Which one of the following elements will remain as liquid inside pure boiling water?
Li
Ga
Cs
Br
The Correct Option is B
Solution and Explanation
Element State at Boiling Water Temperature Analysis
Boiling water has a temperature of \( 100^\circ\text{C} \). We need to find an element that has a melting point below \( 100^\circ\text{C} \) and a boiling point above \( 100^\circ\text{C} \) to remain liquid at this temperature.
- \( \text{Li} \) and \( \text{Cs} \): Alkali metals (Li and Cs) react vigorously with water.
- \( \text{Br}_2 \): Bromine (\( \text{Br}_2 \)) has a boiling point of \( 58.8^\circ\text{C} \), so it would be a gas at \( 100^\circ\text{C} \).
- \( \text{Ga} \): Gallium (\( \text{Ga} \)) has a melting point of \( 29.8^\circ\text{C} \) and a boiling point of \( 2400^\circ\text{C} \). Thus, gallium will remain liquid inside boiling water.
Conclusion:
Gallium will remain liquid inside boiling water.
Top Questions on p -Block Elements
- Why is PCl\(_5\) a known compound, but NCl\(_3\) is not?
- Bihar Board XII - 2025
- Chemistry
- p -Block Elements
- Which acid is the strongest: HF, HCl, HBr, HI?
- AIIMS Paramedical - 2025
- Chemistry
- p -Block Elements
- Oxidation numbers of oxygen in H_2O_2 and H_2O:
- AIIMS Paramedical - 2025
- 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
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
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 Main exam
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: 0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g H₂O. Molar mass of (X) is ___________g mol\(^{-1}\).
- JEE Main - 2025
- Organic Chemistry
- In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _____.
- JEE Main - 2025
- Reaction Mechanisms & Synthesis
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions