CBSE Class 12 Chemistry — NCERT Solutions

Given:

• Mass of benzene = 22 g
• Mass of carbon tetrachloride = 122 g
• Total mass of solution = 22 + 122 = 144 g

Formula:
$$\text{Mass percentage of a component} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$

Mass percentage of benzene:
$$= \frac{22}{144} \times 100 = 15.28\%$$

Mass percentage of CCl₄:
$$= \frac{122}{144} \times 100 = 84.72\%$$

Answer: Mass percentage of benzene = 15.28% and of CCl₄ = 84.72%

Given:

• 30% by mass benzene means 30 g of benzene in 100 g of solution.
• Mass of CCl₄ = 100 − 30 = 70 g

Molar masses:

• Benzene (C₆H₆): $M = 6 \times 12 + 6 \times 1 = 78\,\text{g mol}^{-1}$
• CCl₄: $M = 12 + 4 \times 35.5 = 154\,\text{g mol}^{-1}$

Moles:
$$n_{\text{benzene}} = \frac{30}{78} = 0.3846\,\text{mol}$$
$$n_{\text{CCl}_4} = \frac{70}{154} = 0.4545\,\text{mol}$$

Mole fraction of benzene:
$$x_{\text{benzene}} = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{CCl}_4}} = \frac{0.3846}{0.3846 + 0.4545} = \frac{0.3846}{0.8391} = 0.459$$

Mole fraction of CCl₄:
$$x_{\text{CCl}_4} = 1 - 0.459 = 0.541$$

Answer: Mole fraction of benzene = 0.459 and of CCl₄ = 0.541

Given: Reaction of alcohols with KI to prepare alkyl iodides.

Concept: KI is used with phosphoric acid (H₃PO₄) and not H₂SO₄ for the conversion of alcohols to alkyl iodides.

Explanation:

H₂SO₄ is an oxidising acid. If H₂SO₄ is used along with KI, the following side reactions occur:

$$\text{H}_2\text{SO}_4 + 2\text{KI} \rightarrow \text{K}_2\text{SO}_4 + 2\text{HI}$$

The HI formed is then oxidised by H₂SO₄:

$$\text{H}_2\text{SO}_4 + 2\text{HI} \rightarrow \text{SO}_2 + \text{I}_2 + 2\text{H}_2\text{O}$$

Thus H₂SO₄ oxidises HI (and KI) to I₂, which cannot act as a nucleophile for the substitution reaction. Hence H₂SO₄ is not used; instead, non-oxidising acids like H₃PO₄ are used.

Given: Propane, $\text{C}_3\text{H}_8$; dihalogen derivatives (using Cl as representative halogen).

Concept: Replace two hydrogen atoms of propane with halogen atoms in all possible ways.

The different dihalogen derivatives of propane are:

(i) 1,1-Dichloropropane:
$$\text{CH}_3\text{CH}_2\text{CHCl}_2$$

(ii) 1,2-Dichloropropane:
$$\text{CH}_3\text{CHClCH}_2\text{Cl}$$

(iii) 1,3-Dichloropropane:
$$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{Cl}$$

(iv) 2,2-Dichloropropane:
$$\text{CH}_3\text{CCl}_2\text{CH}_3$$

(v) 1,1-Dichloropropane (gem on C1) is listed above; additionally:

$$\text{ClCH}_2\text{CHClCH}_3 \quad \text{(1,2-dichloropropane)}$$

All four structural isomers:

1. $\text{ClCH}_2\text{CH}_2\text{CH}_2\text{Cl}$ — 1,3-dichloropropane
2. $\text{ClCH}_2\text{CHClCH}_3$ — 1,2-dichloropropane
3. $\text{CH}_3\text{CCl}_2\text{CH}_3$ — 2,2-dichloropropane
4. $\text{CH}_3\text{CH}_2\text{CHCl}_2$ — 1,1-dichloropropane

A metal higher in the reactivity series displaces a metal lower down from its salt solution. So we arrange them from most reactive to least reactive:

• Mg is most reactive
• then Al
• then Zn
• then Fe
• then Cu

So the order of displacement is:

Mg > Al > Zn > Fe > Cu

Reducing power increases as the standard reduction potential becomes more negative. From the given values:

• Ag⁺/Ag = +0.80 V
• Hg²⁺/Hg = +0.79 V
• Cr³⁺/Cr = -0.74 V
• Mg²⁺/Mg = -2.37 V
• K⁺/K = -2.93 V

So the increasing order of reducing power is:

Ag < Hg < Cr < Mg < K.

Given: Various alcohol structures to classify.

Concept: An alcohol is classified based on the number of carbon atoms directly attached to the carbon bearing the –OH group.

