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Explore popular questions from Structure of Atom for NEET. This collection covers Structure of Atom previous year NEET questions hand picked by experienced teachers.

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Structure of Atom

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Q 1. If {tex} n = 6 , {/tex} the correct sequence for filling of electrons will be

{tex} n s \rightarrow ( n - 2 ) f \rightarrow ( n - 1 ) d \rightarrow n p {/tex}

B

{tex} n s \rightarrow ( n - 1 ) d \rightarrow ( n - 2 ) f \rightarrow n p {/tex}

C

{tex} n s \rightarrow ( n - 2 ) f \rightarrow n p \rightarrow ( n - 1 ) d {/tex}

D

{tex} n s \rightarrow n p ( n - 1 ) d \rightarrow ( n - 2 ) f {/tex}

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Q 2. According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

A

{tex} n = 6 {/tex} to {tex} n = 1 {/tex}

B

{tex} n = 5 {/tex} to {tex} n = 4 {/tex}

{tex} n = 6 {/tex} to {tex} n = 5 {/tex}

D

{tex} n = 5 {/tex} to {tex} n = 3 {/tex}

Explanation

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Q 3. A {tex}\mathrm {0.66 kg}{/tex} ball is moving with a speed of {tex}100\ \mathrm { m } / \mathrm { s } {/tex} The associated wavelength will be {tex} \left( h = 6.6 \times 10 ^ { - 34 } \mathrm { J\ s } \right) {/tex}

A

{tex} 6.6 \times 10 ^ { - 32 } \mathrm { m } {/tex}

B

{tex} 6.6 \times 10 ^ { - 34 } \mathrm { m } {/tex}

{tex} 1.0 \times 10 ^ { - 35 } \mathrm { m } {/tex}

D

{tex} 1.0 \times 10 ^ { - 32 } \mathrm { m } {/tex}

Explanation

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Q 4. Maximum number of electrons in a subshell of an atom is determined by the following

A

{tex} 2l + 1 {/tex}

B

{tex} 4l - 2 {/tex}

C

{tex}2 n ^ { 2 } {/tex}

{tex} 4l + 2 {/tex}

Explanation

Total number of subshells ={tex} (2l+1) {/tex} ∴ Maximum number of electrons in a subshell = {tex} 2(2l+1) = 4l+2 {/tex}

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Q 5. Which of the following is not permissible arrangement of electrons in an atom?

A

{tex} n = 5 , l = 3 , m = 0 , s = + 1 / 2 {/tex}

{tex} n = 3 , l = 2 , m = - 3 , s = - 1 / 2 {/tex}

C

{tex} n = 3 , l = 2 , m = - 2 , s = - 1 / 2 {/tex}

D

{tex} n = 4 , l = 0 , m = 0 , s = - 1 / 2 {/tex}

Explanation

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Q 6. Which one is the wrong statement?

A

The uncertainty principle is {tex} \Delta E \times \Delta t \geq \frac { h } { 4 \pi } {/tex}

B

Half filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement.

The energy of 2{tex} s {/tex}-orbital is less than the energy of 2{tex} p {/tex}-orbital in case of hydrogen like atoms.

D

de-Broglie's wavelength is given by {tex} \lambda = \frac { h } { m v } , {/tex} where {tex} m = {/tex} mass of the particle, {tex} v = {/tex} group velocity of the particle.

Explanation

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Q 7. How many electrons can fit in the orbital for which {tex} n = 3 {/tex} and {tex} l = 1 ? {/tex}

A

2

6

C

10

D

14

Explanation

Total 6 electrons can fit for n=3 and l=1 If the value of third shell is 3 or n=3 with sub-shell value l=1. This means each orbital will contain two electrons. There exist three p orbitals. Each electron will combine with three orbitals; finally, you will find six electrons will fit into the orbit.

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Q 8. Which of the following pairs of {tex} d {/tex}-orbitals will have electron density along the axes?

A

{tex} d _ { z ^ { 2 } , } d _ { x z } {/tex}

B

{tex} d _ { x z } , d _ { y z } {/tex}

{tex} d _ { z ^ { 2 } , } d _ { x ^ { 2 } - y ^ { 2 }} {/tex}

D

{tex} d _ { x y } , d _ { x ^ { 2 } - y ^ { 2 } } {/tex}

Explanation

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Q 9. Two electrons occupying the same orbital are distinguished by

A

azimuthal quantum number

spin quantum number

C

principal quantum number

D

magnetic quantum number.

Explanation

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Q 10. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?

A

{tex}4s\ 3s \ 3p \ 3d {/tex}

B

{tex} 3s \ 3p \ 3d \ 4s {/tex}

{tex}3s \ 3p \ 4s \ 3d {/tex}

D

{tex} 3s \ 4s \ 3p \ 3d {/tex}

Explanation

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Q 11. The number of {tex} d {/tex}-electrons in {tex} \mathrm { Fe } ^ { 2 + } ( Z = 26 ) {/tex} is not equal to the number of electrons in which one of the following?

A

{tex} d {/tex}-electrons in {tex} \mathrm { Fe } ( Z = 26 ) {/tex}

B

{tex} p {/tex}-electrons in {tex} \mathrm{Ne}( Z = 10 ) {/tex}

C

{tex}s{/tex}-electrons in {tex} \mathrm { Mg } ( Z = 12 ) {/tex}

{tex} p {/tex}-electrons in {tex}\mathrm{Cl} ( Z = 17 ) {/tex}

Explanation

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Q 12. The angular momentum of electron in {tex}'d'{/tex} orbital is equal to

A

2{tex} \sqrt { 3 } \ h {/tex}

B

6 {tex}h{/tex}

{tex} \sqrt { 6 }\ h {/tex}

D

{tex} \sqrt { 2 }\ h{/tex}

Explanation

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Q 13. What is the maximum number of orbitals that can be identified with the following quantum numbers?{tex}\mathrm{ [ n = 3 , l = 1 , m _ { l } = 0 ]}{/tex}

1

B

2

C

3

D

4

Explanation

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Q 14. Calculate the energy in joule corresponding to light of wavelength {tex}45 \mathrm { nm } {/tex} . (Planck's constant, {tex} h = 6.63 \times 10 ^ { - 34 } \mathrm { J } \mathrm s {/tex} , speed of light, {tex} \left. c = 3 \times 10 ^ { 8 } \mathrm { m } \ \mathrm { s } ^ { - 1 } \right) {/tex}

A

{tex} 6.67 \times 10 ^ { 15 } {/tex}

B

{tex} 6.67 \times 10 ^ { 11 } {/tex}

C

{tex} 4.42 \times 10 ^ { - 15 } {/tex}

{tex} 4.42 \times 10 ^ { - 18 } {/tex}

Explanation

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Q 15. {tex} \mathrm { Be } ^ { 2 + } {/tex} is isoelectronic with which of the following ions?

A

{tex} \mathrm { H } ^ { + } {/tex}

{tex} \mathrm {Li}^+ {/tex}

C

{tex} \mathrm { Na } ^ { + } {/tex}

D

{tex} \mathrm { Mg } ^ { 2 + } {/tex}

Explanation

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Q 16. Based on equation {tex} E = - 2.178 \times 10 ^ { - 18 } \mathrm { J } \left( \frac { Z ^ { 2 } } { n ^ { 2 } } \right) {/tex}, certain conclusions are written. Which of them is not correct?

A

Equation can be used to calculate the change in energy when the electron changes orbit.

For {tex} n = 1 , {/tex} the electron has a more negative energy than it does for {tex} n = 6 {/tex} which means that the electron is more loosely bound in the smallest allowed orbit.

C

The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.

D

Larger the value of {tex} n , {/tex} the larger is the orbit radius.

Explanation

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Q 17. The value of Planck's constant is {tex} 6.63 \times 10 ^ { - 34 } \mathrm { J\ s } {/tex} . The speed of light is {tex} 3 \times 10 ^ { 17 } \mathrm { nm }\ \mathrm { s } ^ { - 1 } {/tex} . Which value is closest to the wavelength in nanometer of a quantum of light with frequency of {tex} 6 \times 10 ^ { 15 } \ \mathrm { s } ^ { - 1 } ? {/tex}

50

B

75

C

10

D

25

Explanation

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Q 18. The outer electronic configuration of {tex} \mathrm { Gd }{/tex} (At. No. 64 ) is

A

4{tex} f ^ { 5 } 5 d ^ { 4 } 6 s ^ { 1 } {/tex}

4{tex} f ^ { 7 } 5 d ^ { 1 } 6 s ^ { 2 } {/tex}

C

4{tex} f ^ { 3 } 5 d ^ { 5 } 6 s ^ { 2 } {/tex}

D

4{tex} f ^ { 4 } 5 d ^ { 5 } 6 s ^ { 1 } {/tex}

Explanation

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Q 19. According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as {tex} \left( h = 6.62 \times 10 ^ { - 27 } \text { ergs, } c = 3 \times 10 ^ { 10 } \mathrm { cm }\ \mathrm { s } ^ { - 1 } \text { , } \right. {/tex} {tex} \left. N _ { A } = 6.02 \times 10 ^ { - 23 } \mathrm { mol } ^ { - 1 } \right) {/tex}

{tex} \frac { 1.196 \times 10 ^ { 8 } } { \lambda } {/tex}

B

{tex} \frac { 2.859 \times 10 ^ { 5 } } { 2 } {/tex}

C

{tex} \frac { 2.859 \times 10 ^ { 16 } } { \lambda } {/tex}

D

{tex} \frac { 1.196 \times 10 ^ { 16 } } { \lambda } {/tex}

Explanation


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Q 20. Maximum number of electrons in a subshell with {tex} l = 3 {/tex} and {tex} n = 4 {/tex} is

14

B

16

C

10

D

12

Explanation

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Q 21. The correct set of four quantum numbers for the valence electron of rubidium atom {tex} ( Z = 37 ) {/tex} is

A

{tex} 5,1,1 , + 1 / 2 {/tex}

B

{tex} 6,0,0 , + 1 / 2 {/tex}

{tex} 5,0,0 , + 1 / 2 {/tex}

D

{tex} 5,1,0 , + 1 / 2 {/tex}

Explanation

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Q 22. The total number of atomic orbitals in fourth energy level of an atom is

A

8

16

C

32

D

4

Explanation

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Q 23. The energies {tex} E _ { 1 } {/tex} and {tex} E _ { 2 } {/tex} of two radiations are {tex}25 eV{/tex} and {tex}50 eV{/tex} respectively. The relation between their wavelengths {tex} i . e . , \lambda _ { 1 } {/tex} and {tex} \lambda _ { 2 } {/tex} will be

A

{tex} \lambda _ { 1 } = \lambda _ { 2 } {/tex}

{tex} \lambda _ { 1 } = 2 \lambda _ { 2 } {/tex}

C

{tex} \lambda _ { 1 } = 4 \lambda _ { 2 } {/tex}

D

{tex} \lambda _ { 1 } = \frac { 1 } { 2 } \lambda _ { 2 } {/tex}

Explanation