Solved Example: Classification of Element and Periodicity | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Solved Example on Classification of Elements and 
Periodicity 
JEE Mains 
Q1:  When the following five anions are arranged in order of decreasing ionic radius, the correct 
sequence is: 
(A) ????
?? -
, ?? -, ???? , ?? ?? -
, ?? -
 
(B) ?? -
, ????
?? -
, ?? ?? -
, ????
?? , ?? -
 
(C) ????
?? -
, ?? -
, ????
-
, ?? -
, ?? ?? -
 
(D) ?? -
, ????
?? -
, ????
-
, ?? ?? -
, ?? -
 
Ans:  (D) I- will have the largest ionic radius increase further, we know that ionic radius will down the 
group and decrease across a period. And for isoelectronic species, the ionic radius will increase with 
increase in negative charge 
 
 
Q2:  Which of the following is wrong regarding the stability of the ions ???? , ???? and ???? 
(A) ????
?? +
< ????
?? +
< ????
?? +
 
(B) ????
?? +
> ????
?? +
> ????
?? +
 
(C) ????
?? +
> ????
?? +
 
(D) ????
?? +
< ????
?? +
 
Ans: (D) In group 13, the stability of +2 oxidation state will increase and stability of +4 oxidation state 
will decrease down the group due to inert pair effect, which is negligible in Ge and most significant in 
Pb. 
 
 
Q3:  The law of triads is not applicable on 
(A) ???? , ????
I 
 
(B) ???? , ?? , ???? 
(C) ?? , ???? , ???? 
(D) ???? , ???? , ???? 
Ans: (B) We check the atomic radius of all the elements in the four options. 
For option (B), we find that the atomic mass of the middle elements is not equal to the average of 
atomic mass of other 2 elements. 
 
 
Page 2


JEE Solved Example on Classification of Elements and 
Periodicity 
JEE Mains 
Q1:  When the following five anions are arranged in order of decreasing ionic radius, the correct 
sequence is: 
(A) ????
?? -
, ?? -, ???? , ?? ?? -
, ?? -
 
(B) ?? -
, ????
?? -
, ?? ?? -
, ????
?? , ?? -
 
(C) ????
?? -
, ?? -
, ????
-
, ?? -
, ?? ?? -
 
(D) ?? -
, ????
?? -
, ????
-
, ?? ?? -
, ?? -
 
Ans:  (D) I- will have the largest ionic radius increase further, we know that ionic radius will down the 
group and decrease across a period. And for isoelectronic species, the ionic radius will increase with 
increase in negative charge 
 
 
Q2:  Which of the following is wrong regarding the stability of the ions ???? , ???? and ???? 
(A) ????
?? +
< ????
?? +
< ????
?? +
 
(B) ????
?? +
> ????
?? +
> ????
?? +
 
(C) ????
?? +
> ????
?? +
 
(D) ????
?? +
< ????
?? +
 
Ans: (D) In group 13, the stability of +2 oxidation state will increase and stability of +4 oxidation state 
will decrease down the group due to inert pair effect, which is negligible in Ge and most significant in 
Pb. 
 
 
Q3:  The law of triads is not applicable on 
(A) ???? , ????
I 
 
(B) ???? , ?? , ???? 
(C) ?? , ???? , ???? 
(D) ???? , ???? , ???? 
Ans: (B) We check the atomic radius of all the elements in the four options. 
For option (B), we find that the atomic mass of the middle elements is not equal to the average of 
atomic mass of other 2 elements. 
 
 
 
Q4:  The atomic volume was choosen as the basic of periodic classification of elements by 
(A) Niels Bohr 
(B) Mendeleev 
(C) Lothar meyer 
(D) Newlands 
Ans: (C) Lothar Meyer used atomic volume for plotting Lothar Mayer's volume curves. 
 
Q5:  The electronic configuration ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? is for: 
(A) f-block element 
(B) d-block element 
(C) p-block element 
(D) s-block element 
Ans: (D) Since, here s orbital is filled last, it is a s -block element 
 
Q6: The majority of gaseous elements in the periodic table are placed 
(A) At bottom left hand side 
(B) At top right hand side 
(C) Below the main table 
(D) Along side ?? block elements 
Ans: (B) Majority of low atomic number non metallic elements are gases which are placed at top 
right hand side. 
 
 
Q7:  Fluorine has the highest electronegativity among the ?? ?? ?? ?? ?? ?? group on the Pauling scale, but 
the electron affinity of fluorine is less than that of chlorine because: 
(A) The atomic number of fluorine is less than that of chlorine 
(B) Fluorine being the first member of the family behaves in an unusual manner 
(C) Chlorine can accommodate an electron better than fluorine by utilizing its vacant 3d orbital 
(D) Small size, high electron density and an increased electron repulsion makes addition of an 
electron to fluorine less favourable than that in the case of chlorine. 
Ans: (D) All the reasons given are correct. 
 
 
 
Page 3


JEE Solved Example on Classification of Elements and 
Periodicity 
JEE Mains 
Q1:  When the following five anions are arranged in order of decreasing ionic radius, the correct 
sequence is: 
(A) ????
?? -
, ?? -, ???? , ?? ?? -
, ?? -
 
(B) ?? -
, ????
?? -
, ?? ?? -
, ????
?? , ?? -
 
(C) ????
?? -
, ?? -
, ????
-
, ?? -
, ?? ?? -
 
(D) ?? -
, ????
?? -
, ????
-
, ?? ?? -
, ?? -
 
Ans:  (D) I- will have the largest ionic radius increase further, we know that ionic radius will down the 
group and decrease across a period. And for isoelectronic species, the ionic radius will increase with 
increase in negative charge 
 
 
Q2:  Which of the following is wrong regarding the stability of the ions ???? , ???? and ???? 
(A) ????
?? +
< ????
?? +
< ????
?? +
 
(B) ????
?? +
> ????
?? +
> ????
?? +
 
(C) ????
?? +
> ????
?? +
 
(D) ????
?? +
< ????
?? +
 
Ans: (D) In group 13, the stability of +2 oxidation state will increase and stability of +4 oxidation state 
will decrease down the group due to inert pair effect, which is negligible in Ge and most significant in 
Pb. 
 
 
Q3:  The law of triads is not applicable on 
(A) ???? , ????
I 
 
(B) ???? , ?? , ???? 
(C) ?? , ???? , ???? 
(D) ???? , ???? , ???? 
Ans: (B) We check the atomic radius of all the elements in the four options. 
For option (B), we find that the atomic mass of the middle elements is not equal to the average of 
atomic mass of other 2 elements. 
 
 
 
Q4:  The atomic volume was choosen as the basic of periodic classification of elements by 
(A) Niels Bohr 
(B) Mendeleev 
(C) Lothar meyer 
(D) Newlands 
Ans: (C) Lothar Meyer used atomic volume for plotting Lothar Mayer's volume curves. 
 
Q5:  The electronic configuration ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? is for: 
(A) f-block element 
(B) d-block element 
(C) p-block element 
(D) s-block element 
Ans: (D) Since, here s orbital is filled last, it is a s -block element 
 
Q6: The majority of gaseous elements in the periodic table are placed 
(A) At bottom left hand side 
(B) At top right hand side 
(C) Below the main table 
(D) Along side ?? block elements 
Ans: (B) Majority of low atomic number non metallic elements are gases which are placed at top 
right hand side. 
 
 
Q7:  Fluorine has the highest electronegativity among the ?? ?? ?? ?? ?? ?? group on the Pauling scale, but 
the electron affinity of fluorine is less than that of chlorine because: 
(A) The atomic number of fluorine is less than that of chlorine 
(B) Fluorine being the first member of the family behaves in an unusual manner 
(C) Chlorine can accommodate an electron better than fluorine by utilizing its vacant 3d orbital 
(D) Small size, high electron density and an increased electron repulsion makes addition of an 
electron to fluorine less favourable than that in the case of chlorine. 
Ans: (D) All the reasons given are correct. 
 
 
 
 
Q8:  The last electron in each normal elements of period is filled in 
(A) The same energy sublevel 
(B) The same enrgy level 
(C) The same orbital 
(D) Relation between ?? ?? and ?? ?? is uncertain 
Ans: (B) The last electron in each period is filled in np orbital which is the same shell, i.e. same 
energy level. 
 
