Q.1. Within each pair of elements F and Cl, S and Se, and Li and Na, respectively, the elements that release more energy upon an electron gain are 
(a) Cl, Se and Na
(b) Cl, S and Li
(c) F, S and Li
(d) F, Se and Na
Solution: The electron gain enthalpy of Cl is more than that of F. Thus the pair that is Cl, S and Li releases more energy than F, S and Li. The electron gain enthalpy of S and Li is more than Se and Na respectively.
Q.2. The first ionization energy (in kJ mol–1) of Na, Mg, Al and Si respectively, are 
(a) 496, 737, 577, 786
(b) 496, 577, 737, 786
(c) 786, 737, 577, 496
(d) 496, 577, 786, 737
Solution: Moving from left to right in a period, the size of the element decreases and the value of first ionization enthalpy increases. Thus, the ionization energy of Na, Mg, Al and Si should be 496, 577, 737 and 786 kJ mol–1 respectively but due to the entry of the d-orbital in 3rd period the effective nuclear charge of Al increases. So the ionization energy of Al is lower than Mg.
Hence, the correct order of first ionization energy of Na, Mg, Al, and Si is 496, 737, 577, and 786 kJ mol–1 respectively.
Q.3. The increasing order of the atomic radii of the following elements is 
(a) (II) < (III) < (IV) < (I) < (V)
(b) (IV) < (III) < (II) < (I) < (V)
(c) (III) < (II) < (I) < (IV) < (V)
(d) (I) < (II) < (III) < (IV) < (V)
Solution: Moving left to right along a period, the size of the elements decreases and moving down the group, the size of the elements increases. Thus, Br is the biggest in size since it belongs to 4th period followed by Cl which belongs to 3rd period. Fluorine is the smallest element in the periodic table, and carbon comes before O in 2nd period. Hence, the correct increasing order of atomic radii is: (III), (II), (I), (IV), (V).
Q.4. B has a smaller first ionization enthalpy than Be. Consider the following statements:
(I) It is easier to remove 2p electron than 2s electron.
(II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be.
(III) 2s electron has more penetration power than 2p electron.
(IV) Atomic radius of B is more than Be.
(Atomic number B = 5, Be = 4)
The correct statements are 
(a) (I), (II) and (IV)
(b) (II), (III) and (IV)
(c) (I), (II) and (III)
(d) (I), (III) and (IV)
Solution: The electronic configuration of B (Z = 5) is 1s22s22p1 and Be (Z = 4) is 1s22s22p0. Thus, it is easier to remove an electron from B than Be. The 2p orbital of B is more shielded than 2s orbital of Be. s-orbital has more penetration power than p-orbital.
Q.5. The acidic, basic and amphoteric oxides, respectively, are 
(a) Na2O, SO3, Al2O3
(b) Cl2O, CaO, P4O10
(c) N2O3, Li2O, Al2O3
(d) MgO, Cl2O, Al2O3
Solution: Non-metals of p-block elements form acidic oxide and metals of p-block elements form amphoteric oxide since they have properties in between metals and non-metals. Oxide of metals of s-block elements are basic in nature.
Acidic oxide: N2O3
Basic oxide: Li2O
Amphoteric oxide: Al2O3
Q.6. If the magnetic moment of a dioxygen species is 1.73 BM, it may be 
Solution: The molecular electronic configuration of the following species is
Q.7. In general, the properties that decrease and increase down a group in the periodic table, respectively, are: 
(a) Atomic radius and electronegativity.
(b) Electron gain enthalpy and electronegativity.
(c) Electronegativity and atomic radius.
(d) Electronegativity and electron gain enthalpy.
Solution: Generally, electronegativity decreases down the group as the size increases. This can also be formulated as:
Electronegativity = 1/ size
Q.8. Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and +3 oxidation states. This is due to: 
(a) Inert pair effect
(b) Diagonal relationship
(c) Lattice effect
(d) Lanthanoid contraction
So, Due to the inert pair effect, thallium exists in more than one oxidation state. Also, for thallium + 1 oxidation state is more stable than +3 oxidation state.
Q.9. When the first electron gain enthalpy(ΔegH) of oxygen is - 141 kJ/mol, its second electron gain enthalpy is: 
(a) A more negative value than the first
(b) Almost the same as that of the first
(c) Negative, but less negative than the first
(d) A positive value
Solution: The second electron gain enthalpy of oxygen is positive as energy has to be added for the addition of another electron.
Q.10. The electronegativity of aluminium is similar to: 
Solution: Be and Al shows diagonal relationship due to which these two elements have similar electronegativity.
Q.11. The 71st electron of an element X with an atomic number of 71 enters into the orbital: 
∴ Orbital occupied by last e- is 5d.
