Madhya Pradesh Board 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

Given/Concept: The standard electrode potential is always measured relative to the Standard Hydrogen Electrode (SHE), whose potential is taken as zero.

Method:

1. Set up a galvanic cell by connecting the Mg²⁺|Mg half-cell with the Standard Hydrogen Electrode (SHE).

• The cell is: $\text{Mg(s)} | \text{Mg}^{2+}(1\,\text{M}) || \text{H}^+(1\,\text{M}) | \text{H}_2(1\,\text{bar}) | \text{Pt(s)}$

2. Maintain all species at unit activity (1 M concentration for ions, 1 bar pressure for gases, 298 K).
3. Measure the EMF of the cell using a voltmeter.
4. Since Mg is a stronger reducing agent than H₂, Mg acts as the anode and SHE acts as the cathode.
5. The measured cell potential gives:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
$$E^\circ_{\text{cell}} = E^\circ_{\text{SHE}} - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$
$$E^\circ_{\text{cell}} = 0 - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$

The experimentally measured value is $E^\circ_{\text{cell}} = +2.37\,\text{V}$, so $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37\,\text{V}$.

Given: Standard electrode potentials:

• $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$
• $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$

Concept: A spontaneous reaction occurs when the cell EMF is positive, i.e., the metal with lower (more negative) electrode potential displaces the metal with higher electrode potential from its salt solution.

Working:
If copper sulphate is stored in a zinc pot, the following redox reaction would occur:
$$\text{Zn(s)} + \text{CuSO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{Cu(s)}$$

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34 - (-0.76) = +1.10\,\text{V}$$

Since $E^\circ_{\text{cell}} > 0$, the reaction is spontaneous. Zinc will dissolve and copper will be deposited.

Conclusion: No, copper sulphate solution cannot be stored in a zinc pot because zinc is more reactive than copper and will displace copper from the copper sulphate solution, corroding the zinc pot.

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:

Given:

• Initial concentration, $[R]_1 = 0.03\,\text{M}$
• Final concentration, $[R]_2 = 0.02\,\text{M}$
• Time interval, $\Delta t = 25\,\text{min}$

Formula:
$$\text{Average rate} = -\frac{\Delta[R]}{\Delta t} = -\frac{[R]_2 - [R]_1}{\Delta t}$$

In minutes:
$$\text{Average rate} = -\frac{(0.02 - 0.03)\,\text{mol L}^{-1}}{25\,\text{min}} = -\frac{-0.01}{25}\,\text{mol L}^{-1}\text{min}^{-1}$$
$$= 4 \times 10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}$$

In seconds (converting: $25\,\text{min} = 25 \times 60 = 1500\,\text{s}$):
$$\text{Average rate} = -\frac{(0.02 - 0.03)}{1500}\,\text{mol L}^{-1}\text{s}^{-1} = \frac{0.01}{1500}$$
$$= 6.66 \times 10^{-6}\,\text{mol L}^{-1}\text{s}^{-1}$$

Answer: Average rate $= 4 \times 10^{-4}\,\text{mol L}^{-1}\text{min}^{-1} = 6.66 \times 10^{-6}\,\text{mol L}^{-1}\text{s}^{-1}$

(i) α-Methoxypropionaldehyde
α-carbon is C-2 of propionaldehyde (propanal). A methoxy (–OCH₃) group is attached at C-2.
$$\text{CH}_3-\underset{\displaystyle|}{{\overset{\displaystyle\text{OCH}_3}{\text{CH}}}}-\text{CHO}$$
Structure: CH₃–CH(OCH₃)–CHO

(ii) 3-Hydroxybutanal
Butanal with –OH at C-3.
$$\text{CH}_3-\underset{\displaystyle|}{{\overset{\displaystyle\text{OH}}{\text{CH}}}}-\text{CH}_2-\text{CHO}$$
Structure: CH₃–CH(OH)–CH₂–CHO

(iii) 2-Hydroxycyclopentane carbaldehyde
A cyclopentane ring with –CHO at C-1 and –OH at C-2.
Structure: Cyclopentane ring with –CHO substituent at C-1 and –OH at C-2 (both on adjacent carbons of the ring).

