Question:
Match List I with List IILIST I LIST II A Chlorophyll 1 \(Na _2 CO _3\) B Soda ash 2 \(CaSO _4\) C Dentistry, Ornamental work 3 \(Mg ^{2+}\) D Used in white washing 4 \(Ca ( OH )_2\)
Match List I with List II
| LIST I | LIST II | ||
| A | Chlorophyll | 1 | \(Na _2 CO _3\) |
| B | Soda ash | 2 | \(CaSO _4\) |
| C | Dentistry, Ornamental work | 3 | \(Mg ^{2+}\) |
| D | Used in white washing | 4 | \(Ca ( OH )_2\) |
Updated On: Mar 19, 2025
- A-III, B-I, C-II, D-IV
- A-II, B-III, C-IV, D-I
- A-III, B-IV, C-I, D-II
- A-II, B-I, C-III, D-IV
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
To correctly match List I with List II, let's analyze each item:
A. Chlorophyll: - Chlorophyll contains magnesium (\( \text{Mg}^{2+} \)) as its central metal ion. - Match: A – III (\( \text{Mg}^{2+} \)).
B. Soda ash: - Soda ash is chemically sodium carbonate (\( \text{Na}_2\text{CO}_3 \)). - Match: B – I (\( \text{Na}_2\text{CO}_3 \)).
C. Dentistry, Ornamental work: - Calcium sulfate (\( \text{CaSO}_4 \)) is used in dentistry and ornamental work, especially in plaster form. - Match: C – II (\( \text{CaSO}_4 \)).
D. Used in white washing: - Slaked lime (\( \text{Ca(OH)}_2 \)) is commonly used in white washing. - Match: D – IV \(\text{Ca}(OH)_2\)
Conclusion: The correct matches are: \[ \text{A – III, B – I, C – II, D – IV} \]
Was this answer helpful?
0
0
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
View More Questions
Questions Asked in JEE Main exam
- There is a weak base 'B' having $\text{pK}_b = 5.691$ of molarity 0.02M. When 0.02M HCl solution has been added, then pH of resultant buffer solution has been found to be 9. Take total volume of resultant buffer solution to be 100 ml. Find the value of 'x' & 'y', where 'x' is volume of HCl solution in ml & 'y' is volume of 'B' solution in ml. Given $\log(5) = 0.691$
- JEE Main - 2026
- Ionic Equilibrium In Solution
- Internal energy of a gas is given as U = 3nRT. 1 mole of He gas takes 126 J of heat and its temperature rises by 4°C. Atmospheric pressure is \(P_0 = 10^5\) Pa and area of piston is 17 cm\(^2\). Find distance moved by piston.
- JEE Main - 2026
- Thermodynamics
- If the arithmetic mean of \(\dfrac{1}{a}\) and \(\dfrac{1}{b}\) is \(\dfrac{5}{16}\) and \(a,\,4,\,\alpha,\,b\) are in increasing A.P., then both the roots of the equation \[ \alpha x^2-ax+2(\alpha-2b)=0 \] lie between:
- JEE Main - 2026
- Quadratic Equations
- What will be the number of significant figures in the summation of \(0.153,\; 153.2\) and \(153.2\)?
- JEE Main - 2026
- Units, Dimensions and Measurements
- In an adiabatic process, the temperature reduces to \( \frac{1}{4} \)th and volume increases to 8 times. Find the adiabatic constant of the gas.
- JEE Main - 2026
- Thermodynamics
View More Questions
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