NEB Class 12 Science Chemistry Question Paper 2077 Nepal

1. Hybrid orbitals are formed by mixing of atomic orbitals of nearly equal energy belonging to the same atom; the number of hybrid orbitals formed equals the number of atomic orbitals mixed.
2. Hybrid orbitals are equivalent in energy and shape, are directed in space to give minimum repulsion (definite geometry), and form stronger sigma bonds than pure atomic orbitals due to greater directional overlap.

(i) Primary standard solution: A solution prepared by directly dissolving an accurately weighed amount of a primary standard substance (one that is pure, stable, non-hygroscopic, with high equivalent weight, e.g. oxalic acid, $Na_2CO_3$) in a known volume; its concentration is known exactly without further standardisation.

(ii) Acidimetry: A volumetric (titrimetric) method of determining the strength (concentration) of an acid by titrating it against a standard solution of a base.

Blue vitriol is hydrated copper(II) sulphate, $CuSO_4\cdot5H_2O$. On heating it loses water of crystallisation in steps:

On strong heating (above ~$720^\circ C$) anhydrous $CuSO_4$ decomposes: $2CuSO_4 \rightarrow 2CuO + 2SO_2 + O_2$.

(i) Aldol Condensation: Two molecules of acetaldehyde combine in the presence of dilute alkali:

which on heating loses water to give but-2-enal.

(ii) Rosenmund's reduction: Acid chloride is reduced to aldehyde with $H_2$ over Pd poisoned with $BaSO_4$:

Primary amine: $CH_3CH_2CH_2NH_2$ — propan-1-amine (1-propanamine). (Also $CH_3CH(NH_2)CH_3$ = propan-2-amine.)

Secondary amine: $CH_3CH_2-NH-CH_3$ — N-methylethanamine (N-methylethylamine).

(i) Titration error: The difference between the actual equivalence point and the observed end point of a titration (e.g. due to the indicator changing colour slightly before or after the true equivalence point).

(ii) Unknown solution: The solution whose concentration (strength) is to be determined by titrating it against a standard solution of known concentration.

Mixing calculation: Let $x$ L of $10\,M$ and $(1-x)$ L of $3\,M$ HCl give $1$ L of $6\,M$.

Moles balance: $10x+3(1-x)=6\times1\Rightarrow 7x+3=6\Rightarrow 7x=3\Rightarrow x=\dfrac{3}{7}\,\text{L}\approx 0.4286\,\text{L}$.

So $\approx 0.429\,\text{L}$ of $10\,M$ HCl and $\approx 0.571\,\text{L}$ of $3\,M$ HCl.

Enthalpy of combustion: The enthalpy change when one mole of a substance is completely burnt (oxidised) in excess oxygen under standard conditions.

Combustion of glucose: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$.

$\Delta H_{comb}=[6\,\Delta H_f(CO_2)+6\,\Delta H_f(H_2O)]-[\Delta H_f(C_6H_{12}O_6)+6\,\Delta H_f(O_2)]$.

$=[6(-395)+6(-269.4)]-[1169+0]$

$=[-2370-1616.4]-1169=-3986.4-1169=-5155.4\ \text{KJ mol}^{-1}$.

(Note: the printed heat of formation of glucose, $+1169$, appears to have a sign error; with the values as given, $\Delta H_{comb}\approx -5155.4\ \text{KJ mol}^{-1}$.)

(i) From 1,1,1-trichloroethane: Hydrolysis with aqueous alkali then acidification: $CH_3CCl_3 + 2H_2O \rightarrow CH_3COOH + 3HCl$ (the $-CCl_3$ becomes $-COOH$).

(ii) From methyl magnesium iodide: $CH_3MgI + CO_2 \rightarrow CH_3COOMgI \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)I$.

(iii) From ethanenitrile (acetonitrile): $CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$ (acid hydrolysis).

Ethanoic acid → methanoic acid: This is a step-down (descent) of the series. One route: $CH_3COOH \xrightarrow{NH_3,\Delta} CH_3CONH_2 \xrightarrow{Br_2/KOH (\text{Hofmann})} CH_3NH_2$; then $CH_3NH_2 \xrightarrow{HNO_2} CH_3OH \xrightarrow{[O]} HCOOH$ (methanol oxidised to methanoic acid).

Structure & IUPAC of $C_4H_{10}O$: $CH_3CH_2CH_2CH_2OH$ — butan-1-ol. (Other isomers: butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol, plus ethers like diethyl ether.)

Victor Meyer's method (1° vs 2° alcohol): Treat alcohol with HI to form alkyl iodide, then with AgNO₂ to form nitroalkane, then with $HNO_2$ (nitrous acid) and finally NaOH:

• Primary (propan-1-ol) → nitroalkane gives red colour with NaOH (nitrolic acid).
• Secondary (propan-2-ol) → gives blue colour (pseudonitrole). (Tertiary gives no colour.)

(i) Oxo-process (hydroformylation): $CH_2=CH_2 + CO + H_2 \xrightarrow{Co\ catalyst} CH_3CH_2CHO$ (alkene + CO + H₂ → aldehyde).

(ii) Baeyer's test: Alkenes decolourise cold dilute alkaline $KMnO_4$ giving a diol: $CH_2=CH_2 + [O] + H_2O \xrightarrow{KMnO_4} CH_2(OH)CH_2(OH)$ (test for unsaturation).

Propan-2-ol → propan-1-ol: $CH_3CH(OH)CH_3 \xrightarrow{conc.\ H_2SO_4,\Delta} CH_3CH=CH_2 \xrightarrow{HBr\ (\text{peroxide, anti-Markovnikov})} CH_3CH_2CH_2Br \xrightarrow{aq.\ KOH} CH_3CH_2CH_2OH$.

(i) Rate law equation: An experimentally determined expression relating reaction rate to the molar concentrations of reactants raised to appropriate powers, e.g. $\text{rate}=k[A]^x[B]^y$.

(ii) Half-life: The time required for the concentration of a reactant to fall to half of its initial value.

Order vs molecularity: Order is the experimentally determined sum of powers of concentration terms in the rate law and may be fractional or zero; molecularity is the number of reacting species that collide in an elementary step, is always a whole number, and is a theoretical concept for elementary reactions.

Order from data (rate $=k[A]^x[B]^y$): Comparing Expt 1 and 4 ($[B]$ const): $\dfrac{2.4\times10^{-2}}{6.0\times10^{-3}}=4=(0.4/0.1)^x=4^x\Rightarrow x=1$. The printed values for B-dependence are inconsistent (Expt 2 vs 3 show rate decreasing with increasing [B]); taking the commonly intended data, order in B = 2 and overall rate $=k[A][B]^2$ with $k$ from Expt 1: $6.0\times10^{-3}=k(0.1)(0.1)^2\Rightarrow k=6$. Then at $[A]=0.5,[B]=0.2$: rate $=6(0.5)(0.2)^2=6(0.5)(0.04)=0.12\ \text{mol L}^{-1}\text{s}^{-1}$.