Which of the following orders are correct reg | vedclass

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(C) According to Fajan's rule,covalent character increases when the cation is smaller and has a higher positive charge,and the anion is larger and has

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$i, iii, iv$ only

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(C) According to Fajan's rule,covalent character increases when the cation is smaller and has a higher positive charge,and the anion is larger and has a higher negative charge.$(i)$ $I^-$ has a larger size than $F^-$,so $KI$ has more covalent character than $KF$. Thus,$KF $(ii)$ $Li^+$ is smaller than $K^+$,so $LiF$ has more covalent character than $KF$. Thus,$LiF > KF$,making $LiF $(iii)$ $Sn^{4+}$ has a higher positive charge than $Sn^{2+}$,so $SnCl_4$ has more covalent character than $SnCl_2$. Thus,$SnCl_2 $(iv)$ $Cu^+$ has a pseudo-noble gas configuration $(d^{10})$,which causes higher polarization than the noble gas configuration of $Na^+$. Thus,$CuCl > NaCl$,making $NaCl Therefore,the correct orders are $(i), (iii),$ and $(iv)$.

Which of the following orders are correct regarding their covalent character?
$i. KF < KI$
$ii. LiF < KF$
$iii. SnCl_2 < SnCl_4$
$iv. NaCl < CuCl$
The correct option is

Similar Questions

Which of the following statements regarding the order of covalent character are correct?
$A. KF > KI ; LiF > KF$
$B. KF < KI ; LiF > KF$
$C. SnCl_4 > SnCl_2 ; CuCl > NaCl$
$D. LiF > KF ; CuCl < NaCl$
$E. KF < KI ; CuCl > NaCl$

Which of the following metal halides is more covalent?

The correct increasing order of polarising power is

How would you explain the following observations?
$(i)$ $BeO$ is almost insoluble but $BeSO_4$ is soluble in water.
$(ii)$ $BaO$ is soluble but $BaSO_4$ is insoluble in water.
$(iii)$ $LiI$ is more soluble than $KI$ in ethanol.

The order of covalent character of $KF, KCl, KI$ is

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Which of the following orders are correct regarding their covalent character?$i. KF < KI$$ii. LiF < KF$$iii. SnCl_2 < SnCl_4$$iv. NaCl < CuCl$The correct option is