JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 14

Multiple Choice Questions (MCQs) for JEE Chemistry: Classification Of Elements and Periodicity In Properties

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Practice Classification Of Elements and Periodicity In Properties MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: The energy released for the process X(g) + e⁻ → X⁻(g) is minimum and maximum respectively for (a) F (b) Cl (c) N (d) B:
(a) c & a
(b) d & b
(c) a & b
(d) c & b
Show Answer & Explanation

Answer: (d) c & b

Explanation:
1. Nitrogen has a half filled 2p³ shell and has almost zero or slightly positive EA, so minimum energy is released for N (c).
2. Cl releases the maximum (349), so (c) and (b).

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 2: Which of the following electronic configuration is expected to have highest electron affinity:
(a) 2s² 2p⁰
(b) 2s² 2p²
(c) 2s² 2p³
(d) 2s² 2p¹
Show Answer & Explanation

Answer: (b) 2s² 2p²

Explanation:
1. 2s² 2p² (carbon) has the highest EA (122) since it can reach the half filled 2p³ level.
2. 2s² 2p⁰ (Be), 2s² 2p³ (N) have almost none and B has 27.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 3: Consider (i) O(g) + e⁻ → O⁻(g); ΔH₁, (ii) F(g) + e⁻ → F⁻(g); ΔH₂, (iii) Cl(g) + e⁻ → Cl⁻(g); ΔH₃, (iv) Na(g) → Na⁺(g) + e⁻; ΔH₄. The incorrect statement is:
(a) ΔH₁ and ΔH₂ is less negative than ΔH₃
(b) ΔH₂ is more negative than ΔH₁
(c) ΔH₂ , ΔH₃ are negative while ΔH₁ is positive
(d) ΔH₁ , ΔH₂ and ΔH₃ are negative while ΔH₄ is positive
Show Answer & Explanation

Answer: (c) ΔH₂ , ΔH₃ are negative while ΔH₁ is positive

Explanation:
1. Statement 3 is wrong: O⁺ e is exothermic (delta H1 is negative at -141 kJ per mole).
2. Statement 1 and 2 are right and 4 is also true.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 4: The correct set of decreasing o rder of electronegativity is:
(a) Li, H, Na
(b) Na, H, Li
(c) H, Li, Na
(d) Li, Na, H
Show Answer & Explanation

Answer: (c) H, Li, Na

Explanation:
1. Electronegativity: H 2.1 > Li 1.0 > Na 0.9.
2. So H, Li, Na.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 5: Polarity of a bond can be explained by:
(a) Electron affinity
(b) Ionisation potential
(c) Electronegativity
(d) All of the above
Show Answer & Explanation

Answer: (c) Electronegativity

Explanation:
1. Polarity arises from a difference in the electronegativity of bonded atoms.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 6: Electronegativity values for elements are useful in predicting:
(a) Bond energy of a molecule
(b) Polarity of a bond
(c) Nature of an oxide
(d) All
Show Answer & Explanation

Answer: (d) All

Explanation:
1. Electronegativity predicts bond polarity, bond energy and the acid or base nature of an oxide, so all of these.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 7: Th e pair with minimum dif fere nce in electronegativity is:
(a) F, Cl
(b) C,H
(c) P, H
(d) Na, Cs
Show Answer & Explanation

Answer: (c) P, H

Explanation:
1. The EN differences are F-Cl 1.0, C-H 0.4, P-H 0.0 and Na-Cs 0.2.
2. P and H (both 2.1) differ least.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 8: Least electronegative element is:
(a) I
(b) Br
(c) C
(d) Cs
Show Answer & Explanation

Answer: (d) Cs

Explanation:
1. Cs is the least electronegative element listed (0.7).
2. I 2.5, Br 2.8 and C 2.5 are higher.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 9: The nomenclature of ICl is iodine chloride because:
(a) Size of I < Size of Cl
(b) Atomic number of I > Atomic number of Cl
(c) E.N. of I < E.N. of Cl
(d) E. A. of I < E. A. of Cl
Show Answer & Explanation

