\because$ in presence of strong field ligand, pairing of electrons occurs so in this complex no unpaired electron is present and

it is low spin complex.

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For a solution formed by mixing liquids $L$ and $M$, the vapour pressure of $L$ plotted against the mole fraction of $M$ in solution is shown in the following figure. Here $x_{L}$ and $x_M$ represent mole fractions of $L$ and $M$, respectively, in the solution. The correct statement(s) applicable to this system is(are)

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$[D]$ The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed from $x_{L}=0$ to $x_{L}=1$

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