Periodic Table

Periodic Table DPP-01

1. Which of the following set of elements obey Newland's octave rule –

(A) Na, K, Rb
(B) F, Cl, Br
(C) Be, Mg, Ca
(D) B, Al, Ga

2. Which of the following is/are Dobereiner triad :-

(a) P, As, Sb (b) Cu, Ag, Au (c) Fe, Co, Ni (d) S, Se, Te
Correct answer is :-
(A) a and b
(B) b and c
(C) a and d
(D) All

3. Which is not anomalous pair of elements in the Mendeleev's periodic table:-

(A) Ar and K
(B) Co and Ni
(C) Te and I
(D) Al and Si

4. Which are correct match :-

(a) Eka silicon – Be (b) Eka aluminium – Ga
(c) Eka manganese – Tc (d) Eka scandium – B
(A) b, c
(B) a, b, d
(C) a, d
(D) All

5. Which of the following set of atomic numbers represent representative element

(A) 5, 13, 30, 53
(B) 11, 33, 58, 84
(C) 5, 17, 31, 54
(D) 9, 31, 53, 83
6. Which statement is wrong for the long form of periodic table :-
(A) Number of periods are 7 and groups 18
(B) No. of valence shell electrons in a period are same
(C) IIIrd B group contains 32 elements
(D) Lanthanides and actinides are placed in same group

7. Which of the following statement is false :-

(A) Elements of ns2np6 electronic configuration lies in 1st to 6th period
(B) Typical elements lies in 3rd period
(C) The seventh period will accommodate thirty two elements
(D) Boron and silicon are diagonally similar

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Periodic Table

8. From atomic number 58 to 71, elements are placed in ;-

(A) 5th period and III A group
(B) 6th period and III B group
(C) separate period and group
(D) 7th period and IV B group

9. The IUPAC name of the element which is placed just after 105Db is the periodic table, will be :-
(A) Unnilpentium
(B) Ununnilium
(C) Unnilhexium
(D) Unnilquadium

10. The element that was present in Ascending position in Lothar – Meyer curve is ..................
(A) Cl
(B) Ca
(C) Si
(D) Mn

11. Atomic weight of an element ‘X’ is 32 and that of element Z is 127. At wt. of their intermediate element
Y, as per Dobereiner triad will be :
(A) 171
(B) 32
(C) 79.5
(D) 85.5

12. The element which occupies peak of Lothar Meyer Curve is :

(A) V
(B) Se
(C) K
(D) La

13. Which of the following pairs shows diagonal relationship ?

(A) Li and Mg
(B) Na and K
(C) Zn and Cd
(D) Li and Be

14. Which element are not called transition element ?

(A) Fe, CO, Ni
(B) W, Cu, Au
(C) Zn, Hg, Cd
(D) Sc, V, Mn

15. Lanthanoids belong to :

(A) Vth period
(B) VIIth period
(C) IIIrd group
(D) IVth group

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Answer C C D A D B A B C A C C A C C

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Periodic Table

Solutions DPP-01
1. Ans. C
Be, Mg, Ca obeys Newland's octave rule.

2. Ans. C
Dobereiner triad is followed by -
(a) P, As, Sb (d) S, Se, Te

3. Ans. D
In option (D) Al and Si is not anomalous pair of elements in the Mendeleev's periodic table.
Atomic mass of Si > Atomic mass of Al

4. Ans. A
Eka silicon – Ge
Eka aluminium – Ga
Eka manganese – Tc
Eka Boron - Sc

5. Ans. D
Representative elements are elements of s and p-block
Atomic No. Electronic configuration
9 [He] 2s2 3p5
10
31 [Ar] 4s2 3d 4p1
53 [Kr] 5s24 d 5p5
10

10 14
83 [Xe] 6s25 d 4f 6p2

6. Ans. B
→ Number of valence shell electrons are same in groups.
→ IIIrd B groups contains 4 transition, 14 Lanthanides & 14 actinides

7. Ans. A
ns𝟐 np𝟔 → 18th group elements general configuration.

8. Ans. B
These elements are Lanthanides which are placed in 6th period and III B group.

9. Ans. C
106 → Unnilhexium

10. Ans. A
Halogens occupy Ascending positions in Lothar – Meyer curve.

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Periodic Table

11. Ans. C
X Y Z
32 A 127
32 + 137 159
A= = = 79.5
2 2

12. Ans. C
Alkali metals occupy top positions at Lothar Meyer curve.

13. Ans. A
Li and Mg show diagonal relationship

14. Ans. C
Zn, Cd, Hg is not transition elements.

15. Ans. C
Lanthanoids belong to III group and VI period.

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Periodic Table

Periodic Table DPP-02

1. Effective nuclear charge in a group generally :-

(A) Increase Down the group
(B) Decrease down the group
(C) Remains constant
(D) First increase than remains constant

2. In Sodium atom the screening is due to:-

(A) 3s 2 , 3p6
(B) 2s1
(C) 1s 2 , 2s 2 , 2p6
(D) 1s 2 , 2s 2

3. Find correct order of penetration power

(A) ns > nd > np > nf
(B) nf > nd > np > ns
(C) ns > np > nd > nf
(D) nf > np > nd > ns

4. Screening effect is not observed in:-

(A) He+
(B) Li+2
(C) H
(D) All of these

5. Value of 𝐙𝐞𝐟𝐟 on Last 𝐞– of N

(A) 3.5
(B) 3.8
(C) 3.9
(D) 3.25

6. If 𝐙𝐞𝐟𝐟 of Li is X then 𝐙𝐞𝐟𝐟 of Boron will be:

(A) 2X
(B) X + 1.95
(C) X + 0.65
(D) X + 1.3

7. Correct order of 𝐙𝐞𝐟𝐟 of different ions of an element (M) is:-

(A) M+ < M < M–
(B) M+ > M > M–
(C) M+ < M > M–
(D) M– > M+ > M

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Periodic Table

8. Correct order of 𝐙𝐞𝐟𝐟 is in following:-

(A) Mg +2 > Na+ > F – > O–2 > N –3
(B) Mg +2 > Na+ > N –3 > O–2 > N –3
(C) Mg +2 < Na+ < F – > O–2 > N –3
(D) Mg +2 < Na+ < F – < O–2 < N –3

9. Value of 𝛔 for last 𝐞– of 𝐂𝐫 +𝟑

(A) 12.75
(B) 5.3
(C) 12.3
(D) 11.25

10. Zeff for 4s electron in 30Zn is :

(A) 1.65
(B) 4
(C) 12.85
(D) 4.35

11. What is the value of effective nuclear charge for last electron of yttrium (Y) ?
(A) 36
(B) 3
(C) 39
(D) 9

12. Which of following subshell electrons apply maximum screening effect ?

(A) s
(B) p
(C) d
(D) f

13. Find the shielding constant () by the first electron enter in H- ?
(A) .70
(B) 1
(C) .35
(D) .30

14. Maximum Number of electrons in sulphur which has same value of 𝐙𝐞𝐟𝐟

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Answer D C C D C D B A C D B A D B

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Periodic Table

Solutions DPP-02
1. Ans. D
Effective nuclear change in a group first increases then remains constant.

2. Ans. C

These electrons will show shielding effect electron present in 3s.

3. Ans. C
Correct order of penetration power is:
ns > np > nd > nf

4. Ans. D
H, He+ , Li+2 all are single electron species. So, there is no other electron available for shielding. That's why
screening effect is not observed in all of these.

5. Ans. C
M = (1s 2 ) (2s 2 2p3 )
Zeff = Z – σ
= 7 – (2 × 0.85 + 4 × 0.35)
= 7 – (3.1)
= 3.9

6. Ans. D
We know from Left to right in periodic table Z (atomic no.) increase by 1, σ increase by 0.35.
So, Zeff = Z – σ increase by 0.65 (applicable only for s & p block)
Li Be B

x + 0.65 x + 0.65 + 0.65

B  x + 1.3

7. Ans. B
Z
Zeff ∝
e
For M, M+, M– value of Z (Atomic number) remains same.
So, on increasing Number of electrons, Zeff decreases
M+ > M > M –

8. Ans. A
Z
Zeff ∝
e
 Given series is isoelectronic series.
So, No. of electron is same, but atomic number (Z) is different.
So, Zeff ∝ Z

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Periodic Table

That why on increasing 'Z', 'Zeff ' also increases.

Mg +2 > Na+ > F – > O–2 > N –3
Z = 12 11 9 8 7

9. Ans. C
Cr = 1s 2 2s 2 2p6 3s 2 3p6 4s1 3d5
Cr +3 = 1s 2 2s 2 2p6 3s 2 3p6 3d3
(1s 2 ) (2s 2 2p6 ) (3s 2 3p6 ) (3d3 )
σ = 1 × 2 + 0.85 × 8 + 0.35 × 10
σ = 12.3

10. Ans. D
Electronic configuration of 30Zn : 1s2 2s2 2p6 3s2 3p6 4s2 3d10
Slater configuration : (1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s2)
 = (1 × 0.35) + (18 × 0.85) + (10 × 1) = 25.65
zeff = z – 
= 30 – 25.65 = 4.35

11. Ans. B
Electronic configuration of 30Zn : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1
Slater configuration : (1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s2 4p6) (4d1) (5s2)
 = (1 × 36) = 36
zeff = z – 
= 39 – 36 = 3

12. Ans. A
Subshell closest to nucleus will apply maximum screening effect.

13. Ans. D
H = 1s1
H- = 1s1
 = 0.30 X 1 = 0.30
Zeff = Z -  = 1 – 0.30 = 0.70
Zeff = 0.70
So  is = 0.30
Ans. is (4)

14. Ans. 8
S → 1s 2 2s 2 2p6 3s 2 3p4

8 electrons which has same value of Zeff .

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Periodic Table

Periodic Table DPP-03

1. The correct order of increasing atomic size of element N, F, Si & P :-

(A) N < F < Si < P
(B) F > N < P < Si
(C) F < N < P < Si
(D) F < N < Si < P

2. Arrange the following in order of increasing atomic radii Na, Si, Al, Ar :-
(A) Na < Si < Al < Ar
(B) Si < Al < Na < Ar
(C) Ar < Al < Si < Na
(D) Na < Al < Si < Ar

3. X-X bond length is 1.00Å and C-C bond length is 1.54Å. If electronegativities of X and C are 3.0 and 2.0
respectively then C-X bond length is likely to be ?
(A) 1.11
(B) 2.1Å
(C) 1.18Å
(D) 2.4Å

4. Arrange the elements in increasing order of atomic radius Na, Rb, K, Mg :-

(A) Na, K, Mg, Rb
(B) K, Na, Mg, Rb
(C) Mg, Na, K, Rb
(D) Rb, K, Mg, Na

5. For the element X, student Mansi Measured its radius as 102 nm, student Rohit as 203 nm and Ankur as
100 nm. Using same-apparatus. Their teacher explained that measurement were correct by saying that
recorded value by three students were :-
(A) crystal, vander wall's and covalent radii
(B) Covalent, Crystal and vander wall's radii
(C) Vander wall's, ionic and covalent radii
(D) No is correct.

6. Which of the following ion has largest size :-

(A) F –
(B) Al+3
(C) Cs +
(D) O–2

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Periodic Table

7. In which of the following pair radius of second species is smaller than that of first specie.
(A) Li, Na
(B) Na+ , F –
(C) N –3 , Al+3
(D) Mn+7 , Mn+4

8. Which of the following order of atomic ionic radius is not correct :-

(A) I– > I > I+
(B) Mg +2 > Na+ > F –
(C) P +5 < P +3
(D) Li > Be > B

9. Which of the following order of radii is correct:-

(A) Li < Be < Mg
(B) H + < Li+ < H –
(C) O < F < Ne
(D) Na+ > F – > O–2

10. Consider the isoelectronic series, 𝐊 + , 𝐒 𝟐− , 𝐂𝐥– 𝐚𝐧𝐝 𝐂𝐚+𝟐 , the radii of the ions decrease as :-
(A) Ca+2 > K + > Cl– > S –2
(B) Cl– > S 2− > K + > Ca+2
(C) S 2 – > Cl– > K + > Ca+2
(D) K + > Ca+2 > S 2− > Cl–

11. Which of the following pair has approximately the same atomic radii ?
(A) Zn and Hf
(B) Al and Ga
(C) (a) and (b) both
(D) Al and Mg

12. The covalent and Vander Waals radii of nitrogen respectively are :
(A) 0.37 Å, 1.2Å
(B) 0.37 Å, 3.7Å
(C) 1.2 Å, 1.2Å
(D) 1.2 Å, 0.37Å

13. The ionic radius of a cation is always :

(A) less than atomic radius
(B) more than atomic radius
(C) equal to atomic radius
(D) cannot be predicted

14. The ionic radii of N3, O2 and F are respectively given by :

(A) 1.36, 1.40, 1.71
(B) 1.36, 1.71, 1.40
(C) 1.71, 1.40, 1.36
(D) 1.71, 1.36, 1.40

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Periodic Table

15. Choose the correct order of ionic radius for the following species :
(A) Cl−  I −  Te2 −  Ar +
(B) Te2 −  I −  Cl−  Ar +
(C) I −  Te2−  Cl−  Ar +
(D) I −  Cl−  Ar +  Te2 −

16. The correct sequence which shows decreasing order of the ionic radii of the element is :
(A) O2−  F−  Na+  Mg2+  Al3+
(B) Al3+  Mg2+  Na+  F−  O2−
(C) Na+  Mg2+  Al3+  O2−  F−
(D) Na+  F−  Mg2+  O2−  Al3+

17. Ionic radii are

(A) Inversely proportional to square of effective nuclear charge
(B) directly proportional to effective nuclear charge
(C) directly proportional to square of effective nuclear charge
(D) Inversely proportional to effective nuclear charge

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Answer C B C C A C C B B C C A A C B A D

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Periodic Table

Solutions DPP-03
1. Ans. C
F < N < P < Si
Generally, Left to right in a period size decreases and down the group size increases.

2. Ans. B
Si < Al < Na < Ar (Due to Vander Waal radius)

3. Ans. C
rc +x = rc + rx – 0.09∆x (Here x = difference in electronegativity i.e. 1)
1.00 1.54
= + – 0.09 (1)
2 2
= 1.27 – 0.09 = 1.18Å
So, C-X bond length = 1.18 Å

4. Ans. C
Mg < Na < K < Rb
Left to right in a period atomic radius decreases and top to bottom in a group atomic radius increases.

5. Ans. A
Covalent radii < Crystal radii < Vander waal's radi

6. Ans. C
Cs + ion has largest size due to large value of 'n'.

7. Ans. C
(i) Li < Na (n ↑ , atomic size ↑)
(ii) Na+ < F – (Zeff For isoelectronic, Z ↑ , size ↓)
(iii) N –3 > Al+3 (Zeff For isoelectronic, Z ↑ , size ↓)
1
(iv) Mn+7 < Mn+4 (size ∝ )
Zeff

8. Ans. B
1
size ∝
Zeff
In option (B) correct order is
F – > Na+ > Mg +2 (For isoelectronic, Z ↑ , size ↑ )

9. Ans. B
1
Size ∝ ∝ Number of shell
Zeff
(A) Li > Be < Mg
(B) H + < Li+ < H –
(C) Ne > O > F
(D) O–2 > F – > Na+ (For iso electronic Z ↑, size ↓)

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Periodic Table

10. Ans. C
For isoelectronic species.
Z ↑, size ↓
S 2– > Cl– > K + > Ca+2

11. Ans. C
Zr and Hf same atomic radii due to lanthanide contraction.
Al and Ga same atomic radii due to lanthanide contraction.

12. Ans. A
Covalent < Vander walls
0.37 Å 1.2 Å

13. Ans. A
Cation has a smaller number of electrons than neutral atom. So, in cation Zeff is more hence, ionic radius of
cation is always less than atomic radius.

14. Ans. C
Order of size (isoelectronic species)
N-3 O-2 F-
10e– 10e– 10e-
Size order : F- < O-2 < N-3

15. Ans. B
For isoelectronic species
Size order : Anion > Natural > Cation

16. Ans. A
For isoelectronic species.
Size  Anion > Neutral > Cation

17. Ans. D
1
Ionic radii 
Z eff

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Periodic Table

Periodic Table DPP-04

1. Decreasing order of size of ions is :

(A) Br – > S –2 > Cl– > N –3
(B) N –3 > S –2 > Cl– > Br –
(C) Br – > Cl– > S –2 > N –3
(D) N –3 – > Cl– > S –2 > Br

2. The correct order of size would be:-

(A) Ni < Pd ≈ Pt
(B) Pd < Pt < Ni
(C) Pt > Ni > Pd
(D) Pd > Pt > Ni

3. Four elements P, Q, R, S have ground state electronic configuration as :-

𝐏 → [𝐍𝐞]𝟑𝐬 𝟐 𝟑𝐩𝟑 ; 𝐐 → [𝐍𝐞]𝟑𝐬 𝟐 𝟑𝐩𝟏 ; 𝐑 → [𝐀𝐫]𝟑𝐝𝟏𝟎 𝟒𝐬 𝟐 𝟒𝐩𝟑 ; 𝐒 → [𝐀𝐫]𝟑𝐝𝟏𝟎 𝟒𝐬 𝟐 𝟒𝐩𝟏
Comment which of the following option represent the correct order of true (T) & False (F) statement.
(i) Size of P < Size of Q
(ii) Size of R < Size of S
(iii) Size of P < Size of R (Appreciable difference)
(iv) Size of Q < Size of S (Appreciable difference)
(A) TTTT
(B) TTTF
(C) FFTT
(D) TTFF

4. Correct order of ionic radii is

(A) Ti4+ < Mn7+
(B) 37 Cl– < 35 Cl–
(C) K + > Cl–
(D) P 3+ > P 5+

5. The size of the following species increases in the order :-

(A) Mg 2+ < Na+ < F –
(B) F – < Na+ < Mg 2+
(C) Mg 2+ < F – < Na+
(D) Na+ < F – < Mg 2+

6. In the following smallest ionic radius of Cr would be of :

(A) K2CrO4
(B) CrO2
(C) CrF3
(D) CrCl3

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Periodic Table

7. Correct order of Atomic radius is :

(A) Mn+2  Mn+5  Mn+7
(B) Mn+5  Mn+2  Mn+7
(C) Mn+7  Mn+5  Mn+2
(D) Mn+5  Mn+7  Mn+2

8. From the given set of species, point out the species from each set having least atomic radius :-
(1) O2 ,F− ,Na+ (2) Ni, Cu, Zn (3) Li, Be, Mg (4) He, Li+, H-
(A) O-2, Cu, Li, H-
(B) Na+, Ni, Be, Li+
(C) F-, Zn, Mg, He
(D) Na+, Cu, Be, He

