Question

The atomic number of three elements A, B, C are 11,14 and 17 respectively. State the group of which these elements belongs in the m9dern periodic table. Write the formula of the compound formed when the element B reacts with C

Answer 1

A: 2,8,1 ...Since valence electrons is 1,so it will b found in 1st grp

B:2,8,4... Since no. of valence electrons is 4,so (4+10=14) 14th grp

C:2,8,7...17th grp

Now valency of B is 4,n that of C is 1,so covalent bond will be formed with formula BC4...

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Answer 2

Element A belongs to group 1, element B belongs to group 4 and element C belongs to group 7. Formula of compound formed when the element B reacts with C is $SiCl_{4}$.

Explanation:

• Element A has atomic number 11 and its electronic distribution is 2, 8, 1. As it contains only one valence electron hence, it belongs to group 1 of the modern periodic table. This element is sodium.

• Element B has atomic number 14 and its electronic distribution is 2, 8, 4. This is silicon and since, it contains 4 valence electrons hence, it belongs to group 4 of the periodic table. Element C has atomic number 17 and its electronic distribution is 2, 8, 7.

• Since, there are 7 valence electrons present in element C. So, it belongs to group 7 of the periodic table. This element is chlorine. A compound formed by element B and C is as follows.

               $Si + 2Cl_{2} \rightarrow SiCl_{4}$

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