NCERT Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Multiple Choice Questions

NCERT Solutions For Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Multiple Choice Questions

Question 1. An element belongs to Group 15 and the third period of the periodic table. Its electronic configuration will be

1. ls²2s²2p²
2. ls²2s²2p⁴
3. ls²2s²2p⁶3s²3p³
4. ls²2s²2p⁶3s²3p²

Answer: 3. ls²2s²2p⁶3s²3p³

The electronic configuration of the valence shell of group 15 elements in the periodic table is ns²np³ where n is the tire period number.

Therefore, the element located in the third period has the electronic configuration ls²2s²2p⁶3s²3p³

Questions 2. Which one ofthe following has the lowest ionization energy

1. ls²2s²2p⁶
2. ls²2s²2p⁵
3. ls²2s²2p⁶3s¹
4. ls²2s²2p³

Answer: 2. ls²2s²2p⁵

The electronic configuration ls²2s²2p⁶3s¹ is that of an alkali metal. In a certain period of the periodic table, the ionization potential of alkali metals is the lowest.

Question 3. If 2nd the ionization 1st ionization energy of the atom-atom is — is 13.6 eV, then the

1. 27.2 eV
2. 40.8 eV
3. 54.4 eV
4. 108.8eV

Answer: 3. 54.4 eV

The first ionization energy of the -atom is calculated as:

⇒ \(\frac{2 \pi^2 m Z^2 e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=13.6 \mathrm{eV}\)

The second ionization energy of the He atom is = 13.6 X Z2 eV = 13.6 X 22 eV = 54.4 eV.

Questions 4. The stable bivalency of Pb and trivalency of Bi is

1. Due to d contraction in pb and bi
2. Due to relativistic contraction of the 6s -orbitals of pb
3. And bi, leading to an inert pair effect
4. Due to the screening effect
5. Due to the attainment of noble gas configuration.

Answer: 2. Due to relativistic contraction of the 6s -orbitals of pb

Due to the relativistic contraction of 6s -orbital, an inert pair effect is observed as a result of which, the lower oxidation states of elements attain stability on moving down a group in the periodic table. Thus lead shows stable bivalency while bismuth shows stable trivalency

Question 5. Which of the following is correct?

1. Radius of ca²⁺ < cl- < s²
2. The radius of cl- <s^2-ca²⁺
3. Radius of s^2- = cl- = ca²⁺
4. Radius of s^2- cl- < ca²⁺

Answer: 1. Radius of ca²⁺ < cl- < s²⁺

The number of electrons in Ca²⁺, S^2- and CI^–= 18, i.e., all three ions are isoelectronic.

For any ion, ionic radius oc(e/Z) [where, e = electron number and Z = atomic number]. Therefore, the correct order of the radii ofthe given three ions is—Ca²⁺ < Cl^– < S^2-

Question 6. For BCl₃, AlCl₃, and GaCl₃, the increasing order of ionic character is

1. BCl₃ < AlCl₃ < GaCl₃
2. GaCl₃ < AlCl₃ < BCl₃
3. BCl₃ <GaCl₃ <AlCl₃
4. AlCl₃ < BCl₃ < GaCl₃

Answer: 3. BCl₃ <GaCl₃ <AlCl₃

The ionic character of a compound depends on the polarising power of the cation present in the compound. The greater the magnitude of polarising power, the lesser the ionic character of the cation.

Polarisingpower of B³⁺, Al³⁺, and Ga³⁺ follows the order: Al³⁺ < Ga³⁺ < B³⁺, Thus, the order of increasing ionic character is— BCl₃ < GaCl₃ < AICI₃.

Question 7. The hydrides of the first elements in groups 15-17, namely NH₃, H₃O, and HF respectively show abnormally high values for ting and boiling points. This is due to

1. Small sizes of n, 0 and
2. The ability to form extensive intermolecular-bonding
3. The ability to form extensive intramolecular h-bonding
4. Effective van der waals interaction

Answer: 2. The ability to form extensive intramolecular-bonding

The ionization potential values of, O, and F are quite high and thus, they form intermolecular H-bonding in the compounds, NH₃, H₃O, and HF. As a result, these compounds show abnormally high values for melting and boiling points.

