Classification of Elements and Periodicity in Properties — DPP 1

20 NEET questions from Classification of Elements and Periodicity in Properties, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

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1
The correct order of second ionisation potential for the given elements is:
  1. A $\mathrm { C } > \mathrm { N } > \mathrm { O } > \mathrm { F }$
  2. B $\mathrm { O } > \mathrm { N } > \mathrm { F } > \mathrm { C }$
  3. C $\mathrm { O } > \mathrm { F } > \mathrm { N } > \mathrm { C }$
  4. D $\mathrm { F } > \mathrm { o } > \mathrm { N } > \mathrm { C }$
Show answer & explanation

Correct answer: C

C$^{+}$ : 1s$^{2}$2s$^{2}$2p$^{1}$
N$^{+}$ : 1s$^{2}$2s$^{2}$2p$^{2}$
O$^{+}$ : 1s$^{2}$2s$^{2}$2p$^{3}$
F$^{+}$ : 1s$^{2}$2s$^{2}$2p$^{4}$
Generally, IE$_{2}$ increase along a period but IE$_{2}$ of O+ is more than F+ due to extra half filled p subshell.
Therefore, option (3) is the correct answer.
2
Which one of the following is not the correct statement?
  1. A $\mathrm { IE } _ { 1 }$ of He is the maximum among all elements.
  2. B $\Delta _ { \mathrm { eg } } \mathrm { H } _ { 1 }$ for noble gases is zero/positive.
  3. C Electronegativity is maximum for fluorine.
  4. D $\mathrm { IE } _ { 1 }$ for nitrogen is less than that of oxygen.
Show answer & explanation

Correct answer: D

IE$_{1}$ of N is greater than IE$_{1}$ of O because of the stable halffiled electronic confiuration of N.
Therefore, option (4) is the correct answer
3
Amphoteric behaviour is shown by the oxides of:
  1. A Na and Mg
  2. B Pb and Ba
  3. C Cr and Mg
  4. D Ga and Al
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Correct answer: D

4
The element having highest electronegativity;
  1. A has period number 3 and group number 16.
  2. B has electronic configuration $[ \mathrm { Ne } ] 3 \mathrm {s} ^ { 2 } 3 \mathrm { p } ^ { 3 }$.
  3. C has 7 valence electrons and belongs to same period as that of Li .
  4. D has 6 electrons in their p-subshell and belongs to same period as that of Mg .
Show answer

Correct answer: C

5
The atomic volume was chosen as the basis of the periodic classification of elements by;
  1. A Niels Bohr
  2. B Mendeleev
  3. C Lothar Meyer's
  4. D Newlands
Show answer

Correct answer: C

6
Which among $\mathrm { P } , \mathrm { S } , \mathrm { Cl }$ and F will have the most negative electron gain enthalpy and which is the least negative?
  1. A $\mathrm { F } , \mathrm { Cl }$
  2. B $\mathrm { Cl } , \mathrm { F }$
  3. C $\mathrm { S } , \mathrm { Cl }$
  4. D $\mathrm { Cl } , \mathrm { P }$
Show answer

Correct answer: D

7
The successive ionisation energies for an unknown element are:
IE$_{1}$ = 899 kJ/mol
IE$_{2}$ = 1757 kJ/mol
IE$_{3}$ = 14,847 kJ/mol
IE$_{4}$ = 17,948 kJ/mol
To which family in the periodic table does the unknown element most likely belong?
  1. A Carbon family
  2. B Boron family
  3. C Alkaline earth metal family
  4. D Nitrogen family
Show answer & explanation

Correct answer: C

There is a large jump in ionisation energy from the second to the third one. Alkaline earth metals have the electronic confiuration [noble gas]ns$^{2}$; the third electron is to be removed from the highly stable electronic confiuration i.e., [noble gas] which will require very high energy.
Therefore, option (3) is the correct answer
8
The correct order of the ionic radii of $\mathrm { O } ^ { 2 - } , \mathrm { N } ^ { 3 - } , \mathrm { F } ^ { - }$, $\mathrm { Mg } ^ { 2 + } , \mathrm { Na } ^ { + }$and $\mathrm { Al } ^ { 3 + }$ is;
  1. A $\mathrm { N } ^ { 3 - } < \mathrm { O } ^ { 2 - } < \mathrm { F } ^ { - } < \mathrm { Na } ^ { + } < \mathrm { Mg } ^ { 2 + } < \mathrm { Al } ^ { 3 + }$
  2. B $\mathrm { Al } ^ { 3 + } < \mathrm { Na } ^ { + } < \mathrm { Mg } < \mathrm { O } ^ { 2 - } < \mathrm { F } ^ { - } < \mathrm { N } ^ { 3 - }$
  3. C $\mathrm { Al } ^ { 3 + } < \mathrm { Mg } ^ { 2 + } < \mathrm { Na } ^ { + } < \mathrm { F } ^ { - } < \mathrm { O } ^ { 2 - } < \mathrm { N } ^ { 3 - }$
  4. D $\mathrm { N } ^ { 3 - } < \mathrm { F } ^ { - } < \mathrm { O } ^ { 2 - } < \mathrm { Mg } ^ { 2 + } < \mathrm { Na } ^ { + } < \mathrm { Al } ^ { 3 + }$
Show answer

Correct answer: C

9
How many periods and vertical columns are there in the long form of the periodic table?
  1. A 8,12
  2. B 6,8
  3. C 7,18
  4. D 6,18
Show answer

Correct answer: C

10
Fluorine has the highest electronegativity among the $\mathrm { ns } ^ { 2 } \mathrm { np } ^ { 5 }$ group on the Pauling scale, but the electron affinity of fluorine is less than that of chlorine because:
  1. A the atomic number of fluorine is less than that of chlorine.
  2. B fluorine being the first member of the family behaves unusually.
  3. C chlorine can accommodate an electron better than fluorine by utilising its vacant 3 d orbital.
  4. D small size, high electron density and an increased electron repulsion make the addition of an electron to fluorine less favourable than that in the case of chlorine.
Show answer & explanation

Correct answer: D

There is more interelectronic repulsion in the 2p subshell of florine than in chlorine (3p). So extra electrons will be added easily in the 3p subshell of chlorine as compared to the 2p subshell of florine.
Therefore, option (4) is the correct answer.
11
The electronegavitity values of $\mathrm { C } , \mathrm { N } , \mathrm { O }$ and F on Pauling scale:
  1. A increase from carbon to fluorine.
  2. B decrease from carbon to fluorine.
  3. C increase up to oxygen and is minimum at fluorine.
  4. D is minimum at nitrogen and then increase continuously.
Show answer

Correct answer: A

12
Which of the following relation is correct concerning the first (I) and second (II) ionisation energies of sodium and magnesium?
  1. A $\mathrm { I } _ { \mathrm { Mg } } = \mathrm { II } _ { \mathrm { Na } }$
  2. B $\mathrm { I } _ { \mathrm { Na } } > \mathrm { I } _ { \mathrm { Mg } }$
  3. C $\mathrm { II } _ { \mathrm { Mg } } > \mathrm { II } _ { \mathrm { Na } }$
  4. D $\mathrm { II } _ { \mathrm { Na } } > \mathrm { II } _ { \mathrm { Mg } }$
Show answer & explanation

Correct answer: D

The second ionisation energy of sodium is greater than that of Mg. In the case of sodium ion (i.e. Na⁺) the electron removal is from the stable inert gas confiuration (1s$^{2}$ 2s$^{2}$ 2p$^{6}$) which requires much higher energy.
Therefore, option (4) is the correct answer
13
Among the elements Ca, Mg, P and Cl , the order of increasing atomic radii is:
  1. A $\mathrm { Mg } < \mathrm { Ca } < \mathrm { Cl } < \mathrm { P }$
  2. B $\mathrm { Cl } < \mathrm { P } < \mathrm { Mg } < \mathrm { Ca }$
  3. C $\mathrm { P } < \mathrm { Cl } < \mathrm { Ca } < \mathrm { Mg }$
  4. D $\mathrm { Ca } < \mathrm { Mg } < \mathrm { P } < \mathrm { Cl }$
Show answer

Correct answer: B

14
The most electronegative element in the periodic table is $\_\_\_\_$ -
  1. A nitrogen
  2. B oxygen
  3. C chlorine
  4. D fluorine
Show answer & explanation

