Periodic Classification of Elements

NCERT InText Solution

Question: 1. Did Dobereiner's Triads also exist in the columns of Newlands' Octaves?

Answer: Yes, Dobereiner's Triads exist in the columns of Newlands' Octaves.

In Newlands' Octaves Lithium (Li), Sodium (Na), and Potassium (K) exists as same these elements are in Doereiner's Triads.

Question: 2. What were the limitations of Dobereiner's classification?

Answer: Dobereiner could identify only three triads among the elements known at that time. Because of this Dobereiner's Classification of element was not found to be useful.

Question: 3. What were the limitations of Newlands' Law of Octaves?

Answer: Newlands' Octaves was applicable only upto calcium (Ca), as after calcium (Ca) every eighth element did not possess the properties similar to that of the first.

At the time of Newlands only 56 elements were known. After discovery of several new elements, their properties did not fit into the Law of Octaves.

Newlands placed cobalt (Co) and nickel (Ni) in the same group as they have similar properties, but kept iron (Fe) very far from these elements while iron, cobalt and nickel have similar properties.

Cobalt (Co), and nickel (Ni) are placed in same column as chlorine (Cl), fluorine (F) and Bromine, while cobalt and nickel has completely different properties than those of chlorine, fluorine and bromine.

Hence, Newlands' Law Octaves work well with lighter elements only.

Question: 4. Use Mendeleev's Periodic Table to predict the formulae for the oxides of the following elements:

K, C, Al, Si, Ba

Answer:

(a) K is in 1^st group and formula of oxides for 1^st group is R₂O

Thus, formula of oxide of K is K₂O

(b) C is in 4^th group and formula of oxides for 4^th group is RO₂

Thus, formula of oxide of C is CO₂

(c) Al is in the 3^rd group and formula of oxides for 3^rd group is R₂O₃

Thus, formula of oxide of Al(Aluminium) is Al₂O₃

(d) Si is in 4^th group and formula of oxides for 4^th group is RO₂

Thus, formula of oxide of Si(Silicon) is SiO₂

(e) Ba is in the 2^nd group and formula of oxides for 2^nd group is RO

Thus, formula of oxide of Ba(Barium) is BaO

Question: 5. Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table? (any two)

Answer: Scandium and Germanium

Question: 6. What were the criteria used by Mendeleev in creating his Periodic Table?

Answer: Criteria used by Mendeleev in creating his Periodic Table:

(a) He selected hydrogen and oxygen as they are very reactive and formed compounds with most elements. The formulae of the hydrides and oxides formed by an element were treated as one of the basic properties of an element for its classification.

(b) He took atomic masses of element as basic property and arranged elements in increasing order of their atomic mass in creating Periodic Table.

Question: 7. Why do you think the noble gases are placed in a separate group?

Answer: Noble gases are not reactive and they do not react with oxygen or hydrogen, thus they are placed in a separate group.

Question: 8. How could the Modern Periodic Table remove various anomalies of Mendeleev Periodic Table?

Answer: Anomalies of Mendeleev's Periodic Table were removed in following ways in the Modern Periodic Table:

(a) Hydrogen placed in 1^st group with elements having valence electrons equal to 1(one)

(b) As elements are arranged in the increasing order of their atomic number, thus there were no confusion left for the place of isotopes of elements.

(c) Many elements were did not placed in the correct order of their increasing atomic masses in Mendeleev Periodic Table, such as Cobalt having greater atomic mass than nickel was placed before Nickel.

Since, in Modern Periodic Table elements were placed in the increasing order of their atomic numbers, thus these anomalies get corrected.

Question: 9. Name two elements you would expect to show chemical reactions similar in magnesium. What is the basis for your choice?

Answer: Be (Beryllium) and Ca (Calcium) are two more elements which show chemical reactions similar in Mg (Magnesium).

These two elements Be (Beryllium) and Ca (Calcium) show chemical reactions similar to Mg (Magnesium) because they all have equal number of valence electrons, i.e. valency. And because of this they all are placed in similar 2^nd group.

Question: 10. Name

(a) three elements have a single electron in their outermost shell

Answer: H (Hydrogen), Li (Lithium), Na (Sodium).

These elements have single electron in their outermost shell and they are placed in 1^st group.

(b) two elements that have two electrons in their outermost shells.

Answer: Be (Beryllium) and Mg (Magnesium)

These elements have two electrons in their outermost shell and because of that they are placed in 2^nd group.

(c) three elements with filled outermost shells.

Answer: He (Helium), Ne (Neon) and Ar (Argon)

All elements having filled outermost shells are kept in 18^th group in the Modern Periodic Table. All such elements are inert gases. Helium, Neon and Argon are inert gases.

Question: 11.

(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements.

Answer: Lithium, sodium, potassium have one electrons present in their outermost shell.

(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have common?

Answer: Helium and Neon are inert gases. There atoms have completely filled outermost shell.

Question: 12. In the Modern Periodic Table, which are the metals among the first ten elements?

Answer: Lithium (Li) and Beryllium (Be) are the metals among the first ten elements.

Question: 13. By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristics?

Ga, Ge, As, Se, Be

Answer: Be (Beryllium) have the maximum metallic characteristics.

Because metallic character decreases by moving from left to right in a Periodic Table. (Be) Beryllium is in the second group while Ga(Gallium) is in the 13^th group, Ge (Germanium) is in the 14^th group, As (Arsenic) is in the 15^th group and Se(Selenium) is in the 16^th group.

Thus, Be (Beryllium) have the maximum metallic characters among the given elements as it has lowest group number.