Q9:  The greater stability of the lower oxidation state in heavier ?? block metals in the consequence 
of 
(A) Electronic transition within ?? -orbitals 
(B) Electronic transition from ?? to ?? -orbitals 
(C) Inert pair effect 
(D) Expansion of octet 
Ans: (C) This is a consequence of inert pair effect, according to which in heavier p -block elements, 
the s orbitals become very close to nucleus and thus become relatively inert and are not removed. 
 
 
Q10:  Oxidation number of p -block elements is [Excluding inert gases] 
(A) Equal to group number 
(B) Group number +2 
(C) Between the range [Group no...(Group no. 8)] 
(D) Number of unpaired electrons in the valence shell 
Ans: (C) p-block elements can display a range of oxidation states due to availability of electrons in 
both of ?? and ?? orbitals and also possibility of attaining noble gas configuration by adding electrons. 
 
 
 
Q11:  The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is: 
(A) ?? > ?? > ?? > ?? 
(B) ?? > ?? > ?? > ?? 
(C) ?? > ?? > ?? > ?? 
(D) ?? > ?? > ?? > ?? 
Ans: (C) We note that order of (IE) is F > N > O > C 
Here (IE) of N > (IE) of O because N has a stable halffilled orbital configuration and also IE increases 
across a period. 
Page 4


JEE Solved Example on Classification of Elements and 
Periodicity 
JEE Mains 
Q1:  When the following five anions are arranged in order of decreasing ionic radius, the correct 
sequence is: 
(A) ????
?? -
, ?? -, ???? , ?? ?? -
, ?? -
 
(B) ?? -
, ????
?? -
, ?? ?? -
, ????
?? , ?? -
 
(C) ????
?? -
, ?? -
, ????
-
, ?? -
, ?? ?? -
 
(D) ?? -
, ????
?? -
, ????
-
, ?? ?? -
, ?? -
 
Ans:  (D) I- will have the largest ionic radius increase further, we know that ionic radius will down the 
group and decrease across a period. And for isoelectronic species, the ionic radius will increase with 
increase in negative charge 
 
 
Q2:  Which of the following is wrong regarding the stability of the ions ???? , ???? and ???? 
(A) ????
?? +
< ????
?? +
< ????
?? +
 
(B) ????
?? +
> ????
?? +
> ????
?? +
 
(C) ????
?? +
> ????
?? +
 
(D) ????
?? +
< ????
?? +
 
Ans: (D) In group 13, the stability of +2 oxidation state will increase and stability of +4 oxidation state 
will decrease down the group due to inert pair effect, which is negligible in Ge and most significant in 
Pb. 
 
 
Q3:  The law of triads is not applicable on 
(A) ???? , ????
I 
 
(B) ???? , ?? , ???? 
(C) ?? , ???? , ???? 
(D) ???? , ???? , ???? 
Ans: (B) We check the atomic radius of all the elements in the four options. 
For option (B), we find that the atomic mass of the middle elements is not equal to the average of 
atomic mass of other 2 elements. 
 
 
 
Q4:  The atomic volume was choosen as the basic of periodic classification of elements by 
(A) Niels Bohr 
(B) Mendeleev 
(C) Lothar meyer 
(D) Newlands 
Ans: (C) Lothar Meyer used atomic volume for plotting Lothar Mayer's volume curves. 
 
Q5:  The electronic configuration ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? is for: 
(A) f-block element 
(B) d-block element 
(C) p-block element 
(D) s-block element 
Ans: (D) Since, here s orbital is filled last, it is a s -block element 
 
Q6: The majority of gaseous elements in the periodic table are placed 
(A) At bottom left hand side 
(B) At top right hand side 
(C) Below the main table 
(D) Along side ?? block elements 
Ans: (B) Majority of low atomic number non metallic elements are gases which are placed at top 
right hand side. 
 
