Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

4 months ago

Class 11th Chemistry Classification of Elements and Periodicity in Properties

The pair, in which ions are isoelectronic with Al³⁺ is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Au + NaCN + O₂ → Na [Au (CN)₂]

$Z n + N a [A u {(C N)}_{2}] \to N a_{2} [Z n {(C N)}_{4}] + A u$

$A i s {[A u {(C N)}_{2}]}^{-} a n d B i s {[Z n {(C N)}_{4}]}^{- 2}$

0 0

New answer posted

4 months ago

Class 11th Chemistry Chemical Bonding and Molecular Structure

Bonding in which of the following diatonic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

(A) NO

(B) N₂

(C) O₂

(D) C₂

(E) B₂

Choose the most appropriate answer from the options given below

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Bond strength $\propto$ Bond order

$N_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2}$

$O_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{* 1}$

$C_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2}$

$B_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{1} \equiv π_{2 p y}^{1}$

NO Number of electron = 7 + 8 = 15

B.O. Similar to $N_{2}^{-}$

$N_{2}^{-} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{*}$

B.O. of N₂ = 3B.O of C₂ = $\frac{8 - 4}{2} = 2$

Removal of e form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e in orbital.

0 0

New answer posted

4 months ago

Class 11th Chemistry

What are major differences between covalent and ionic bonds?

0 Follower 3 Views

Pallavi Arora

Beginner-Level 5

The name of covalent and ionic bonds tell their formation story in small detail. Co- means sharing or coexisting, so whenever there is bond fromation due to sharing of valence electron, it is known as covalent bond. Similerly, ionic bonds suggest that when a bond formation takes place due to ions, which attract each other, it is considered as ionic bond. Ionic bonds are also known as electrovalent bonds. There are some major differences between Electrovalent and covalent bonds. Chekc the table below:

Aspect	Covalent Bond	Ionic Bond
Reason of Formation	sharing of electrons between atoms	transfer of electrons from one atom to another
Elements	Usually non-metals with non-metals	Usually metals with non-metals
Strength	Moderate	the strongest
Nature	Directional Bond & Poor conductors of electricity	Non-directional Bond & Conducts electricity in aqueous state
Electronegativity Difference	Small (generally < 1.7)	Large (generally > 1.7)
Physical State	Usually Gases, liquids, and soft solids	Hard crystalline solids
Melting & Boiling Points	low to moderate	high
Solubility	Non-polar solvents	Polar solvents like water
Examples	H? , O? , CH? , HCl, NH?	NaCl, KBr, CaCl? , MgO

For more detail related to the chemical bonding in class 11 chemistry read our notes.

Chemical Bonding and Molecular Structure

0 0

New answer posted

4 months ago

Class 11th Physics Motion in Straight Line

A force on an object of mass 100g is $(10 \hat{i} + 5 \hat{j}) N .$ The position of that object at t = 2s is $(a \hat{i} + b \hat{j}) m$ after starting from rest. The value of $\frac{a}{b}$ will be

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

$\vec{F} = 10 \hat{i} + 5 \hat{j}$

$\vec{a} = \frac{10}{m} \hat{i} + \frac{5}{m} \hat{j}$

$= \frac{10}{0.1 k g} \hat{i} + \frac{5}{0.1} \hat{j}$

$\vec{a} = 100 \hat{i} + 50 \hat{j}$

$a_{x} = 100, a_{y} = 50$

$S_{x} = u t + \frac{1}{2} a t^{2}$

$= 0 * 2 + \frac{1}{2} * 100 * 2 * 2$

Sx = 200 m

$\frac{a}{b} = \frac{200}{100} = 2$

0 0

New answer posted

4 months ago

Class 11th Physics Waves

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20cm. The length of the open organ pipe is……….cm.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$H = \frac{λ}{4}$

$\Rightarrow 20 = \frac{λ}{4}$

$λ = 80 c m$

Now first overtone for open organ pipe

$L_{2} = 4 \frac{λ}{4} = λ = 80 c m$

0 0

New answer posted

4 months ago

Class 11th Physics Thermodynamics

A steam engine intakes 50g of steam at $100 ° C$ per minute and cools it down to $20 ° C .$ If latent heat of vaporization of steam is 540 cal g^-¹, then the heat rejected by the steam engine per minute is………….* 10³ cal.

(Given : specific heat capacity of water : 1 cal g^-¹ ⁰C^-¹)

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

$Q_{1} = \underset{w a t e r}{20 ° C} \to 100 ° s t e a m$

$Q_{1} = m c Δ T + m L$

$= m [c Δ T + L]$

$= 31000 = 31 * 1 0^{3} c a l$

$Q_{2} = \frac{293}{373} * 31 * 1 0^{3}$

$= 24.35 * 1 0^{3} c a l$

0 0

New answer posted

4 months ago

Class 11th Physics Mechanical Properties of Fluids

The velocity of upper layer of water in a river is 36kmh^-¹. Shearing stress between horizontal layers of water is 10^-³ Nm^-². Depth of the river is…….m. (Co-efficient of viscosity of water is 10^-² Pa.s)

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

$τ = η \frac{v}{h}$

Given

$τ = 1 0^{- 3} N / m^{2}$

(shear stress)

h =?

v = 36 km/hr = 10 m/sec

$1 0^{- 3} = \frac{1 0^{- 2} * 10}{h}$

h = 100 m

0 0

New answer posted

4 months ago

Class 11th Work, Energy and Power

A 0.5kg block moving at a speed of 12ms^-¹ compresses a spring through a distance 30cm when its speed is halved. The spring constant of the spring will be…………Nm^-¹.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

0 0

New answer posted

4 months ago

Class 11th Physics

A uniform chain of 6m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is………….m.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = $μ N$

$\Rightarrow λ x g = μ λ (L - x) g$

$x = 0.5 (6 - x)$

$\Rightarrow x = 3 - 0.5 x$

$1.5 x = 3$

x = 2m

0 0

New question posted

4 months ago

Class 11th Maths

Let lx – 2y = µ be a tangent to the hyperbola $a^{2} x^{2} - y^{2} = b^{2} .$ Then ${(\frac{λ}{a})}^{2} - {(\frac{μ}{b})}^{2}$ is equal to

0 Follower 3 Views

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
679k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience