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In the following diagram for the first three periods of the periodic table, five elements have been represented by the letters `a,b,c,d` and `e` (which are not their chemical symbols): (i) Select the letter which represents a halogen. (ii) Select the letter which represents a noble gas. (iii) What type of bond is formed between `a` and `b`? (iv) What type of bond is formed between `c` and `b`? (v) Which element will form a divalent anion? |
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Answer» (i) The halogens are placed in group 17. Now, in the above given table the element present in group 17 is b. Thus, the letter b represents a halogen. (ii) The noble gases are placed in group 18 of the periodic table. In the above given table, the element placed in group 18 is e. Thus, the letter e represents a noble gas. (iii) We have now to find the type of bond formed betwen the elements a and b. Now, if we look at the above given table, we find that element a is placed non-metal reacts with another non-metal, the covalent bonds are formed. So, the bond between a and b is covalent bond. (iv) And now we have to find out the type of bond between the elements c and b. The element c is in group 2 so it is a metal. The element b is in group 17 so it is a non metal. We know that when a metal reacts with a non -metal, then an ionic bond is formed. Thus, the bond formed between the elements c and b will be ionic bond. (v) We will now find out that element which forms a divalent anion. A divalent means a negative ion having 2 units of negative charge. A divalent anion is formed by a non-metal atom having 6 valence electrons (so that it can accept 2 more electrons to complete the octet and form a divalent anion). Thus a divalent anion will be formed by a non metal element of group 16 because it will have 6 valence electrons in its atom. Now, in the above given table, the elemented has been placed in group 16 of the periodic table, so element will form a divalent anion. |
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