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Consider the following transformations
(I) $Xe{{F}_{6}}+NaF\to N{{a}^{+}}{{[Xe{{F}_{7}}]}^{-}}$
(II) $2PC{{l}_{5}}(s)\to {{[PC{{l}_{4}}]}^{+}}{{[PC{{l}_{5}}]}^{-}}$
(III) ${{[Al{{({{H}_{2}}O)}_{6}}]}^{3+}}+{{H}_{2}}O\to {{[Al{{({{H}_{2}}O)}_{5}}OH]}^{2+}}+{{H}_{3}}{{O}^{+}}$

Possible transformation is:
(A) I, II, III
(B) I, III
(C) I, II
(D) II, III

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Last updated date: 20th Jun 2024
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Answer
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Hint: One of the basic fundamental rules to check whether a transformation or a reaction is possible or not is by checking whether the elements in the reactant and products are balanced or not.

Complete step by step solution:
-In the first option, the reaction is between Xenon fluoride and Sodium fluoride which gives an unstable compound. Xenon fluoride here acts as a fluoride donor and gives a balanced equation.

$Xe{{F}_{6}}+NaF\to N{{a}^{+}}{{[Xe{{F}_{7}}]}^{-}}$

-In the second option, the phosphorus pentachloride is undergoing self ionization in the polar solutions. The reaction is also balanced. This is a reversible reaction. In dilute solutions, the reaction occurs in the forward direction, but at higher concentrations, the reaction in the backward direction becomes more prevalent. The \[PC{{l}_{5}}\]in the reactant is in $s{{p}^{3}}d$hybridization, while ${{[PC{{l}_{4}}]}^{+}}$and ${{[PC{{l}_{5}}]}^{-}}$in the products are in $s{{p}^{3}}$ and $s{{p}^{3}}{{d}^{2}}$ hybridization respectively.

$2PC{{l}_{5}}(s)\to {{[PC{{l}_{4}}]}^{+}}{{[PC{{l}_{5}}]}^{-}}$

-In the third option, Aluminium Hexa-aqua complex ion on dissociation leads to the formation of the conjugate base that is ${{[Al{{({{H}_{2}}O)}_{5}}OH]}^{2+}}$and a proton which combines with water molecule forming a hydronium ion. $A{{l}^{3+}}$being highly charged and small attracts water molecule and binds via dative covalent bonds and it also attracts the lone pair of electrons from the O-H bond and the water molecule splits into a proton and a hydroxide ion.

${{[Al{{({{H}_{2}}O)}_{6}}]}^{3+}}+{{H}_{2}}O\to {{[Al{{({{H}_{2}}O)}_{5}}OH]}^{2+}}+{{H}_{3}}{{O}^{+}}$

So, option (A) is correct.

Note: $Xe{{F}_{6}}$is solid at room and it readily sublimes into yellow vapours. It can be prepared by heating $Xe{{F}_{2}}$ at about $300{}^\circ C$and 60 atm pressure, but with the use of $Ni{{F}_{2}}$as a catalyst, the temperature can be lowered to $120{}^\circ C$. $PC{{l}_{5}}$is one of the most important phosphorus chlorides which is a colourless, water-sensitive, and moisture-sensitive solid chemical compound. It is used as a chlorinating reagent. It is highly corrosive and reacts violently with water, this is a very dangerous substance.