JEE 2011 :: General Questions :: Chemistry

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 Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in these reactions as given in Column II.

Answer:

Explanation:

Match the transformation in Column I with appropriate options in Column II

Answer:

Explanation:

(A)  $CO_{2}(s) \rightarrow  CO_{2}(g)$

It is just a phase transition (sublimation) as no chemical change has occurred. Sublimation is always endothermic. Product is gas, more disordered, hence $\triangle$ s  is positive

(B) $CaCO_{3}(s) \rightarrow   CaO(s)+CO_{2}(g)$

It is a chemical decomposition, not a phase change. Thermal decomposition occurs at the expense of energy, hence endothermic. Product contains a gaseous species, hence  $\triangle S>0$

(C) $2H \rightarrow H_{2}(g)$

A new H-H covalent bond is formed, hence, $\triangle H <0.$ Also, the product is less disordered than the reactant, $\triangle S  <0$

(D) Allohopes are considered as different phase, hence $P_{(white,solid)} \rightarrow  P_{red,solid)}$ transition as well as allotropic change.

Also. red phosphorus is more ordered than white phosphorus, $\triangle$ S< 0.

The total number of contributing structures showing hyperconjugation (involving C-H bonds) for the following carbocation is

Answer:

Explanation:

These are total 6$\alpha$ -H to sp² carbon and they all can participate in hyperconjugation.

The volume (in mL) of 0.1 M $AgNO_{3}$  required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [$Cr(H_{2}O)_{15}Cl]Cl_{2}$, as silver chloride is close to

Answer:

Explanation:

mmol of complex = 30 x 0.01 = 0.3 Also, 1 mole of complex $[Cr(H_{2}O)_{5}Cl]Cl_{2}$ gives only two moles of chloride ion when dissolved in solution

$[Cr(H_{2}O)_{5}Cl)Cl_{2} \rightarrow   [Cr(H_{2}O)_{5}Cl]^{2+}+2Cl^{-}$

$\Rightarrow$  mmol of Cl^- ion produced from its 0.3 mmol = 0.6

Hence, 0.6 mmol of Ag⁺ would be required for precipitition

$\Rightarrow$  0.60 mmol  of $Ag^{+}$ = 0.1M x V (in mL)

$\Rightarrow$     V=6mL 

The number of hexagonal faces that are present in a truncated octahedron is

Answer:

Explanation:

The truncated octahedron is the l4-faced Archimedean solid, with 14 total faces. 6 squares and 8 regular hexagons. The truncated octahedron is formed by removing the six right square pyramids one from each point of a regular octahedron as

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics