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Mock Test Series for JEE Main & Advanced 2022

JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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Question 1: 2nd electron affinity for halogens is zero – explain.
Solution: After addition of one electron in halogen atom formation of halide ion takes place which has fully filled configuration (ns2np6) and thus possesses no tendency to gain one more electron.

Question 2: NaOH behaves as a base while Zn(OH)2 is amphoteric why?
Solution: In NaOH the bond electronegativity difference between Na and oxygen is greater than between H and O and therefore it is the Na–O bond that breaks releasing OH. But in case of Zn—O—H bond the difference of electronegativity of Zn—O and O—H are almost same. So there is equal probability that the bond breaks in both ways leading to an amphoteric behavior.

Question 3: Why does xenon react with fluorine whereas neon does not?
Solution: Xe has lower ionisation energy than Ne. The valence electrons in Xe (n = 5) are much farther from the nucleus than those of Ne (n=2) and much less tightly held by the nucleus; they are more willing to be shared that those in neon. Also xenon has empty 5d orbital which can help to accommodate the bonding pairs of electrons, while neon has all the valence orbitals filled.

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