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6 months ago

Class 11th Maths

Let O be the origin and A be the point z₁ = 1 + 2i. If B is the point z₂, Re(z₂) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

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Raj Pandey

Contributor-Level 9

$\frac{z_{2} - 0}{(1 + 2 i) - 0} = | \frac{O B}{O A} | e^{\frac{i π}{4}} \Rightarrow z_{2} = (1 + 2 i) (1 + i) = - 1 + 3 i ∴ a r g z_{2} = π - t a n^{- 1} 3 a n d | z_{2} | = \sqrt[]{10}$

$z_{1} - 2 z_{2} = 3 - 4 i ∴ a r g (z_{1} - 2 z_{2}) = - t a n^{- 1} \frac{4}{3} \Rightarrow | z_{1} - 2 z_{2} | = | 2 + 4 i + 1 - 3 i | = \sqrt[]{10}$

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6 months ago

Class 11th Ncert Solutions Chemistry Class 11th

On complete combustion of 0.492g of an organic compound containing C, H and O, 0.7938 g of CO₂ and 0.4428 g of H₂O was produced. The % composition of oxygen in the compounds is____________.

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Raj Pandey

Contributor-Level 9

% of C in organic compound

$= \frac{12}{14} * \frac{W C O_{2}}{W . O . C .} * 100$

$= \frac{952.56}{21.648} = 44 %$

$% o f H = \frac{2}{18} * \frac{W H_{2} O}{W . O . C .} * 100$

= $\frac{20}{18} * \frac{0.4428}{0.492} * 100$

= $\frac{88.56}{8.856} = 10 %$

= 100 – 54 = 46%

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Class 11th Redox Reactions

The disproportionation of $M n O_{4}^{2 -}$ in acidic medium resultant in the formation of two manganese compounds A and B. If the oxidation state of Mn is B is smaller than that of A, then the spin-only magnetic moment $(μ)$ value of B in BM is___________.

(Nearest integer)

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Raj Pandey

Contributor-Level 9

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

...more

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

$= \sqrt[]{n (n + 2)} = \sqrt[]{3 (3 + 2)} = \sqrt[]{15}$

$= 3.87 \approx 4$

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6 months ago

Class 11th Ncert Solutions Chemistry Class 11th

The number of interhalogens from the following having square pyramidal structure is:

CIF₃, IF₇, BrF₅, BrF₃, I₂Cl₆, IF₅, ClF, ClF₅

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Raj Pandey

Contributor-Level 9

$C l F_{3}$ ® T-shaped (sp³d)

IF₇ ® Pentagonal bipyramidal (sp³d³)

BrF₅ ® Square pyramidal (sp³d²)

BrF₃ ® T-Shaped (sp³d)

I₂Cl₆ ® Triangular bipyramidal (sp³d)

$I F_{5} \to$ Square Pyramidal (sp³d²)

ClF ® (sp³)

ClF₅ ® Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

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6 months ago

Class 11th Chemistry Chemical Bonding and Molecular Structure

The number of paramagnetic species among the following is

B₂, Li₂, C₂, $C_{2}^{-}$ , $O_{2}^{- 2}$ , $O_{2}^{+}$ and $H e_{2}^{+}$ .

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Raj Pandey

Contributor-Level 9

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

$\begin{array}{l} L i_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} \to D i a m a g n e t i c \\ C_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} \to D i a m a g n e t i c \\ C_{2}^{-} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{1} \to P a r a m a g n e t i c \end{array}$

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

$O_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{0} \to$ Paramagnetic

$H e_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 1} \to$ Paramagnetic

$∴$ Paramagnetic molecules are $= B_{2}, C_{2}^{-}, O_{2}^{+}, H e_{2}^{+}$

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6 months ago

Class 11th Chemistry

In the given reaction,

X + Y + 3Z XYZ₃

If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ₃. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ₃ is_____________g.

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Raj Pandey

Contributor-Level 9

x + y + 3z = xyz₃

1 mole 1 mole 0.05 mole

$∴ n_{x} = 1$

$n_{y} = 1$

$n_{z} = \frac{0.05}{3} = 0.0167$ here z is limiting reagent.

$? \frac{0.05}{3}$ mole z gives 1 mole xyz₃

mass of xyz₃ = n * molecular mass

= $\frac{0.05}{3} * (10 + 20 + 3 * 30) a . m . u .$

= 0.5 * 4 = 2g

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New answer posted

6 months ago

Class 11th Ncert Solutions Chemistry Class 11th

The geometry around boron in the product 'B' formed from the following reaction is

$B F_{3} + N a H \overset{450 K}{\to} A + N a F$

A + NMe₃ $\overset{}{\to} B$

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Raj Pandey

Contributor-Level 9

$2 B F_{3} + 6 N a H \to B_{2} H_{6} + 6 N a F$

$B_{2} H_{6} + 2 \overset{?}{N} {(M e)}_{3} \to 2 H_{3} B \leftarrow : N {(M e)}_{3}$

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6 months ago

Class 11th Ncert Solutions Chemistry Class 11th

Choose the correct order of density of the alkali meals:

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Raj Pandey

Contributor-Level 9

Density of metal a very specific and depends upon many factors.

Li ® 0.53 gm/cm³

Na ® 0.97 gm/cm³

K ® 0.86 gm/cm³

Rb ®1.53 gm/cm³

Cs ® 1.90 gm/cm³

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6 months ago

Class 11th Redox Reactions

The reaction of H₂O₂ with potassium permanganate in acidic medium leads to the formation of mainly.

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Raj Pandey

Contributor-Level 9

$2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}$

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Class 11th Chemical Equilibrium

20 mL of 0.1 M NH₄Oh is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given K_b (NH₄OH) = 1 * 10^-5, log2 = 0.30, log3 = 0.48, log5 = 0.69, log7 = 0.84, log11 = 1.04

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Raj Pandey

Contributor-Level 9

$[N H_{4} C l] = \frac{2}{60} = \frac{1}{30} M$

$p H = 7 - \frac{1}{2} P K_{b} - \frac{1}{2} l o g C$

$= 7 - \frac{5}{2} - \frac{1}{2} l o g (\frac{1}{30}) = 5.24$

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