Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve: Overview, Questions, Preparation

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Ans:

Fehling's test is positive when compound 'B' is used. It gives an iodoform test and proves that it is an aldehyde. Because it fails Fehling's test, compound 'C' is a ketone.

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Ans: The aromatic compound 'A' does not give Tollen's reagent test, it is not an aromatic aldehyde. As it responds to an iodoform test called methyl ketone.

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Ans: Functional isomers

Compound I will react faster with HCN due to less steric hindrance and greater positive charge on carbon atom of carbonyl group. Two methyl groups increase electron density on carbonyl carbon in compounds II hence the rate of nucleophilic attack is less.
Mechanism of the reaction:

 

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Ans: Liquid 'A' reduces ammoniacal silver nitrate and 'B' is a ketone which forms white crystalline solid on treatment with hydrogen sulphite

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Butanol is a four-carbon aldehyde, with the aldehyde group on the fourth carbon. It has a functional group and a carbonyl group that interacts dipole-dipole. Because butanol possesses a polar O-H bond, it exhibits intermolecular H bonding, which is not feasible in butanal due to the lack of a polar bond. The boiling point of butanol is higher than that of butan-1-ol.

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(i) 3-Phenylprop-2-ene-1-al.

(ii) Cyclohexanecarbaldehyde

(iii) 3-Oxopentan-1-al

(iv) IUPAC name: But-2-enal

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Ans:

A: Ethane-1, 2-dial

B: Benzene-1, 4-carbaldehyde

C: 3-Bromobenzaldehyde

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Benzal chloride can be made by chlorinating toluene in the presence of sunshine and then using the hydrolysis process to obtain benzaldehyde. This method can be used to make benzaldehyde in a commercial setting.

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The electrophile formed when benzene reacts with benzoyl chloride in the presence of anhydrous AlCl₃ is benzoylinium cation, and the resultant product is benzophenone. Friedel Crafts acylation reaction is the name for this reaction.

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ClCH₂COOH > FCH₂COOH > C₆H₅CH₂COOH > , CH₃COOH > CH₃CH₂OH

When compared to other alcohols, CH₃CH₂OH is the least acidic. Because of the halogen, FCH₂COOH is quite acidic. It is either an electron donating group or an electron receiving group in acidic strength. Acidic strength increases as the number of electron withdrawing groups increases. The acidic strength of electron donating groups diminishes.

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It's referred to as cross Aldol Condensation.Cross aldol condensation occurs when two separate aldehydes or ketones combine to form a single molecule.

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When alkaline KMnO₄ oxidises compound "B" with compound "A," acid is formed.

When compound "A" is reduced with LiAlH₄, it transforms into compound "B." The letter "B" stands for a type of alcoholic beverage.

When compounds "A" and "B" are heated with conc. H2SO4, compound "C" is formed, which has a fruity odour.

 

 

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NO₂CH₂COOH  >  FCH₂COOH  > C₆H₅COOH

To begin, determine whether it has an acidic strength associated with it, and whether it is an electron donating or electron withdrawing group. Acidic strength increases as the number of electron withdrawing groups increases. The acidic strength reduces as the electron donating group increases. In decreasing order, the most influential acidic strength is −NO₂> −F> −C₆H₅.

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Solution: Carbonyl group is polar in nature. Due to larger electronegativity of oxygen as compared to carbon, carbon acquires partial positive charge while O acquires partial negative charge.

Because of slight positive charge on C atom, it is attacked by nucleophiles and, therefore, undergoes nucleophilic addition reaction.
Ethylenic double bond is a non-polar bond and is a source of electrons. Therefore, it is attacked by electrophiles and undergoes electrophilic addition reactions.

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The presence of a lone pair of electrons on an atom of the OH group diminishes the electrophilic character of C through resonance; carboxylic acids contain a carbonyl group. As a result, the carbonyl atom's partial positive charge is lowered.

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Although all of them have a hydrogen atom connected to an oxygen atom (—O—H), carboxylic acids are more acidic than alcohols or phenols. This can be explained by the conjugate base's stability after  H+ ions have been removed from the acid and phenol.

This resonance hybrid may be presented as:-

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Due to the development of polysubstitution products, ethylbenzene is usually made through acetylation of benzene followed by reduction and direct alkylation.

Due to resonance, the benzene ring must be deactivated. The electrophilic attacking agent is rarely produced by the benzene ring.

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Yes, both reactions look to be the same, but they are not. To convert benzene to benzaldehyde, the Gatterman-Koch reaction is performed. Benzene is treated with an acid chloride in the presence of anhydrous AlCl₃ in acylation processes. The Gatterman-Koch reaction is used to make HCOCl in situ by reacting CO with HCl gas in the presence of anhydrous AlCl₃.

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Correct option: B

CH₃−CH₂−C≡CH + H₂O CH₃−CH₂−CO−CH₃

Butan-2-one is produced when butyne is hydrated in the presence of Hg ions as a catalyst. The ketone is the product. After the triple bond is broken, the OH group attaches to the secondary carbocation.

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Correct option: B or (ii)

In nucleophilic addition processes, the reactivity of aldehydes and ketones are steric factors. The carbonyl group with the highest positive charge will be the most reactive, and if these groups have a positive inductive impact, the attack tendency will be reduced.

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Correct option: C

Because the phenoxide ion produced is stabilised by resonance, phenol is more acidic than ethanol. When it comes to ethanol, resonance does not help to stabilise the phenoxide ion. The acidity of chloroacetic acid is higher than that of acetic acid. Alcohols and phenols are less acidic than carboxylic acids.

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Correct option: B

When phenol reacts with acid chloride in the presence of an alkali, the alkali is pyridine, and the acid chloride is benzoyl chloride, compound C₆H₅OCOC₆H₅ is formed. During this process, pyridine is utilised to neutralise 4HCl.

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Correct option: C

Acetones belong to the ketonic functional category. To distinguish between aldehyde and ketone, utilise Fehling's solution. Carbonyl groups are found in sodium hydrogen sulphite, phenyl hydrazine, and Grignard reagent. Ketones and aromatic aldehydes do not react with Fehling's solution.

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Correct options: D

Aldehydes without a - hydrogen atom are the subject of Cannizaro's reaction.

Option D, CH₃CHO has 3 alpha  hydrogen atoms and it does not give Cannizaro's reaction 

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Correct option: B

In KOH adds to the compound

Due to the production of a stable conjugate base after hydronium is removed from phenol, it is more stable than alcohol.

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Ans: Correct option: D

Tautomers are structural isomers that exist in two or more interconvertible structures of a single chemical molecule. Tautomerism transforms 'A' into acetone.

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Correct option: B

The position of a functional group in the same carbon chain differs in positional isomerism. The functional group in the chemical or the unsaturated double or triple bond in the compound can help us figure it out.

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Correct option: C

_I2 and NaOH It is an iodoform reaction involving methyl ketones, the solution is the correct reagent. Because the reactants and product have double bonds, double bonds do not work throughout the reaction. There are also ketone and carboxylic groups present. The ketonic group is oxidised under the carboxylic group during the process. 

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Correct option: B

When a primary alcohol, such as Butan-1-ol, reacts with a powerful oxidising agent, such as KMnO₄, it produces carboxylic acid, however when a secondary alcohol, such as Butan-2-ol, is oxidised under the pressure of KMnO₄, it produces ketone. As a result, option B is right.

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Correct option: A

The Clemmensen Reduction method is used to reduce a carbonyl group to CH2. Amalgams are a type of reducing agent that has been around for a long time. In the Clemmensen reduction, hydrogen chloride acts as an acid, protonating the ketone and transferring the electrons from the zinc amalgam to the carbon atoms

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Correct options: B and D

 Consider the following product CH₃ -CHO C and O are broken into positive and negative bonds. C should have one condition: hydrogen must be available; if it isn't, aldol condensation will not occur. When two products are mixed with the availability of - hydrogen, the positive and negative bonds between carbon and hydrogen are exchanged, and water (H₂0) is eliminated, resulting in aldol condensation.

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Correct options: B and C

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Solution:  (b, d)
Clemmensen reduction is used to convert cyclohexanone into cyclohexane and benzophenone into diphenyl methane .

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Correct option: A and C

Grignard Reaction is the addition of an organomagnesium halide to form a tertiary or secondary alcohol.

Canizzaro's Reaction involves the base induced disproportionation of two molecules. Aldol Condensation is an organic reaction when an enolate ion reacts with a carbonyl compound to form beta hydroxy ketone with dehydration to give a conjugated enone.

The HVZ reaction involves alpha bromination of carboxylic acids and this reacts with chlorine or bromine in presence of a small amount of red phosphorus to give alpha halo carboxylic acids. 

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Correct options: A and B

A: Benzophenone can be obtained by Friedel-craft acylation reaction 

B: Benzophenone can also be obtained by the reaction between benzoyl chloride and diphenyl cadmium.

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Correct options: A and B 

The given compound is a planar molecule with sp2 hybridised carbon. Carbon atoms are attacked by nucleophiles. If the nucleophile approaches from the front, the A and B molecules shift to the front and back positions, respectively, and the carbon becomes tetrahedral. Another alternative is that A and B molecules will be located above and below the plane, respectively. As a result, the choices C and D aren't represented as planar molecules.

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Ans: A

The smallest aldehyde is formaldehyde. It features a triangular planar shape with two bond pairs and no lone pairings. C and O share one pair of electrons in their double bond. As a result, the C atom is sp² hybridised. As a result, both Assertion and Reason are true.

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Ans: E

The two reactions produce different products for different reasons. One involves the oxidation of aldehydes, while the other involves the reduction of carboxylic acids. Aldehydes are easily oxidised by mild oxidising agents, hence compounds containing  - CHO are rapidly oxidised.

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Ans: D

Because of the presence of electron withdrawing carbonyl groups, the alpha hydrogen atom in carbonyl compounds is acidic. In nature, hydrogen is very acidic.

Because the cation is released in the form of H⁺ , the anion created after the loss of the -hydrogen atom is resonance stabilised.

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Ans: C

There is no alpha hydrogen atom in the Cannizaro process. Formaldehyde and aromatic aldehydes do not contain alpha hydrogen. It proceeds through the Cannizaro reaction. Formaldehyde is the most reactive of all aldehydes.

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Ans: D

The silver mirror test can be used to determine Tollen's. [Ag (NH₃)₂] + OH ^- . Only aldehydes, not ketones, react with Tollen's reagent to create silver.

A silver mirror test is not given with this affirmative test.

The carbonyl group is present in both aldehyde and ketone.

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 Ans: (i) →  (d); (ii) →  (e); (iii) →  (a); (iv) →  (b); (v) →  (c)

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Ans: (i) →  (b); (ii) →  (e); (iii) →  (d); (iv) →  (a); (v) →  (c)

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Ans: (i) →  (c); (ii) →  (d); (iii) →  (a); (iv) →  (b).

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Ans: (i) — (e);  (ii) — (d);  (iii) — (a);  (iv) — (b);  (v) — (f); (vi) — (c)

Sol. (a) In extraction of iron lime stone is added on a flux

$\begin{array}{rr}& {\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3?\mathrm{C}\mathrm{a}\mathrm{O}+{\mathrm{C}\mathrm{O}}_2\\ & {\mathrm{C}\mathrm{O}}_2+{\mathrm{S}\mathrm{i}\mathrm{O}}_2?{\mathrm{C}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{O}}_3\end{array}$

Sol. Reactivity towards ${\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3$  depend on stability of carbocation formed

Sol. Balmer series lies in the visible region.

Sol. ${\mathrm{C}\mathrm{r}}^{2+}$ and ${\mathrm{F}\mathrm{e}}^{2+}$  both have 4 unpaired electrons.

Sol. $2\mathrm{P}\mathrm{b}{\left({\mathrm{N}\mathrm{O}}_3\right)}_2\stackrel{\mathrm{\Delta }}{?}2\mathrm{P}\mathrm{b}\mathrm{O}+2{\mathrm{N}\mathrm{O}}_2+{\mathrm{O}}_2$

$2{\mathrm{N}\mathrm{O}}_2?{\mathrm{N}}_2{\mathrm{O}}_4$ colourless

${\mathrm{N}}_2{\mathrm{O}}_4+2\mathrm{N}\mathrm{O}?2{\mathrm{N}}_2{\mathrm{O}}_3$

Oxidation state in ${\mathrm{N}}_2{\mathrm{O}}_3$  is +3

Kindly go through the solution

Sol. ${\mathrm{O}}^{2-}>{\mathrm{F}}^{-}>{\mathrm{N}\mathrm{a}}^{+}>{\mathrm{M}\mathrm{g}}^{2+}$ . Greater the Zeff, smaller is size.

Sol. A - B has the lowest potential energy and it means it has stronger bond.

Kindly go through the solution 

Sol.

Sol. At equilibrium rate of forward and backward reaction becomes equal.

Kindly go through the solution

Sol.

