Class 11th

Identify the compounds A, X and Z in the following reactions :

i) A + 2HCl + 5H₂O → 2NaCl + X i )X HBO₂ Z

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Kindly go through the resolution

ii)

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Explain the following :

(i) Gallium has higher ionisation enthalpy than aluminium (ii) Boron does not exist as B³⁺ ion (iii) Aluminium forms [AlF₆] ^3– ion but boron does not form [BF₆] ^3– ion (iv) PbX₂ is more stable than PbX_4._{(v) P}_{b⁴⁺ acts as an oxidising agent but Sn²⁺ acts as a reducing agent.}_(vi)_{Electron gain enthalpy of chlorine is more negative as compared to fluorine}_(vii)_{Tl (NO}₃_{)3 acts as an oxidising agent. &nbs}

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Class 11th Systems of Particles and Rotational Motion

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This is a Short Answers Type Questions as classified in NCERT Exemplar

(i) When moving down the periodic table within a group, ionization enthalpy generally decreases due to increased atomic radius. However, in the case of gallium (Ga) and aluminum (Al), gallium experiences a higher effective nuclear charge due to less shielding from its inner electrons, resulting in a higher ionization enthalpy compared to aluminum.

(ii) As boron is smaller in size and the sum of its first three ionization enthalpies i.e. ΔH₁+ΔH₂+ΔH₃ is very large so boron does not allow to lose its all three valence electrons and exist as +3

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(iii) As compared to boron, aluminium is larger in size and has vacant d-orbital through which it can expand its coordination number from +4 to +6. Can easily acquire the sp³d² hybridisation State. On the other hand, boron is smaller in size with no d-orbitals and can expand its coordination number maximum up to +4 so boron can maximum the form BF₄⁻but not BF₆²⁻.

(iv) Lead is extra stable in its +2 oxidation state than in +4 oxidation state due to inert pair effect.

(v) Due to the inert Pair effect, lead is more stable in its +2 oxidation state than in +4. So if it is present in +4 oxidation state in any compound it can readily reduce itself by gaining two electrons. Thus acting as an oxidising agent.

Reaction: Pb⁴⁺+2e⁻→Pb²⁺

While for Sn2+, +4 oxidation state is extra stable than +2 oxidation state so in any compound where acid is present in +2 oxidation state it loses two more electron and oxidises itself to +4 oxidation state. Thus acting as a reducing agent.

Reaction: Sn²⁺→Sn⁴⁺+2e⁻

(vi) Electronic configuration of F & Clare:

In fluorine the new electron added will be accommodated in 2p sub-shell while in chlorine the new electron goes to 3p subshell. The size of the 2p subshell is smaller than the size of 3p sub shell. So, the added electron in 2p sub shell experiences strong inter-electronic repulsion which is comparatively less in the case of 3p sub shell of chlorine. The added electron doesn't feel much nuclear attraction in fluorine, that's why fluorine doesn't easily gain new electrons. Therefore, fluorine has less negative electron gain enthalpy than chlorine.

(vii) When we move down the group in group 13 due to the inert pair effect for Thallium, +3 oxidation state is less stable than its +1 oxidation state. In Tl (NO₃)₃ the central atom Tl exist in +3 oxidation state which is not much stable so it readily gains two electrons to form compounds in which Thallium is in +1 oxidation state that is TINO₃.

(viii) As we move down the group in group 14 size of element increases, electronegativity decreases which results in the decrease of M-M born strength hence the catenation decreases

(ix) Complete hydrolysis of BF₃does not occur, rather it gets partially hydrolysed. BF₃ reacts with water to form boric acid and HF the HF further reacts with H₃BO₃ and prevents Hydrolysation of BF₃.

(x) In group 14, as we move down the size of elements increases and electronegativity decreases. Silicon does not have sp² state and does not form sp² double bond like carbon does in graphite. Rather silicon exists in sp³ hybridisation state which forms 3-D structure more like diamond.

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The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

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no $\sum_{i} F_{i} \neq 0$

The sum of torques about a certain point O $\sum_{i} r_{i} * F_{i} = 0$

The sum of torques about any other O $\sum_{i} r_{i} F_{i} = 0$

The sum of torques about any other point O'

$\sum_{i} (r_{i} - a) * F_{i} = \sum_{i} r_{i} * F_{i} - a * \sum_{i} F_{i}$

Here the second term need not vanish.

