
Which of the following compounds will not dissolve in aqueous \[NaOH\]?
A.
B.
C.
D.
Answer
557.4k+ views
Hint: Determine the solubility of unknowns in water, \[\% \] sodium hydroxide saturated sodium bicarbonate and \[\% \] hydrochloric acid. If the unknown is insoluble in water, but does dissolve in \[5\% \] sodium hydroxide solution then your unknown probably contains an acidic functional group that is deprotonated by the sodium hydroxide producing an ionic compound.
Complete answer:
\[NaOH\] is a strong base. We need strong acids which have acidification and have acidic \[H - \] ions. Here we react to the reaction between strong acid and strong base. Both react and form the conjugate base and contain stability of the compound.
We see option step by step:
In the option A, the reaction forms:
In the option B, the reaction forms:
${O^ - }$
In the option C, the reaction forms:
${O^ - }$
In the option D, the reaction forms:
Option (A) contains resonance and it becomes stable and as it is followed in option (B). here option (C) is unstable compound and in option (D) Sulphur has \[-ve\] charge which is easily dispersed then it contains stability.
Hence, option (C) is the correct answer.
Note: The stability of a complex particle in solution is determined by the nature of the central atom and the ligands. The most important characteristics of the central atom, determining the stability of the complex compound, are the degree of oxidation the dimensions, and the electronic structure. In the case of complexes with monatomic ligands, stability is dependent on the same characteristics in the ligand.
Complete answer:
\[NaOH\] is a strong base. We need strong acids which have acidification and have acidic \[H - \] ions. Here we react to the reaction between strong acid and strong base. Both react and form the conjugate base and contain stability of the compound.
We see option step by step:
In the option A, the reaction forms:
In the option B, the reaction forms:
${O^ - }$
In the option C, the reaction forms:
${O^ - }$
In the option D, the reaction forms:
Option (A) contains resonance and it becomes stable and as it is followed in option (B). here option (C) is unstable compound and in option (D) Sulphur has \[-ve\] charge which is easily dispersed then it contains stability.
Hence, option (C) is the correct answer.
Note: The stability of a complex particle in solution is determined by the nature of the central atom and the ligands. The most important characteristics of the central atom, determining the stability of the complex compound, are the degree of oxidation the dimensions, and the electronic structure. In the case of complexes with monatomic ligands, stability is dependent on the same characteristics in the ligand.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

