A) Teflon \[{{[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]}^{0}}\]
B) Neoprene \[{{[Co(en)C{{l}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}\]
C) Terylene
D) Nylon 66 \[{{[Co{{(en)}_{3}}]}^{3+}}\]
Correct Answer: B
Solution :
Teflon is a polymer of tetrafluoroethylene. \[=k[A][B]\] Neoprene is a polymer of chloroprene (2-chloro buta-1, 3-diene). \[N_{2}^{-}<{{N}_{2}}<N_{2}^{2-}\] \[N_{2}^{2-}<N_{2}^{-}<{{N}_{2}}\] Terylene is a polymer of terepthalic acid and ethylene glycol. Nylon-66 is a polymer of hexamethylene diamine and adipic acid, \[{{N}_{2}}<N_{2}^{2-}<N_{2}^{-}\] \[N_{2}^{-}<N_{2}^{2-}<{{N}_{2}}\]You need to login to perform this action.
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