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Science Answer the questions.[Che-3M]

Periodic Classification of Elements · CBSE STD 10 (CBSE - English Medium). 5 questions.

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Answer the questions.[Che-3M]

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Question 13 Marks
Write the formula of the product formed when the element A (atomic number 19) combines with the element B (atomic number 17). Draw its electronic dot structure. What is the nature of the bond formed?
Answer
Electronic configuration of element A(Z = 19) is 2, 8, 8, 1.
Electronic configuration of element B(Z = 17) is 2, 8, 7.
The element A has one valence electron while the element B has seven electrons in its valence shell. One electron gets transferred from the atom of element A to the atom of element B. As a result of electron transfer, an ionic bond is formed.

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Question 23 Marks
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.
Answer
Eka-silicon is identified as germanium (Ge) which is placed in group 4 of Mendeleev's periodic table. The valency of germanium is 4 so the chemical formula for chloride must be GeCl4.
Eka-aluminium was later identified as gallium (Ga). It is placed in group 3 of the Mendeleev's periodic table. Hence the valence of gallium is 3 and the formula of chloride would be GaCl3.
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Question 33 Marks
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as? Give one example of such a set of elements.
Answer
This is an example of Dobereiner's triads which is a group of three elements. In Dobereiner's triads the atomic masses of elements are arranged in such a way that the atomic mass of middle element is almost the mean of atomic masses of first and third elements.
For example; Ca, Sr and Ba is one of the Dobereiner's triads. Here, atomic masses of Ca, Sr and Ba are 40, 88 and 137 respectively. Atomic mass of Sr that is 88 is almost the mean of atomic masses of Ca and Ba.
Atomic mass of $\text{Sr}=\frac{40+137}{2}=88.5$
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Question 43 Marks
If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride?
Answer
If an element is placed in group 14, then it has 4 electrons in its outermost orbit. Hence, the formula of its chloride is ECl4. Since the element can make compound by sharing of electrons, so its compound will have covalent bond.
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Question 53 Marks
“Hydrogen occupies a unique position in Modern Periodic Table”. Justify the statement.
Answer
The position of the element hydrogen is still not clear even in the Modern Periodic Table.
  • In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron. However, it is not a metal whereas alkali-metals are metallic in nature.
  • In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well.
It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form. Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.
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