Q.12. The correct order of the atomic radii of C, Cs, Al, and S is: 
(a) C < S < Al < Cs
(b) S < C < Cs < Al
(c) S < C < Al < Cs
(d) C < S < Cs < Al
Solution: On going down the group, size increases while going from left to right in a period size decreases, so order is C < S < Al < Cs.
Q.13. The correct option with respect to the Pauling electronegativity values of the elements is: 
(a) Te > Se
(c) Si < Al
(d) P > S
Solution: Correct order of electronegativity values of the elements is
Si > Al; S > P; Se > Te; Ge > Ga.
Q.14. The relative stability of +1 oxidation state of group 13 elements follows the order: 
(a) Al < Ga < Tl < In
(b) Tl < In < Ga < Al
(c) Ga < Al < In < Tl
(d) Al < Ga < In < Tl
Solution: Due to inert pair effect, the stability of +1 oxidation state increases down the group.
Thus, correct order of stability is Al < Ga < In < Tl
Q.15. The element with Z = 120 (not yet discovered) will be an/a: 
(a) Inner-transition metal
(b) Alkaline earth metal
(c) Alkali metal
(d) Transition metal
Solution: Elements with Z = 120 will belong to alkaline earth metals.
Its electronic configuration may be represented as [Og] 8s2.
Q.16. The size of the iso-electronic species Cl-, Ar and Ca2+ is affected by: 
(a) Azimuthal quantum number of valence shell
(b) Electron-electron interaction in the outer orbitals
(c) Principal quantum number of valence shell
(d) Nuclear charge
Solution: Iso-electronic species differ in size due to different effective nuclear charge.
Q.17. The IUPAC symbol for the element with atomic number 119 would be: 
Solution: Symbol for 1 is u and for 9 is e.
∴ UPAC symbol for 119 is uue.
Q.18. The correct statements among I to III regarding group 13 element oxides are, 
(I) Boron trioxide is acidic.
(II) Oxides of aluminium and gallium are amphoteric.
(III) Oxides of indium and thallium are basic.
(a) (I) and (II) only
(b) (I), (II) and (III)
(c) (I) and (III) only
(d) (II) and (III) only
(I) B2O3 - Acidic Oxide
(II) Al2O3 & Ga2O3 - Amphoteric Oxide
(III) In2O3 & Tl2O - Basic Oxide
Q.19. In comparison to boron, beryllium has: 
(a) Lesser nuclear charge and lesser first ionisation enthalpy.
(b) Greater nuclear charge and lesser first ionisation enthalpy.
(c) Greater nuclear charge and greater first ionisation enthalpy.
(d) Lesser nuclear charge and greater first ionisation enthalpy.
Solution: Nuclear charge : B > Be
Be = 1s2 2s2 (more stable)
B = 1s2 2s2 2p1
∴ Ionisation energy of Be is greater than B due to ns2 outer electronic configuration.
Q.20. For Na+, Mg2+, F- and O2-; the correct order of increasing ionic radii is: 
(a) Na+ < Mg2+< F- < O2-
(b) Mg2 < Na+ < F- < O2-
(c) Mg2+ < O2- < Na+< F-
(d) O2- < F- < Na+< Mg2+
Solution: Isoelectronic series: Mg2+ < Na+ < F- < O2-
When negative charge increase, the radius of ion increases.
Q.21. The group having isoelectronic species is 
(a) O2–, F–, Na+, Mg2+
(b) O–, F–, Na, Mg+
(c) O2–, F–, Na, Mg2+
(d) O–, F–, Na+, Mg2+
Solution: Mg2+, Na+, O2– and F– all have 10 electrons each.
Q.22. Consider the following ionization enthalpies of two elements ‘A’ and ‘B’. 
Which of the following statements is correct?
(a) Both ‘A’ and ‘B’ belong to group-1 where ‘A’ comes below ‘B’.
(b) Both ‘A’ and ‘B’ belong to group-2 where ‘A’ comes below ‘B’.
(c) Both ‘A’ and ‘B’ belong to group-1 where ‘B’ comes below ‘A’.
(d) Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’.
Solution: Generally, the ionization enthalpies or energy increases from left to right in a period and decreases from top to bottom in a group. Several factors such as atomic radius, nuclear charge, shielding effect are responsible for change of ionization enthalpies.
Here, Ist ionization enthalpy of A and B is greater than group I (Li 520 kJmol–1 to Cs 374 kJmol–1), which means element A and B belong to group –2 and all three given ionization enthalpy values are less for element B means B will come below A.
Q.23. The electronic configuration with the highest ionization enthalpy is: 
(a) [Ar] 3d10 4s2 4p3
(b) [Ne] 3s2 3p1
(c) [Ne] 3s2 3p3
(d) [Ne] 3s2 3p2
Solution: S1< P1< S2< P2< P4< P3< P5< P6(IE order)