Structure of Atom — Page 11 | JEE Chemistry

Q101

Given below are two statements : Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II) : If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is

\geqslant \sqrt{\dfrac{h}{\pi}} \times \dfrac{1}{2 m}

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

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Both Statement I and Statement II are false

D Statement I is true but Statement II is false

Correct Answer

Option B

Solution

Statement I : Correct This statement is describing the uncertainty principle.

It states that it is impossible to measure both the position and the momentum of an object.

Statement II : Correct.

If the uncertainty in the measurement of position ( $\Delta x$ ) and the uncertainty in the measurement of momentum ( $\Delta p$ ) are equal for an electron, then the uncertainty in the measurement of velocity ( $\Delta v$ ) is greater than or equal to $\sqrt{\dfrac{h}{\pi}}\dfrac{1}{2m}$ This is derived from the Heisenberg uncertainty principle,

\Delta x\,.\,\Delta {p_x} \ge {h \over {4\pi }}

Given,

\Delta x = \Delta p

Substitute this in

\Delta x\,.\,\Delta p \ge {h \over {4\pi }}

as

\Delta p\,.\,\Delta p \ge {h \over {4\pi }}

\Delta {p^2} \ge {h \over {4\pi }}

\Delta {p^2} \ge {h \over {4\pi }}

\left\{ \begin{array}{ll}The\,formula\,for\,momentum\,(P)\,is\,P = mV \\ For\,uncerta{\mathop{\rm \int}} y\,\Delta P = m\Delta v \end{array} \right.

So,

m\Delta v \ge \sqrt {{h \over {4\pi }}}

\Delta v \ge \sqrt {{h \over {4\pi }}} \times {1 \over m}

\Delta v \ge {1 \over 2} > \sqrt {{h \over \pi }} > {1 \over m}

\Delta v \ge \sqrt {{h \over \pi }} {1 \over {2m}}

Both the statements are true (correct). So answer is (2) Both statement I and statement II are true.

Q102

Consider the hypothetical situation where the azimuthal quantum number,

l

, takes values 0, 1, 2, ....., n + 1, where n is the principal quantum number. Then, the element with atomic number :

A 13 has a half-filled valence subshell

B 9 is the first alkali metal

C 8 is the first noble gas

D 6 has a 2p-valence subshell

Correct Answer

Option A

Solution

Under hypothetical situation, the value of l is greater than n which varies from 0 to n + 1.

For n = 1, l = 0, 1, 2 n = 2, l = 0, 1, 2, 3 Elements follow the following electronic configuration 1s 1p 1d 2s 2p 2d 2f Atomic number (Z) = 9 1s2 1p6 1d1 Atomic number = 6 1s2 1p4 Atomic number 8 1s2 1p6 Atomic number 13 1s2 1p6 1d5 (half filled)

Q103

The correct statement about probability density (except at infinite distance from nucleus) is :

A It can be zero for 1s orbital

B It can be zero for 3p orbital

C It can never be zero for 2s orbital

D It can negative for 2p orbital

Correct Answer

Option B

Solution

$\phi$ 2 (probability density) can be zero for 3p orbital other than infinite distance. It has one radial node.

Q104

The orbital having two radial as well as two angular nodes is :

A 3p

B 5d

C 4d

D 4f

Correct Answer

Option B

Solution

Number of radial nodes = (n – l – 1) Number of angular nodes = l for 5d; n = 5, l = 2 5d orbital has two radial nodes and two angular nodes

Q105

In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields? (A) n = 1, l = 0, m = 0 (B) n = 2, l = 0, m = 0 (C) n = 2, l = 1, m = 1 (D) n = 3, l = 2, m = 1 (E) n = 3, l = 2, m = 0

A (D) and (E)

B (C) and (D)

C (B) and (C)

D (A) and (B)

Correct Answer

Option A

Solution

As here is no electric and magnetic field so ignore m to calculate the energy of orbital.

As here atom is multi-electron so ( n + l ) rule is applicable here.

This rule says those orbitals which have more value of ( n + l ) will have more energy.

In (A), n + l = 1 + 0 = 1 In (B), n + l = 2 + 0 = 2 In (C), n + l = 2 + 1 = 3 In (D), n + l = 3 + 2 = 5 In (E), n + l = 3 + 2 = 5 So (D) and (E) will be of equal energy.

Q106

Compare the energies of following sets of quantum numbers for multielectron system. (A)

\mathrm{n}=4,1=1

(B)

\mathrm{n}=4,1=2

(C)

\mathrm{n}=3, \mathrm{l}=1

(D)

\mathrm{n}=3,1=2

(E)

\mathrm{n}=4,1=0

Choose the correct answer from the options given below :

A

(\mathrm{E})>(\mathrm{C})>(\mathrm{A})>(\mathrm{D})>(\mathrm{B})

B

(\mathrm{B})>(\mathrm{A})>(\mathrm{C})>(\mathrm{E})>(\mathrm{D})

C

(\mathrm{C})<(\mathrm{E})<(\mathrm{D})<(\mathrm{A})<(\mathrm{B})

D

(\mathrm{E})<(\mathrm{C})<(\mathrm{D})<(\mathrm{A})<(\mathrm{B})

Correct Answer

Option C

Solution

To compare the energies of electrons in a multielectron system, we must consider both the principal quantum number (

\mathrm{n}

) and the azimuthal quantum number (

\mathrm{l}

).

In multielectron atoms, the energy levels are influenced by electron-electron repulsions, which modify the energy ordering compared to the hydrogen atom.