9. The correct order of radii is :

(A) N < Be < B
(B) F- < O2- < N3-
(C) Na < Li < K
(D) Fe3+ < Fe2+ < Fe4+

10. Which of the following sequence is correct for decreasing order of ionic radius ?
(A) Se-2, I-, Br-, O-2, F-
(B) I-, Se-2, O-2, Br-, F-
(C) Se-2, I-, Br-, F-, O-2
(D) I-, Se-2, Br-, O-2, F-

11. Select the correct order of ionic radius :

(A) Fe3+  Fe2 +  Fe4 +
(B) Ti3+  Ti2+  Ti4 +
(C) Fe4 +  Fe3+  Fe2 +
(D) Ti2+  Ti4 +  Ti3+

12. The set representing the correct order of ionic radius is :

(A) Li+>Be2+>Na+ >Mg2+
(B) Na+ >Mg2+> Be2+> Li+
(C) Li+>Na+ >Mg2+>Be2+
(D) Na+ >Mg2+> Li+> Be2+

13. Which of the following order of atomic radius are correct :-

(A) Li < Be < Na
(B) Ni < Cu < Zn
(C) Ti > V > Cr
(D) Ti > Zr ≈ Hf

14. Size of H– is smaller than how many elements among these: H, Li, Be, B, C, N, O, F, F –

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Answer A A B D A A C B B D C D B,C 0

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Periodic Table

Solutions DPP-04
1. Ans. A
N–3 O–2 F–
S–2 Cl–
Br–
size ∝ no. ofshell(n)
So,

2. Ans. A

3. Ans. B
(i) Size of P3 < Q L → R in periodic table Zeff ↑ A. R. ↓
3p 3p1
(ii) Size of R3 < S1 L → R in periodic table Zeff ↑ A. R. ↓
4p 4p

(iii) Size of P3 < R3 size ∝ no. of shell (n)

3p 4p
(iv) Size of Q < S1 not appreciable difference due to d contraction (Al ≈ Ga)
3p1 4p

4. Ans. D
1
(size ∝ )
Zeff

Ti+4 > Mn+7 ; K + < Cl– {For iso electronic ; σ is constant, so when Z increases then Z eff increases and size
decreases.
37 −
Cl = 35 Cl– (e– & P are same)
1
P 3+ > P 5+ (size ∝ ).
Zeff

5. Ans. A
For iso electronic species
1
(size ∝ )
Zeff

Mg 2+ < Na+ < F −

6. Ans. A
K2CrO4 = K+ + CrO−42

CrO2 = Cr +2 + O2−2
CrF3 = Cr +3 + F−
CrCl3 = Cr +3 + Cl−
As positive charge increases, ionic radius decreases.

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Periodic Table

7. Ans. C
Mn+7  Mn+5  Mn+2

8. Ans. B
(1) O2 ,F− ,Na+ → More Zeff, small size.
(2) Ni, Cu, Zn → Ni has more Zeff, so small size.
(3) Li, Be, Mg → Be has more Zeff, small size.
(4) He, Li+, H– → Li+ has more Zeff, small size.

9. Ans. B
For Isoelectronic series, ionic radii depends on Z/e- value [where Z→ Atomic number and e- is number of
electrons]
Z
Higher the value of smaller the ionic radii.
e−

10. Ans. D
No. of shells  Atomic Radius
- ve charge  size
I − Se−2  Br − O −2  F−
  (No. of shells)
5 4 2

11. Ans. C
Positive charge  Ionic Radius 
So, Ans is Fe4 +  Fe3+  Fe2 +

12. Ans. D

13. Ans. B,C

14. Ans. 0
H– has greatest size among given species.

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Periodic Table

Periodic Table DPP-05

1. Factor that has no effect on the ionisation potential:

(A) atomic size
(B) atomic mass
(C) Effective nuclear charge
(D) Shielding effect

2. (a) 𝐌(𝐠)
–
→ 𝐌(𝐠) (b) 𝐌(𝐠) → 𝐌(𝐠)
+
(c) 𝐌(𝐠)
+ +𝟐
→ 𝐌(𝐠) (d) 𝐌(𝐠)
+𝟐 +𝟑
→ 𝐌(𝐠)
Minimum and maximum Ionisation Potential would be of :-
(A) a, d
(B) b, c
(C) c, d
(D) d, a

3. Correct order of first ionisation energy is :-

(a) Li < B < Be < C (b) O < N < F (c) Be < N < Ne
(A) a, b
(B) b, c
(C) a, c
(D) a, b, c

4. Successive ionisation energies of an element 'X' are given below (in kcal/mol)
𝐈𝐏𝟏 𝐈𝐏𝟐 𝐈𝐏𝟑 𝐈𝐏𝟒
165 195 556 595
Electronic configuration of the element 'X' is:-
(A) 1s 2 , 2s 2 2p6 , 3s 2 3p2
(B) 1s 2 , 2s1
(C) 1s 2 , 2s 2 2p2
(D) 1s 2 , 2s 2 2p6 , 3s 2

5. Out of 𝐍𝐚+ , 𝐌𝐠 +𝟐 , 𝐎–𝟐 𝐚𝐧𝐝 𝐍 –𝟑 , the pair of species showing minimum and maximum IP would be.
(A) Na+ , Mg +2
(B) Mg +2 , N –3
(C) N –3 , Mg +2
(D) O–2 , N –3

6. The maximum second ionisation energy is of : -

(A) Mn
(B) Sc
(C) Cr
(D) Ti

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Periodic Table

7. The strongest reducing agent among the following is :-

(A) Na
(B) Mg
(C) Al
(D) K

8. Write correct order of non-metallic characteristics :-

(A) F < Cl < Br < I
(B) F < Cl > Br > I
(C) F > Cl > Br > I
(D) F > Br > I > Cl

9. Correct order of I.P is :

(A) Zn > Mn
(B) Xe > Ne
(C) Sc > Mn
(D) Na > As

10. Select the correct order of 2nd I.P of C.N.O.F :

(A) C > N > O > F
(B) F > O > N > C
(C) O > F > N > C
(D) C > O > N > F
11. Choose the correct order of ionization energy for the following species :
(A) Sc > La > Y
(B) Sc > Y  La
(C) Sc > Y > La
(D) Sc < Y > La

12. The decreasing order of the ionization potential of the following elements is :
(A) Ne >Cl > P > S > Al > Mg
(B) Ne > Cl > P > S > Mg > Al
(C) Ne > Cl > S > P > Mg > Al
(D) Ne > Cl > S > P > Al > Mg

13. The third ionisation energy is maximum for :

(A) Nitrogen
(B) Phosphorous
(C) Aluminium
(D) Boron

14. Select the incorrect order of I.E.:-

(A) Cl– > Cl > Cl+
(B) Cl+ > Cl > Cl–
(C) Cl > Cl+ > Cl–
(D) Cl– > Cl+ > Cl

15. Consider ionization of an element "E" :

+𝐈𝐏𝟏 +𝐈𝐏𝟐 +𝐈𝐏𝟑 +𝐈𝐏𝟒 +𝐈𝐏𝟓
+ 𝟐+ 𝟑+ 𝟒+ ∗
𝐄(𝐠) → 𝐄(𝐠) → 𝐄(𝐠) → 𝐄(𝐠) → 𝐄(𝐠) → 𝐄(𝐠)
If E+5 does not have any electron. Find number of electrons in valence shell of element "E".

