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Question

(A) $(1)$ Free radicals are electrically neutral species containing an odd number of electrons (unpaired electron). Thus,$(1)-(B)-(E)$.$(2)$ Electroph

Vedclass · Accepted Answer

$(1)-(B)-(E), (2)-(D)-(F), (3)-(A)-(C)$

Vedclass · Answer

(A) $(1)$ Free radicals are electrically neutral species containing an odd number of electrons (unpaired electron). Thus,$(1)-(B)-(E)$.$(2)$ Electrophiles are electron-deficient species,often having an incomplete octet,and act as Lewis acids. Thus,$(2)-(D)-(F)$.$(3)$ Nucleophiles are electron-rich species with a complete octet (lone pairs) and act as Lewis bases. Thus,$(3)-(A)-(C)$.Therefore,the correct matching is $(1)-(B)-(E), (2)-(D)-(F), (3)-(A)-(C)$.

Column-$A$	Column-$B$
$(1)$ Free radical	$(A)$ Lewis base
$(2)$ Electrophile	$(B)$ Electrically neutral
$(3)$ Nucleophile	$(C)$ Octet complete
	$(D)$ Lewis acid
	$(E)$ Odd number of electrons
	$(F)$ Incomplete octet

Match the items in Column-$A$ with the appropriate items in Column-$B$:
Column-$A$ Column-$B$
$(1)$ Free radical $(A)$ Lewis base
$(2)$ Electrophile $(B)$ Electrically neutral
$(3)$ Nucleophile $(C)$ Octet complete
$(D)$ Lewis acid
$(E)$ Odd number of electrons
$(F)$ Incomplete octet

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Match the items in Column-$A$ with the appropriate items in Column-$B$:Column-$A$Column-$B$$(1)$ Free radical$(A)$ Lewis base$(2)$ Electrophile$(B)$ Electrically neutral$(3)$ Nucleophile$(C)$ Octet complete$(D)$ Lewis acid$(E)$ Odd number of electrons$(F)$ Incomplete octet