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Chemistry
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An increase in equivalent conductance of a strong electrolyte with dilution is mainly due toA strong electrolyte is almost completely ionised at all concentrations, so dilution does not increase the number of ions.
In a concentrated solution the ions are close together and strong inter-ionic attractions slow them down.
On dilution the ions move farther apart, the inter-ionic attractions decrease and the ions move more freely, i.e., their mobility increases.
This is why the equivalent (molar) conductance of a strong electrolyte rises only slightly and linearly with $\sqrt{c}$ on dilution (Debye–Hückel–Onsager equation: $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$).
For a weak electrolyte, in contrast, the increase is mainly due to an increase in the number of ions.
Hence the increase is mainly due to an increase in the ionic mobility of ions. -
25.3 g of sodium carbonate, $Na_2CO_3$, is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $Na^+$, and carbonate ions, $CO_3^{2-}$, are respectively (Molar mass of $Na_2CO_3$ = 106 g $mol^{-1}$)Moles of $Na_2CO_3$ = $\frac{25.3}{106} = 0.2387$ mol
Molarity of $Na_2CO_3$ = $\frac{0.2387}{250}\times1000 = 0.955$ M
Complete dissociation: $Na_2CO_3 \rightarrow 2Na^+ + CO_3^{2-}$
$[Na^+] = 2\times0.955 = 1.910$ M
$[CO_3^{2-}] = 0.955$ M -
Property of the alkaline earth metals that increases with their atomic number isDown group 2 the atomic size increases, so ionisation energy and electronegativity both decrease.
Hydroxides: the lattice enthalpy decreases down the group more rapidly than the hydration enthalpy (the small $OH^-$ ion makes lattice enthalpy very sensitive to cation size), so the solubility increases: $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.
Sulphates: the lattice enthalpy stays nearly constant (large $SO_4^{2-}$ ion) while the hydration enthalpy falls, so the solubility decreases: $BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.
Hence the property that increases with atomic number is the solubility of their hydroxides in water. -
Which of the following pairs has the same size?Zr (4d series) and Hf (5d series) belong to the same group; normally the size should increase from Zr to Hf.
But the 14 lanthanoids come between them. The 4f electrons shield the nuclear charge poorly, so the atomic and ionic radii decrease steadily across the lanthanoids (lanthanoid contraction).
This contraction cancels the expected increase in size from the 4d to the 5d series.
As a result $Zr^{4+}$ and $Hf^{4+}$ have almost identical radii (about 72 pm and 71 pm), and the two elements have very similar properties.
$Ti^{4+}$ (3d series) is clearly smaller than $Zr^{4+}$, and $Fe^{2+}$ and $Ni^{2+}$ differ because of increasing nuclear charge across the period.
Hence the pair with the same size is $Zr^{4+}$, $Hf^{4+}$. -
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?A non-volatile solute lowers the vapour pressure of the solvent; the lowering is proportional to the mole fraction of solute particles (Raoult's law).
Adding water dilutes the solution, so the mole fraction of solute particles decreases, the mole fraction of water increases and the vapour pressure rises.
Adding NaCl or $Na_2SO_4$ increases the number of solute particles, which lowers the vapour pressure further.
Adding more 1.00 molal KI solution leaves the concentration, and hence the vapour pressure, unchanged.
Hence the vapour pressure increases on addition of water. -
For the reaction $N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g)$ the value of rate of disappearance of $N_2O_5$ is given as $6.25\times10^{-3}$ mol $L^{-1}s^{-1}$. The rate of formation of $NO_2$ and $O_2$ is given respectively asFor $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt}$
$\frac{d[NO_2]}{dt} = 2\times\left(-\frac{d[N_2O_5]}{dt}\right) = 2\times6.25\times10^{-3}$
$\frac{d[NO_2]}{dt} = 1.25\times10^{-2}$ mol $L^{-1}s^{-1}$
$\frac{d[O_2]}{dt} = \frac{1}{2}\times\left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2}\times6.25\times10^{-3}$
$\frac{d[O_2]}{dt} = 3.125\times10^{-3}$ mol $L^{-1}s^{-1}$ -
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at $25^\circ C$. The value of standard Gibbs energy, $\Delta G^\circ$, will be (F = 96500 C $mol^{-1}$)Cell reaction: $Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$
Copper loses two electrons, so n = 2.