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Answer B A D D C C D C A C C B A A,C,D 3

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Periodic Table

Solutions DPP-05
1. Ans. B
Ionisation potential does not depend on atomic mass.

2. Ans. A
Successive Ionisation Potential of an element is always greater than previous one

(a) M(−g) ⎯⎯⎯⎯

→M(g)
( )
IP1 M−

(b) M(g) ⎯⎯⎯⎯

→M(+g)
( )
IP2 M−

( )
IP3 M−
→M2(g+)
(c) M(+g) ⎯⎯⎯⎯

(d) M2(g+) ⎯⎯⎯⎯

( )
IP4 M−
→ M3(g+)

IP4 > IP3 > IP2 > IP1

Minimum IP → IP1 → a
Maximum IP → IP4 → d

3. Ans. D
Li Be B C N O F Ne
L → R Z eff  
2s1 2s2 2p1 2p2 2p3 2p 4 2p5 2p6  I.E.  
 
 2 1 
s2 >p1 p3 >P 4 (half filled) s >p  
3 4

p >p  

Li < B < Be < C < O < N < F < Ne → I. E.1

4. Ans. D
165, 195, 556, 595
Here we get large Jump in I.E. value after removal of second electron so, it is clear that removal of third electron
requires more energy. Hence, element 'X' should have 2 electron in its outer most shell.
So, electronic configuration of X → 1s2 2s2 2p6

5. Ans. C
N –3 O–2 Na+ Mg +2 → isoelectronic series
Z 7 8 10 12
e 10 10 10 10
As value of Z increases, then Zeff increases which cause increase in IP.
N 3− < O2− < Na+ < Mg 2+
N 3− minimum
Mg 2+ maximum

6. Ans. C
Sc Ti V Cr Mn
4s 2 3d1 4s 2 3d2 4s 2 3d3 4s1 3d5 4s 2 3d5

↓ – e–
4s1 3d1 4s1 3d2 4s1 3d3 4s 0 3d5 4s1 3d5
↓
Due to half filled subshell more energy required to
remove an electron from this.

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Periodic Table

7. Ans. D
1
We know that Reducing power ∝
I.E

Na Mg Al
K
I.E. → Mg > Al > Na > K
——

Minimum I.E
Maximum Reducing power. Hence, strong reducing agent.

8. Ans. C

F > Cl > Br > I

9. Ans. A
(A) IP of Zn > Mn
d10 d5
(B) IP of Xe > Ne
(C) IP of Sc > Mn
d1 d5
(D) IP of Na > As
s1 p3

10. Ans. C
Second Ionisation potential order : Be < C < B < N < F < O < Ne < Li

11. Ans. C
1
I.P 
size
12. Ans. B
Mg > Al P > S Noble gases having
s2 p1 p3 p4 maximum due to ns2np6
Stable stable stable configuration

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Periodic Table

13. Ans. A
I.P.
B N

Al P

N+2 P +2 Al+2 B+2

2p1 3p1 3s1 2s1

14. Ans. A,C,D

Cl+ Cl Cl– → isoprotonic
Z 17 17 17
e 16 17 18
Z
↑ Zeff ↑ IE ↑
e
So, Cl+ > Cl > Cl− → Correct
But incorrect will be → A, C, D

15. Ans. 3
E5+ contains zero electron. So, total electrons in E = 5
Electronic configuration = 1s 2 2s2 2p1
Valence shell electrons = 3

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Periodic Table

Periodic Table DPP-06

1. Process in which maximum energy is released :-

(A) O → O–2
(B) Mg + → Mg +2
(C) Cl → Cl–
(D) F → F –

2. (𝐠) ; ∆𝐇𝐞𝐠 = 𝟕𝟒𝟒. 𝟕 𝐊𝐉/𝐦𝐨𝐥𝐞. The positive value of ∆𝐇𝐞𝐠 is due to :-

𝐎(𝐠) + 𝟐𝐞– → 𝐎𝟐−
(A) Energy is released to add 1 e– to O–1
(B) Energy is required to add 1 e– to O–1
(C) Energy is needed to add 1e– to O
(D) None of the above is correct

3. Electron affinity is a :-
(A) Relative strength to attract the shared electron pair
(B) Necessary energy required to remove the electron from the ultimate orbit
(C) Energy released when an electron is added to the outermost shell
(D) Energy released when an electron is added to the inner shell

4. Which of the following electronic configuration is expected to have highest electron affinity :-
(A) [He]2s 2 2p0
(B) [He]2s 2 2p2
(C) [He]2s 2 2p3
(D) [He]2s 2 2p1
5. Select the correct code in terms of true (T) and false (F), respectively.
(i) Process of adding electron to an atom may be exothermic or endothermic.
(ii) In most of the elements energy is released when an electron is added to the atom.
(iii) Halogens have maximum ∆𝐇𝐞𝐠 in their respective period because of their stable electronic
configuration.
(iv) Electron affinity order S < O
(v) ∆𝐇𝐞𝐠 is maximum positive for inert gases in their respective period
(A) T F F T F
(B) F T T F T
(C) T T F F T
(D) F F F T F

6. Correct order of ∆𝐇𝐞𝐠 :-

(A) F + > Cl+ > Cl > F
(B) F + > Cl > Cl+ > F
(C) F + > F > Cl+ > Cl
(D) Cl > F > F + > Cl+

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Periodic Table

7. If ∆𝐇𝐞𝐠 of
(Oxygen) O = a KJ/mol 𝐎+ = b KJ/mol
(Sulphur) S = c KJ/mol 𝐒 + = d KJ/mol
Then find correct order of ∆𝐇𝐞𝐠 ?
(A) a>c>b>d
(B) a>c>d>b
(C) b>d>c>a
(D) b>d>a>c

