Chemistry

Cumene (isopropylbenzene) $\xrightarrow[(ii)\ H_2O,\ H^+]{(i)\ O_2}$ Phenol, so (i) matches (i) $O_2$; (ii) $H_2O/H^+$. $CH_3COOH \xrightarrow[(ii)\ H_2,\ catalyst]{(i)\ CH_3OH,\ H^+} CH_3CH_2OH$ (esterification followed by hydrogenolysis), so (ii) matches (i) $CH_3OH, H^+$; (ii) $H_2$, catalyst. $CH_3CH_2CH_2OH \xrightarrow[(ii)\ H^+/H_2O]{(i)\ conc.\ H_2SO_4,\ \Delta} CH_3-CH(OH)-CH_3$ (dehydration to propene, then Markovnikov hydration), so (iii) matches (i) conc. $H_2SO_4, \Delta$; (ii) $H^+/H_2O$. Benzene $\xrightarrow[(ii)\ NaOH,\ \Delta;\ (iii)\ H^+]{(i)\ Oleum}$ Phenol (sulphonation, alkali fusion, acidification), so (iv) matches (i) oleum; (ii) NaOH, $\Delta$; (iii) $H^+$. Correct match: i-b, ii-c, iii-d, iv-a.

$C_2H_6 \xrightarrow[UV\ light]{Cl_2} CH_3CH_2Cl$ (X) $CH_3CH_2Cl \xrightarrow{NH_3} CH_3CH_2NH_2$ (Y) $CH_3CH_2NH_2 \xrightarrow[(ii)\ H_2O]{(i)\ NaNO_2 + HCl} CH_3CH_2OH$ (Z) So the major product Z is $C_2H_5OH$.

$BiO(OH)_{(s)} \rightleftharpoons BiO^+_{(aq)} + OH^-_{(aq)}$; at equilibrium $[BiO^+] = [OH^-] = S$ (S = solubility) $K = K_{sp} = S^2 = 4 \times 10^{-10}$ (AB type) $S = 2 \times 10^{-5}$ M, so $[OH^-] = 2 \times 10^{-5}$ M $pOH = 5 - \log 2 = 5 - 0.3010 = 4.699$ $pH = 14 - 4.699 = 9.301$

$CO_2(g) + C(s) \rightarrow 2CO(g)$ Initial (t = 0): $CO_2$ = 1 L, CO = 0 After reaction: $CO_2 = (1 - x)$, CO = $2x$ $V_{total} = (1 - x) + 2x = 1 + x = 1.4 \Rightarrow x = 0.4$ $V_{CO} = 2x = 0.8\ dm^3$ and $V_{CO_2} = 1 - x = 0.6\ dm^3$

n = 2, l = 1 $\rightarrow$ 2p n = 4, l = 0 $\rightarrow$ 4s n = 5, l = 3 $\rightarrow$ 5f n = 3, l = 2 $\rightarrow$ 3d Correct match: i-b, ii-c, iii-d, iv-a.

Benzene $+ 6Cl_2 \xrightarrow[dark,\ cold]{Anhydr.\ AlCl_3} C_6Cl_6$ (X, hexachlorobenzene – electrophilic substitution of all six H) Benzene $+ 3Cl_2 \xrightarrow[500K]{UV} C_6H_6Cl_6$ (Y, benzene hexachloride – free-radical addition) Number of chlorine atoms in X and Y are 6 and 6 respectively.

$CH_3CH_2CH_2OH \xrightarrow{PCl_5} CH_3CH_2CH_2Cl + POCl_3 + HCl$, so X = $POCl_3$ $CH_3CH_2CH_2Cl \xrightarrow[\Delta]{alc.\ KOH} CH_3-CH=CH_2$ (Y) $CH_3-CH=CH_2 \xrightarrow[(C_6H_5CO)_2O_2]{HBr} CH_3CH_2CH_2Br$ (Z, anti-Markovnikov/peroxide effect) So X = $POCl_3$ and Z = $CH_3CH_2CH_2-Br$.

$V_2O_5$: preparation of $H_2SO_4$ from $SO_2$ (Contact process) Fe: $N_2 + 3H_2 \rightarrow 2NH_3$ (Haber process) $PdCl_2$: oxidation of ethyne to ethanal (Wacker process) Ni complex: polymerisation of alkynes Correct match: i-c, ii-a, iii-d, iv-b.

In $ClF_3$, Cl has 3 bond pairs and 2 lone pairs, so hybridisation is $sp^3d$. The two lone pairs occupy equatorial positions, giving a T-shaped geometry. So $ClF_3$ is T-shaped with two lone pairs on the central Cl atom.