Question 8. Decreasing the basic character of K₂O, BaO, CaO, and MgO is

1. K₂O > BaO > CaO > MgO
2. K₂O > CaO > BaO > MgO
3. MgO > BaO > CaO > K₂O
4. MgO > CaO > BaO > K₂O

Answer: 1. K₂O > BaO > CaO > MgO Alkali metal oxides are highly basic. Down the group in the periodic table, the ionization potential of the alkaline earth metals decreases.

Question 9. Amongst Be, B, Mg, and A₁ the second ionization potential is maximum for

1. B
2. Be
3. Mg
4. Al

Answer: 1. B

The electronic configuration of B (5) is ls²2s²2p¹. The second ionization potential of B is the maximum because an electron has to be removed from the filled 2s orbital which will require a high amount of energy.

Question 10. An element X belongs to the period and fifteenth group of the periodic table. Which of the following statements is true

1. It has a filled s -s-orbital and a partially filled d -d-orbital
2. It has filled s -and p -orbital and a partly filled d -orbital
3. It has filled s -and p orbitals and a half-filled d -orbital.
4. It has a half-filled p -p-orbital and filled s and d – d-orbital

Answer: 4. It has a half-filled p-orbital and filled s and d – orbital

The element X is positioned at the 4th period and 15th group. Hence the element is As. (Atomic mass = 33) Electronic configuration ofthe element: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²43³ Thus the element has a half-filled p-orbital and a filled s  -and d – orbital

Question 11. Which of the following atoms should have the highest 1st electron affinity—

1. F
2. O
3. N
4. C

Answer: 1. F

Decreasing order of electron affinity: F > O > N > C.

Question 12. Which of the set of oxides are arranged in the proper order of basic, amphoteric acidic

1. SO₂,P₂O₅,CO
2. BaO, Al₂O₃, SO₂
3. CaO, SiO₂> Al₂O₃
4. CO₂, Al₂O₃, CO

Answer: 2. BaO→basic, Al₂O₃ amphoteric, SO₂ → acidic

Question 13. Which ofthe following orders presents a correct sequence of the increasing basic nature ofthe given oxides—

1. Al₂O₃ < MgO < Na₂O < K₂O
2. MgO < K₂O < Al₂O₃ < Na₂O
3. Na₂O < K₂O < MgO < Al₂O₃
4. K₂O < Na₂O < Al₂O₃ < MgO

Answer: 1. Al₂O₃ < MgO < Na₂O < K₂O

With the increase in the value of the electropositivity of metals, the basic nature of their oxides also increases.

Question 14. The increasing order of the ionic radii of the given isoelectronic species is —

1. S^2- , Cl^2- Ca²⁺ , K
2. Ca²⁺ , K⁺ , Cl^– S^2-
3. K⁺ , S^2- , Ca²⁺ , Cl^–
4. Cl-, Ca²⁺, K²⁺, S²⁺

Answer: 2. Ca²⁺ , K⁺ , Cl^– S^2-

For any atom Orion, ionic radius (e/Z) [e=number of electrons and Z= atomic number]. Since the given ions are isoelectronic their electron number is also the same.

It means the higher the atomic number lower the ionic radius of the ion. Therefore, the correct order of the ionic radii ofthe ions is Ca²⁺, K⁺, Cl^– S^2-

Question 15. The first ionization potential of Na is 5.1 eV. The value of electron-gain enthalpy of Na⁺ will be

1. +2.55ev
2. -2.55ev
3. -5.1ev
4. -10.ev

Answer: 3. -5.1ev

For any element, the value of the first ionization potential and that of the electron-gain enthalpy of its unipositive ion. However, for electron gain enthalpy energy is liberated while for ionization potential energy is absorbed. Hence, they have the same magnitude but opposite sign.

Question 16. Which of the given represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se, and Ar

1. Ca<Ba<S<Se<Ar
2. Ca < S < Ba < Se < Ar
3. S < Se < Ca < Ba < Ar
4. Ba < Ca < Se < S < Ar

Answer: 4. Ba < Ca < Se < S < Ar

On moving from left to right across a period in the periodic table, the values of ionization enthalpy of elements increase, while moving down a group it decreases.

