JEE 2011 :: General Questions :: Chemistry

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Geometrical shapes of the complexes formed by the reaction of $Ni^{2+}$ with $Cl^{-}$,  $CN^{-}$ and $H_{2}O$ , respectively, are 

Answer:

Explanation:

 $Ni^{2+}+4CN^{-} \rightarrow  [Ni(CN)_{4}]^{2-}$ . Here $Ni^{2+}$  has d⁸ configuration with CN^- as strong ligand.

$d^{8}$  configuration in strong ligand field  gives dsp²-hybridisation  , hence  square planar geometry 

$Ni^{2+} +4Cl^{-} \rightarrow  [NiCl_{4}]^{2-}$

 Here  $Ni^{2+}$ has $d^{8}$ -configuration with $Cl^{-}$ as weak  ligand 

 d⁸-configuration in weak ligand field gives sp³ hybridisation, hence  tetrahedral geometry  Ni²⁺ with H₂O forms 

  $[Ni(H_{2}O)_{6}]^{2+}$ complex  and H₂O is a weak  ligand 

 Therefore , $Ni(H_{2}O)^{2+}$  has octahedral  geometry 

$AgNO_{3}(aq)$ was added to an aqueous KCI solution gradually and the conductivity of the solution was measured. The plot of conductance (A) versus the volume of $AgNO_{3}$ is

Answer:

Explanation:

As $AgNO_{3}$ is added to the solution, KCI will be displayed according to the following reaction

 $AgNO_{3}(aq) +KCl(aq) \rightarrow   AgCl(s)+KNO_{3}(aq)$

For every mole of KCI displaced from the solution, one mole of KNO₃ comes into the solution resulting in almost constant conductivity. As the end point is reached, added AgNO₃ remain in solution increasing ionic concenkation, hence conductivity increases

Bombardment of aluminium by α-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are

Answer:

Explanation:

(i) $^{27}_{13}Al+^{4}_{2}He (\alpha) \rightarrow ^{30}_{14}Si+^{1}_{1}X\Rightarrow X $  is proton $^{1}_{1}H$

(ii)$^{27}_{13}Al+^{4}_{2}He (\alpha) \rightarrow ^{30}_{15}P+^{1}_{0}Y\Rightarrow Y $  is neutron $^{1}_{0}n$

  $^{30}_{15}P \rightarrow ^{30}_{14}Si+^{0}_{1}Z$

$\Rightarrow$       Z is positron  $_{+1}^{0} e$

• General Questions - Physics
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