Cell reaction: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$ ; $P_{H_2}$ = 2 atm, $[H^+] = 2 \times 10^{-2}$ M $E_{OP} = 0 - \frac{0.059}{2} \log \frac{(2 \times 10^{-2})^2}{2}$ $= -\frac{0.059}{2}[\log(2 \times 10^{-4})]$ $= -\frac{0.059}{2}[0.3 - 4]$ $= -\frac{0.059}{2}[-3.7]$ $= +0.109$ V

Unit of k = $mol^{1-n}\,L^{n-1}\,time^{-1}$ Zero order (n = 0): $mol\,L^{-1}\,s^{-1}$ First order (n = 1): $s^{-1}$ Second order (n = 2): $mol^{-1}\,L\,s^{-1}$ Third order (n = 3): $mol^{-2}\,L^2\,s^{-1}$ Correct match: i-d, ii-c, iii-a, iv-b.

$_{22}Ti \rightarrow [Ar]\,4s^2\,3d^2$ ; $Ti^{2+} \rightarrow [Ar]\,3d^2$ Number of unpaired electrons, n = 2 Spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.84$ BM

Benzene $\xrightarrow[AlCl_3]{CH_3Cl}$ Toluene (W) (Friedel-Crafts alkylation) Toluene $\xrightarrow[dil.\ H_2SO_4,\ warm]{dil.\ HNO_3}$ o-nitrotoluene (X) + p-nitrotoluene (Y) X and Y are positional isomers with different boiling points, so they can be separated by fractional distillation.

Power rating of bulb = 150 W Energy converted into light per second $= 150 \times \frac{8}{100} = 12$ J Number of photons emitted per second $= \frac{Total\ energy}{Energy\ of\ one\ photon} = \frac{12}{4.42 \times 10^{-19}}$ $= 2.71 \times 10^{19}$

Acetate + dil. $H_2SO_4 \rightarrow CH_3COOH$ Acetic acid vapours have the smell of vinegar and turn blue litmus red, so the anion is acetate, $CH_3COO^-$.

$R-C \equiv N \rightarrow R-CH_2-NH_2$ can be carried out by: (i) $LiAlH_4$; (ii) $H_2O$ $H_2/Ni$ (catalytic hydrogenation) $Na(Hg)/C_2H_5OH$ (Mendius reduction) $Sn + HCl$ reduces nitro compounds, and $Br_2/aq.\ NaOH$ is the Hofmann bromamide reagent for amides – neither reduces nitriles to primary amines. So the correct reagents are I, III and IV only.

Oxygen shows variable oxidation states, not only $-2$: $H_2O$: $O^{2-}$, $-2$ $H_2O_2$: $O_2^{2-}$, $-1$ $KO_2$: $O_2^{-}$, $-\frac{1}{2}$ $O_2$: 0 $OF_2$: $+2$ $O_2F_2$: $+1$ So the statement 'Oxygen exhibits only $-2$ oxidation state' is incorrect.

$Ce \rightarrow [_{54}Xe]\,6s^2\,5d^1\,4f^1$ $Ce^{4+} \rightarrow [_{54}Xe]\,4f^0$ After losing one more electron (from $Ce^{3+}$), cerium acquires the stable $4f^0$ (noble gas) configuration, hence it shows +4 oxidation state.

During Lassaigne's test, the organic compound is fused with sodium metal, so the elements present in the compound (N, S, halogens) are converted from covalent form to ionic form (NaCN, $Na_2S$, NaX).

Molecular formula of urea = $NH_2CONH_2$; atomicity w.r.t. H = 4 Moles of urea $= \frac{5.4}{60}$ Moles of H $= \frac{5.4}{60} \times 4$ No. of H-atoms $= \frac{5.4}{60} \times 4 \times 6.022 \times 10^{23} = 2.168 \times 10^{23}$

Metamers have the same molecular formula but different alkyl groups on either side of a polyvalent functional group. $CH_3OCH_2CH_2CH_3$ (methyl propyl ether) and $CH_3CH_2OCH_2CH_3$ (diethyl ether) are metamers.

Both N and O are 2nd period elements, so they form only p$\pi$-p$\pi$ bonds but not d$\pi$-p$\pi$ bonds due to absence of d-orbitals. So 'Nitrogen can form d$\pi$-p$\pi$ bond with oxygen' is the incorrect statement.

$\underset{(colourless)}{HPh} \rightleftharpoons H^+ + \underset{(pink)}{Ph^-}$ In acidic medium, due to common ion effect, the equilibrium lies to the left, so the solution is colourless. In alkaline medium, ionisation of HPh increases, so the concentration of $Ph^-$ ions increases, which imparts pink colour to the solution. Note: ALLEN's key gives 'colourless to pink' (oxalic acid in flask, NaOH from burette); the official key used in this CSV gives 'pink to colourless' (NaOH in flask).

$C_2H_4$ ($H_2C=CH_2$): 5 $\sigma$ bonds, 1 $\pi$ bond $C_2H_2$ ($H-C \equiv C-H$): 3 $\sigma$ bonds, 2 $\pi$ bonds $CH_4$: 4 $\sigma$ bonds $NH_3$: 3 $\sigma$ bonds, 1 lone pair Correct match: i-d, ii-a, iii-c, iv-b.

Heat absorbed by the system, q = +500 J Work done by the system, w = $-200$ J $\Delta U = q + w = 500 - 200 = 300$ J