Correct answer: D

The most electronegative element in the periodic table is Fluorine.
Therefore, option (4) is the correct answer.
15
The first ionisation energy of $\mathrm { Li } , \mathrm { Be } , \mathrm { B }$ and C are in the order:
  1. A $\mathrm { Li } < \mathrm { Be } > \mathrm { B } < \mathrm { C }$
  2. B $\mathrm { Li } > \mathrm { Be } > \mathrm { B } > \mathrm { C }$
  3. C $\mathrm { Li } < \mathrm { Be } < \mathrm { B } > \mathrm { C }$
  4. D $\mathrm { Li } > \mathrm { Be } > \mathrm { B } < \mathrm { C }$
Show answer & explanation

Correct answer: A

Across the period (i.e. 2$^\mathrm{nd}$ period) the size of the atom decreases and the nuclear charge increases. So generally the ionisation energy increases. However, the ionisation energy of Be is greater than B because of the more penetration power of the 2s$^{2}$ subshell electrons of Be as compared to that of the 2p$^{1}$ subshell electron of B.
Therefore, option (1) is the correct answer
16
Which electronic configuration of neutral atoms will have the highest first ionisation energy?
  1. A $1 \mathrm {s} ^ { 2 } 2 \mathrm {s} ^ { 2 } 2 \mathrm { p } ^ { 4 }$
  2. B $1 \mathrm {s} ^ { 2 } 2 \mathrm {s} ^ { 2 } 2 \mathrm { p } ^ { 3 }$
  3. C $1 \mathrm {s} ^ { 2 } 2 \mathrm {s} ^ { 2 } 2 \mathrm { p } ^ { 2 }$
  4. D $1 \mathrm {s} ^ { 2 } 2 \mathrm {s} ^ { 2 } 2 \mathrm { p } ^ { 1 }$
Show answer

Correct answer: B

17
The first ionization energy of magnesium is smaller as compared to that of elements X and Y , but higher than that of Z . The elements $\mathrm { X } , \mathrm { Y }$ and Z, respectively, are;
  1. A chlorine, lithium and sodium.
  2. B argon, lithium and sodium.
  3. C argon, chlorine and sodium.
  4. D neon, sodium and chlorine.
Show answer

Correct answer: C

18
Elements X, Y and Z have atomic numbers 19, 37 and 55 respectively. Which of the following statements is true?
  1. A Their ionisation potential would increase with the increasing atomic number.
  2. B 'Y' would have an ionisation potential in between those of ' X ' and ' Z '.
  3. C 'Z' would have the highest ionisation potential.
  4. D 'Y' would have the highest ionisation potential.
Show answer & explanation

Correct answer: B

X = $_{19}$K ; Y = $_{37}$Rb and Z = $_{55}$Cs
Ionisation potential decreases down the group.
Therefore, option (2) is the correct answer
19
Which represents alkali metals (i.e. group I metals) based on $\mathrm { IE } _ { 1 }$ and $\mathrm { IE } _ { 2 }$ values?
  1. A X - IE$_{1}$ = 100; IE$_{2}$ = 110
  2. B Y - IE$_{1}$ = 95; IE$_{2}$ = 1200
  3. C Z - IE$_{1}$ = 195; IE$_{2}$ = 500
  4. D M - IE$_{1}$ = 200; IE$_{2}$ = 250
Show answer & explanation

Correct answer: C

In alkali metals, removal of the fist electron is easier because after removal of this electron, the atom acquires extra stable confiuration but the second electron is to be removed from ns$^{2}$np$^{6}$ confiuration i.e. stable noble gas confiuration. So IE$_{2}$ is quite higher.
Therefore, option (3) is the correct answer.
20
The first ionisation potential of Al is smaller than that of Mg because:
  1. A the atomic size of $\mathrm { Al } > \mathrm { Mg }$.
  2. B the atomic size of $\mathrm { Al } < \mathrm { Mg }$.
  3. C Al has one unpaired electron in 3p orbital.
  4. D Mg has incompletely filled 3 s orbital.
Show answer & explanation

Correct answer: C

The penetration of p subshell electrons is less than s subshell electrons. Thus 3s electron of magnesium is closer to the nucleus as compared to the 3p electron of aluminium.
Therefore, option (3) is the correct answer

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