The question asks us to identify the amphoteric oxides among the given options of group 14 element oxides. Amphoteric oxides are those that can react both as acids and bases. To solve this, we need to consider the properties of each given oxide.
Based on the above analysis, the oxides that are amphoteric from the given group 14 elements are \(SnO_2\) and \(PbO_2\). Therefore, the correct answer is:
\(SnO_2, PbO_2\)
These oxides (\(SnO_2\) and \(PbO_2\)) can react with both acids, such as hydrochloric acid, and bases like sodium hydroxide, exhibiting their amphoteric behavior.
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: \(\left[\text{Ni}(\text{NH}_3)_5\right]\text{Br}\)
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.
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $