Q1. Write the general outer electronic configuration of s, p, d, and f block elements. Also assign the position of elements having outer electronic configuration as
i) ns2 np4 for n=3
ii) (n-2) f7 (n-1) d1 ns2 for n=6, in the periodic table.
Solution
The general outer configuration of s block elements is ns1-2 p block = ns2 p1-6 d block= (n-1) d1-10ns0-2 f block = (n-2) f1-14 (n-1) d0-1,ns2
i) ns2 np4 for n=3 Ans) In this case the element belongs to third period of the periodic table as n=3. The group number of the element is 16
ii) (n-2)f7 (n-1) d1 ns2 for n=6 Ans) In this case the element belongs to the sixth period of the periodic table. The group is lanthanoid.
Q2. The elements are classified into four blocks depending on:
- 1) Size of atoms of element
- 2) Colour of elements of group
- 3) Mass of elements of group
- 4) The type of orbitals that are being filled
Solution
The elements are classified into four blocks depending on type of orbital being filled i.e., s, p, d and f blocks.
Q3. What is periodicity? Discuss its cause.
Solution
The periodical repetition of similar properties of the elements after certain regular intervals when the elements are arranged in order of increasing atomic number is called periodicity. This property is due to the recurrence of similar valence shell configuration after regular intervals.
Q4. The fourth period involves the filling of
- 1) 4s, 3d, and 4p orbitals.
- 2) 3s, 4d, and 4p orbitals.
- 3) 4s, 4d, and 4p orbitals.
- 4) 5s, 4d, and 4p orbitals.
Solution
The fourth period involves the filling of 4s, 3d, and 4p orbitals.
Q5. Identify the element belonging to third period and 17th group of the periodic table.
- 1) Silicon
- 2) Bromine
- 3) Chlorine
- 4) Sulphur
Solution
Chlorine belongs to third period and 17th group of the periodic table. Its atomic number is 17 and its configuration is 2, 8, 7.
Q6. Newland arranged elements in increasing order of atomic weights and noted that every eighth element had properties similar to:
- 1) Third element
- 2) Fourth element
- 3) Second element
- 4) First element
Solution
Newland profounded the Law of Octaves but it was true only for elements upto calcium.
Q7. Why the first ionization enthalpy of carbon is more than that of boron but the reverse happens in the case of second ionisation enthalpy? Explain.
Solution
The first ionisation enthalpy of carbon is more than that of boron because the effective nuclear charge of carbon is more and its atomic size is smaller as compared to boron. The second ionisation enthalpy of carbon is less than that of boron because after losing one electron the boron atom changes into B+ which has stable electronic configuration. As a result more energy is required to remove an electron from B+ than from C+.
Q8. Why do we call the d block elements as transition metals?
Solution
In the periodic table the left hand side of the d- block elements is occupied by the reactive metals of group 1 and 2 and towards the right hand side the less reactive elements of groups 13 and 14 are present. The d-block elements form a bridge or transition between the two so they are known as the transition metals.
Q9. Calculate the amount of energy produced when 7.1g of chlorine, in the form of chlorine atoms, is converted to Cl– ions in a gaseous state? Given that the electron gain enthalpy of chlorine is 3.7 eV.
Solution
Energy produced during conversion of 1 mol of Cl (g) to Cl–= Energy produced during conversion of 0.2 mol of Cl (g) to Cl– =
Q10. The systemic name for an element with atomic number (Z = 118) is:
- 1) Ununhexium
- 2) Ununoctium
- 3) Unbinilium
- 4) Unnilunium
Solution
The systematic name for an element with atomic number (Z = 118) is ununoctium.
Q11. The transition elements have characteristic electronic configuration which can be represented as:
- 1) (n-2) d1 – 10 (n-1) s2 p6 d1 or 2 ns1
- 2) (n-1)s2 p6 d 10 ns2 np6 nd1-10
- 3) (n-2) d1 – 10 (n-1) p6 ns2
- 4) (n-1) d1-10 ns0-2
Solution
In transition elements, there is filling of d orbitals in the penultimate shell.
