Order of Covalent bond;
$\mathrm{A.~KF > KI ; LiF > KF}$
$\mathrm{B.~KF < KI ; LiF > KF}$
$\mathrm{C.~SnCl_4 > SnCl_2 ; CuCl > NaCl}$
$\mathrm{D.~LiF > KF ; CuCl < NaCl}$
$\mathrm{E.~KF < KI ; CuCl > NaCl}$
Choose the correct answer from the options given below :
Solution
B is correct $\mathrm{KF}<\mathrm{KI}$; $\mathrm{LiF}>\mathrm{KF}$
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$\mathrm{C}$ is correct $\underset{+4}{\mathrm{SnCl}_{4}}>\underset{+2}{\mathrm{SnCl}_{2}} ; \mathrm{CuCl}>\mathrm{NaCl}$
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$\mathrm{E}$ is correct $\mathrm{KF}<\mathrm{KI}$; $\mathrm{CuCl}>\mathrm{NaCl}$
About this question
Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding
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