Answer: (c) E.N. of I < E.N. of Cl

Explanation:
1. In ICl, the less electronegative I is written first and named first.
2. The more electronegative Cl is written second as chloride.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 10: Among the following least and most polar bonds are respectively :- (a) C⁻ I (b) N⁻ O (c) C⁻ F (d) P⁻ F:
(a) d and c
(b) a and d
(c) b and d
(d) b and c
Show Answer & Explanation

Answer: (b) a and d

Explanation:
1. C-I (2.5 - 2.5 = 0) is the least polar and P-F (4.0 - 2.1 = 1.9) is the most polar.
2. N-O is 0.5 and C-F is 1.5.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 11: If the ionisation potential is IP, electron affinity is EA and electronegativity is X then which of the following relation is correct:
(a) 2X – EA – IP = 0
(b) 2EA – X⁻ IP = 0
(c) 2IP – X⁻ EA = 0
(d) All of the above
Show Answer & Explanation

Answer: (a) 2X – EA – IP = 0

Explanation:
1. Mulliken: X = (IP + EA)/2, so 2X - IP - EA = 0.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 12: The properties which are not common to both groups 1 and 17 elements in the periodic table are:
(a) Electropositive character increases down the groups
(b) Reactivity decreases from top to bottom in these groups
(c) Atomic radii increases as the atomic number increases
(d) Electronegativity decreases on moving down the group
Show Answer & Explanation

Answer: (b) Reactivity decreases from top to bottom in these groups

Explanation:
1. Reactivity increases down group 1 but decreases down group 17, so this property is not common.
2. The other three properties are common to both groups.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 13: Electronegativity of an element can be measured using:
(a) Pauling's scale
(b) Mulliken's scale
(c) Both
(d) None
Show Answer & Explanation

Answer: (c) Both

Explanation:
1. Both the Pauling scale and the Mulliken scale measure electronegativity.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 14: As we proceed across the period in periodic table, we find there is a decrease in:
(a) Ionisation energy
(b) Electron affinity
(c) Electronegativity
(d) Atomic radii
Show Answer & Explanation

Answer: (d) Atomic radii

Explanation:
1. Atomic radius decreases across a period as Zeff increases.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 15: Which compound strongly absorb CO₂ ?
(a) BeO
(b) K₂O
(c) H₃PO₄
(d) P₄O₆
Show Answer & Explanation

Answer: (b) K₂O

Explanation:
1. K2O is a strongly basic oxide and absorbs the acidic gas CO2.
2. BeO is amphoteric and H3PO4 and P4O6 are acidic.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 16: The electronegativities of the following elements: H, O, F, S and Cl increase in the order:
(a) H < O < F < S < Cl
(b) Cl < H < O < F < S
(c) H < S < O < Cl < F
(d) H < S < Cl < O < F
Show Answer & Explanation

Answer: (d) H < S < Cl < O < F

Explanation:
1. Electronegativities: H 2.1 < S 2.5 < Cl 3.0 < O 3.5 < F 4.0.

Teacher's Note:
1. Use Pauling values to judge bond polarity. Electronegativity rises across a period and falls down a group.
2. Revise the trend with two or three examples.

Question 17: Which of the following is the most basic oxide?
(a) SeO₂
(b) Al₂O₃
(c) Sb₂O₃
(d) Bi₂O₃
Show Answer & Explanation

Answer: (d) Bi₂O₃

Explanation:
1. Basic character of oxides increases down a group.
2. SeO2 is acidic, Al2O3 and Sb2O3 are amphoteric, and Bi2O3 is the most basic.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 18: Identify the correct order of the size of the following?
(a) Ca²⁺ < K⁺ < Ar < Cl⁻ (b) Ar < Ca²⁺ < K⁺ < Cl⁻ (c) Ca²⁺ < Ar (d) Ca²⁺ < K⁺ < Ar < S²⁻ Show Answer & Explanation

