JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 05

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

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Multiple Choice Questions

Question 1: Among the following elements (electronic configurations given), the one with the highest ionisation energy is
(a) [Ne] 3s2 3p3
(b) [Ne] 3s2 3p4
(c) [Ne] 3s2 3p5
(d) [Ar] 3d10 4s2 4p2
Show Answer & Explanation

Answer: (c) [Ne] 3s2 3p5

Explanation:
1. These are P, S, Cl and Ge; Cl is the right-most element of period 3 (excluding Ar) and has the highest IE (1251 kJ/mol).

Teacher's Note:
1. IE increases across a period and decreases down a group.

Question 2: Out of Na+, Mg2+, O2- and N3-, the pair showing minimum and maximum IP is
(a) Na+, Mg2+
(b) Mg2+, N3-
(c) N3-, Mg2+
(d) O2-, N3-
Show Answer & Explanation

Answer: (c) N3-, Mg2+

Explanation:
1. These ions are isoelectronic (10 electrons). N3- has the lowest nuclear charge, so its electron is held loosest (minimum IP); Mg2+ has the highest charge (maximum IP).

Teacher's Note:
1. In an isoelectronic series, IP rises with Z.

Question 3: The element having the highest ionisation potential among C, N, O and Si, P, S is
(a) P
(b) N
(c) S
(d) O
Show Answer & Explanation

Answer: (b) N

Explanation:
1. Along a period IE increases, and moving up a group it increases, so N (1402 kJ/mol) is higher than P, and higher than O (1314).

Teacher's Note:
1. N has a half-filled 2p subshell.

Question 4: The strongest reducing agent among the following is
(a) Na
(b) Mg
(c) Al
(d) K
Show Answer & Explanation

Answer: (d) K

Explanation:
1. Reducing power increases with ease of losing electrons (lowest IE). K has the lowest IE and highest reducing character.

Teacher's Note:
1. Metals lower in group 1 reduce more strongly.

Question 5: The following graph shows variation of IP with atomic number in the second period (Li to Ne). The value of IP for Na (Z = 11) will be
(a) Above Ne
(b) Below Ne but above O
(c) Below Li
(d) Between N and O
Show Answer & Explanation

Answer: (c) Below Li

Explanation:
1. Na is the first element of the third period: larger size and lower IE than Li (496 against 520 kJ/mol), so it is below Li.

Teacher's Note:
1. IE decreases down a group.

Question 6: Consider A(g) to A+(g) + e- (IE = A1), A+(g) to A2+(g) + e- (A2) and A2+(g) to A3+(g) + e- (A3). The correct order of IE is
(a) A1 > A2 > A3
(b) A1 = A2 = A3
(c) A1 < A2 < A3
(d) A3 = A2 < A1
Show Answer & Explanation

Answer: (c) A1 < A2 < A3

Explanation:
1. Successive ionisation energies always increase: IE1 < IE2 < IE3.

Teacher's Note:
1. Each electron is removed from a more positive ion.

Question 7: IE1, IE2 and IE3 of an element are 10 eV, 15 eV and 45 eV respectively. The most stable oxidation state of the element will be
(a) +1
(b) +2
(c) +3
(d) +4
Show Answer & Explanation

Answer: (b) +2

Explanation:
1. The big jump from IE2 to IE3 means the third electron comes from a stable core.
2. So the element prefers +2.

Teacher's Note:
1. Look for the largest jump between successive IEs.

Question 8: In the process Cl(g) + e- to Cl-(g), delta H is
(a) Positive
(b) Negative
(c) Zero
(d) None
Show Answer & Explanation

Answer: (b) Negative

Explanation:
1. Energy is released when chlorine gains an electron: delta H = -349 kJ/mol.

Teacher's Note:
1. The first electron affinity of halogens is exothermic.

Question 9: The process in which maximum energy is released is
(a) O to O2-
(b) Mg+ to Mg2+
(c) Cl to Cl-
(d) F to F-
Show Answer & Explanation

Answer: (c) Cl to Cl-

Explanation:
1. Cl to Cl- releases 349 kJ/mol, the largest. O to O2- and Mg+ to Mg2+ absorb energy.
2. F gains 328 kJ/mol.

Teacher's Note:
1. Cl has the highest electron gain enthalpy.

Question 10: The element of which atomic number has the highest electron affinity?
(a) 35
(b) 17
(c) 9
(d) 53
Show Answer & Explanation

Answer: (b) 17

Explanation:
1. Cl (Z = 17) has the highest electron gain enthalpy among halogens (-349 kJ/mol), higher than F because of the small size and electron repulsion in F.

Teacher's Note:
1. The order of EA: Cl > F > Br > I.

Question 11: Which of the following statements about electron affinity is correct?
(a) That of carbon is greater than oxygen
(b) That of fluorine is less than iodine
(c) That of fluorine is less than chlorine
(d) That of sulphur is less than oxygen
Show Answer & Explanation

Answer: (c) That of fluorine is less than chlorine

Explanation:
1. EA (kJ/mol): Cl 349, F 328, I 295: fluorine is less than chlorine.
2. C (122) is less than O (141), and S (200) is more than O.

Teacher's Note:
1. F is anomalous due to its small size.

Question 12: Which of the following elements will form the most stable bivalent anion?
(a) Fluorine
(b) Oxygen
(c) Chlorine
(d) Nitrogen
Show Answer & Explanation

Answer: (b) Oxygen

Explanation:
1. Oxygen needs two electrons to complete its octet and forms O2- most readily among these.

