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• advanced 2013 - Chemistry 2013
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Consider the following complex ions P, Q , and R

$p= [FeF_{6}]^{3-},Q=[V(H_{2}O)_{6}]^{2+} and R= [Fe(H_{2}O)_{6}]^{2+}$

 the correct order of the complex ions. According in their spin-only magnetic moment values (in BM) IS

Answer:

Explanation:

Plan spin only magnetic moment $=\sqrt{N(N+2)}$ BM, when N is the number of unpaired elecrons. In presence of weak ligand (as H₂O, Cl^-, F^-) there is no pairing of electrons, and electrons donated by ligands are filled in outer vacant orbitals 

 In presence of strong ligand

(as CN^-, CO, NH₃,en) electrons are paired and electrons from ligands are filled in available inner orbitals

The compound that  does not liberate CO₂, on treatment with aqueous sodium bicarbonate  solution is 

Answer:

Explanation:

 Plan    $NaHCO_{3}\rightleftharpoons Na^{+}+HCO_{3}^{-}$

 $HCO_{3}^{-}$   is decomposed by acid releasing CO₂

 $HCO_{3}^{-}+H^{+}\rightarrow H_{2}O+CO_{2}$

 If acid is stronger than $HCO_{3}^{-}$     then CO₂ is released 

 Phenol is less acidic and thus does not liberate CO₂   and NaHCO₃

KI in acetone undergoes S_N 2 reaction with each P, Q, R and S. the rates of the reaction varies as

Answer:

Explanation:

 Plan  Acetone is an aprotic solvent and can dissolve both the nucleophile and the substrate  and thus S_N   2 reaction is favoured, also

 Thus, reactivity order is 

 S >P>R>Q

The standard enthalpies of formation of CO₂(g), H₂O(l)  and glucose (s)   at 25° C are -400 kJ/mol, -300kJ/mol and -1300kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° C  is 

Answer:

Explanation:

 Plan   $\triangle _{c}H^{0}$   (standard heat of combustion ) is the standard enthalpy change when one mole of the substance is completely oxidised.

  Also standard heat of formation ( $\triangle _{f}H^{0}$ ) can be taken as the standard  of their substance.

   $H^{0}_{H_{2}O}=\triangle_{f}H^{0}(H_{2}O)=-300kJmol^{-1}$

 $H^{0}_{glucose }=\triangle_{f}H^{0}(glucose)=-1300kJmol^{-1}$

      $H^{0}_{O_{2} }=\triangle_{f}H^{0}(O_{2})=0.00$

  $C_{6}H_{12}O_{6}(s)+6O_{2}(g)\rightarrow 6CO_{2}(g)+6H_{2}O(l)$

  $\triangle_{c}H^{0} (glucose)$

                 = $6[\triangle_{f}H^{0}(CO_{2})+\triangle_{f}H^{0} (H_{2}O)]$

       -  $[\triangle_{f}H^{0}(C_{6}H_{12}O_{6})+6\triangle_{f}H^{0} O_{2})]$

    =6[-400-300]-[-1300+6 X0]

   =-2900 kJ mol^-1

 Molar mass of C₆H₁₂O₆= 180 g mol^-1

 Thus, standard heat of combustion of glucose per gram

  =   $\frac{-2900}{180}=-16.11kJ g^{-1}$

• advanced 2013 - Physics 2013
• advanced 2013 - Chemistry 2013
• advanced 2013 - Mathematics 2013