8. Ionisation energy of F- is 320 kJmol-1. The electron gain enthalpy of fluorine would be :
(A) -320 KJ/mol
(B) -160 KJ/mol
(C) +320 KJ/mol
(D) +160 KJ/mol

9. Correct order of decreasing tendency to form anion :

(A) Cl, F, S, O
(B) S, O, P, N
(C) Si, C, P, N
(D) Cl, S, Si, P

10. Which of the following order is incorrect ?

(A) Be < B < Li < C (electron affinity)
(B) B < Be < C < N (ionization potential)
(C) B < C < O < F (electron affinity)
(D) Ne < O < N < F (electron affinity)

11. The formation of O(g)

2-
starting from O(g) is endothermic by 639 kJ mol-1. If electron gain enthalpy of O(g)

is -141 KJ mol-1 the second electron gain enthalpy of oxygen would be :

(A) -780 kJ mol-1
(B) +780 kJ mol-1
(C) -498 kJ mol-1
(D) -498 kJ mol-1

12. Correct electron gain enthalpy order is :

(A) N < O < Cl < Al
(B) O < N < Al < Cl
(C) N < Al < O < Cl
(D) Cl < N < O < Al

13. The element having very high ionization enthalpy but zero electron gain enthalpy is :
(A) H
(B) F
(C) He
(D) Be

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Periodic Table

14. The following of the oxide ion O2- (g) requires first an exothermic and then an endothermic step as shown
below :
O(g) + e− → O− (g); H = − 142kJmol−1
O(g) + e− → O− (g); H = 884kJmol−1
(A) O- ion has comparatively larger size than oxygen atom
(B) Oxygen has high electron affinity
(C) O- ion will tend to resist the addition of another electron\
(D) Oxygen is more electronegative

15. The Incorrect order of property of the following is :

(1) Basic nature Na2O > MgO > Al2O3 > SiO2 (2) 2nd I.P. Na > S > P > Si
(3) Electron affinity O > S > Se (4) Size I- < I < I+
(A) 1, 3, 4
(B) 2, 3, 4
(C) 3, 4
(D) 1, 2, 3

16. Consider the following sequence of reaction

If electronic configuration of element A is [𝐍𝐞]𝟑𝐬 𝟏 , then which of the following order is correct regarding
given enthalpies ?
(A) |∆H4 | = |∆H5 |
(B) |∆H2 | > |∆H1 |
(C) |∆H2 | > |∆H3 |
(D) |∆H1 | = |∆H6 |

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Answer

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Periodic Table

Solutions DPP-06
1. Ans. C
EA of Cl > EA of F

2. Ans. B
O(g) +e- ⎯⎯
→ O-(g) ; ΔH=-x
O-(g) +e- ⎯⎯ 2-
→ O(g) ; ΔH=+y → Energy is required to add one electron to O1-
O(g) +2e- ⎯⎯
→ O-2
(g) ; ΔH= y - x =positive → Because y > x

3. Ans. C
It is amount of energy released when an electron is added to the outermost shell of an isolated gaseous atom.

4. Ans. B
When electron is added to [He]2s22p2 then the configuration formed is very stable [He] 2s22p3. Thus, energy
released for this addition of clectron will be maximum.

5. Ans. C
(i) Electron gain Enthalpy may be exothermic or endothermic.
(ii) Addition of first electron is exothermic process for most of the elements.
(iii) Halogens have maximum ∆Heg in their respective period as ∆Heg increase across of the period Due to
more Zeff and not due to stable configuration.
(iv) S > O; E A (3rd period > 2nd period)
(v) ∆Heg = positive for noble gases due to their stable configuration.

6. Ans. A
(A) ∆Heg order
F < F + } Positive charge More attraction
Cl < Cl+
1
I.E. of F > I.E. of Cl I.E. 
no.of shell
& |I. E of F| = |ΔHeg of F + |
|I. E of Cl| = |ΔHeg of Cl+ |
So, ∆Heg of F + > ∆Heg of Cl+
∆Heg of Cl > ∆Heg of F
Correct order of ∆Heg ⟹ F + > Cl+ > Cl > F

7. Ans. D
ΔHeg of O < O+
} Positive charge → More Attraction → more ΔHeg
ΔHeg of S < S +
∆Heg of S > O
I.E. of O > I.E. of S
|I. E. of O| = |ΔHeg of O+ |
|I. E. of O| = |ΔHeg of O+ |
|I. E. of S| = |ΔHeg of S + |
 Heg of O+ > ∆Heg of S+
Then correct order
O+ > S + > S > O (∆Heg )
b>d>c>a

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Periodic Table

8. Ans. A
IE of F- = - Heg of F
So ans is -320 KJ/mol

9. Ans. D
Across the period, electron affinity increases. Also, electron affinity of fully filled and half-filled configuration is
minimum. So, correct order is : Cl > S > Si > P

10. Ans. D
Li Be B C N O F
1s 2s
2 1
1s 2s
2 2
1s 2s 2p
2 2 1
1s 2s 2p
2 2 2
1s 2s 2p
2 2 3
1s 2s 2p
2 2 4
1s 2s 2p5
2 2

Li →gaining one e–, it completed full filled configuration while boron does not therefore Li has more electrons
affinity then B.
Be → Full filled, E. A – almost Zero
B → I.P. of Be is more as compare to B due to full filled configuration
C → gaining one e– it is completed half-filled configuration
N → half-filled easily not accept the electron
F → gaining one e– it is completed full filled configuration

11. Ans. B
Given:
O (g) + 2e– ⎯→ O2– (g) Heg1 = 639 kJmol–1
O (g) + e– ⎯→ O– (g) Heg2 = –141 kJmol–1
Subtracting: O– (g) + e– ⎯→ O2– (g) Heg1 = 639 + 141 = 780 kJmol–1

12. Ans. C
Electron gain enthalpy of N < O and Al < Cl also N < Al as nitrogen is half filled.
We know
Heg O > B and B > Al
So, Heg O > Al
Also, Heg Cl > O
Hence, overall order : N < Al < O < Cl

13. Ans. C
He, having very high Ionisation enthalpy but zero electron gain enthalpy because 1s is completely filed.

14. Ans. C
2nd EA is always endothermic process due to repulsion between negatively charged electron and O – ion.

15. Ans. C
1) Basic nature decreases across the period.
2) Correct second ionisation potential order : Na > S > P > Si

16. Ans. A,B,D

(A) |∆H4 | = |∆H6 | [(ΔHeg )A = (ΔHIE )A− ]
(B), (C) → ∆H3 > ∆H2 > ∆H1 [Successive I.E. always higher than previous I.E.]
(D) |∆H1 | = |∆H6 | [(ΔHI.E. )A = (ΔHeg )A+ ]