Sol. $E_{ext}>1.1\mathrm{V}$ , cell reaction is reversed
i.e. $\mathrm{C}\mathrm{u}?{\mathrm{C}\mathrm{u}}^{2+}$ anode
${\mathrm{Z}\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}?\mathrm{Z}\mathrm{n}$ cathode

Both contains $-{\mathrm{N}\mathrm{H}}_2$  group

Sol. (i) Foam - Froth (e)

(ii) Gel - Jellies (c)
(iii) Aerosol - smoke (a)
(iv) Emulsion - milk (f)

Sol. $\mathrm{U}$ and $\mathrm{H}$ are temperature dependent

${\mathrm{C}}_{\mathrm{P},\mathrm{m}}-{\mathrm{C}}_{\mathrm{V},\mathrm{m}}=\mathrm{R}$  (for 1 mole of ideal gas)

$\mathrm{d}\mathrm{U}={\mathrm{C}}_{\mathrm{V}}\mathrm{d}\mathrm{t}$

Sol. $4\mathrm{L}\mathrm{i}+{\mathrm{O}}_2?2{\mathrm{L}\mathrm{i}}_2\mathrm{O}\mathrm{}$ oxide

$2\mathrm{N}\mathrm{a}+{\mathrm{O}}_2?{\mathrm{N}\mathrm{a}}_2{\mathrm{O}}_2$ peroxide

$\mathrm{K}+{\mathrm{O}}_2?{\mathrm{K}\mathrm{O}}_2\mathrm{}$ superoxide

Sol. $Z=101$ belong to actinoids

104 belong to group 4

Sol. 

$P_T=X_A\left(P_A^0-P_B^0\right)+P_B^0$

ATQ

$550=\frac{1}{4}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+4{\mathrm{P}}_{\mathrm{B}}^0$

$560=\frac{1}{5}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+5{\mathrm{P}}_{\mathrm{B}}^0$

${\mathrm{P}}_{\mathrm{A}}^0+3{\mathrm{P}}_{\mathrm{B}}^0=2200$

$\frac{{\mathrm{P}}_{\mathrm{A}}^0\pm 4{\mathrm{P}}_{\mathrm{B}}^0=2800}{{\mathrm{P}}_{\mathrm{B}}^0=600}$

${\mathrm{P}}_{\mathrm{A}}^0=400\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

Sol. first order reaction

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{{\mathrm{a}}_0-\mathrm{x}}$

$\mathrm{K}=\frac{2.303}{90}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{0.25{\mathrm{a}}_0}$

$=0.0154$

$\begin{array}{rr}& t=60\mathrm{\%}=\frac{2.303}{K}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{a_0}{a_0}\dots .\left(2\right)\\ & =\frac{2.303}{0.0154}\times (1-0.602)=59.51\mathrm{m}\mathrm{i}\mathrm{n}s\approx 60\end{array}$

${\mathrm{N}}_2+3{\mathrm{H}}_2?2{\mathrm{N}\mathrm{H}}_3$

$\begin{array}{ll}\frac{2.8\times {10}^3}{28}& \frac{1\times {10}^3}{2}\\ \frac{0.1\times {10}^3\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{R}}& 0.5\times {10}^3\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

$\text{Mass of}{\mathrm{N}\mathrm{H}}_3\text{produced}=0.2\times {10}^3\times 17=3.4\mathrm{k}\mathrm{g}=3400\mathrm{g}$

Kindly go through the solution

eq of ${\mathrm{H}}_2{\mathrm{O}}_2=$ eq of ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$

$\begin{array}{rr}& =\frac{0.316}{158}*5=0.01\\ & =\frac{0.01*17}{0.2}*100=85\mathrm{\%}\end{array}$

(a) In extraction of iron lime stone is added on a flux

$\begin{array}{rr}& {\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3?\mathrm{C}\mathrm{a}\mathrm{O}+{\mathrm{C}\mathrm{O}}_2\\ & {\mathrm{C}\mathrm{O}}_2+{\mathrm{S}\mathrm{i}\mathrm{O}}_2?{\mathrm{C}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{O}}_3\end{array}$

(b) Extraction of silver

$\mathrm{A}\mathrm{g}+\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{N}?{\left. [\begin{array}{l}\text{anionic complex}\\ \mathrm{A}\mathrm{g}(\mathrm{C}\mathrm{N}{)}_2\end{array}\right]}^{-}$

(c) Nickel is purified by Mond's process

(d) $\mathrm{Z}\mathrm{r}$ and $\mathrm{T}\mathrm{i}$ are purified by Van Arkel method by converting to volatile iodide.

Reactivity towards ${\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3$ depend on stability of carbocation formed

Balmer series lies in the visible region.

${\mathrm{C}\mathrm{r}}^{2+}$ and ${\mathrm{F}\mathrm{e}}^{2+}$ both have 4 unpaired electrons.

