Experimentally it was found that a metal oxide has formula
$ M_{0.98}O. $ Metal $M$, present as $ M^{2+} $ and $ M^{3+} $
in its oxide.Fraction of the metal which exists as $ M^{3+} $
would be
- $ 7.01 \% $
- $ 4.08 \% $
- $ 6.05 \% $
- $ 5.08 \% $
The Correct Option is B
Solution and Explanation
Then the fraction of metal as $M^{2+} = (0.98 - x)$
$\therefore 3x + 2(0.98 -x ) = 2$
$x + 1.96 =2$
$x = 0.04$
$\therefore \%$ of $M^{3+}=\frac{0.04}{0.98}\times 100$
$=4.08\%$
Top Questions on The solid state
- Which of the following material is not a semiconductor.
- JEE Main - 2024
- Chemistry
- The solid state
- In a metal deficient oxide sample, MXY2O4(M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M2+ ions present in M is \(\frac{1}{3}\), the value of X is______.
- JEE Advanced - 2024
- Chemistry
- The solid state
A cubic solid is made up of two elements $X$ and $Y$ Atoms of $X$ are present on every alternate corner and one at the enter of cube $Y$ is at $\frac{1}{3} td$ of the total faces The empirical formula of the compound is
- JEE Main - 2023
- Chemistry
- The solid state
- A metal 𝑀 forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _______ $×10^{21}$. (Nearest integer).
( Given $N_A=6.02×10^{23}$)- JEE Main - 2023
- Chemistry
- The solid state
- Match List I with List II
List-I
List-II
(A) Hexagonal (I) ∝ ≠ β ≠ γ ≠ 90° (B) Orthorhombic (II) ∝ = γ = 90°, β ≠ 90° (C) Triclinic (III) ∝ = β = 90°, γ = 120° (D) Monoclinic (IV) ∝ = β = γ = 90°
Choose the correct answer from the options given below:- CUET (UG) - 2023
- Chemistry
- The solid state
Questions Asked in JEE Main exam
- Let \[\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}, \quad \text{and} \quad \vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k} \]be three vectors such that \[\vec{b} \times \vec{a} = \vec{c} \times \vec{a}.\]If the angle between the vector $\vec{c}$ and the vector $3\hat{i} + 4\hat{j} + \hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan^2 \theta$ is:
- JEE Main - 2024
- Vector Algebra
- 10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO\(_2\)(g) and 50 mL of water vapour. The total number of carbon and hydrogen atoms in the hydrocarbon is ______ .
- JEE Main - 2024
- Hydrocarbons
- If each term of a geometric progression \( a_1, a_2, a_3, \dots \) with \( a_1 = \frac{1}{8} \) and \( a_2 \neq a_1 \), is the arithmetic mean of the next two terms and \( S_n = a_1 + a_2 + \dots + a_n \), then \( S_{20} - S_{18} \) is equal to
- JEE Main - 2024
- Arithmetic Mean
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
- JEE Main - 2024
- speed and velocity
- $\textbf{Choose the correct statements about the hydrides of group 15 elements.}$
A. The stability of the hydrides decreases in the order \(\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3\)
B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
D. The basicity of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
Choose the most appropriate from the option given below:- JEE Main - 2024
- p -Block Elements
Concepts Used:
Solid State
Solids are substances that are featured by a definite shape, volume, and high density. In the solid-state, the composed particles are arranged in several manners. Solid-state, in simple terms, means "no moving parts." Thus solid-state electronic devices are the ones inclusive of solid components that don’t change their position. Solid is a state of matter where the composed particles are arranged close to each other. The composed particles can be either atoms, molecules, or ions.
Types of Solids:
Based on the nature of the order that is present in the arrangement of their constituent particles solids can be divided into two types;
- Amorphous solids behave the same as super cool liquids due to the arrangement of constituent particles in short-range order. They are isotropic and have a broad melting point (range is about greater than 5°C).
- Crystalline solids have a fixed shape and the constituent particles are arranged in a long-range order.
SUBSCRIBE TO OUR NEWS LETTER