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Periodic Table

Periodic Table DPP-07

1. Select the group which shows increasing order of EN

(A) Ca, Sr, Ba
(B) Si, C, Cl
(C) F, Cl, Br
(D) Li, Na, K

2. Increasing order of electronegativity is :-

(A) Bi < P < S < Cl
(B) P < Bi < S < Cl
(C) S < Bi < P < Cl
(D) Cl < S < Bi < P

3. The correct set of decreasing order of electronegativity is :-

(A) Li, H, Na
(B) Na, H, Li
(C) H, Li, Na
(D) Li, Na, H

4. 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

5. 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

6. % Ionic character of compound AB is

ENA = 1.8 ENB = 2.6
(A) 42.42%
(B) 24.24%
(C) 15.04%
(D) 17.08%

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Periodic Table

7. Calculate % ionic character in most ionic compound Cs-F

(A) 100%
(B) 93.225%
(C) 97.225%
(D) 98.225%
(ENs = 0.7 ENF = 4)

8. The values of electronegativity of atoms A and B are 1.20 and 4.0 respectively. The percentage of ionic
character of A–B bond is :
(A) 50%
(B) 72.24%
(C) 55.3%
(D) 43%

9. The electronegativity of A is 0.7 and that of B is 4.0. The bond formed between is :
(A) Electrovalent
(B) Covalent
(C) Metallic
(D) None

10. Which of the following is most electronegative ?

(A) Lead
(B) Silicon
(C) Carbon
(D) Tin

11. Compound XY is predominantly ionic as X+Y– if

(A) (IE)x < (IE)y
(B) (EA)x < (EA)y
(C) (EN)x < (EN)y
(D) (IE)x < (IE)y

12. 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

13. The pair with minimum difference in electronegativity is :

(A) F, Cl
(B) C, H
(C) P, H
(D) Na, Cs

14. Which of the following relation is correct ?

(A) 2 IP – EA – EN = 0
(B) 2 IP – EN + EA = 0
(C) 2 EN – IP – EA = 0
(D) EN – IP – EA = 0

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Periodic Table

Passage 15 to 19
Ionization energies of five elements in Kcal/mol are given below

15. Which elements is a noble gas ?

(A) P
(B) T
(C) R
(D) S

16. Which element form stable unipositive ion ?

(A) P
(B) Q
(C) R
(D) S

17. The element having most stable oxidation state +2 is ?

(A) Q
(B) R
(C) S
(D) T

18. Which is a more E.N element ?

(A) P
(B) Q
(C) R
(D) S

19. Element which is most electropositive ?

(A) R
(B) T
(C) Q
(D) S

20. I.P. of an element is equal to 14.2 eV and EA is equal to 2.6 eV, then find electronegativity according to
Pauling scale.

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Answer B A C D B C B B A C C C C C B B C A C 3

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Periodic Table

Solutions DPP-07
1. Ans. B
(EN)C > (EN)Si ; (EN)Cl >(EN)C

2. Ans. A
Order of EN : Bi < P < S < Cl

3. Ans. C
EN order H > Li > Na

4. Ans. D
F > O > Cl > S > H (From Pauling scale)

5. Ans. B
As electronegativity difference increases then bond polarity increases.
C – I = |2.5 – 2.55| = 0.05 minimum bond polarity
P – F = |2.1 – 4| = 1.9 Maximum bond polarity

6. Ans. C
% Ionic character = 16|∆EN| + 3.5 (∆EN)2
 ∆EN = 2.6 – 1.8 = 0.8
= 16 |(0.8)| + 3.5 (0.8)2
= 15.04%

7. Ans. B
EN = 4 – 0.7 = 3.3
% ionic character = 16 |∆EN| + 3.5 (∆EN)2
= 16 |3.3| + 3.5(3.3)2
= 93.225%

8. Ans. B
%Ionic character = 16 + 3.52
 = | X A − XB |
 = | 4 − 1.20 |
 = 2.8
%I.C. = 16 x 2.8 + 3.5 x (2.8)2

= 44.8 + 27.44
= 72.24%

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Periodic Table

9. Ans. A
EN = 4 – 0.7 = 3.3
as EN > 2.1
So, electrovalent bond.

10. Ans. C
Carbon is more Electronegative among given elements.

11. Ans. C
X+Y- is ionic
So (EN)x < (EN)y

12. Ans. C
Element with more electronegativity is written in the end. So, iodine chloride.

13. Ans. C
(A) F, Cl  EN = | 4 – 3.16 | = 0.84
(B) C, H  EN = | 2.55 – 2.2 | = 0.35
(C) P, H  EN = | 2.19 – 2.2 | = 0.01
(D) Na, Cl  EN = | 0.93 – 0.79 | = 0.14
EN for P-H is minimum among the given pairs.

14. Ans. C
IP + EA
EN =
2
So, 2 EN – IP – EA = 0

15. Ans. B
Ist IE of noble gas is very high

16. Ans. B
The element with low value of first IE and very high value of second IE forms stable unipositive ion.

17. Ans. C
The element having low value of IE for removal of initial two electrons and very large IE value for removal of
third electron, will form stable oxidation state +2.

18. Ans. A
The element with highest first ionisation energy among the given is more electronegative.

19. Ans. C
Most electropositive : Lowest 1st IE

20. Ans. 3
IP + EA 14.2 + 2.6 16.8
XP = = = =3
5.6 5.6 5.6

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Periodic Table

Periodic Table DPP-08

1. Correct Order of acidic Character ?

(A) ClOH < IOH < BrOH
(B) ClOH < BrOH < IOH
(C) IOH < BrOH < ClOH
(D) None of these

2. Which oxide of Manganese is most acidic in nature ?

(A) MnO
(B) Mn2 O7
(C) Mn2 O3
(D) MnO2

3. In periodic table, the basic character of oxides ?

(A) Increase from left to right and decreases from top to bottom
(B) decreases from right to left and increases from top to bottom
(C) decreases from left to right and increases from top to bottom
(D) Increases from left to right and decreases from top to bottom