 
Q7:  Fluorine has the highest electronegativity among the ?? ?? ?? ?? ?? ?? group on the Pauling scale, but 
the electron affinity of fluorine is less than that of chlorine because: 
(A) The atomic number of fluorine is less than that of chlorine 
(B) Fluorine being the first member of the family behaves in an unusual manner 
(C) Chlorine can accommodate an electron better than fluorine by utilizing its vacant 3d orbital 
(D) Small size, high electron density and an increased electron repulsion makes addition of an 
electron to fluorine less favourable than that in the case of chlorine. 
Ans: (D) All the reasons given are correct. 
 
 
 
 
Q8:  The last electron in each normal elements of period is filled in 
(A) The same energy sublevel 
(B) The same enrgy level 
(C) The same orbital 
(D) Relation between ?? ?? and ?? ?? is uncertain 
Ans: (B) The last electron in each period is filled in np orbital which is the same shell, i.e. same 
energy level. 
 
Q9:  The greater stability of the lower oxidation state in heavier ?? block metals in the consequence 
of 
(A) Electronic transition within ?? -orbitals 
(B) Electronic transition from ?? to ?? -orbitals 
(C) Inert pair effect 
(D) Expansion of octet 
Ans: (C) This is a consequence of inert pair effect, according to which in heavier p -block elements, 
the s orbitals become very close to nucleus and thus become relatively inert and are not removed. 
 
 
Q10:  Oxidation number of p -block elements is [Excluding inert gases] 
(A) Equal to group number 
(B) Group number +2 
(C) Between the range [Group no...(Group no. 8)] 
(D) Number of unpaired electrons in the valence shell 
Ans: (C) p-block elements can display a range of oxidation states due to availability of electrons in 
both of ?? and ?? orbitals and also possibility of attaining noble gas configuration by adding electrons. 
 
 
 
Q11:  The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is: 
(A) ?? > ?? > ?? > ?? 
(B) ?? > ?? > ?? > ?? 
(C) ?? > ?? > ?? > ?? 
(D) ?? > ?? > ?? > ?? 
Ans: (C) We note that order of (IE) is F > N > O > C 
Here (IE) of N > (IE) of O because N has a stable halffilled orbital configuration and also IE increases 
across a period. 
But after removal of an electron O will have a stable, half -filled orbital configuration and therefore, 
its IE will be more than F
+
. Hence order O > F > N > C 
Q12:  Which statement is wrong: 
(A) ?? nd 
 ionization energy shows jump in alkali metals 
(B) ?? nd 
 electron affinity for halogens is zero 
(C) Maximum electron affinity exists for ?? 
(D) Maximum ionization energy exists for ???? 
Ans: (C) Electron affinity is higher for Cl due to its low electron density. 
(IE)
2
 of alkali metals shows a jump because they have stable noble gas configuration in +1 oxidation 
state 
(IE)
2
 of halogens is zero because halogens have noble gas configuration in -1 oxidation state 
He has maximum ionisation energy due to its small size and fully filled orbital configuration. 
Q13 Which of the following is the configuration of second excited state of the element 
isoelectronic with ?? ?? or ?? or ????
?? 
(A) [???? ]?? ?? ?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ?? 
?? 
(B) [???? ]?? ?? ?? 
(C) [???? ]?? ?? ?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ????
?? ?? ?? ????
?? 
(D) [???? ]?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ????
?? 
Ans: (C) The element isoelectronic with O
2
 is sulphur (s). Its configuration in group state 
[Ne]3 s
2
3p
4
3 d
0
 Outer shell configuration in ground state 
 
 
Q14:  Metallic radii of transition elements 
(A) First increase, then decrease periodically 
(B) First decrease, then remain almost constant 
(C) First increase, then remaining almost constant 
(D) First increase, then increase periodically 
Ans:  (B) Metallic radii of transition elements first decreases due to increase in effective nuclear 
charge, but after since the electrons are filled in (n – 1)d orbital, after a few elements, the nuclear 
charge on ns electrons remains almost constant due to extra stability. 
Page 5


JEE Solved Example on Classification of Elements and 
Periodicity 
JEE Mains 
Q1:  When the following five anions are arranged in order of decreasing ionic radius, the correct 
sequence is: 
(A) ????
?? -
, ?? -, ???? , ?? ?? -
, ?? -
 
