Structure of Atom — Page 12 | JEE Chemistry

Q111

Arrange the following orbitals in decreasing order of energy. A.

\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0

B.

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

C.

\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0

D.

\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1

The correct option for the order is :

A

\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}

B

\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}

C

\mathrm{D}>\mathrm{B}>\mathrm{A}>\mathrm{C}

D

\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}

Correct Answer

Option D

Solution

(A) n = 3; l = 0; m = 0 ; 3s orbital (B) n = 4; l = 0; m = 0 ; 4s orbital (C) n = 3; l = 1; m = 0 ; 3p orbital (D) n = 3; l = 2; m = 0 ; 3d orbital As per Hund’s ruleo of energy is given by (n + l) value.

If value of (n + l) remains same then energy is given by n only.

Q112

The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is

A 8.51 × 105 J mol−1

B 6.56 × 105 J mol−1

C 7.56 × 105 J mol−1

D 9.84 × 105 J mol−1

Correct Answer

Option D

Solution

Note : 1 eV/atom = 96.485 $\times$ 103 J/mol $\therefore$ 13.6 eV/atom = 13.6 $\times$ 96.485 $\times$ 103 J/mol = 1.312 × 106 J mol−1 Energy required to excite the electron from n1 to n2 is

\Delta E = 13.6 \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

= 1.312 × 106 × 1

\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)

= 1.312 ×

{3 \over 4}

× 106 = 9.84 × 105 J mol−1

Q113

Given below are two statements : Statement (I) : A spectral line will be observed for a

2 p_x \rightarrow 2 p_y

transition. Statement (II) :

2 \mathrm{p}_x

and

2 \mathrm{p}_y

are degenerate orbitals. 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 false

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

Let us analyze the two statements in the context of atomic orbitals and electronic transitions: Statement (I) “A spectral line will be observed for a $2 p_x \rightarrow 2 p_y$ transition.”

For an emission (or absorption) line to be observed, there must be a difference in energy between the initial and final states.

In a typical hydrogen-like or many-electron atom (without additional external fields or splitting effects), the three $2p$ orbitals ( $2 p_x$ , $2 p_y$ , $2 p_z$ ) are degenerate—i.e., they all have the same energy.

Consequently, a transition from $2 p_x$ to $2 p_y$ (both having the same energy) would involve no energy change.

Hence, no photon is emitted or absorbed for this “transition.”

So you would not observe a spectral line for such a transition.

Therefore, Statement (I) is false.

Statement (II) “ $2 p_x$ and $2 p_y$ are degenerate orbitals.”

Orbitals within the same subshell (e.g., $2p$ subshell) are typically degenerate (same energy) in an isolated atom (especially a hydrogen-like atom).

Thus, $2 p_x$ and $2 p_y$ (and $2 p_z$ ) do indeed have the same energy.

Therefore, Statement (II) is true.

Conclusion Statement (I): False Statement (II): True Hence, the correct choice is: Option A: Statement I is false but Statement II is true.

Q114

The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is :

A 5.92

B 6.93

C 3.87

D 4.90

Correct Answer

Option A

Solution

In transition metal contains d orbital, and in d orbital maximum no of unpaired electron possible = 5.

Spin only magnetic moment,

{\mu _{spin}} = \sqrt {n\left( {n + 2} \right)}

B. M here n $=$ Number of unpaired electrons. $\therefore$

{\mu _{spin}} = \sqrt {5\left( {5 + 2} \right)}

B. M $=$ 5.92 B. M

Q115

The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 meters per second is approximately

A 10-31 meters

B 10-16 meters

C 10-25 meters

D 10-33 meters

Correct Answer

Option D

Solution

As per de-Broglie wavelength of a particle is Wavelength ( $\lambda$ ) =

{h \over {mv}}

Where h is plank's constant, m is mass of the particle and v is the velocity of the particle

