Question:
Consider the oxides of group 14 elements \(SiO_2,GeO_2,SnO_2,PbO_2,CO\) and \(GeO\). The amphoteric oxides are
Consider the oxides of group 14 elements \(SiO_2,GeO_2,SnO_2,PbO_2,CO\) and \(GeO\). The amphoteric oxides are
Updated On: Nov 20, 2024
- \(GeO, GeO_2\)
- \(SiO_2, GeO_2\)
- \(SnO_2, PbO_2\)
- \(SnO_2, CO\)
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
\(SnO_2\) and \(PbO_2\) are amphoteric.
Was this answer helpful?
0
0
Top Questions on GROUP 14 ELEMENTS
- How many types of oxides do Carbon family form?
- JKCET - 2024
- Chemistry
- GROUP 14 ELEMENTS
- Given below are two statements:
Statement I: The electronegativity of group 14 elements from Si to Pb gradually decreases.
Statement II: Group 14 contains non-metallic, metallic, as well as metalloid elements.
In the light of the above statements, choose the most appropriate from the options given below:- JEE Main - 2024
- Chemistry
- GROUP 14 ELEMENTS
- Evaluate the following statements related to group 14 elements for their correctness.
(A) Covalent radius decreases down the group from C to Pb in a regular manner.
(B) Electronegativity decreases from C to Pb down the group gradually.
(C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d orbitals.
(D) Heavier elements do not form ppi-ppi bonds.
(E) Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given below:- JEE Main - 2024
- Chemistry
- GROUP 14 ELEMENTS
Which one of the following complex ions has geometrical isomers?
Option 1: \(\left[\text{Co}(\text{Cl})_2(\text{en})_2\right]^+\)
Option 2: \(\left[\text{Cr}(\text{NH}_3)_4(\text{en})\right]^{3+}\)
Option 3: \(\left[\text{Co}(\text{en})_3\right]^{3+}\)
Option 4: \(\left[\text{Ni}(\text{NH}_3)_5\right]\text{Br}\)
- TS EAMCET - 2024
- Chemistry
- GROUP 14 ELEMENTS
Given are two statements regarding the properties of carbon in Group 14 of the periodic table:
Statement-I: Carbon has the highest catenation power in group 14 elements.
Statement-II: Carbon has small size and high electronegativity compared to other elements of group 14.- AP EAMCET - 2024
- Chemistry
- GROUP 14 ELEMENTS
View More Questions
Questions Asked in JEE Main exam
- Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:
- JEE Main - 2025
- permutations and combinations
- Let $ A = [a_{ij}] $ be a $3 \times 3 $ matrix such that: $$ A = \begin{bmatrix} 0 & 0 & 4 \\ 1 & 0 & 0 \\ 0 & 1 & 3 \\ \end{bmatrix}, A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix}. $$ Then $ a_{23} $ equals:
- JEE Main - 2025
- Matrix Operations
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- At the interface between two materials having refractive indices \( n_1 \) and \( n_2 \), the critical angle for reflection of an EM wave is \( \theta_c \). The \( n_1 \) material is replaced by another material having refractive index \( n_3 \), such that the critical angle at the interface between \( n_1 \) and \( n_3 \) materials is \( \theta_{c3} \). If \( n_1 > n_2 > n_3 \), \( \frac{n_2}{n_3} = \frac{2}{5} \), and \( \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} \), then \( \theta_{c1} \) is:
- JEE Main - 2025
- Refraction of Light
- Let the lines \[ 3x - 4y - \alpha = 0, \quad 8x - 11y - 33 = 0, \quad 2x - 3y + \lambda = 0 \] be concurrent. If the image of the point \( (1, 2) \) in the line \[ 2x - 3y + \lambda = 0 \text{ is } \left( \frac{57}{13}, \frac{-40}{13} \right), \text{ then } |\alpha \lambda| \text{ is equal to:} \]
- JEE Main - 2025
- Matrices and Determinants
View More Questions