Identify (A) and (B) in the following sequence of reactions.
$SnC{{l}_{2}}+2NaOH\xrightarrow{{}}(A)(\text{white ppt)}\xrightarrow[(excess)]{NaOH}B$
A. $Sn{{(OH)}_{2}},N{{a}_{2}}Sn{{O}_{3}}$
B. $Sn{{(OH)}_{2}},N{{a}_{2}}Sn{{O}_{2}}$
C. $Sn{{(OH)}_{2}},N{{a}_{2}}Sn{{(OH)}_{6}}$
D. $Sn{{(OH)}_{2}},\text{no effect}$

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Hint: This question is asked from the simple inorganic reaction. You can do this if you know the precipitation reactions. The NaOH in excess means there is enough NaOH to react with the reactant to its last possible form.

Complete answer:
In the first step, the reaction is in between $\text{SnC}{{\text{l}}_{\text{2}}}$ and two moles of NaOH. This is a precipitation reaction, where an insoluble salt is formed and precipitates out of the solvent as the two reactants react. Here, the two $\text{O}{{\text{H}}^{\text{-}}}$ will replace the two chloride ions to form the white precipitate which is$Sn{{(OH)}_{2}}$. The salt $NaCl$ is more stable than $SnC{{l}_{2}}$ and hence the $O{{H}^{-}}$ ions displace the $C{{l}^{-}}$ ions to form $NaCl$. The reaction is as follows:
$SnC{{l}_{2}}+2NaOH\to Sn{{(OH)}_{2}}(\text{White ppt)}$
In the next step, $Sn{{(OH)}_{2}}$ will react with the excess of $\text{O}{{\text{H}}^{\text{-}}}$from the NaOH. From study of coordination compounds, we know that the coordination number of $Sn$ is 6. So, in the excess of $\text{O}{{\text{H}}^{\text{-}}}$, all the six coordination bonds will be formed with the$\text{O}{{\text{H}}^{\text{-}}}$ ions acting as ligands. So, the resultant product will be${{\left[ Sn{{(OH)}_{6}} \right]}^{2-}}$. The $N{{a}^{+}}$ ions will form an ionic bond with the complex to neutralize the charge. The reaction is as follows:
$Sn{{(OH)}_{2}}+\text{excess }O{{H}^{-}}={{\left[ Sn{{(OH)}_{6}} \right]}^{2-}}$
So, the overall reaction will be $SnC{{l}_{2}}+2NaOH\to Sn{{(OH)}_{2}}+\text{excess NaOH}\to \text{N}{{\text{a}}_{2}}Sn{{\left( OH \right)}_{6}}$

Therefore, the correct option is ‘C. $Sn{{(OH)}_{2}},N{{a}_{2}}Sn{{(OH)}_{6}}$’

Additional Information:
- $SnC{{l}_{2}}$(Tin Chloride) is also known as stannous chloride is a white crystalline solid. It can form stable dehydrates. But in the aqueous solution, it tends to undergo hydrolysis.
- $SnC{{l}_{2}}$ is most commonly used as a reducing agent and in electrolytic bathing for tin-plating.

Note: The $SnC{{l}_{2}}$ molecule in the gas phase is a bent structure. Because $SnC{{l}_{2}}$ has a lone pair of electrons. $SnC{{l}_{2}}$ is mainly used as a reducing agent for metal plating like gold and silver. Remember that an excess of reagents will usually cause a reaction if the molecules have extra lone pairs and empty orbitals through which they can form bonds. The reaction may not have occurred if $SnC{{l}_{2}}$ was present in excess.