\therefore \lambda

=

{{6.63 \times {{10}^{ - 34}}} \over {0.06 \times 10}}

=

{{6.63 \times {{10}^{ - 34}} \times 10} \over {0.6 \times 10}}

=

{{6.63 \times {{10}^{ - 33}}} \over 6}

\cong 10{}^{ - 33}

Q116

The de Broglie wavelength of an electron in the 4th Bohr orbit is :

A 2

\pi

a0

B 6

\pi

a0

C 8

\pi

a0

D 4

\pi

a0

Correct Answer

Option C

Solution

According to Bohr’s model

{r_n} = {{{n^2}} \over Z} \times {a_0}

Also 2 $\pi$

{r_n}

= n $\lambda$ $\Rightarrow$ 2 $\pi$

{{{n^2}} \over Z} \times {a_0}

= n $\lambda$ $\Rightarrow$ $\lambda$ =

2\pi \times {n \over Z} \times {a_0}

For n = 4 and Z = 1 $\Rightarrow$ $\lambda$ = 8 $\pi$

{a_0}

Q117

For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number

\left( {\overline v } \right)

against

\left( {{1 \over {{n^2}}}} \right)

will be (The Rydberg constant, RH is in wave number unit)

A Linear with intercept

-

RH

B Non linear

C Linear with slope RH

D Linear with slope

-

RH

Correct Answer

Option C

Solution

We know,

\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]

here ni $=$ 8 and nf $=$ n Z $=$ 1 for hydrogen $\therefore$

\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]

So, graph is $\to$ So, graph will be linear with slope RH.

Q118

The difference between the radii of 3rd and 4th orbits of Li2+ is R1 . The difference between the radii of 3rd and 4th orbits of He+ is

\Delta

R2 . Ratio

\Delta

R1 :

\Delta

R2 is :

A 8 : 3

B 3 : 2

C 2 : 3

D 3 : 8

Correct Answer

Option C

Solution

Rn = a0

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

{{\Delta {R_1}} \over {\Delta {R_2}}} = {{{{\left( {{r_4} - {r_3}} \right)}_{L{i^{2 + }}}}} \over {{{\left( {{r_4} - {r_3}} \right)}_{H{e^ + }}}}}

=

{{{{{4^2}} \over 3} - {{{3^2}} \over 3}} \over {{{{4^2}} \over 2} - {{{4^2}} \over 2}}}

=

{{7/3} \over {7/2}} = {2 \over 3}

Q119

Identify the incorrect statement from the following.

A A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit.

B An orbital is the one electron wave function

(\psi)

in an atom.

C The existence of Bohr's orbits is supported by hydrogen spectrum.

D Atomic orbital is characterised by the quantum numbers

\mathrm{n}

and

l

only.

Correct Answer

Option D

Solution

Atomic orbital is characterised by the quantum numbers $\mathrm{n}, l$ and $\mathrm{m}$ . Hence option D is incorrect.

Q120

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose? [RH = 1

\times

105 cm–1, h = 6.6

\times

10–34 Js, c = 3

\times

108 ms–1]

A Balmer,

\infty

\to

2

B Paschen, 5

\to

3

C Paschen,

\infty

\to

3

D Lyman,

\infty

\to

1

Correct Answer

Option C

Solution

Given, RH = 1 $\times$ 105 cm–1 $\Rightarrow$

{1 \over {{R_H}}}

= 10-5 cm $\Rightarrow$

{1 \over {{R_H}}}

= 10-7 cm $\times$ 100 $\Rightarrow$

{1 \over {{R_H}}}

= 100 nm We know,

{1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)

$\Rightarrow$ $\lambda$ =

{1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

[For H atom Z = 1] $\Rightarrow$

\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

$\Rightarrow$ $\lambda$ =

{{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

Given, $\lambda$ = 900 nm $\therefore$

{{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

= 900 $\Rightarrow$

{\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}

By checking each options you can see when nL = 3 and nH = $\infty$ then

{\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}

$\therefore$ Option C is correct.