Question 17. Ionic radii (A) of N^3- , O^2- and F^–

1. 1.71, 1.40, and 1.36
2. 1.36, 1.40, and 1.71
3. 1.71, 1.36 and 1.40
4. 1.36, 1.71 and 1.40

Answer: 2. 1.36, 1.40 and 1.71

The boiling point of Xe is the highest because the boiling point increases down the group from He to Rn due to an increase in van der Waals force of attraction as the size of the atom increases.

Question 19. Which of the following atoms has the highest first ionization energy

1. Rb
2. Na
3. K
4. Sc

Answer: 4. Sc

Question 20. The group having isoelectronic species is-

1. O^2- , F^– , Na⁺ Mg²⁺
2. O^– , F^– ,Na⁺   Mg²⁺
3. O^2- F^– , Na, Mg²⁺
4. O^– F^–, Na⁺, Mg²⁺

Answer: 1. O^2-, F^–, Na⁺ Mg²⁺

Question 21. Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the incorrect one, is

1. Both form nitrides
2. Nitrates of both li and mg yield nO₂ and O₂ on heating
3. Both form soluble carbonates
4. Both form soluble bicarbonates

Answer: 3. Both form soluble carbonates

Despite having a diagonal relationship, Li and Mg differ in carbonate salts. Magnesium forms basic carbonate [3MgCO₃, Mg(OH)₂, 2H₂O] but lithium like other alkali metals form normal carbonate [Li2CO₃] salts. Hence, O-2, F+, and Mg²⁺ are isoelectronic species.

Question 22. What is the value of electron-gain enthalpy of Na⁺ if IEX

1. -5.1 eV
2. -10.2 eV
3. +2.55 eV
4. +10.2 eV

Answer: 1. -5.1 eV

Question 23. In which of the following arrangements, the given sequence is not according to the indicated against it

1. HO < H₂S < H₂Se < H₂Te: increasing pKa values
2. NH₃ < PH₃ < ASH₃ < SbH₃: increasing acidity
3. CO₂<SiO₂<SnO₂<PbO₂: increasing oxidising power
4. HF < HC1 < HBr <HI: increasing acidic strength

Answer: 1. HO < H₂S < H₂Se < H₂Te: increasing pKa values

On moving down a group in the periodic table, the covalent character of the hydrides of the elements increases and so their acidity gradually increases. Therefore, the acidic character of the hydrides of the elements of group VA and VILA follows the order— NH₃< PH₃ < AsH₃ < SbH₃ and HF < HCI < HBr < HI On the other hand, the oxidizing power of the oxides in option follows the trend:

CO₂ < SiO₂ < SnO₂ < PbO₂

Question 24. Identify the wrong statement among the following 

1. The atomic radius of the elements increases as one moves down the first group ofthe periodic table
2. The atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table
3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius
4. Amongst isoelectronic species, the greater the negative charge on the anion, the larger the the ionic radius

Answer: 3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius

On moving from left to right across a period. The proton number and magnitude of the nuclear charge of elements increase gradually leading to a corresponding decrease in the atomic radii of the elements.

Again, moving down a group, with an increase in the atomic number of elements, their atomic radii gradually decrease due to the addition of new shells. For isoelectronic ions, with a decrease in the positive charge of cations and with an increase in the negative charge of anions, ionic radii increase.

Question 25. The reason for lanthanoid contraction is

1. Negligible screening effect of f-orbitals
2. Increasing nuclear charge
3. Decreasingnuclear charge
4. Decreasing screening effect

Answer: 1. Negligible screening effect of f-orbitals.

Question 26. Be²⁺ is isoelectronic with which ofthe following ions?

1. H⁺
2. Li⁺
3. Na⁺
4. Mg2⁺

Answer: 2. Li⁺

⇒ \(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Ion } & \mathrm{Be}^{2+} & \mathrm{H}^{+} & \mathrm{Li}^{+} & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} \\
\hline \text { No. of electron } & 2 & 0 & 2 & 10 & 10 \\
\hline
\end{array}\)

Question 27. Which of the following orders of ionic radii is correctly represented?