(iv) 4-Oxopentanal
A five-carbon chain with an aldehyde (–CHO) at C-1 and a keto (=O) group at C-4.
$$\text{OHC}-\text{CH}_2-\text{CH}_2-\underset{\displaystyle\|}{\overset{\displaystyle}{\text{C}}}-\text{CH}_3$$
Structure: OHC–CH₂–CH₂–CO–CH₃

(v) Di-sec. butyl ketone
sec-Butyl group = CH₃CH₂CH(CH₃)–
Two sec-butyl groups on either side of the carbonyl.
$$\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)-\overset{\displaystyle O}{\overset{\displaystyle\|}{\text{C}}}-\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$$
Structure: (CH₃CH₂CHCH₃)–CO–(CHCH₃CH₂CH₃)

(vi) 4-Fluoroacetophenone
Acetophenone (methyl phenyl ketone) with –F at the para position of the benzene ring.
$$\text{F}-\text{C}_6\text{H}_4-\overset{\displaystyle O}{\overset{\displaystyle\|}{\text{C}}}-\text{CH}_3 \quad (\text{F at para position})$$
Structure: p-F–C₆H₄–CO–CH₃

Given: Silver (Ag, Z = 47) has ground state electronic configuration [Kr] $4d^{10}$ $5s^1$.

Concept: A transition element is defined as one which has an incompletely filled $d$ orbital in its ground state OR in any of its commonly occurring oxidation states.

Working:
Although Ag has completely filled $4d^{10}$ orbitals in its ground state, it can exhibit a +2 oxidation state (Ag²⁺). In the +2 state, the electronic configuration becomes [Kr] $4d^9$, which has an incompletely filled $d$ orbital.

Conclusion: Since Ag can exist in the +2 oxidation state with an incompletely filled $d$ subshell, it qualifies as a transition element.

Concept: An amine is classified based on the number of hydrogen atoms of NH₃ replaced by alkyl/aryl groups.

• Primary (1°): one H replaced → R–NH₂
• Secondary (2°): two H replaced → R₂NH
• Tertiary (3°): three H replaced → R₃N

(i) The structure shown in the image is that of a cyclic secondary amine (cyclohexylamine type) or an N-substituted compound. Based on standard NCERT context, image (i) represents a compound where nitrogen bears two carbon substituents and one H → Secondary amine.

(ii) The structure shown in image (ii) represents a compound where nitrogen bears three carbon substituents and no H → Tertiary amine.

(iii) $(C_2H_5)_2CHNH_2$
The nitrogen atom is bonded to two H atoms and one carbon group (the $–CH(C_2H_5)_2$ group). Since only one H of NH₃ is replaced by an alkyl group, this is a Primary amine (1°).

(iv) $(C_2H_5)_2NH$
The nitrogen atom is bonded to two ethyl groups and one H atom. Two H atoms of NH₃ are replaced → Secondary amine (2°).

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

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

Given: Glucose/sucrose vs. cyclohexane/benzene — all are six-membered ring compounds, yet solubility in water differs.

Concept: 'Like dissolves like' — polar solutes dissolve in polar solvents (water), non-polar solutes do not.

Explanation:

• Glucose and sucrose contain a large number of –OH (hydroxyl) groups in their structures. These –OH groups form hydrogen bonds with water molecules (H-bonding between O–H of solute and O–H of water). This strong interaction makes them readily soluble in water.
• Cyclohexane and benzene are non-polar hydrocarbons. They have no –OH or any other polar group capable of forming hydrogen bonds with water. Hence they are insoluble in water.

Conclusion: The presence of multiple –OH groups in glucose and sucrose enables extensive hydrogen bonding with water, making them water-soluble, whereas the non-polar nature of cyclohexane and benzene prevents any such interaction.