Structure of Atom

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New answer posted•a year ago

Structure of Atom Class 11th

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

           Assertion A: Energy of 2s orbital of hydrogen atom is greater that that of 2s orbital of lithium.

           Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic. 

           **In the light of the above statements, choose the correct answer from the options given below:**

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Energy of orbitals in same subshell decrease on increasing the atomic number.

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New answer posted•a year ago

Structure of Atom Class 11th

Given below are the quantum numbers for 4 electrons.

A. $n = 3, l = 2, m_{l} = 1, m_{s} = + 1 / 2$ 

B. $n = 4, l = 1, m_{l} = 0, m_{s} = + 1 / 2$ 

C. $n = 4, l = 2, m_{l} = - 2, m_{s} = - 1 / 2$ 

D. $n = 3, l = 1, m_{l} = - 1, m_{s} = + 1 / 2$ 

The correct order of increasing energy is

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Sequence of sub-energy level decided by the rule of $(n + l) .$

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New answer posted•a year ago

Structure of Atom Class 11th

Among the following species

$N_{2}, N_{2}^{+}, N_{2}^{-}, N_{2}^{2 -}, O_{2}, O_{2}^{+}, O_{2}^{-}, O_{2}^{2 -}$

The number of species showing diamagnesim is __________.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

In N₂ & $O_{2}^{2 -} \to$ no. of unpaired electron = 0

In other species ® no. of unpaired electron $\neq$ 0

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New answer posted•a year ago

Structure of Atom Class 11th

The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a₀ (Where a₀ = Bohr radius 52.9 pm) is__________ km s^-¹.

(Given: Mass of electron = 9.1 × 10^-³¹ kg, Planck’s constant h = 6.63 × 10^-³⁴ Js)

60 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$E_{1_{H}} = - 2.2 \times 1 0^{- 18} J$

$L i \to L i^{+ 2} + 2 e^{-}, n = 2$

$E_{L i^{+ 2}} = E_{1 H} \times \frac{Z^{2}}{n^{2}} = - 2.2 \times 1 0^{- 18} \times \frac{3^{2}}{2^{2}}$

$λ = 4 \times 1 0^{- 8}$ m

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New answer posted•a year ago

Structure of Atom Class 11th

The longest wavelength of light that can be used for the ionization of lithium atom (Li) in its ground state is x × 10^-⁸m. The value of x is_________. (Nearest Integer)

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$E_{1_{H}} = - 2.2 \times 1 0^{- 18} J$

$L i \to L i^{+ 2} + 2 e^{-}, n = 2$

$E_{L i^{+ 2}} = E_{1 H} \times \frac{Z^{2}}{n^{2}} = - 2.2 \times 1 0^{- 18} \times \frac{3^{2}}{2^{2}}$

$λ = 4 \times 1 0^{- 8}$ m

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New answer posted•a year ago

Structure of Atom Class 11th

The energy of one mole of photons of radiation of wavelength 300nm is (Given: h = 6.63 × 10³⁴ Js,N_A = 6.02 × 10²³ mol^-1 c = 3 × 10⁸ ms^-1)

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Energy of 1 photon = $\frac{h c}{λ}$

Energy of 1 mole of photon = $N_{A} \times \frac{h c}{λ}$ $_{} \frac{}{}$

$= \frac{6.022 \times 1 0^{23} \times 6.63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{300 \times 1 0^{- 9}}$

= 0.399 × 106 J = 399 KJ

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New answer posted•a year ago

Structure of Atom Class 11th

Consider the following statements:

(A) The principal quantum number ‘n’ is a positive integer with values of ‘n’ = 1, 2, 3,….

(B) The azimuthal quantum number $' l'$ for a given ‘n’ (principal quantum number) can have values as  $' l' = 0, 1, 2, . . . n$

(C) Magnetic orbital quantum number $' m_{l}'$ for a particular $' l'$ (azimuthal quantum number) has $(2 l + 1)$ values.

(D)  $\pm 1 / 2$ are two possible orientations of electron spin.

(E) For $l = 5,$ there will be a total of 9 orbitals

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$•$ Values of principal quantum number, n = 1, 2, 3, .

$•$ Values of azimuthal quantum number, $l = 0, 1, 2 . . . . . . . . . . ., n - 1$

$•$ Number of values of magnetic quantum number are $2 l + 1 .$

$•$ Values of spin quantum number are $\pm \frac{1}{2} .$

$•$ Number of orbitals for particular value of $l a r e 2 l + 1, s o f o r l = 5,$ no. of orbitals = 11.

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New answer posted•a year ago

Structure of Atom Class 11th

A 0.5 percent solution of potassium choride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^-¹ and molar mass of KCl is 74.6 g mol^-¹)

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

0.5 % KCl solution has molality (m) = $\frac{0.5 \times 1000}{74.5 \times 99.5}$

$K C l_{(a q)} ? K_{(a g)}^{+} + C l_{(a q)}^{-}$

1 - α α α

And I = $(1 - α + α + α) = 1 + α$

$i = \frac{Δ T f}{k f \times m} = \frac{0.24 \times 74.5 \times 99.5}{1.8 \times 0.5 \times 1000} = 1 + α$

1.976 = 1 + α

$∴ α = 0.976$

% = 97.6%

the nearest 98.

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New answer posted•a year ago

Structure of Atom Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Match List I with List II

List I

Enzyme

List II

Conversion of

A. Invertase

I. Starch into maltose

B. Zymase

II. Maltose into glucose

C. Diastase

III. Glucose into ethanol

D. Maltase

IV. Cane sugar into glucose

Choose the most appropriate answer from the options given below:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Invertase Cane sugar to glucose and fructose

Zymase Glucose to ethanol

Diastase Starch to Maltose

Maltase Maltose to glucose

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New answer posted•a year ago

Structure of Atom Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Which of the following statements are correct?

(A) The electronic configuration of Cr is [Ar] 3d⁵4s¹.

(B) The magnetic quantum number may a negative value.

(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.

(D) The total number of nodes are given by (n-2)

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

(A) Cr = [Ar]3d⁵4s¹

(B) m = $- l t o + l$

(C) According to Aufbau principle, orbital are filled in order of their increasing energies.

(D) Total nodes = n – 1

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List I Enzyme	List II Conversion of
A. Invertase	I. Starch into maltose
B. Zymase	II. Maltose into glucose
C. Diastase	III. Glucose into ethanol
D. Maltase	IV. Cane sugar into glucose

Structure of Atom

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