Which of the following is the correct increas | vedclass

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(A) $(1)$ For alkali metal bicarbonates,the solubility increases down the group due to the decrease in lattice energy being more significant than the

Vedclass · Accepted Answer

Solubility in $H_2O: NaHCO_3 < KHCO_3 < RbHCO_3 < CsHCO_3$

Vedclass · Answer

(A) $(1)$ For alkali metal bicarbonates,the solubility increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$NaHCO_3 $(2)$ Boiling point order for group $15$ hydrides is $PH_3 $(3)$ According to Fajan's rule,as the charge on the cation increases,polarization increases,which increases the covalent character and decreases the ionic character. Thus,$Fe_2O_3$ has more covalent character than $FeO$.$(4)$ $IE_2$ of $Na$ $([Ne] 3s^1)$ is very high because the second electron is removed from a stable noble gas configuration,whereas $Mg$ $([Ne] 3s^2)$ has a lower $IE_2$ as it achieves a stable configuration after removing one electron. Thus,$Mg < Na$.

Which of the following is the correct increasing order for the mentioned property?

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Assertion : $K$,$Rb$ and $Cs$ (all belonging to group $1$) can also form superoxides.
Reason : The ionic radii of $K$,$Rb$ and $Cs$ show the following trend $Cs^{+} < Rb^{+} < K^{+}$.

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