Atomic Structure - Worksheet

The following questions have appeared in JEE - Main / AIEEE. The Year in which the question appeared is mentioned after the question.

For each question 4 options are given out of which only one is correct.

The correct set of four quantum numbers for the valence electrons of Rubidium atom (z = 37) is:

A) \(5, 1, 1, +\dfrac{1}{2}\)

B) \(5, 0, 1, +\dfrac{1}{2}\)

C) \(5, 0, 0, +\dfrac{1}{2}\)

D) \(5, 1, 0, +\dfrac{1}{2}\)

[2014]

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Energy of an electron is given by \(E = -2.178 \times 10^{-18} (\dfrac{Z^2}{n^2})J\). Wavelength of light required to excite an electron in an hydrogen atom from level n =1 to n = 2 will be

(\(h = 6.62 \times 10^{34}Js and c = 3.0 \times 10^8 m/s\))

A) \(2.816 \times 10^{-7}m\)

B) \(6.500 \times 10^{-7}m\)

C) \(8.500 \times 10^{-7}m\)

D) \(1.214 \times 10^{-7}m\)

[2013]

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The electrons identified by quantum numbers n and l (1) n = 4, l = 1 (2) n = 4, l = 0 (3) n = 3, l = 2 (4) n = 4, l = 1 can be placed in the increasing order of energy as

A) (3) < (4) < (2) < (1)

B) (4) < (2) < (3) < (1)

C) (2) < (4) < (1) < (3)

D) (1) < (3) < (2) < (4)

[2012]

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The frequency of light emitted for the transition n = 4 to n= 2 of \(He^+\) is equal to the transition in H atom corresponding to which of the following?

A) n = 4 to n= 3

B) n = 3 to n= 1

C) n = 2 to n= 1

D) n = 3 to n= 2

[2011]

Basic Concepts of Chemistry, Mole Concept & Stoichiometry - Worksheet

The following questions have appeared in JEE - Main / AIEEE. The Year in which the question appeared is mentioned after the question.

For the estimation of nitrogen, 1.4g of organic compound was digested by Kjeldahl metod and the evolved ammonia was absorbed in 60mL of 0.1M sulphuric acid. The unreacted acid required 20mL of 0.1M sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

A) 3%

B) 5%

C) 6%

D) 10%

[2014]

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The molarity of a solution obtained by mixing 750mL of 0.5M HCl with 250mL of 2M HCl will be

A) 1.00M

B) 1.75M

C) 0.975M

D) 0.875M

[2013]

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A gaseous hydrocarbon gives upon combustion 0.72g of water and 3.08g of carbon dioxide. The empirical formula of the hydrocarbon is

A) \(C_3H_4\)

B) \(C_6H_5\)

C) \(C_7H_8\)

D) \(C_2H_4\)

[2013]

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Consider the following reaction:

\[xMnO_4^- + yC_2O_4^{2-} + zH^+ \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O\]

The values of x, y and z in the reaction are, respectively

A) 2, 5 and 8

B) 2, 5 and 16

C) 5, 2 and 8

D) 5, 2 and 16

[2013]

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The mass of potassium dichromate crystals required to oxidise 750mL of 0.6M Mohr’s salt solution is (Molar mass of \(K_2Cr_2O_7\) = 294, Mohr’s Salt = 392)

A) 2.2g

B) 0.49g

C) 0.45g

D) 22.05g

[2011]

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What is the best description of the change that occurs when \(Na_2O\) is dissolved in water?

A) Oxidation Number of oxygen increases

B) Oxidation Number of sodium decreases

C) Oxide ion accepts a pair of electrons

D) Oxide ion donates a pair of electrons

[2011]

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29.5mg of an organic compound containing nitrogen awas digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20mL of 0.1M HCl solution. The excess of the acid required 15mL of 0.1M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

A) 59.0%

B) 47.4%

C) 23.7%

D) 29.5%

[2010]

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The amount of oxalic acid present in a solution can be determined by its titration with \(KMnO_4\) solution in the presence of \(H_2SO_4\). The titration gives unsatisfactory result when carried out in presence of HCl, because

A) HCl gets oxidised by oxalic acid to chlorine

B) HCl furnishes \(H^+\) ions in addition to those from oxalic acid

C) HCl reduces permanganate ion to \(Mn^{2+}\)

D) HCl oxidises oxalic acid to carbon dioxide and water

[2008]

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Which of the following statements are true for the reaction \(2Al (s) + 6HCl (aq) \rightarrow 2Al^{3+}(aq) + 6Cl^- (aq) + 3H_2 (g)\)?

