The reaction $Pb \left( NO _3\right)_2$ and $NaCl$ in water produces a precipitate that dissolves upon the addition of $HCl$ of appropriate concentration The dissolution of the precipitate is due to the formation of
- $PbCl _2$
- $PbCl _4$
- $\left[ PbCl _4\right]^{2-}$
- $\left[ PbCl _6\right]^{2-}$
The Correct Option is C
Approach Solution - 1
We are given the following reaction in aqueous solution:
\[ \text{Pb(NO}_3\text{)}_2 + \text{NaCl} \rightarrow \text{PbCl}_2 (\text{precipitate}) + 2 \text{NaNO}_3 \]
When the precipitate, \( \text{PbCl}_2 \), is formed and then dissolves upon the addition of HCl, this dissolution is due to the formation of a soluble complex ion.
Step 1: Precipitation of \( \text{PbCl}_2 \)
When \( \text{Pb(NO}_3\text{)}_2 \) and \( \text{NaCl} \) are mixed in water, a precipitation reaction occurs: \[ \text{Pb}^{2+} + 2 \text{Cl}^- \rightarrow \text{PbCl}_2 (s) \] Here, lead(II) chloride (\( \text{PbCl}_2 \)) forms as a white precipitate.
Step 2: Dissolution of the Precipitate
Upon adding excess HCl, the \( \text{PbCl}_2 \) precipitate dissolves, forming a soluble lead(II) chloride complex. This occurs because the chloride ions from HCl coordinate with lead ions to form a complex. The formation of the complex ion can be represented as: \[ \text{PbCl}_2 (s) + 2 \text{Cl}^- \rightleftharpoons [\text{PbCl}_4]^{2-} \] The \( [\text{PbCl}_4]^{2-} \) complex is soluble in water, which leads to the dissolution of the precipitate.
Step 3: Conclusion
The dissolution of \( \text{PbCl}_2 \) is due to the formation of the soluble \( [\text{PbCl}_4]^{2-} \) complex ion. Therefore, the correct answer is:
The correct option is C: \( [\text{PbCl}_4]^{2-} \)
Approach Solution -2
When lead ions (Pb2+) react with chloride ions (Cl−), a white precipitate of lead(II) chloride (PbCl2) forms.
This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion
So the correct answer option is (C)
Top Questions on p -Block Elements
- For \( P \), the incorrect statement is:
- JEE Main - 2026
- Chemistry
- p -Block Elements
- Two elements of p-block can form following halides XF$_3$ and YF$_3$. XF$_3$ can act as Lewis acid while YF$_3$ can act as Lewis base. Then hybridization of ‘X’ and ‘Y’ in XF$_3$ and YF$_3$ is respectively.
- JEE Main - 2026
- Chemistry
- p -Block Elements
- An element of p-block forms a species of type EH$_4^{+}$, which when passed through a basic solution of K$_2$[HgI$_4$], forms a brown ppt. Select the correct option:
- JEE Main - 2026
- Chemistry
- p -Block Elements
- Statement-I: An element ‘X’ of P-block forms a hydride H–X, which has the longest bond length, then element ‘X’ will have the shortest covalent radius.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.
- JEE Main - 2026
- Chemistry
- p -Block Elements
Given below are two statements.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Chemistry
- p -Block Elements
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties
Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
- Group 13 Elements: Boron family
- Group 14 Elements: Carbon family
- Group 15 Elements: Nitrogen family
- Group 16 Elements: Oxygen family
- Group 17 Elements: Fluorine family
- Group 18 Elements: Neon family