Solution

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Solution

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Solution

a. [Co(NH₃)₅(CO₃)]Cl
b. K₂[Ni(CN)₄]

Solution

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Solution

Solution

(a) Because sulphur readily gets oxidized itself to more stable +6 state.

(b) Because of absence of d-orbital in Fluorine.
(c) Because size increases from Helium to Radon. Also, dispersion or van der Waal forces increase from Helium to Radon.
or
(a) MnO₂+ 4HCl —>MnCl₂ + Cl₂ + 2H₂O

(b) XeF₆+ KF —>K⁺[XeF₇]⁻

(c) I⁻ (aq) + H⁺ (aq) + O₂(g) —>2I₂(s) + 2H₂O(l)

Solution

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Solution

(a) If the surface area of a reactant is reduced, less particles are exposed to the other reactant. there is a lesser chance of particles colliding thus rate of reaction decreases when surface area of the reactant is reduced.

(b) Catalysts typically speed up a reaction by reducing the activation energy. The presence of a catalyst increases the reaction rate (in both forward and reverse reactions) by providing an alternative pathway with lower activation energy.

(c) With increase in temperature rate of reaction , increase due to increase in number of molecules having threshold energy.

Solution

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(a) (i) Add Ice cold (NaNO₂ + HCl) followed by phenol or β-Naphthol to both the compounds.
Aniline forms orange red dye while ethylamine doesn’t.
ii) Add CHCl₃ and KOH (alc.) to both the compounds.
Aniline gives foul smelling isocyanide while N-Methylaniline doesn’t.(b) CH₃- CH₂- CH₂ - CH₂– OH (butanol)
CH₃-CH₂-CH₂- CH₂- NH₂ (butanamine)
CH₃- CH₂- CH₂- CH₃ (butane)
The hydrogen bonding is strongest in butanol thus the decreasing order of boiling point is:
Butanol > Butanamine > Butane

Solution

(a) Because the – CHO group in glucose is involved in hemiacetal formation and thus is not free i.e., due to cyclic structure of glucose -CHO group is not free.

(b) The two strands of DNA are bound together by hydrogen bonds between the purine base on one strand and the pyrimidine base on the other strand, and vice versa. The only possible pairings in DNA are G (guanine) and C (cytosine) through three H-bonds (i.e., C≡≡G)) and A (adenine) and T (thymine) through two H-bonds (i.e., A = T). This is due to the different sizes and geometries of the bases. The sequence of bases in one strand fixes the sequence of bases in the other strand due to this base-pairing concept. The two strands are thus complimentary rather than identical.

(c) Starch consists of amylose and amylopectin.
● Amylose is a long unbranched chain with a-D(+)-glucose units that are held together with C1–C4 glycosidic linkage.
●Amylopectin is a highly branched chain polymer of a-D-glucose units in which glycosidic linkage is present between C1–C4 atoms and branching occurs through C1–C6 glycosidic linkage.
Cellulose is a straight chain polysaccharide composed only of b-D-glucose units which are joined by glycosidic linkage between C1 of one glucose units and C4 of next glucose units.