A) 11.2L of \(H_2(g)\) at STP is produced for every mole of \(HCl(aq)\) consumed

B) 6L of \(HCl(aq)\) is consumed for every 3L of \(H_2(g)\) produced

C) 33.6L of \(H_2(g)\) is produced regardless of temperature and pressure for every mole of Al that reacts

D) 67.2L of \(H_2(g)\) at STP is produced for every mole of Al that reacts

[2007]

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How many moles of Magnesium Phosphate, \(Mg_3(PO_4)_2\) will contain 0.25 mole of oxygen atoms?

A) \(2 \times 10^{-2}\)

B) \(3.125 \times 10^{-2}\)

C) \(1.25 \times 10^{-2}\)

D) \(2.5 \times 10^{-2}\)

[2006]

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Density of a 2.05M solution of acetic acid in water is 1.02g/mL. The molality of the solution is

A) 1.14 mol/kg

B) 3.28 mol/kg

C) 2.28 mol/kg

D) 0.44 mol/kg

[2006]

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Which of the following chemical reactions depicts the oxidising behaviour of \(H_2SO_4\)?

A) \(2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O\)

B) \(Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O\)

C) \(NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\)

D) \(2PCl_5 + H_2SO_4 \rightarrow2POCl_3 + SO_2Cl_2 + 2HCl\)

[2006]

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If we consider that \(\frac{1}{6}\), in place of \(\frac{1}{12}\), mass of carbon atom is taken to be relative atomic mass unit, the mass of one mole of a substance will

A) decrease twice

B) increase two fold

C) remain unchanged

D) be a function of the molecular mass of the substance

[2005]

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What volume of hydrogen gas at 273K and 1 atm pressure will be consumed in obtaining 21.6g of elemental boron from the reduction of \(BCl_3\) by \(H_2\) (atomic mass of B = 10.8)?

A) 89.6L

B) 67.2L

C) 44.8L

D) 22.4L

[2003]

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With increase of temperature, which of the following will change?

A) Molality

B) Weight fraction of solute

C) Fraction of solute present in water

D) Mole Fraction

[2002]

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The number of atoms in 558.5g of Fe (atomic mass of Fe = 55.85 g/mol) is

A) twice that in 60g of Carbon

B) \(6.02 \times 10^{22}\)

C) half that in 8g of He

D) \(558.5 \times 6.02 \times 10^{23}\)

[2002]

Question on General Organic Chemistry

thinkingchemistry:

Question:

Is a Tertiary Carbocation always more stable than a primary carbocation?

Line of Thinking:

In general any charged species will be unstable compared to an uncharged species. For example \(CH_3OH\) will be more stable that \(CH_3O^-\) under similar conditions. Greater the charge on the species more unstable it will be.

Can you answer the question based on the above Line of Thinking (LoT)?

Is a Tertiary Carbocation always more stable than a Primary Carbocation?

How to distinguish between Carbon dioxide and Sulphur dioxide?

While performing Salt Analysis, the action of heat on a salt with or without an acid will release gases that can be used to identify the anion of the salt. For example, carbonate salts will release \(CO_2\) and bisulphite salts release \(SO_2\). These gases are then identified with appropriate tests.

\(CO_2\) when passed through lime water makes it turbid (milky) and if the passage of \(CO_2\) is prolonged the turbidity disappears and the solution becomes clear again.

\(CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O\)

\(CaCO_3\downarrow + CO_2 \rightarrow Ca(HCO_3)_2\)

\(CaCO_3\) is a precipitate and is responsible for the turbidity of the solution.

\(Ca(HCO_3)_2\) is soluble in water and is responsible in making the solution clear again.

\(SO_2\) when passed through lime water makes it turbid (milky) and if the passage of \(SO_2\) is prolonged the turbidity disappears and the solution becomes clear again.

\(SO_2 + Ca(OH)_2 \rightarrow CaSO_3\downarrow + H_2O\)

\(CaSO_3\downarrow + SO_2 \rightarrow Ca(HSO_3)_2\)

\(CaSO_3\) is a precipitate and is responsible for the turbidity of the solution.

\(Ca(HSO_3)_2\) is soluble in water and is responsible in making the solution clear again.

Visually, the same visual thing is observed when \(CO_2\) and \(SO_2\) is passed through lime water. What is the test to distinguish between the two, then?

Here is the reaction to distinguish between them. The gas is passed into to an aqueous solution of acidified Potassium Dichromate. If the colour of the solution (which is originally orange) turns to green, then the gas is \(SO_2\).

Chromium in its +6 oxidation state and in the dichromate (\(Cr_2O_7^{2-}\)) form is orange in colour and in its +3 oxidation state is green in colour. The change in colour in the above reaction shows that \(Cr_2O_7^{2-}\) ion is reduced to the \(Cr^{3+}\) ionic state. This also means that the compound that reacted with \(Cr_2O_7^{2-}\) ion should have been oxidised.