1. H→ H+→H
2. F→ O₂→Na⁺
3. Na⁺ → F^–→ O^2-
4. Al³⁺ > Mg²⁺ > N₃

Answer: 2. F→ O₂→Na⁺

None of the options is correct; An atom loses an electron (s) to form a cation. Thus, the radius of the formed cation is less than the parent atom.

Again an atom gains electron (s) to form an anion. So, the radius of the formed anion is greater than the parent atom. Therefore, H^– > H > H⁺. Now, for isoelectronic species, with an increase in their atomic number, their ionic radii decrease. Therefore, the correct orders of ionic radii are O^2-→F¯ > Na+ and N^3- > Mg²⁺ > Al³⁺

Question 28. The species Ar, K⁺, and Ca²⁺ contain the same number of electrons. In which order do their radii increase

1. Ca²⁺ <K+<Ar
2. K+ < Ar < Ca²⁺
3. Ar < K+ < Ca²⁺
4. Ca²⁺ < Ar < K⁺

Answer: 1. Ca²⁺ <K+<Ar

For isoelectronic species atomic/ionic radii decrease with the increase of nuclear charge. Hence increasing order atomic/ionic radii  Ca²⁺ < igK+ < 18Ar.

Question 29. Because of lanthanide contraction, which of the following pairs of elements have nearly the same atomic radii (Numbers in the parenthesis are atomic numbers)

1. Zr(40) and Hf(72)
2. Zr(40) and Ta(73)
3. Ti(22) and Zr(40)
4. Zr(40) and Nb(41)

Answer: 1. Zr(40) and Hf(72)

Zr and Hf have the same atomic radii due to the lanthanide contraction.

Question 30. In which ofthe following options the order of arrangement does not agree with the variation of property indicated against it

1. I < Br < Cl < F (increasing electron-gain enthalpy)
2. Li<Na<K<Rb (increasingmetallic radius)
3. Al³⁺ < Mg^2+_ < Na⁺ < F^–(increasing ionic size)
4. B<C<N<0 (increasing first ionization enthalpy)

Answer: 1. I < Br < Cl < F (increasing electron-gain enthalpy)

The increasing order of negative electron gain enthalpy; I < Br < F < Cl and that of first ionization energy: B < C < O < N.

Question 31. The element Z =114 has been discovered recently. It will belong to which of the following family/groups and electronic configuration—

1. Carbon family, [rn]5/I4⁶d¹⁰7s²7p²
2. Oxygen family, [rn]5/14⁶d¹⁰7s²7p⁴
3. Nitrogen family, [rn]5/14⁶d¹⁰7s²7p⁶
4. Halogen family, [rn]5/I4⁶d¹⁰7s²7p⁵

Answer: 1. Carbon family, [rn]5/i4⁶d¹⁰7s²7p²

The electronic configuration of the element having atomic mass 114 is [Rn]5f¹⁴ 6d¹⁰7s²7p² The outer electronic configuration of the element is the same as that of carbon. So the element should belong to the carbon family.

Question 32. Among CaH₂, BeH₂, BaH₂ the order of ionic character

1. BeH₂<CaH₂<BaH₂
2. CaH₂ < BeH₂ < BaH₂
3. BeH₂ < HaH₂ < CaH₂
4. BaH₂ < BeH₂ < CaH₂

Answer: BeH₂<CaH₂<BaH₂

Down a group metallic character increases. Thus ionic character of the metal hydride increases down the group. Hence least ionic compound is BeH₂.

Question 33. Which ofthe following oxides is most acidic

1. MgO
2. BeO
3. BaO
4. CaO

Answer: 2. BeO

Down a group, the metallic character of the elements increases. Hence down the group basic character of the metallic oxide increases. Thus BeO has the most acidic character. It is an amphoteric oxide whereas others are basic oxides.

Question 34. The first ionization enthalpy of Na, Mg, and Si are 496, 737, and 776 kj.mol^-1 respectively. What will be the first ionization enthalpy potential of a kj mol^-1

1. >766 kj – mol^-1
2. >496 and < 737 k).mol^-1
3. >737 and < 766 kj.mol^-1
4. >496 kj- mol^-1

Answer: 2. >496 and < 737 k).mol^-1

The ionization enthalpy of A1 is lower than that of Mg as the 3p1 electron of A1 is easier to remove than to remove an electron from the fully-filled 3s -orbital of Mg.