Q12. The anomalous behavior of first element of s and p block elements of each group as compared to other group members is due to following reasons:
- 1) Large charge/radius ratio
- 2) All of the above
- 3) High electronegativity and non availability of d-orbitals in their valence shell
- 4) Small size of atom
Solution
The anomalous behaviour of first element of s and p block elements of each group as compared to other group members is due to following reasons: Small size of atom, large charge/radius ratio, high electronegativity and non availability of d-orbitals in their valence shell.
Q13. What are magic numbers?
Solution
The physical and chemical properties of an element repeat themselves after the interval of 2, 8, 18 or 32 in their atomic number. So the numbers 2, 8, 18 and 32 are known as magic numbers.
Q14. The ionic radii (in Å) of N3¯, O2¯ and F¯ are respectively:
- 1) 1.36, 1.40 and 1.71
- 2) 1.71, 1.40 and 1.36
- 3) 1.71, 1.36 and 1.40
- 4) 1.36, 1.71 and 1.40
Solution
N3¯, O2¯ and F¯ are isoelectronic. But due to the difference in their nuclear charges, their ionic radii follow the order N3¯ > O2¯ > F¯. Only option 3 matches this order.
Q15. What is the relation between the number of elements present in each period and the number of atomic orbitals?
Solution
The number of elements in each period is twice the number of atomic orbitals available in the energy level.
Q16. What is the relationship between diagonal relationship of an element and the polarizing power of an element?
Solution
On moving along a period, the ionic charge increases while ionic size decreases so polarizing power increases. On moving down the group the ionic size increases and hence polarizing power decreases. On moving diagonally these two effects cancel each other to some extent and hence the properties remain similar.
Q17. The sixth period of the periodic table should have 32 elements. Explain this sentence on the basis of quantum numbers.
Solution
In the modern periodic table, each period starts with the filling of a new principal energy level. Thus, the sixth period begins with the filling of 6s orbital and continues till the filling of seventh energy level. According to the Aufbau rule the sub shell which follows 6s are 4f, 5d, 6p, and 7s. Therefore in the sixth period, electrons can be filled in only 6s, 4f, 5d and 6p sub shells.
Nows subshell has two, p subshell has 3, d-subshell has 5 and f-subshell has 7 orbitals. Hence in all, there are 16 orbitals that can be filled in this period which at the maximum can accommodate 32 electrons, and therefore, the sixth period has 32 elements.
Q18. Among the following groups, select the species which has the smallest radius. Also give appropriate reasons to support your answer.
- i) O, O–, O2-
- ii) K+, Sr2+, Ar
- iii) Si, P, Cl
Solution
i) Among O, O–, O2- the species O has the smallest radii because the radius of the anion is always larger than the radius of the atom from which it is formed. O–and O2- are anions of oxygen.
ii) In K+ and Ar the outer most shell is third where as in Sr2+ it is fourth. Out of K+, Sr2+ and Ar, K+ has smaller size because it has greater nuclear charge.
iii) Among Si, P and Cl, Cl has the smallest radius. Si, P and Cl belong to the same period. In a period the atomic radius decreases with increase in atomic number due to increase in the effective nuclear charge.
Q19. What were the main draw backs of the Mendeleev’s periodic table?
Solution
The main draw backs of the Mendeleev’s periodic table are as follows:
- The position of hydrogen in the periodic table was not certain, it was placed in first as well as eighth group.
- Some elements of higher atomic weight were placed before the elements with lower atomic weight.
- Lanthanides and actinides were not given a proper position.
- Isotopes of elements were placed in the same position in the table.
- The noble gases were not given any space in the periodic table.
- Some chemically similar elements were placed in different groups and some dissimilar elements were placed in the same group.
Q20. Differentiate between lanthanide and actinide series.
Solution
Lanthanoid series Actinoid series
- The sixth period of the periodic table is known as the lanthanoid series.
- The seventh period of the periodic table is known as the actinoid series.
- It involves the filling of 6s orbital.
- It involves the filling of the 7s orbital.
- It contains 32 elements.