The general rule, called the "n+l" rule or Madelung rule, states that the energy of an electron in a multielectron atom is primarily determined by the sum of the principal quantum number (

\mathrm{n}

) and the azimuthal quantum number (

\mathrm{l}

). An orbital with a lower

(\mathrm{n} + \mathrm{l})

value has lower energy. If two orbitals have the same

(\mathrm{n} + \mathrm{l})

value, the orbital with the lower

\mathrm{n}

value has lower energy. Now, let's analyze each set of quantum numbers: (A)

\mathrm{n}=4, \mathrm{l}=1

\mathrm{n} + \mathrm{l} = 4 + 1 = 5

(B)

\mathrm{n}=4, \mathrm{l}=2

\mathrm{n} + \mathrm{l} = 4 + 2 = 6

(C)

\mathrm{n}=3, \mathrm{l}=1

\mathrm{n} + \mathrm{l} = 3 + 1 = 4

(D)

\mathrm{n}=3, \mathrm{l}=2

\mathrm{n} + \mathrm{l} = 3 + 2 = 5

(E)

\mathrm{n}=4, \mathrm{l}=0

\mathrm{n} + \mathrm{l} = 4 + 0 = 4

Based on the

\mathrm{n} + \mathrm{l}

values, the order of increasing energy is: (C) = (E) Among (C) and (E), (C) has lower

\mathrm{n}

value, so it is lower in energy: (C) Among (A) and (D), (D) has lower

\mathrm{n}

value, so it is lower in energy: Thus, the correct order is: (C) The correct answer is Option C: $$ (\mathrm{C})

Q107

The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 x 108 ms–1 and NA = 6.02 x 1023 mol–1)

A 594 nm

B 640 nm

C 700 nm

D 494 nm

Correct Answer

Option D

Solution

Energy required to break one Cl2 molecule =

{{242 \times {{10}^3}} \over {6.02 \times {{10}^{23}}}}

J As E =

{{hc} \over \lambda }

So

\lambda = {{hc} \over E}

=

{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.02 \times {{10}^{23}}} \over {242 \times {{10}^3}}}

= 494 $\times$ 10-9 m = 494 nm

Q108

For electrons in ' 2 s ' and ' 2 p ' orbitals, the orbital angular momentum values, respectively are :

A 0 and

\sqrt{6} \dfrac{h}{2 \pi}

B 0 and

\sqrt{2} \dfrac{h}{2 \pi}

C

\sqrt{2} \dfrac{h}{2 \pi}

and 0

D

\dfrac{h}{2 \pi}

and

\sqrt{2} \dfrac{h}{2 \pi}

Correct Answer

Option B

Solution

The orbital angular momentum for electrons in orbitals is given by the formula: $\text{Orbital angular momentum} = \sqrt{\ell(\ell+1)} \dfrac{\mathrm{h}}{2 \pi}$ For the 2s orbital, the quantum number $\ell$ is 0: Orbital angular momentum = $\sqrt{0(0+1)} \dfrac{\mathrm{h}}{2 \pi} = 0$ For the 2p orbital, the quantum number $\ell$ is 1: Orbital angular momentum = $\sqrt{1(1+1)} \dfrac{\mathrm{h}}{2 \pi} = \sqrt{2} \dfrac{\mathrm{h}}{2 \pi}$

Q109

Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect ? (The Bohr radius is represented by a0)

A The probability density of finding the electron is maximum at the nucleus.

B The electron can be found at a distance 2a0 from the nucleus

C The total energy of the electron is maximum when it is at a distance a0 from the nucleus.

D The magnitude of potential energy is double that of its kinetic energy on an average.

Correct Answer

Option C

Solution

Probablity density =

{\psi ^2}

From the graph you can see probablity density is maximum at the nucleus where r = 0.

$\therefore$ Option A is correct.

(B) For hydrogen atom, electron can be found at a any distance distance from the nucleus.

$\therefore$ Option B is correct.

(C) Distance a0 from the nucleus = First orbital $\therefore$ n = 1 at first orbital As Total energy, En = -13.6 $\times$

{{{Z^2}} \over {{n^2}}}

eV $\therefore$ Total energy for hydrogen at n = 1 is, E1 = -13.6 $\times$

{{{1^2}} \over {{1^2}}}

eV = -13.6 eV So total energy is minimum at distance a0 from the nucleus.

$\therefore$ Option C is incorrect.

(D) Total energy, En = -13.6 $\times$

{{{Z^2}} \over {{n^2}}}

eV Kinetic energy, K.E = 13.6 $\times$

{{{Z^2}} \over {{n^2}}}

eV Potential energy, P.E = -27.2 $\times$

{{{Z^2}} \over {{n^2}}}

eV $\therefore$ Magnitude of potential energy is double that of its kinetic energy. $\therefore$ Option D is correct.

Q110

Consider the following pairs of electrons (A) (a) n = 3,

l

= 1, m1 = 1, ms = +

{1 \over 2}

(b) n = 3, 1 = 2, m1 = 1, ms = +

{1 \over 2}

(B) (a) n = 3,

l

= 2, m1 =

-

2, ms =

-

{1 \over 2}

(b) n = 3,

l

= 2, m1 =

-

1, ms =

-

{1 \over 2}

(C) (a) n = 4,

l

= 2, m1 = 2, ms = +

{1 \over 2}

(b) n = 3,

l

= 2, m1 = 2, ms = +

{1 \over 2}

The pairs of electrons present in degenerate orbitals is/are :

A Only (A)

B Only (B)

C Only (C)

D (B) and (C)

Correct Answer

Option B

Solution

For degenerate orbitals, only the value of m must be different.

The value of (n + l) must be the same.

Hence, the pair of electrons with quantum numbers given in (B) are degenerate.