• Primary (1°): –OH on a carbon attached to only one other carbon (or no carbon).
• Secondary (2°): –OH on a carbon attached to two other carbons.
• Tertiary (3°): –OH on a carbon attached to three other carbons.

(i) $\mathrm{CH_3-\underset{CH_3}{\overset{CH_3}{C}}-CH_2OH}$

The –OH group is on $-CH_2-$, which is attached to only one carbon (the quaternary carbon). Hence it is a Primary alcohol.

(ii) $\mathrm{H_2C=CH-CH_2OH}$

The –OH group is on $-CH_2-$, which is attached to only one carbon (the vinylic $CH$). Hence it is a Primary alcohol (also an allylic alcohol).

(iii) $\mathrm{CH_3-CH_2-CH_2-OH}$

The –OH group is on the terminal $-CH_2-$, attached to only one carbon. Hence it is a Primary alcohol.

(iv) The structure represents a secondary alcohol where –OH is on a carbon bearing two other carbon groups (e.g., $\mathrm{CH_3-\underset{OH}{CH}-CH_3}$). Hence it is a Secondary alcohol.

(v) The structure $\mathrm{CH_2-CH-CH_3}$ with –OH on the middle carbon represents a Secondary alcohol.

(vi) $\mathrm{CH=CH-\underset{OH}{C}-}$ where the carbon bearing –OH is attached to three carbons. Hence it is a Tertiary alcohol (also an allylic alcohol).

Summary:

Order is the sum of powers in the rate law. The dimensions of $k$ depend on the order.

(i) Rate $=k[\mathrm{NO}]^2$

• Order $=2$
• Since Rate $=$ concentration/time,

$$k=\frac{\text{rate}}{[\mathrm{NO}]^2}$$
So units of $k$ are
$$\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^2} =\mathrm{L\,mol^{-1}\,s^{-1}}$$

(ii) Rate $=k[\mathrm{H_2O_2}][\mathrm{I^-}]$

• Order $=1+1=2$
• Units of $k$ are again

$$\mathrm{L\,mol^{-1}\,s^{-1}}$$

(iii) Rate $=k[\mathrm{CH_3CHO}]^{3/2}$

• Order $=\frac{3}{2}$
• Units of $k$:

$$k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{3/2}} =\mathrm{L^{1/2}\,mol^{-1/2}\,s^{-1}}$$

(iv) Rate $=k[\mathrm{C_2H_5Cl}]$

• Order $=1$
• Units of $k$:

$$k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol\,L^{-1}}}=\mathrm{s^{-1}}$$

(i) Cyanohydrin: Addition product formed when HCN adds to an aldehyde or ketone. Example:

$$\mathrm{RCHO + HCN \rightarrow RCH(OH)CN}$$

(ii) Acetal: Compound formed when an aldehyde reacts with two molecules of alcohol in presence of dry HCl. Example:

$$\mathrm{RCHO + 2ROH \xrightarrow{H^+} RCH(OR)_2 + H_2O}$$

(iii) Semicarbazone: Derivative formed when a carbonyl compound reacts with semicarbazide. Example:

$$\mathrm{R_2C=O + H_2N-NHCONH_2 \rightarrow R_2C=NNHCONH_2 + H_2O}$$

(iv) Aldol: \beta-hydroxy aldehyde or \beta-hydroxy ketone formed by aldol reaction. Example: two molecules of ethanal give aldol:

$$2\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}$$

(v) Hemiacetal: Product formed when one molecule of alcohol adds to an aldehyde. Example:

$$\mathrm{RCHO + ROH \rightleftharpoons RCH(OH)OR}$$

(vi) Oxime: Derivative formed when a carbonyl compound reacts with hydroxylamine. Example:

$$\mathrm{R_2C=O + NH_2OH \rightarrow R_2C=NOH + H_2O}$$

(vii) Ketal: Cyclic or acyclic derivative formed when a ketone reacts with alcohols; with ethylene glycol it gives a cyclic ketal. Example:

$$\mathrm{R_2C=O + 2ROH \xrightarrow{H^+} R_2C(OR)_2 + H_2O}$$

(viii) Imine: Compound containing the group $>C=NH$ formed by reaction of a carbonyl compound with ammonia. Example:

$$\mathrm{R_2C=O + NH_3 \rightarrow R_2C=NH + H_2O}$$

(ix) 2,4-DNP-derivative: Yellow, orange or red derivative formed when a carbonyl compound reacts with 2,4-dinitrophenylhydrazine. Example:

$$\mathrm{R_2C=O + H_2NNHC_6H_3(NO_2)_2 \rightarrow R_2C=NNHC_6H_3(NO_2)_2 + H_2O}$$

(x) Schiff's base: A substituted imine formed when a carbonyl compound reacts with a primary amine. Example:

$$\mathrm{R_2C=O + R'NH_2 \rightarrow R_2C=NR' + H_2O}$$

Remove electrons first from $ns$ orbital and then from $(n-1)d$ or $4f/5f$ orbitals as applicable.

