JEE Main On 8 April 2017 Question 14
Question: The pair of compounds having metals in their highest oxidation state is : [JEE Online 08-04-2017]
Options:
A) $ Mn{O_2}andCr{O_2}C{l_2} $
B) $ {{[\text{ Fe(CN}{{)}_6}]}^{\text{3-}}}and,{{[\text{ Cu(CN}{{)}_4}]}^{\text{2-}}} $
C) $ {{[NICl _4]}^{2-}}and{{[CoCl _4]}^{2-}} $
D) $ {{[\text{ FeC}{l_4}]}^{-}}andC{o_2}{O_3} $
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Answer:
Correct Answer: A
Solution:
Calculation of oxidation states of the given compounds:
A) $\mathrm{MnO}_2$ and $\mathrm{CrO}_2 \mathrm{Cl}_2$ :
- $\mathrm{MnO}_2$ :
$$ x+2(-2)=0 \Rightarrow x=+4 $$
- $\mathrm{CrO}_2 \mathrm{Cl}_2$ :
$$ x+2(-2)+2(-1)=0 \Rightarrow x=+6 $$
B) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$ :
- $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ :
$$ x+6(-1)=-3 \Rightarrow x=+3 $$
- $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$ :
$$ x+4(-1)=-2 \Rightarrow x=+2 $$
C) $\left[\mathrm{NiCl}_4\right]^{2-}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$ :
- $\left[\mathrm{NiCl}_4\right]^{2-}$ :
$$ x+4(-1)=-2 \Rightarrow x=+2 $$
- $\left[\mathrm{CoCl}_4\right]^{2-}$ :
$$ x+4(-1)=-2 \Rightarrow x=+2 $$
D) $\left[\mathrm{FeCl}_4\right]^{-}$and $\mathrm{Co}_2 \mathrm{O}_3$ :
- $\left[\mathrm{FeCl}_4\right]^{-}$:
$$ x+4(-1)=-1 \Rightarrow x=+3 $$
- $\mathrm{Co}_2 \mathrm{O}_3$ :
$$ 2 x+3(-2)=0 \Rightarrow 2 x=6 \Rightarrow x=+3 $$
The pair with metals in their highest oxidation state is A$) \mathrm{CrO}_2 \mathrm{Cl}_2$ with Cr in +6 .
BETA