- It is incomplete.
- Due to the filling up of 4f orbitals, it constitutes the first f transition series.
- Due to the filling up of 5f orbitals, it constitutes the second f transition series.
Q21. Why noble gases have largest positive electron gain enthalpies in their respective periods? Explain.
Solution
The noble gases have a stable electronic configuration so they have largest positive electron gain enthalpies in their respective periods.
Q22. Which of the following elements has the largest ionisation enthalpy?
- 1) F
- 2) N
- 3) C
- 4) O
Solution
Ionization enthalpy decreases down the group on moving from top to bottom in a periodic table.
Q23. Explain briefly the factors responsible for the difference in the behavior of the first member of the group in the s and p block as compared to the other members of the same group.
Solution
The factors responsible for the behavior of the first member of group in the s and p block as compared t the other members of the same group are:
(i) Small atomic size of the first element.
(ii) Large charge / radius ratio.
(iii) High electro negativity.
(iv) Absence of d-orbitals in the valence shell of the first element.
(v) Ability to form p-p
multiplpe bonds.
Q24. Differentiate between Mendeleev’s Periodic Law and Modern Periodic Law.
Solution
Mendeleev’s periodic law Modern periodic law
- It states that the properties of elements are a periodic function of their atomic weights.
- It states that the properties of elements are periodic functions of their atomic numbers.
- Mendeleev’s periodic table is based on Mendeleev’s periodic law.
- The Long-form of periodic table is based on the modern periodic law.
Q25. How can you predict the group, period and block of an element explain?
Solution
To know the group, period and block of an element first of all we write the electronic configuration of an element. The period is represented by the principal quantum number of the valence shell. Block is predicted on the basis of sub shell which receives the last electron. Group is predicted on the basis of the number of electrons in the penultimate or the outer most shell.
Q26. The electronic configuration of an element is (n-1) d1 ns2 here n = 4. Find the position of the element in the periodic table?
Solution
The electronic configuration of the element is 3d14s2. So, the element belongs to d-block, fourth period and group 3 of the periodic table.
Q27.
i) Write the correct order of the chemical reactivity of F, Cl, O and N on the basis of their oxidizing power.
ii) Explain the cause of diagonal relationship.
Solution
i) The oxidizing character increases across a period i.e. on moving from left to right. So, among F, O and N, oxidizing power decreases in the order F > O > N. However, with in a group, oxidizing power decreases from top to bottom. Thus, F is a stronger oxidizing agent than Cl. So, the overall decreasing order of oxidizing power is: F > Cl > O > N.
ii) In the periodic table, along the period electro negativity increases and electropositivity increases along a group. As a result the two effects tend to cancel each other in moving diagonally from top left to bottom right. Therefore, the elements are diagonally related in this way tend to have similar properties.
Q28. What is the atomic volume of germanium, if its atomic mass is 72.6 and its density is 5.47g cm-3
Solution

Q29. Differentiate between metals and non-metals. On the basis of atomic number and position in the periodic table arrange the following elements in increasing order of metallic character. Si, Be, Mg, P.
Solution
Metals Non-metal
i) They have a strong tendency to lose electrons to form cations.
i) They have a strong tendency to accept electrons to form anions.
ii) They are highly electropositive.
ii) They are highly electronegative.
iii) They are good reducing agents.
iii) They are good oxidizing agents.
iv) Their oxides are of basic or amphoteric in nature.
iv) Their oxides are acidic or neutral. Metallic character increases down a group and decreases along a period as we move from left to right so the increasing order of metallic character is: P < Si < Be < Mg < Na.
Q30. In which period of the periodic table are the man made radioactive elements present? Why are the 4f and 5f transition series placed separately in the periodic table?
Solution
The man made radioactive elements are present in sixth period of the periodic table. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification.
Q31. Explain with the help of chemical reaction that MgO is basic oxide while P4O10 is an acidic oxide
Solution
MgO reacts with aqueous HCl to form salt and water thus exhibiting its basic character P4O10 reacts with water to form phosphoric acid
The solution turns blue litmus red.