4. Which of the following does not act as an amphoteric oxide ?

(A) Al2 O3
(B) ZnO
(C) BeO
(D) NO

5. Least basic oxide is :-

(A) Fe2O3
(B) FeO
(C) BaO
(D) Na2O

6. Which of the following sequence correctly represent the decreasing acid nature of oxides ?
(A) Li2O > BeO > B2O3 > CO2 > N2O3
(B) N2O3>CO2>B2O3>BeO>Li2O
(C) CO2 > N2O3 > B2O3 > Li2O
(D) B2O3 > CO2 > N2O3 > Li2O > BeO

7. The most basic among these hydroxides is :

(A) Be(OH)2
(B) Mg(OH)2
(C) Ca(OH)2
(D) Ba(OH)2

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Periodic Table

8. Which one of the following oxides is neutral ?

(A) CO
(B) SnO2
(C) ZnO
(D) SiO2

9. An element ‘X’ which occurs in the first short period has an outer electronic structure s 2p1. What is the
formula and acid -base character of its oxides :
(A) XO3, basic
(B) X2O3 basic
(C) X2O3 acidic
(D) XO2 acidic

10. Correct basic nature order is :

(A) Na2O < K2O < MgO < Al2O3
(B) Al2O3 < MgO < Na2O < K2O
(C) K2O < Na2O < Al2O3
(D) MgO < K2O < Al2O3 < Na2O

11. Which is incorrect order ?

(A) H2O < H2S < H2Se < H2Te (acidic nature)
(B) H2O < H2S < H2Se < H2Te (basic nature)
(C) HF < HCl < HBr < HI (acidic nature)
(D) HF > HCl > HBr > HI (basic nature)

Paragraph for Question 12 to 13

Nature of bond can be predicted on the basis of electronegativity of bonded atoms, greater difference
in electronegativity (X), more will be the polarity of bond, and polar bond are easily broken in polar
solvent like water. For hydroxy acids 𝐗 𝐎 – 𝐗 𝐀 difference predict the nature of oxide formed by the
element A.
|𝐗 𝐎 – 𝐗 𝐀 | > |𝐗 𝐎 – 𝐗 𝐇 | then A – O – H show basic nature (NaOH)
|𝐗 𝐎 – 𝐗 𝐀 | < |𝐗 𝐎 – 𝐗 𝐇 | then A – O – H show acidic nature (H – O – Cl)
With the help of EN values [𝐄𝐍𝐀 = 𝟏. 𝟖, 𝐄𝐍𝐁 = 𝟐. 𝟔, 𝐄𝐍𝐂 = 𝟏. 𝟔, 𝐄𝐍𝐃 = 2.8] answer the following questions
for the compounds HAO, HBO, HCO, HDO.

12. Compounds whose aqueous solution is acidic and order of their acidic strength
(A) AOH, COH ; AOH < COH
(B) HDO, HBO ; HDO > HBO
(C) AOH, COH ; AOH > COH
(D) HDO, HBO ; HDO < HBO
13. Compounds whose aqueous solution is basic and order of their basic strength
(A) AOH, COH ; AOH < COH
(B) HDO, HBO ; HDO > HBO
(C) AOH, COH ; AOH > COH
(D) HDO, HBO ; HDO < HBO

14. Select total number of acidic compounds out of given below.

𝐂𝐬𝐎𝐇, 𝐎𝐂(𝐎𝐇)𝟐 , 𝐒𝐎𝟐 (𝐎𝐇)𝟐 , 𝐒𝐫(𝐎𝐇)𝟐 , 𝐂𝐚(𝐎𝐇)𝟐 , 𝐁𝐚(𝐎𝐇)𝟐 , 𝐁𝐫𝐎𝐇, 𝐍𝐚𝐎𝐇, 𝐎𝟐 𝐍𝐎𝐇

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Periodic Table

Answer Key

Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Answer C B C D A B D A C B B B A 4

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Periodic Table

Solutions DPP-08
1. Ans. C
ClOH > BrOH > IOH (Same group)
Generally, In a group top to bottom electronegativity decreases so Acidic Character decreases.

2. Ans. B
Order of Acidic strength
Mn2 O7 > MnO2 > Mn2 O3 > MnO
more electronegativity  more acidic strength

3. Ans. C

4. Ans. D
Oxides of Zn, Al, Be , Ga, Pb, Sn are amphoteric in nature.

5. Ans. A
Metal Oxides are Basic in nature.
Na2O > BaO > FeO > Fe2O3

6. Ans. B
Across the period, EN increases so acidic strength increases.

7. Ans. D
In M(OH)2
Lower the electronegativity of M, more will be basic strength.

8. Ans. A
CO is neutral oxide.

9. Ans. C
Valence electrons of X = s2p1  Cation formed = X3+
So, oxide formed = X2O3.
Across the period, acidic strength increases, so this oxide will be acidic.

10. Ans. B
Down the group, basic nature increases and across the period, basic nature decreases.
So, Al2O3 < MgO < Na2O < K2O

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Periodic Table

11. Ans. B
Down the group, acidic strength of hydra acids increases.

12. Ans. B
A + O − H → A+ + O̅H (Base)
A − O − H → AO + H +
–
(Acid)
AOH |XO – XA ||XO – XH |
|3.5 – 1.8| |3.5 – 2.1|
1.7 > 1.4 Base
BOH |XO – XB ||XO – XH |
|3.5 – 2.6| 1.4
0.9 < 1.4 Acid
COH |XO – XC ||XO – XH |
1.9 > 1.4 (Base)
DOH |XO – XD ||XO – XH |
0.7 > 1.4 (Acid)
(D-O bond is less polar than B-O)

13. Ans. A
A − O − H → A+ + OH – Base
– ⊕
A − O − H → AO + H Acid
AOH |XO – XA ||XO – XH |
|3.5 – 1.8| |3.5 – 2.1|
1.7 > 1.4 Base
BOH |XO – XB ||XO – XH |
|3.5 – 2.6| 1.4
0.9 < 1.4 Acid
COH |XO – XC ||XO – XH |
1.9 > 1.4 (Base)

DOH |XO – XD ||XO – XH |

0.7 > 1.4 (Acid)
A (C-O bond is more polar than A-O)

14. Ans. 4
Basic → CsOH, Sr(OH)2 , Ca(OH)2 , Ba(OH)2 , NaOH
OC(OH)2 → H2 CO3 → Carbonic Acid (Acid)
SO2 (OH)2 → H2 SO4 → Sulphuric Acid (Acid)
BrOH → Acidic
O2 NOH → HNO3 → Nitric acid (Acid)

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