Out of \(CO_2\) and \(SO_2\), only \(SO_2\) can be oxidized as in \(CO_2\), Carbon is already in its highest oxidation state (+4). The reaction between \(SO_2\) and acidified \(Cr_2O_7^{2-}\) ion is given below.

\(Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O\)

What do you think will the colour change be if acidified \(KMnO_4\) is used instead of acidified \(K_2Cr_2O_7^{2-}\)?

Hint: \(Mn^{7+}\) is Purple in colour, \(Mn^{4+}\) is buff coloured, \(Mn^{2+}\) is pale pink coloued and \(Mn^{6+}\) is green coloured.

The pH Scale

At a given temperature pH of all solutions will be falling within a given range. This range depends on the extent to which water is ionized at that given temperature.

Water, being a weak electrolyte, will be partially ionized. The partial ionization of water can be represented as below;

\[H_2O \rightleftharpoons H^+ + HO^-\]

At \(25^o C\), the degree of dissociation (\(\alpha\)) of water is \(1.8 \times 10^{-9}\). This means that for every mole of water taken \(1.8 \times 10^{-9}\) moles of water is ionized.

The acid dissociation constant of water (\(K_a\)) can be written as,

\(K_a = \frac {[H^+][HO^-]}{[H_2O]} = \frac{C \alpha \times C \alpha}{C(1-\alpha)}\)

where C is the concentration of water and \(\alpha\) is the degree of dissociation of water.

The concentration of water is \(\frac {1000 g/L}{18 g/mol} \approx 55.5 mol/L \)

As \(\alpha \ll 1\), we can neglect it with respect to 1. Neglecting it with respect to 1 and substituting the values of C and \(\alpha\) in the above equation for \(K_a\) we get,

\(K_a = C\alpha^2 = \frac{1000}{18} \times 1.8 \times 1.8 \times 10^{-18} = 1.8 \times 10^{-16}\)

We can now define another constant \(K_w\), the ionic product of water as,

\(K_w = K_a \times [H_2O] = [H^+][HO^-]\)

\(K_w = 1.8 \times 10^{-16} \times \frac{1000}{18} = 10^{-14}\)

\(K_w = 1.8 \times 10^{-14} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-7}\)

\(pH = -log[H^+] = -log 10^{-7} = 7\)

pH of water at \(25^oC\) is 7. This will be the neutral pH as the \([H^+] = [HO^-]\). If the \([H^+] > [HO^-]\), then the value of pH will become less than 7 and if the \([H^+] < [HO^-]\), then the value of pH will become more than 7. To determine the lower and upper limit of the pH value at this temperature and thus the pH scale at this temperature we have to down by 7 units and up by 7 units because 7 will be the mid point of the scale. Thus, the pH scale at \(25^oC\) will be 0 to 14.

Let us say that, at \(125^oC\), \(K_w\) of water is \(1.0 \times 10^{-12}\). What would be the pH of pure water at this temperature and if a solution with pH = 7, is available at this temperature, will it be acidic, basic or neutral?

\(K_w = 1.0 \times 10^{-12} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-6}\)

\(pH = -log[H^+] = -log 10^{-6} = 6\)

pH of water at \(125^oC\) is 6. This will be the neutral pH as the \([H^+] = [HO^-]\). If the \([H^+] > [HO^-]\), then the value of pH will become less than 6 and if the \([H^+] < [HO^-]\), then the value of pH will become more than 6. The pH scale at this temperature will be 0 to 12. Also, the solution whose pH is 7 at this temperature would be basic.

Question on Ionic Equilibrium

thinkingchemistry:

Question:

Saccharin (\(K_a = 2 \times 10^{-12} \)) is a weak acid represented by formula HSaC. A \(4 \times 10^{-4}\) mole amount of saccharin is dissolved in 200mL of water of pH 3. Assuming no change in volume, calculate the concentration of \(SaC^{-1}\) ions in the resulting solution at equilibrium.

Line of Thinking:

At equilibrium, saccharin will dissociate to give \(SaC^{-1}\) ions and \(H^+\) ions. pH of the solution is 3 and hence we will be able to calculate the concentration of \(H^+\) ions in solution. The initial concentration of saccharin can also be calculated from the data given in the question.

Also, \(K_a = \dfrac {[Sac^{-1}]\times[H^+]}{[HSaC]}  = \dfrac{C\alpha \times [H^+]}{C(1 - \alpha)}\) where \(\alpha\) is the degree of dissociation and \(C\) is the initial concentration of saccharin.

Can you calculate the answer based on the above Line of Thinking (LoT)?

Can you solve this question based on the lines of thinking given?