Question 35. Which is correct regarding the size of the atom

1. B < Ne
2. Na > K
3. N < O
4. V > Ti

Answer: 2. The atomic radii of noble gases are the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.

Question 36. An element (X) belongs to the fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of (X)? It has

1. Partially filled d -d-orbitals and filled orbital,
2. Filled s -orbital and filled p orbitals.
3. Filled s -orbital and half-filled p -orbitals.
4. Half-filled d -orbitals and filled s -orbitals.

Answer: 3. filled s -orbital and half-filled p -orbitals.

The electronic configuration of (X) can be written as X: ls²2s²2p⁶3s³3p⁶4s²3d¹⁰4p³ So, element (X) has filled s -and d -orbitals and half-filled p -orbitals.

Question 37. Which of the following transformations produces the maximum amount of energy

1. M-(g)→M(g)
2. M(g)→M-(g)
3. M⁺(g)→M³⁺(g)

Answer: 4. M²⁺+(g)→M³⁺(g)

Question 38. The amount of energy released when 106 atoms of iodine in a vapor state are converted into- ions is 4.8 x 10-13 J. What is the electron affinity of iodine in kJ mol^-1

1. 489
2. 289
3. 259
4. 389

Answer: 2. 289

Question 39. The elements that occupy the peaks of the ionization energy potential curve are

1. Na, K, Rb, Cs
2. Cl, F, Br, I
3. Na, Mg, Cl, I
4. He, Ne, Ar, Kr

Answer: 4. He, Ne, Ar, Kr

Question 40. The electronic configuration ofthe atom having maximum difference second and third ionization enthalpies is

1. ls²2s²2p⁶3s²
2. ls²2s²2p⁶3s²3p¹
3. ls²2s²2p⁶3s²3p²
4. ls²2s²2p⁶3s¹

Answer: 1. ls²2s²2p⁶3s²

Question 41. Identify the least stable ion amongst the following

1. Li^–
2. Be^–
3. B^–
4. C^–

Answer: 2. Be^–

Question 42. If each orbital can accommodate a maximum of four electrons, the number of elements in the third period of the periodic table will be

1. 10
2. 12
3. 14
4. 16

Answer: 4. 16

Question 43. Three elements X, Y, and Z are present in the third short period and their oxides are ionic, amphoteric, and giant molecules respectively. The correct order of atomic numbers of X, Y, and Z is

1. Z< Y<X
2. X<Z<Y
3. Y<Z<X
4. X<Y<Z

Answer: 4. X<Y<Z

Question 44. A gaseous mixture of He, Ne, Ar, and Kr is irradiated with photons of frequency appropriate to ionize Ar. The ion(s) present in the mixture will be

1. Only
2. Ar+ and He+
3. Ar+ and Ne
4. Ar+ and Kr+

Answer: 4. Ar+ and Kr+

Question 45. Boiling point of Kr & Rn are -152°C & -62°C respectively. Then the boiling point of Xe is expected to be

1. -92C
2. -87C
3. -107C
4. 77C

Answer: 3. -107C

Question 46. Which ofthe following is smallest in size

1. Li+ (aq)
2. Na+ (aq)
3. K+ (aq)
4. Rb+ (aq)

Answer: 4. Rb+ (aq)

Question 47. The atomic radius of Li is 1.23 A and the ionic radius of Li+ is 0.76 A. Percentage of the volume occupied by a single valence electron in Li is

1. 35
2. 52.5
3. 76.4
4. 83.72

Answer: 3. 76.4

Question 48. The number of valence electrons in element A is 3 and that in element B is 6. The most probable compound from A and B is

1. A₂B
2. AB₂
3. A₆B₃
4. A₂B₃

Answer: 4. A₂B₃

Question 49. The ionic radius of ‘Cr’ is the minimum in which of the following compounds

1. K₂CrO₄
2. CrF₃
3. CrO₂
4. CrCl₃

Answer: 1. K₂CrO₄

Question 50. The correct order of radii is

1. N < Be < B
2. F^– <O^2- <N^3-
3. Na < Li < K
4. Fe³⁺ <Fe²⁺<Fe⁴⁺

Answer: 2. F^– <O^2- <N^3-