(B) ?? -
, ????
?? -
, ?? ?? -
, ????
?? , ?? -
 
(C) ????
?? -
, ?? -
, ????
-
, ?? -
, ?? ?? -
 
(D) ?? -
, ????
?? -
, ????
-
, ?? ?? -
, ?? -
 
Ans:  (D) I- will have the largest ionic radius increase further, we know that ionic radius will down the 
group and decrease across a period. And for isoelectronic species, the ionic radius will increase with 
increase in negative charge 
 
 
Q2:  Which of the following is wrong regarding the stability of the ions ???? , ???? and ???? 
(A) ????
?? +
< ????
?? +
< ????
?? +
 
(B) ????
?? +
> ????
?? +
> ????
?? +
 
(C) ????
?? +
> ????
?? +
 
(D) ????
?? +
< ????
?? +
 
Ans: (D) In group 13, the stability of +2 oxidation state will increase and stability of +4 oxidation state 
will decrease down the group due to inert pair effect, which is negligible in Ge and most significant in 
Pb. 
 
 
Q3:  The law of triads is not applicable on 
(A) ???? , ????
I 
 
(B) ???? , ?? , ???? 
(C) ?? , ???? , ???? 
(D) ???? , ???? , ???? 
Ans: (B) We check the atomic radius of all the elements in the four options. 
For option (B), we find that the atomic mass of the middle elements is not equal to the average of 
atomic mass of other 2 elements. 
 
 
 
Q4:  The atomic volume was choosen as the basic of periodic classification of elements by 
(A) Niels Bohr 
(B) Mendeleev 
(C) Lothar meyer 
(D) Newlands 
Ans: (C) Lothar Meyer used atomic volume for plotting Lothar Mayer's volume curves. 
 
Q5:  The electronic configuration ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? , ?? ?? ?? , ?? ?? ????
, ?? ?? ?? is for: 
(A) f-block element 
(B) d-block element 
(C) p-block element 
(D) s-block element 
Ans: (D) Since, here s orbital is filled last, it is a s -block element 
 
Q6: The majority of gaseous elements in the periodic table are placed 
(A) At bottom left hand side 
(B) At top right hand side 
(C) Below the main table 
(D) Along side ?? block elements 
Ans: (B) Majority of low atomic number non metallic elements are gases which are placed at top 
right hand side. 
 
 
Q7:  Fluorine has the highest electronegativity among the ?? ?? ?? ?? ?? ?? group on the Pauling scale, but 
the electron affinity of fluorine is less than that of chlorine because: 
(A) The atomic number of fluorine is less than that of chlorine 
(B) Fluorine being the first member of the family behaves in an unusual manner 
(C) Chlorine can accommodate an electron better than fluorine by utilizing its vacant 3d orbital 
(D) Small size, high electron density and an increased electron repulsion makes addition of an 
electron to fluorine less favourable than that in the case of chlorine. 
Ans: (D) All the reasons given are correct. 
 
 
 
 
Q8:  The last electron in each normal elements of period is filled in 
(A) The same energy sublevel 
(B) The same enrgy level 
(C) The same orbital 
(D) Relation between ?? ?? and ?? ?? is uncertain 
Ans: (B) The last electron in each period is filled in np orbital which is the same shell, i.e. same 
energy level. 
 
Q9:  The greater stability of the lower oxidation state in heavier ?? block metals in the consequence 
of 
(A) Electronic transition within ?? -orbitals 
(B) Electronic transition from ?? to ?? -orbitals 
(C) Inert pair effect 
(D) Expansion of octet 
Ans: (C) This is a consequence of inert pair effect, according to which in heavier p -block elements, 
the s orbitals become very close to nucleus and thus become relatively inert and are not removed. 
 
 
Q10:  Oxidation number of p -block elements is [Excluding inert gases] 
(A) Equal to group number 
(B) Group number +2 
(C) Between the range [Group no...(Group no. 8)] 
(D) Number of unpaired electrons in the valence shell 
Ans: (C) p-block elements can display a range of oxidation states due to availability of electrons in 
both of ?? and ?? orbitals and also possibility of attaining noble gas configuration by adding electrons. 
 