• Cr$^{3+}$: Cr = $[Ar]3d^54s^1$; remove 3 electrons $$\rightarrow [Ar]3d^3$
• Cu$^+$: Cu = $[Ar]3d^{10}4s^1$; remove 1 electron $$\rightarrow [Ar]3d^{10}$
• Co$^{2+}$: Co = $[Ar]3d^74s^2$; remove 2 electrons $$\rightarrow [Ar]3d^7$
• Mn$^{2+}$: Mn = $[Ar]3d^54s^2$; remove 2 electrons $$\rightarrow [Ar]3d^5$
• Pm$^{3+}$: Pm = $[Xe]4f^56s^2$; remove 3 electrons $$\rightarrow [Xe]4f^4$
• Ce$^{4+}$: Ce = $[Xe]4f^15d^16s^2$; remove 4 electrons $$\rightarrow [Xe]4f^0 = [Xe]$
• Lu$^{2+}$: Lu = $[Xe]4f^{14}5d^16s^2$; remove 2 electrons $$\rightarrow [Xe]4f^{14}$
• Th$^{4+}$: Th = $[Rn]6d^27s^2$; remove 4 electrons $$\rightarrow [Rn]$

So the configurations are as listed in the final answer.

Classification depends on how many hydrogen atoms of ammonia are replaced by alkyl/aryl groups.

• (i) $(CH_3)_2CHNH_2$: one alkyl group attached through nitrogen, so primary amine.
• (ii) $CH_3(CH_2)_2NH_2$: one alkyl group attached through nitrogen, so primary amine.
• (iii) $CH_3NHCH(CH_3)_2$: nitrogen is attached to two alkyl groups and one hydrogen, so secondary amine.
• (iv) $(CH_3)_3CNH_2$: nitrogen is still attached to only one carbon group and two hydrogens, so primary amine.
• (v) $C_6H_5NHCH_3$: nitrogen is attached to an aryl group and a methyl group, so secondary amine.
• (vi) $(CH_3CH_2)_2NCH_3$: nitrogen is attached to three carbon groups, so tertiary amine.
• (vii) $m$-Br$C_6H_4NH_2$: $-NH_2$ group attached to benzene ring with only one carbon substituent on nitrogen, so primary amine.

Given: Names of coordination compounds. We apply IUPAC rules: write the coordination entity in square brackets with metal last, then counter-ions outside.

(i) tetraamminediaquacobalt(III) chloride

• Central metal: Co(III), i.e., Co³⁺
• Ligands: 4 NH₃ (tetraammine) + 2 H₂O (diaqua)
• Charge on complex ion: +3, so 3 Cl⁻ outside

$$[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3$$

(ii) potassium tetracyanidonickelate(II)

• Central metal: Ni(II), i.e., Ni²⁺ (anionic complex → potassium outside)
• Ligands: 4 CN⁻ (tetracyanido)
• Charge on complex ion: 2−4 = −2, so K₂ outside

$$K_2[\text{Ni}(\text{CN})_4]$$

(iii) tris(ethane-1,2-diamine)chromium(III) chloride

• Central metal: Cr(III), i.e., Cr³⁺
• Ligands: 3 en (tris(ethane-1,2-diamine))
• Charge on complex ion: +3, so 3 Cl⁻ outside

$$[\text{Cr}(\text{en})_3]\text{Cl}_3$$

(iv) amminebromidochloridonitrito-N-platinate(II)

• Central metal: Pt(II), anionic complex
• Ligands: NH₃ (ammine), Br⁻ (bromido), Cl⁻ (chlorido), NO₂⁻ bonded through N (nitrito-N)
• Charge: 2 − 1 − 1 − 1 = −1, so no outer cation shown (the name implies it is an anion; written as the anion)

$$[\text{Pt}(\text{NH}_3)\text{BrCl}(\text{NO}_2)]^-$$

(v) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

• Central metal: Pt(IV), i.e., Pt⁴⁺
• Ligands: 2 Cl⁻ (dichloro) + 2 en (bis(ethane-1,2-diamine))
• Charge on complex ion: 4 − 2 = +2, so 2 NO₃⁻ outside

$$[\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2$$

(vi) iron(III) hexacyanidoferrate(II)

• Two metal centres: Fe³⁺ (cation) and Fe²⁺ (in anionic complex)
• Complex anion: [Fe(CN)₆]⁴⁻
• Charge balance: 3 Fe³⁺ balanced by 4 [Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃

$$\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$$

A monosaccharide is the simplest carbohydrate unit. It cannot be hydrolysed further to give a simpler carbohydrate. It is a polyhydroxy aldehyde or polyhydroxy ketone.