Teacher's Note:
1. Group 16 elements form X2-.

Question 13: The energy released for X(g) + e- to X-(g) is minimum and maximum respectively for (a) F (b) Cl (c) N (d) B
(a) c and a
(b) d and b
(c) a and b
(d) c and b
Show Answer & Explanation

Answer: (d) c and b

Explanation:
1. N has a half-filled stable p subshell and has the least (endothermic) EA; Cl has the maximum EA.

Teacher's Note:
1. Cl > F > B > N in the order of EA.

Question 14: Consider (i) O(g) + e- to O-(g), delta H1; (ii) F(g) + e- to F-(g), delta H2; (iii) Cl(g) + e- to Cl-(g), delta H3; (iv) Na(g) to Na+(g) + e-, delta H4. The incorrect statement is
(a) delta H1 and delta H2 are less negative than delta H3
(b) delta H2 is more negative than delta H1
(c) delta H2 and delta H3 are negative while delta H1 is positive
(d) delta H1, delta H2 and delta H3 are negative while delta H4 is positive
Show Answer & Explanation

Answer: (c) delta H2 and delta H3 are negative while delta H1 is positive

Explanation:
1. delta H1 = -141, delta H2 = -328, delta H3 = -349 and delta H4 = +496 kJ/mol.
2. Hence delta H1 is negative, so statement (3) is incorrect.

Teacher's Note:
1. Remember all first electron gain enthalpies of O, F and Cl are exothermic.

Question 15: In which of the following processes is the least energy required?
(a) F-(g) to F(g) + e-
(b) P-(g) to P(g) + e-
(c) S-(g) to S(g) + e-
(d) Cl-(g) to Cl(g) + e-
Show Answer & Explanation

Answer: (b) P-(g) to P(g) + e-

Explanation:
1. The energy needed equals the electron affinity of the neutral atom: P (72 kJ/mol) is the least; S 200, Cl 349, F 328.

Teacher's Note:
1. Removing an electron from an anion is the reverse of electron gain.

Question 16: The pair with minimum difference in electronegativity is
(a) F, Cl
(b) C, H
(c) P, H
(d) Na, Cs
Show Answer & Explanation

Answer: (c) P, H

Explanation:
1. Pauling values: P = 2.1 and H = 2.1, difference 0.
2. F, Cl 0.8; C, H 0.4; Na, Cs 0.1 (0.9 and 0.7 gives 0.2).

Teacher's Note:
1. P-H bond is almost non-polar.

Question 17: The least electronegative element is
(a) I
(b) Br
(c) C
(d) Cs
Show Answer & Explanation

Answer: (d) Cs

Explanation:
1. Cs has the lowest EN (0.7) among the choices.

Teacher's Note:
1. Electronegativity decreases down a group.

Question 18: ICl is named iodine chloride because
(a) The size of I is less than that of Cl
(b) The atomic number of I is greater than that of Cl
(c) The electronegativity of I is less than that of Cl
(d) The electron affinity of I is less than that of Cl
Show Answer & Explanation

Answer: (c) The electronegativity of I is less than that of Cl

Explanation:
1. In a binary compound the less electronegative element is named first.
2. EN of I (2.5) is less than that of Cl (3.0).

Teacher's Note:
1. The more electronegative atom takes the -ide ending.

Question 19: Among the bonds (a) C-I (b) N-O (c) C-F (d) P-F, the least and most polar are respectively
(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. EN difference: C-I = 0 (2.5 and 2.5), N-O = 0.5, C-F = 1.5, P-F = 1.9.
2. Least: C-I; most: P-F.

Teacher's Note:
1. Compare EN differences from the Pauling table.

Question 20: 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 Pauling (bond energy) and Mulliken (IP and EA) scales are used.

Teacher's Note:
1. Allred-Rochow is a third scale.

Question 21: As we proceed across a period in the periodic table, 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 from left to right; IE, EA and EN generally increase.

Teacher's Note:
1. Same shell, increasing nuclear charge.

Question 22: The electronegativities of 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. Pauling EN: H 2.1, S 2.5, Cl 3.0, O 3.5, F 4.0.

Teacher's Note:
1. Learn F > O > N, Cl > Br > I.

Question 23: Which of the following oxides is different from the other three?
(a) MgO
(b) SnO
(c) PbO
(d) ZnO
Show Answer & Explanation

Answer: (a) MgO

Explanation:
1. MgO is basic, while SnO, PbO and ZnO are amphoteric.

Teacher's Note:
1. Be, Al, Zn, Sn, Pb give amphoteric oxides.

Question 24: Identify the correct order of size of the following
(a) Ca2+ < K+ < Ar < Cl- < S2-
(b) Ar < Ca2+ < K+ < Cl- < S2-
(c) Ca2+ < Ar < K+ < Cl- < S2-
(d) Ca2+ < K+ < Ar < S2- < Cl-
Show Answer & Explanation

Answer: (a) Ca2+ < K+ < Ar < Cl- < S2-

Explanation:
1. All are isoelectronic (18 electrons): size decreases as Z increases.
2. Z: S 16, Cl 17, Ar 18, K 19, Ca 20, so S2- > Cl- > Ar > K+ > Ca2+.

Teacher's Note:
1. Ar being a neutral atom is slightly larger than K+.

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Classification Of Elements and Periodicity In Properties Objective Questions & Solutions for JEE Chemistry

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