$2\mathrm{P}\mathrm{b}{\left({\mathrm{N}\mathrm{O}}_3\right)}_2\stackrel{\mathrm{\Delta }}{?}2\mathrm{P}\mathrm{b}\mathrm{O}+2{\mathrm{N}\mathrm{O}}_2+{\mathrm{O}}_2$

$2{\mathrm{N}\mathrm{O}}_2?{\mathrm{N}}_2{\mathrm{O}}_4$ colourless

${\mathrm{N}}_2{\mathrm{O}}_4+2\mathrm{N}\mathrm{O}?2{\mathrm{N}}_2{\mathrm{O}}_3$

Oxidation state in ${\mathrm{N}}_2{\mathrm{O}}_3$ is +3

${\mathrm{O}}^{2-}>{\mathrm{F}}^{-}>{\mathrm{N}\mathrm{a}}^{+}>{\mathrm{M}\mathrm{g}}^{2+}$ . Greater the Zeff, smaller is size.

A - B has the lowest potential energy and it means it has stronger bond.

Correct IUPAC name is 4-Bromo-2-methylcyclopentane carboxylic acid

Kindly go through the solution

$E_{ext}>1.1\mathrm{V}$ , cell reaction is reversed
i.e. $\mathrm{C}\mathrm{u}?{\mathrm{C}\mathrm{u}}^{2+}$ anode
${\mathrm{Z}\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}?\mathrm{Z}\mathrm{n}$ cathode

Both contains $-{\mathrm{N}\mathrm{H}}_2$ group

(i) Foam - Froth (e)
(ii) Gel - Jellies (c)
(iii) Aerosol - smoke (a)
(iv) Emulsion - milk (f)

$\mathrm{U}$ and $\mathrm{H}$ are temperature dependent

${\mathrm{C}}_{\mathrm{P},\mathrm{m}}-{\mathrm{C}}_{\mathrm{V},\mathrm{m}}=\mathrm{R}$  (for 1 mole of ideal gas)

$\mathrm{d}\mathrm{U}={\mathrm{C}}_{\mathrm{V}}\mathrm{d}\mathrm{t}$

$4\mathrm{L}\mathrm{i}+{\mathrm{O}}_2?2{\mathrm{L}\mathrm{i}}_2\mathrm{O}\mathrm{}$ oxide

$2\mathrm{N}\mathrm{a}+{\mathrm{O}}_2?{\mathrm{N}\mathrm{a}}_2{\mathrm{O}}_2$ peroxide

$\mathrm{K}+{\mathrm{O}}_2?{\mathrm{K}\mathrm{O}}_2\mathrm{}$ superoxide

 $Z=101$ belong to actinoids

104 belong to group 4

$P_T=X_A\left(P_A^0-P_B^0\right)+P_B^0$

ATQ

$550=\frac{1}{4}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+4{\mathrm{P}}_{\mathrm{B}}^0$

$560=\frac{1}{5}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+5{\mathrm{P}}_{\mathrm{B}}^0$

${\mathrm{P}}_{\mathrm{A}}^0+3{\mathrm{P}}_{\mathrm{B}}^0=2200$

$\frac{{\mathrm{P}}_{\mathrm{A}}^0\pm 4{\mathrm{P}}_{\mathrm{B}}^0=2800}{{\mathrm{P}}_{\mathrm{B}}^0=600}$

${\mathrm{P}}_{\mathrm{A}}^0=400\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

First order reaction

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{{\mathrm{a}}_0-\mathrm{x}}$

$\mathrm{K}=\frac{2.303}{90}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{0.25{\mathrm{a}}_0}$

$=0.0154$

$\begin{array}{rr}& t=60\mathrm{\%}=\frac{2.303}{K}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{a_0}{a_0}\dots .\left(2\right)\\ & =\frac{2.303}{0.0154}\times (1-0.602)=59.51\mathrm{m}\mathrm{i}\mathrm{n}s\approx 60\end{array}$

${\mathrm{N}}_2+3{\mathrm{H}}_2?2{\mathrm{N}\mathrm{H}}_3$

$\begin{array}{ll}\frac{2.8\times {10}^3}{28}& \frac{1\times {10}^3}{2}\\ \frac{0.1\times {10}^3\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{R}}& 0.5\times {10}^3\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

$\text{Mass of}{\mathrm{N}\mathrm{H}}_3\text{produced}=0.2\times {10}^3\times 17=3.4\mathrm{k}\mathrm{g}=3400\mathrm{g}$

Kindly go through the solution

eq of ${\mathrm{H}}_2{\mathrm{O}}_2=$ eq of ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$

$\begin{array}{rr}& =\frac{0.316}{158}*5=0.01\\ & =\frac{0.01*17}{0.2}*100=85\mathrm{\%}\end{array}$

Kindly go through the solution

Kindly go through the solution

At equilibrium rate of forward and backward reaction becomes equal.