Answer: (a) Ca²⁺ < K⁺ < Ar < Cl⁻

Explanation:
1. Ca²⁺, K⁺, Ar, Cl⁻ and S²⁻ all have 18 electrons.
2. For equal electrons, size increases as nuclear charge falls: Ca²⁺ < K⁺ < Ar < Cl⁻ < S²⁻.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 19: The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is :
(a) Mn > Cr > Ti > V
(b) Ti > V > Cr > Mn
(c) Cr > Mn > V > Ti
(d) V > Mn > Cr > Ti
Show Answer & Explanation

Answer: (c) Cr > Mn > V > Ti

Explanation:
1. IE2 values (kJ per mole): Ti 1310, V 1410, Cr 1590, Mn 1510.
2. Cr⁺ is 3d⁵, so removing the second electron is hardest.
3. The order is Cr > Mn > V > Ti.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 20: Which of the following oxides is not expected to react with sodium hydroxide ?
(a) BeO
(b) B₂O₃
(c) CaO
(d) SiO₂
Show Answer & Explanation

Answer: (c) CaO

Explanation:
1. BeO is amphoteric, B2O3 and SiO2 are acidic, so all react with NaOH.
2. CaO is basic and does not.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 21: Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy ?
(a) [Ne]3s² 3p¹
(b) [Ne]3s² 3p³
(c) [Ne]3s² 3p²
(d) [Ar]3d¹⁰ 4s² 4p³
Show Answer & Explanation

Answer: (b) [Ne]3s² 3p³

Explanation:
1. Phosphorus (3s² 3p³) has a half filled 3p and the smaller size, so its IE (1012) exceeds that of As (947).
2. Al and Si are lower.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 22: The correct order of the decreasing ionic radii among the following isoelectronic species is:
(a) K⁺ > Ca²⁺ > Cl⁻ > S²⁻
(b) Ca²⁺ > K⁺ > S²⁻ > Cl⁻
(c) Cl⁻ > S²⁻ > Ca²⁺ > K⁺
(d) S²⁻ > Cl⁻ > K⁺ > Ca²⁺
Show Answer & Explanation

Answer: (d) S²⁻ > Cl⁻ > K⁺ > Ca²⁺

Explanation:
1. For isoelectronic ions the one with the lowest nuclear charge is the largest.
2. S²⁻ > Cl⁻ > K⁺ > Ca²⁺.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 23: Which is correct order of IP₁:
(a) Na > Al
(b) Mg > Al
(c) Ga > Ca
(d) Mg > Be
Show Answer & Explanation

Answer: (b) Mg > Al

Explanation:
1. Mg (738) > Al (578), since Mg has a stable filled 3s².
2. Na < Al, Ga < Ca, and Mg < Be.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 24: What is the value of electron gain enthalpy of Na⁺ if IE₁ of Na = 5.1 eV:
(a) +0.2 eV
(b) –5.1 eV
(c) –10.2 eV
(d) +2.55 eV
Show Answer & Explanation

Answer: (b) –5.1 eV

Explanation:
1. Na⁺ gaining an electron is the reverse of ionising Na, so the enthalpy is -IE1 = -5.1 eV.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 25: Which of the following orders of ionic radii is correctly represented ?
(a) H⁻ > H⁺ > H
(b) Na⁺ < F⁻ < O²⁻
(c) F⁻ > O²⁻ > Na⁺
(d) Al³⁺ > Mg²⁺ > N³⁻
Show Answer & Explanation

Answer: (b) Na⁺ < F⁻ < O²⁻

Explanation:
1. Na⁺ (102 pm) < F⁻ (133) < O²⁻ (140) are isoelectronic and fall in this order.
2. H⁻ > H > H⁺ is the right order for hydrogen species, not the one in option 1.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

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