 
 
Q11:  The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is: 
(A) ?? > ?? > ?? > ?? 
(B) ?? > ?? > ?? > ?? 
(C) ?? > ?? > ?? > ?? 
(D) ?? > ?? > ?? > ?? 
Ans: (C) We note that order of (IE) is F > N > O > C 
Here (IE) of N > (IE) of O because N has a stable halffilled orbital configuration and also IE increases 
across a period. 
But after removal of an electron O will have a stable, half -filled orbital configuration and therefore, 
its IE will be more than F
+
. Hence order O > F > N > C 
Q12:  Which statement is wrong: 
(A) ?? nd 
 ionization energy shows jump in alkali metals 
(B) ?? nd 
 electron affinity for halogens is zero 
(C) Maximum electron affinity exists for ?? 
(D) Maximum ionization energy exists for ???? 
Ans: (C) Electron affinity is higher for Cl due to its low electron density. 
(IE)
2
 of alkali metals shows a jump because they have stable noble gas configuration in +1 oxidation 
state 
(IE)
2
 of halogens is zero because halogens have noble gas configuration in -1 oxidation state 
He has maximum ionisation energy due to its small size and fully filled orbital configuration. 
Q13 Which of the following is the configuration of second excited state of the element 
isoelectronic with ?? ?? or ?? or ????
?? 
(A) [???? ]?? ?? ?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ?? 
?? 
(B) [???? ]?? ?? ?? 
(C) [???? ]?? ?? ?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ?? 
?? ?? ?? ????
?? ?? ?? ????
?? 
(D) [???? ]?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ????
?? 
Ans: (C) The element isoelectronic with O
2
 is sulphur (s). Its configuration in group state 
[Ne]3 s
2
3p
4
3 d
0
 Outer shell configuration in ground state 
 
 
Q14:  Metallic radii of transition elements 
(A) First increase, then decrease periodically 
(B) First decrease, then remain almost constant 
(C) First increase, then remaining almost constant 
(D) First increase, then increase periodically 
Ans:  (B) Metallic radii of transition elements first decreases due to increase in effective nuclear 
charge, but after since the electrons are filled in (n – 1)d orbital, after a few elements, the nuclear 
charge on ns electrons remains almost constant due to extra stability. 
 
Q15:  For the formation of a covalent bond, the difference in the value of electro negativities 
should be: 
(A) Equal to or less than 1.7 
(B) More than 1.7 
(C) 1.7 or more 
(D) None of these 
Ans: (A) If difference in electronegativity > 1.7, ionic bond 
 
Q16:  Which one of the following is the correct order of interactions? 
(A) Covalent < hydrogen bonding < Vander Waal's < dipole-dipole 
(B) Van der Waal's < hydrogen bonding < dipole-dipole < covalent 
(C) Van der Waal's < dipole-dipole < hydrogen bonding < covalent 
(D) Dipole-dipole < Van der Waal's < hydrogen 
Ans: (C) In covalent bond, there is actual bond formation therefore they are the strongest of allowed 
by hydrogen bonding. Vander waal’s interactions are least strong because they are developed due to 
momentary dipole interactions. 
Q17:  Properties of the elements of which of the following pairs do not resemble? 
(A) Li and Mg 
(B) Be and ???? 
(C) ???? and ???? 
(D) ?? and ???? 
Ans: (C) The properties of other three pairs resemble due to diagonal relationship between them 
(identical charge 1 size ratio) 
Q18:  Which of the following statement is not true? 
(A) The atoms have no tendency to accept electrons in empty higher energy levels 
(B) The atoms have no tendency to accept electrons in empty high energy sublevels 
(C) The alkali metals have no tendency to accept electrons 
(D) The atoms with exactly half-filled electronic configurations have no tendency to accept 
electrons. 
Ans: (D)The added electron will have no tendency to go to the higher energy level or sublevel as this 
will further increase the energy of the system. Alkali metals have no tendency to accept electrons 
due to low effective nuclear charge. Atoms with exactly hal f filled configuration will have low (but not 
zero) tendency to accept electrons.