$\Delta G^\circ = -nFE^\circ_{cell}$
$\Delta G^\circ = -2\times96500\times0.46$
$\Delta G^\circ = -88780$ J
$\Delta G^\circ \approx -89.0$ kJ -
For an endothermic reaction, energy of activation is $E_a$ and enthalpy of reaction is $\Delta H$ (both of these in kJ/mol). Minimum value of $E_a$ will beIn an endothermic reaction the products lie at a higher energy than the reactants by $\Delta H$.
The reactants must climb to the top of the energy barrier (the activated complex), which lies above the products.
From the energy profile: $E_a(\text{forward}) = \Delta H + E_a(\text{backward})$
$E_a(\text{backward})$ is always a positive quantity.
Therefore $E_a(\text{forward}) > \Delta H$; the activation energy must be more than $\Delta H$. -
Which one of the following ions has electronic configuration $[Ar]3d^6$? (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)When a transition metal forms a cation, the 4s electrons are removed before the 3d electrons.
Co (27): $[Ar]3d^74s^2$; $Co^{3+}$: remove two 4s and one 3d electron, giving $[Ar]3d^6$
Ni (28): $[Ar]3d^84s^2$; $Ni^{3+}$: $[Ar]3d^7$
Mn (25): $[Ar]3d^54s^2$; $Mn^{3+}$: $[Ar]3d^4$
Fe (26): $[Ar]3d^64s^2$; $Fe^{3+}$: $[Ar]3d^5$
Hence the ion with $[Ar]3d^6$ is $Co^{3+}$. -
In which of the following equilibria are $K_c$ and $K_p$ not equal?$K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants. $K_p = K_c$ only when $\Delta n_g = 0$.
$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$: $\Delta n_g = 2 - 2 = 0$
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$: $\Delta n_g = 2 - 2 = 0$
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$: $\Delta n_g = 2 - 2 = 0$
$2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$: carbon is a solid and is not counted, so $\Delta n_g = 2 - 1 = 1$ and $K_p = K_c(RT)$
Hence $K_c$ and $K_p$ are not equal for $2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$.
Note: the worked line in the source PDF treats carbon as a gas and gets $\Delta n_g = 0$, which contradicts its own answer; the equation as printed in the question, with C(s), is used here. -
Which of the following ions will exhibit colour in aqueous solutions?An ion is coloured in aqueous solution when it has unpaired electrons in d (or f) orbitals that can undergo d–d (or f–f) transitions by absorbing visible light.
$Sc^{3+}$: $[Ar]3d^0$, no d electrons, colourless.
$La^{3+}$: $[Xe]4f^0$, no f electrons, colourless.
$Lu^{3+}$: $[Xe]4f^{14}$, completely filled, no unpaired electron, colourless.
$Ti^{3+}$: $[Ar]3d^1$, one unpaired d electron; the $t_{2g} \rightarrow e_g$ transition absorbs visible light, so $[Ti(H_2O)_6]^{3+}$ is purple.
Hence $Ti^{3+}$ exhibits colour. -
Which one of the following compounds is a peroxide?A peroxide contains the peroxide ion $O_2^{2-}$ (an –O–O– linkage) with oxygen in the −1 oxidation state.
$BaO_2$: Ba is always +2, so the two oxygens together carry −2, i.e., each O is −1. It contains $O_2^{2-}$ and is a true peroxide (it gives $H_2O_2$ with dilute acid).
$KO_2$: K is +1, so the unit is $O_2^-$, the superoxide ion (oxidation state $-\frac{1}{2}$).
$NO_2$ and $MnO_2$ are normal dioxides with oxygen in the −2 state (N and Mn are +4).
Hence the peroxide is $BaO_2$. -
In which of the following pairs of molecules/ions, the central atoms have $sp^2$ hybridization?Steric number = number of σ bonds + number of lone pairs on the central atom; steric number 3 means $sp^2$ and 4 means $sp^3$.
