Question

The electronic structures of six elements A To F are given in the table below: Which of the compounds formed are wrong? (a) 1 and 3 (b) 2 and 3 (c) 3 and 4 (d) 2 and 4

Answer 1

B IS THE CORRECT OPTION

BECAUSE ACCORDING TO TABLE ONLY B RATIO IS CORRECT

Answer 2

(d) 2 and 4 cannot be formed.

• When elements combine, they tend to get stabilized by achieving noble gas configuration.
• Compound 1: A needs to lose 1 electron to attain the configuration of helium and B needs two more electrons to complete its octet. Hence, 2 A donates 2 electrons which are accepted by 1 B to form stable A₂B.
• Compound 2: B and C both need to accept 2 and 1 electron respectively. Hence B₂C cannot be formed.
• Compound 3: C needs to accept 1 electron and D can lose 3 electrons. So 3 C atoms can accept 1 electron each from the D atom to form DC₃.
• Compound 4: E needs to accept 2 electrons and F wants to donate 2 electrons. These two will combine to form stable FE and not FE₂.

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