(X), (Y), (Z) are elements in the third period. Oxide of (X) is ionic, that of (Y) is amphoteric, and of (Z) a giant molecule.

(X), (Y) and (Z) will have atomic numbers in the order :

(X) < (Y) < (Z)

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(Z) < (Y) < (X)

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(X) < (Z) < (Y)

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(Y) < (X) < (Z)

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Solution

The correct option is D (X) < (Y) < (Z)
For third period, "s" block will form ionic oxide ( $N a_{2} O, M g O$ ), 13th group forms amphoteric oxide ( $A l_{2} O_{3}$ ) and 14th group element will form giant oxide $(S i O_{2})$ .
So the correct order of atomic numbers is (X) < (Y) < (Z)
Hence option A is correct.

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(X), (Y), (Z) are elements in the third period. Oxide of (X) is ionic, that of (Y) is amphoteric, and of (Z) a giant molecule. (X), (Y) and (Z) will have atomic numbers in the order :

The correct option is D (X) < (Y) < (Z)For third period, "s" block will form ionic oxide (Na2O, MgO), 13th group forms amphoteric oxide (Al2O3) and 14th group element will form giant oxide (SiO2).So the correct order of atomic numbers is (X) < (Y) < (Z)Hence option A is correct.

(X), (Y), (Z) are elements in the third period. Oxide of (X) is ionic, that of (Y) is amphoteric, and of (Z) a giant molecule.

(X), (Y) and (Z) will have atomic numbers in the order :