$BF_3$: 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NO_2^-$: 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NH_2^-$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$NH_3$: 3 σ bonds + 1 lone pair = 4, so $sp^3$.
$H_2O$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
Hence both central atoms are $sp^2$ hybridised in $BF_3$ and $NO_2^-$. -
The correct order of the decreasing ionic radii among the following isoelectronic species isAll four ions have 18 electrons (isoelectronic), but different nuclear charges: $S^{2-}$ (Z = 16), $Cl^-$ (Z = 17), $K^+$ (Z = 19), $Ca^{2+}$ (Z = 20).
For isoelectronic species, the greater the nuclear charge, the more strongly the same number of electrons is pulled in and the smaller the ion.
So the radius decreases as Z increases: $S^{2-}$ (184 pm) > $Cl^-$ (181 pm) > $K^+$ (138 pm) > $Ca^{2+}$ (100 pm).
Decreasing order of ionic radii: $S^{2-} > Cl^- > K^+ > Ca^{2+}$ -
The number of atoms in 0.1 mol of a triatomic gas is ($N_A = 6.02\times10^{23}\ mol^{-1}$)Number of molecules in 0.1 mol = $0.1\times N_A = 0.1\times6.02\times10^{23} = 6.02\times10^{22}$
Each molecule of a triatomic gas contains 3 atoms.
Number of atoms = $3\times6.02\times10^{22}$
$= 1.806\times10^{23}$ -
The existence of two different coloured complexes with the composition of $[Co(NH_3)_4Cl_2]^+$ is due to$[Co(NH_3)_4Cl_2]^+$ is an octahedral complex of the type $[MA_4B_2]$.
The two $Cl^-$ ligands can occupy adjacent positions ($90^\circ$ apart, cis isomer) or opposite positions ($180^\circ$ apart, trans isomer).
These are geometrical isomers; the cis form is violet and the trans form is green.
Ionisation isomerism needs exchange of a ligand with a counter ion, linkage isomerism needs an ambidentate ligand, and coordination isomerism needs both cation and anion to be complexes; none applies to this single complex ion.
Hence the two coloured forms are due to geometrical isomerism. -
Oxidation states of P in $H_4P_2O_5$, $H_4P_2O_6$, $H_4P_2O_7$ are respectivelyTake H = +1, O = −2 and let the oxidation state of P be x.
$H_4P_2O_5$: $4(+1) + 2x + 5(-2) = 0$, so $2x = 6$, x = +3
$H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0$, so $2x = 8$, x = +4
$H_4P_2O_7$: $4(+1) + 2x + 7(-2) = 0$, so $2x = 10$, x = +5
The oxidation states are +3, +4, +5 respectively. -
Which one of the following molecular hydrides acts as a Lewis acid?Molecular hydrides are classified as electron-deficient, electron-precise and electron-rich.
$B_2H_6$ is electron-deficient: it has too few electrons to write a conventional Lewis structure (it has 3-centre–2-electron bonds), so it accepts electron pairs and acts as a Lewis acid, e.g., $B_2H_6 + 2NH_3 \rightarrow 2BH_3\cdot NH_3$ type adducts and $B_2H_6 + 2CO \rightarrow 2BH_3\cdot CO$.
$CH_4$ is electron-precise; it is neither a Lewis acid nor a Lewis base.
$NH_3$ and $H_2O$ are electron-rich, with lone pairs on the central atom; they act as Lewis bases.
Hence the hydride that acts as a Lewis acid is $B_2H_6$. -
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons byWhen higher alkanes are heated to a high temperature in the absence of air (often with a catalyst), the large molecules break down into smaller alkanes and alkenes.
This thermal decomposition is called pyrolysis or cracking, e.g., $C_6H_{14} \xrightarrow{773\ K} C_6H_{12} + H_2$, $C_4H_8 + C_2H_6$, $C_3H_6 + C_2H_4 + CH_4$
Smaller hydrocarbons are gases, so liquid hydrocarbons are converted into a mixture of gaseous hydrocarbons (as in the manufacture of oil gas from kerosene).
Hydrolysis does not affect hydrocarbons, oxidation gives $CO_2$ and water, and distillation only separates the components without breaking them.
Hence the process is cracking. -
Which one of the following is employed as a tranquilizer drug?Tranquilizers are neurologically active drugs used to treat stress, anxiety and mild or severe mental diseases; they relieve tension by inducing a sense of well-being.
Valium (diazepam) is a benzodiazepine tranquilizer; others are chlordiazepoxide, meprobamate, equanil and serotonin-related drugs.
Mifepristone is a synthetic steroid used as an antifertility ('morning after') drug.
Promethazine is an antihistamine, and naproxen is a non-steroidal analgesic (anti-inflammatory).
Hence the tranquilizer is valium. -
Which one of the following does not exhibit the phenomenon of mutarotation?Mutarotation is the change in specific rotation of a sugar solution with time; it needs a free hemiacetal or hemiketal (anomeric –OH) group so that the ring can open and re-close to interconvert the α and β forms.
Fructose (a monosaccharide), maltose and lactose (reducing disaccharides) all have a free anomeric –OH group, so they show mutarotation.
In sucrose, C-1 of α-D-glucose is linked to C-2 of β-D-fructose; both anomeric carbons are used up in the glycosidic bond.
So sucrose has no free hemiacetal or hemiketal group; it is a non-reducing sugar and cannot show mutarotation.
Hence (+) sucrose does not exhibit mutarotation. -
Which of the following structures represents Neoprene polymer?Neoprene (polychloroprene) is a synthetic rubber formed by free radical addition polymerisation of chloroprene, 2-chloro-1,3-butadiene.
$n\ CH_2=C(Cl)-CH=CH_2 \rightarrow -[CH_2-C(Cl)=CH-CH_2]_n-$
Polymerisation occurs by 1,4-addition, leaving one double bond in each repeating unit.
$-[CH(C_6H_5)-CH_2]_n-$ is polystyrene, $-[CH_2-CH(CN)]_n-$ is polyacrylonitrile (orlon) and $-[CH_2-CH(Cl)]_n-$ is polyvinyl chloride (PVC).
Hence neoprene is $-[CH_2-C(Cl)=CH-CH_2]_n-$. -
Which one of the following compounds has the most acidic nature?Acidity depends on the stability of the anion formed after loss of $H^+$.
Phenol gives the phenoxide ion, in which the negative charge on oxygen is delocalised over the benzene ring by resonance; this stabilises the ion and makes phenol acidic ($pK_a \approx 10$).
In benzyl alcohol the –OH is on an $sp^3$ carbon, so the alkoxide ion formed cannot be stabilised by resonance with the ring.
Cyclohexanol and dicyclohexylmethanol are ordinary alcohols; their alkoxide ions are destabilised by the +I effect of the alkyl groups ($pK_a \approx 16$–18).
Hence phenol is the most acidic. -
Which of the following statements about primary amines is 'False'?Alkyl groups release electrons (+I effect) and increase the electron density on nitrogen, so alkyl amines are stronger bases than ammonia (true).
In aryl amines the lone pair on nitrogen is delocalised into the benzene ring, so they are weaker bases than alkyl amines (true).
Primary aliphatic amines react with nitrous acid to give unstable aliphatic diazonium salts, which decompose to alcohols with evolution of nitrogen: $RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O$ (true).
Primary aromatic amines react with nitrous acid at 273–278 K to form arenediazonium salts, which are stable at this temperature: $C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. Phenol is not the product of this reaction.
Hence the false statement is that aryl amines react with nitrous acid to produce phenols. -
Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?An amide treated with bromine and aqueous or ethanolic NaOH gives a primary amine with one carbon atom less (Hoffmann bromamide degradation).
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Acetamide has two carbons; the carbonyl carbon is lost as carbonate and the product is methyl amine.
$PCl_5$ dehydrates acetamide to acetonitrile ($CH_3CN$); soda lime and hot conc. $H_2SO_4$ do not give methyl amine (the latter hydrolyses it to acetic acid).
Hence the reagent is $NaOH-Br_2$.