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Chemistry
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Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii isMg, P and Cl belong to the third period. Across a period the nuclear charge increases while electrons enter the same shell, so the atomic radius decreases from left to right: Mg > P > Cl.
Ca lies just below Mg in group 2. Down a group a new shell is added, so the radius increases: Ca > Mg.
Atomic radii (pm): Cl = 99, P = 110, Mg = 160, Ca = 197.
Increasing order of atomic radii: Cl < P < Mg < Ca -
The pressure exerted by 6.0 g of methane gas in a 0.03 $m^3$ vessel at $129^\circ C$ is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J $K^{-1}mol^{-1}$)Molar mass of $CH_4$ = 12.01 + 4(1.01) = 16.05 g $mol^{-1}$
Moles of methane: $n = \frac{6.0}{16.05} = 0.3738$ mol
T = 129 + 273 = 402 K; V = 0.03 $m^3$
$P = \frac{nRT}{V} = \frac{6.0}{16.05}\times\frac{8.314\times402}{0.03}$
$P = 0.3738\times111408$
P = 41648 Pa -
Among the following which one has the highest cation to anion size ratio?The ratio $\frac{r_{cation}}{r_{anion}}$ is highest when the cation is the largest and the anion is the smallest.
Cation sizes: $Li^+ < Na^+ < Cs^+$, so $Cs^+$ is the largest cation.
Anion sizes: $F^- < I^-$, so $F^-$ is the smallest anion.
The combination of the largest cation with the smallest anion is CsF.
Hence CsF has the highest cation to anion size ratio. -
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will beWork done in expansion: $w = -p_{ext}\Delta V$
Expansion into vacuum is free expansion: there is nothing to push against, so $p_{ext} = 0$.
$w = -0\times\Delta V = 0$
This is true for any amount of gas, so the 'three moles' does not matter.
Work done = zero -
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be ($h = 6.6\times10^{-34}$ J s)de Broglie relation: $\lambda = \frac{h}{mv}$
$\lambda = \frac{6.6\times10^{-34}}{0.66\times100}$
$\lambda = \frac{6.6\times10^{-34}}{66}$
$\lambda = 1.0\times10^{-35}$ m -
In which of the following molecules does the central atom not have $sp^3$ hybridization?Count σ bonds + lone pairs on the central atom.
$CH_4$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$BF_4^-$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$NH_4^+$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$SF_4$: sulphur has 6 valence electrons; 4 are used in bonds and 2 remain as one lone pair. 4 bond pairs + 1 lone pair = 5, so $sp^3d$ (see-saw shape).
Hence the central atom is not $sp^3$ hybridised in $SF_4$. -
For vaporization of water at 1 atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are 40.63 kJ $mol^{-1}$ and 108.8 J $K^{-1}mol^{-1}$, respectively. The temperature when Gibbs energy change ($\Delta G$) for this transformation will be zero is$\Delta G = \Delta H - T\Delta S$
When $\Delta G = 0$: $\Delta H = T\Delta S$
$T = \frac{\Delta H}{\Delta S} = \frac{40.63\times10^3\ J\,mol^{-1}}{108.8\ J\,K^{-1}mol^{-1}}$
T = 373.4 K
(This is the boiling point of water, where liquid and vapour are in equilibrium at 1 atm.) -
Match List-I (substances) with List-II (processes) employed in the manufacture of the substances and select the correct option.
Pick the matching item from Column II for each row.
Column I
- a. Sulphuric acid
- b. Steel
- c. Sodium hydroxide
- d. Ammonia
Column II
- i. Haber's Process
- ii. Bessemer's Process
- iii. Leblanc Process
- iv. Contact Process
Correct answer: a → iv, b → ii, c → iii, d → i
Sulphuric acid is manufactured by the Contact process: $SO_2$ is oxidised to $SO_3$ over $V_2O_5$, which is absorbed in $H_2SO_4$ to give oleum.
Steel is made from pig iron by Bessemer's process, in which air is blown through the molten iron to oxidise the impurities.
Sodium hydroxide was obtained through the Leblanc process (sodium carbonate from the Leblanc process is causticised with lime to give NaOH).
Ammonia is manufactured by Haber's process: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ over an iron catalyst.
Correct match: a-iv, b-ii, c-iii, d-i. -
The rate of the reaction $2NO + Cl_2 \rightarrow 2NOCl$ is given by the rate equation rate = $k[NO]^2[Cl_2]$. The value of the rate constant can be increased byThe rate constant k is characteristic of a reaction at a given temperature; it does not depend on the concentrations of the reactants.
Increasing [NO] or [$Cl_2$] increases the rate of the reaction, but the value of k stays the same.
The dependence of k on temperature is given by the Arrhenius equation: $k = Ae^{-E_a/RT}$
As T increases, $e^{-E_a/RT}$ increases, so k increases (roughly doubling for every 10 K rise).
Hence the rate constant can be increased by increasing the temperature. -
Which of the following oxidation states is the most common among the lanthanoids?The general configuration of the lanthanoids is $[Xe]4f^{1-14}5d^{0-1}6s^2$.
All of them readily lose the two 6s electrons and one 5d or 4f electron to form $Ln^{3+}$ ions.
The +3 state is the most common and most stable oxidation state for every lanthanoid.
+2 and +4 states are shown only by a few elements where an empty, half-filled or completely filled 4f subshell results, e.g., $Ce^{4+}$ ($4f^0$), $Eu^{2+}$ ($4f^7$), $Yb^{2+}$ ($4f^{14}$), $Tb^{4+}$ ($4f^7$); +5 is not shown.
Hence the most common oxidation state is +3. -
Some of the properties of the two species, $NO_3^-$ and $H_3O^+$, are described below. Which one of them is correct?$NO_3^-$: nitrogen has 3 σ bonds and no lone pair (steric number 3), so it is $sp^2$ hybridised and the ion is trigonal planar.
$H_3O^+$: oxygen has 3 σ bonds and one lone pair (steric number 4), so it is $sp^3$ hybridised and the ion is pyramidal.
So the two species differ both in the hybridisation of the central atom ($sp^2$ and $sp^3$) and in shape (planar and pyramidal).
Hence they are dissimilar in hybridization for the central atom, with different structures. -
In the following reaction
$C_6H_5CH_2Br \xrightarrow[2.\ H_3O^+]{1.\ Mg,\ Ether}$ X,
the product 'X' isStep 1: benzyl bromide reacts with magnesium in dry ether to form the Grignard reagent: $C_6H_5CH_2Br + Mg \xrightarrow{ether} C_6H_5CH_2MgBr$
In a Grignard reagent the carbon bonded to magnesium is strongly nucleophilic and basic (carbanion-like).
Step 2: with $H_3O^+$ it takes up a proton to form the hydrocarbon: $C_6H_5CH_2MgBr + H_2O \rightarrow C_6H_5CH_3 + Mg(OH)Br$
So the net change is replacement of Br by H.
Hence X is toluene, $C_6H_5CH_3$.