Kindly go through the solution

Bond strength $\propto$ Bond order

$N_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2$

$O_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*1}$

$C_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2$

$B_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1\equiv {\pi }_{2py}^1$

NO Number of electron = 7 + 8 = 15

B.O. Similar to $N_2^{-}$

$N_2^{-}\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*}$

B.O. of N₂ = 3B.O of C₂ = $\frac{8-4}{2}=2$

Removal of e form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e in orbital.

 $A{\mathcal{l}}^{+3}$ → number of electrons = 10

Br → number of electrons = 36

$B{e}^{+2}$ → number of electrons = 2

$C{\mathcal{l}}^{-}$ → number of electrons = 18

$L{i}^{+}$ → number of electrons = 2

$S^{-2}$ → number of electrons = 18

K⁺ → number of electrons = 18

O^-2 → number of electrons = 10

Mg⁺² → number of electrons = 10

O^-2 & Mg⁺² are isoelectronic with Al⁺³

Au + NaCN + O₂ → Na [Au (CN)₂]

$Zn+Na\left[Au{\left(CN\right)}_2\right]\to N{a}_2\left[Zn{\left(CN\right)}_4\right]+Au$

$Ais{\left[Au{\left(CN\right)}_2\right]}^{-}andBis{\left[Zn{\left(CN\right)}_4\right]}^{-2}$

Number of electrons in

$P{H}_3=15+3=18$

$B_2{H}_6=5\times 2+6=16$

$\begin{array}{l}CC{l}_4=6+17\times 4=6+68=74\\ N{H}_3=7+1\times 3=10\\ LiH=3+1=4\\ BC{l}_3=5+17\times 3=56\end{array}$

$B_2{H}_6\&BC{l}_3$ are e^- deficient molecules. B₂H₆ is dimer of BH₃, both compound has 6e^- only.

Size of hydrated ion $\propto$ $\frac{1}{Ionicmobility}$

Size of hydrated ions $\propto \frac{1}{Ionicsize}$

Size of hydrated ions ; $B{e}^{+2}>M{g}^{+2}>C{a}^{+2}>S{r}^{+2}$

Ionic mobility ; $B{e}^{+2}<M{g}^{+2}<C{a}^{+2}<S{r}^{+2}$

$AgC\mathcal{l}+2N{H}_3\to \underset{\left(Solublecomplex\right)}{\left[Ag{\left(N{H}_3\right)}_2\right]C\mathcal{l}}$

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l}C{e}^{+4}+e^{-}\to C{e}^{+3}{E}^0=1.61V\\ C{e}^{+3}+3e^{-}\to Ce{E}^0=-2.336V\end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

$E_{M{n}^{+3}/M{n}^{+2}}^o=1.51V$

the strongest oxidizing agent have the highest reduction potential. So Mn³⁺ is the strongest oxidizing agent.

Eutrophication of water body results in loss of biodiversity.

Ortho product has intramolecular H- bonding & para product has intermolecular H- bonding. Thus it can be separated by steam distillation due to difference in B.

Molar mass of C₇H₅N₃O₆ = 12 × 7 + 1 × 5 + 14 × 3 + 16 × 6 = 84 + 42 + 96 = 227 g/mol

Number of moles = $\frac{681}{227}$  = 3 mol

$\therefore$ number of N-atoms

$=3\times 6.02\times 10^{23}\times 3=9\times 6.02\times 10^{23}$

$=5418\times 10^{21}$

$\therefore x=5418$ 

$E_{1_H}=-2.2\times 10^{-18}J$

$Li\to L{i}^{+2}+2e^{-},n=2$

$E_{L{i}^{+2}}=E_{1H}\times \frac{Z^2}{n^2}=-2.2\times 10^{-18}\times \frac{3^2}{2^2}$

$\lambda =4\times 10^{-8}$ m

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium ; $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165\times 1000}{-550}K$

$\begin{array}{l}=3\times 100K\\ =300K\end{array}$

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

= 0.00002 × 10³ mm

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2\times 0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}\times 1\right)}^2}{\left(\frac{0.5}{1.5}\times 1\right)}$