Sum of all torques about any point is zero.

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Class 11th Systems of Particles and Rotational Motion

Identify the compounds A, X and Z in the following reactions :

i) A + 2HCl + 5H₂O → 2NaCl + X ii )X HBO₂ Z

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10 months ago

A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig). The sphere is struck horizontally at a height h from the floor. Match the following:

(a) h = R/2 (i) Sphere rolls without slipping with a constant velocity and no loss of energy.

(b) h = R (ii) Sphere spins clockwise, loses energy by friction.

(c) h = 3R/2 (iii) Sphere spins anti-clockwise, loses energy by friction.

(d) h = 7R/5 (iv) Sphere has only a translational motion, looses energy by friction.

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Class 11th System of Particles and Rotational Motion

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Consider a fig in which m is mass of sphere and R is the radius having h height above the floor

The sphere will roll without slipping when w=v/r where v is linear velocity and w is angular velocity .

By conservation of momentum

mv (h-R)=Iw=2/5mR² (v/R)

mv (h-R)=2/5mvR

h-R=2/5R

so h = 7/5R .d $\to i$ sphere will roll here so no loss of energy.

Torque = F (h-R)

For torque=0 h=R sphere will have only translational motion. It would lose energy by friction.

b $\to i v$

the sphere will spin clockwise when t>0 so h>R

so c $\to i i$ and a $\to i i i$

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10 months ago

A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in Fig. Match the following (most appropriate choice):

(a) mg/4 < F mg < /2 (i) Cube will move up.

(b) F > mg/2 (ii) Cube will not exhibit motion.

(c) F > mg (iii) Cube will begin to rotate and slip at A.

(d) F = mg/4 (iv) Normal reaction effectively at a/3 from A, no motion.

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This is a short answer type question as classified in NCERT Exemplar

Moment of force $τ$ ₁= F (a)….anticlockwise

Moment of weight mg $τ$ ₂=mg (a/2)……. (clockwise)

Cube will not exhibit motion then $τ$ _{1
$= τ$}₂

$F * a = m g * \frac{a}{2}$

F=mg/2

Cube will rotate when $τ$ _{1
$> τ$}₂

$F * a > m g * \frac{a}{2}$

F>mg/2

At a/3 from point A then

mg $* \frac{a}{3} = F * a o r F = \frac{m g}{3}$

when F=mg/4 which is less then mg/3 there will be no motion.

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10 months ago

Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction does not proceed. Explain the reason.

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Class 11th Systems of Particles and Rotational Motion

Contributor-Level 10

This is a Short Answers Type Questions as classified in NCERT Exemplar

The following reaction takes place in the test tube

i) When Aluminium is treated with dilute hydrochloric acid.

Al + HCl -à AlCl₃

ii) When Aluminium is treated with dilute sodium hydroxide solution.

Al + NaOH -àNaAlO₂ + H₂

In both the cases, the Hydrogen atoms are evolved. So, when a burning match stick is brought near the test tube we hear a pop sound.

Aluminium does not react with concentrated nitric acid. A "layer" of "Aluminium Oxide" is formed, when aluminium reacts with the nitric acid because "nitric acid" is an oxidising agent. Due t

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The variation of angular position θ , of a point on a rotating rigid body, with time t is shown in Fig. Is the body rotating clock-wise or anti-clockwise?

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This is a short answer type question as classified in NCERT Exemplar

As the slope of graph is positive and positive slope indicates anti clockwise rotation which is traditionally taken as positive.

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When BCl₃ is treated with water, it hydrolyses and forms [B(OH)₄]⁻only whereas AlCl₃ in acidified aqueous solution forms [Al(H₂0)₆]³⁺

Explain what is the hybridisation of boron and aluminium in these species?

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Class 11th System of Particles and Rotational Motion

Contributor-Level 10

This is a Short Answers Type Questions as classified in NCERT Exemplar

BCl₃+3H₂O→B (OH)₃+3HCl

B (OH)₃+2H₂O→ [B (OH)₄]⁻+H₃O⁺

B (OH)₃due to its incomplete octet accepts an electron pair (OH)^- to give has configucation [B (OH)₄]⁻.

Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridization of B in [B (OH)₄]⁻ is sp³

AlCl₃ + 6H₂O [Al (H₂O)₆]³⁺ +3Cl^-

Hence, hybridization of Al is sp³d²

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Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

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