Structure of Atom Questions — JEE Chemistry

Q1

<p>

\text{ Match List I with List II : }

</p> <table class=tg><thead> <tr> <th class=tg-c3ow></th> <th class=tg-c3ow colspan=2>List I<br>(Quantum Numbers)</th> <th class=tg-c3ow></th> <th class=tg-c3ow>List II (Orbital)</th> </tr></thead> <tbody> <tr> <td class=tg-baqh></td> <td class=tg-baqh>'n'</td> <td class=tg-baqh>'I'</td> <td class=tg-baqh></td> <td class=tg-baqh></td> </tr> <tr> <td class=tg-c3ow>A.</td> <td class=tg-c3ow>2</td> <td class=tg-baqh>1</td> <td class=tg-c3ow>I.</td> <td class=tg-c3ow>3d</td> </tr> <tr> <td class=tg-c3ow>B.</td> <td class=tg-c3ow>4</td> <td class=tg-baqh>0</td> <td class=tg-c3ow>II.</td> <td class=tg-c3ow>2p</td> </tr> <tr> <td class=tg-c3ow>C.</td> <td class=tg-c3ow>5</td> <td class=tg-baqh>3</td> <td class=tg-c3ow>III.</td> <td class=tg-c3ow>4s</td> </tr> <tr> <td class=tg-c3ow>D.</td> <td class=tg-c3ow>3</td> <td class=tg-baqh>2</td> <td class=tg-c3ow>IV.</td> <td class=tg-c3ow>5f</td> </tr> </tbody></table>Choose the correct answer from the options given below.

A A-IV, B-II, C-III, D-I

B A-II, B-III, C-I, D-IV

C A-II, B-III, C-IV, D-I

D A-I, B-II, C-III, D-IV

Correct Answer

Option C

Solution

Principal Quantum Number ( $n$ ): This number tells us the principal energy shell.

Q2

The number of radial and angular nodes in 4d orbital are, respectively

A 1 and 2

B 3 and 2

C 1 and 0

D 2 and 1

Correct Answer

Option A

Solution

We know, Radial nodes = n $-$ l $-$ 1 and Angular nodes = l For 4d orbital, n = 4 l = 2 $\therefore$ Radial nodes = 4 $-$ 2 $-$ 1 = 1 Angular nodes = 2

Q3

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is : [Given : The threshold frequency of platinum is 1.3

\times

1015 s

-

1 and h = 6.6

\times

10

-

34 J s.]

A 3.21

\times

10

-

14 J

B 6.24

\times

10

-

16 J

C 8.58

\times

10

-

19 J

D 9.76

\times

10

-

20 J

Correct Answer

Option C

Solution

The minimum energy possessed by photons will be equal to the work function of the metal.

$\therefore$ Emin = h $\nu$ 0 J = 6.6 $\times$ 10 $-$ 34 $\times$ 1.3 $\times$ 1015 = 8.58 $\times$ 10 $-$ 19 J

Q4

Which one of the following groupings represents a collection of isoelectronic species? (At. nos. : Cs : 55, Br : 35)

A N3-, F-, Na+

B Be, Al3+, Cl-

C Ca2+, Cs+, Br

D Na+, Ca2+, Mg2+

Correct Answer

Option A

Solution

Isoelectronic means those species whose electron number are same. .tg .tg Species Atomic Number Electron Number N3- 7 7 + 3 = 10 F- 9 9 + 1 = 10 Na+ 11 11 - 1 = 10 Be 4 4 Al3+ 13 13 - 3 = 10 Cl- 17 17 + 1 = 8 Ca2+ 20 20 - 2 = 18 Cs+ 55 55 - 1 = 54 Br 35 35 Mg2+ 12 12 -2 = 10 $\therefore$ N3-, F-, Na+

Q5

The number of orbitals associated with quantum number n = 5, ms = +

{1 \over 2}

is :

A 11

B 25

C 15

D 50

Correct Answer

Option B

Solution

Total number of orbitals = n2 = (5)2 = 25

Q6

If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been :

A All of the

\alpha

-particles pass through the gold foil without decrease in speed.

B

\alpha

-particles are deflected over a wide range of angles.

C All

\alpha

-particles get bounced back by 180

^\circ

D

\alpha

-particles pass through the gold foil deflected by small angles and with reduced speed.

Correct Answer

Option D

Solution

As in Thompson model, protons are diffused (charge is not centred) $\alpha$ -particles deviate by small angles and due to repulsion from protons, their speed decreases.

Q7

The correct decreasing order of energy for the orbitals having, following set of quantum numbers : (A) n = 3, l = 0, m = 0 (B) n = 4, l = 0, m = 0 (C) n = 3, l = 1, m = 0 (D) n = 3, l = 2, m = 1 is :

A (D) > (B) > (C) > (A)

B (B) > (D) > (C) > (A)

C (C) > (B) > (D) > (A)

D (B) > (C) > (D) > (A)

Correct Answer

Option A

Solution

(A) $\mathrm{n}+\ell=3+0=3$ (B) $\mathrm{n}+\ell=4+0=4$ (C) $\mathrm{n}+\ell=3+1=4$ (D) $\mathrm{n}+\ell=3+2=5$ Higher $\mathrm{n}+\ell$ valuc, higher the encrgy & if same $\mathrm{n}+\ell$ value, then higher $\mathrm{n}$ value, higher the energy.

Thus : $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$ .

Q8

Which of the following sets of quantum numbers represents the highest energy of an atom?

A n = 3, l = 0, m = 0, s = +1/ 2

B n = 3, l = 1, m = 1, s = +1/ 2

C n = 3, l = 2, m = 1, s = +1/ 2

D n = 4, l = 0, m = 0, s = +1/ 2

Correct Answer

Option C

Solution

In which quantum number have highest ( n + l ) value, will represent highest energy of an atom.

In option (C), n + l = 3 + 2 = 5 = maximum.

Q9

The wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097

\times

107 m-1)

A 406 nm

B 192 nm

C 91 nm

D 9.1

\times

10-8 nm

Correct Answer

Option C

Solution

We know Rydberg formula,

{1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

[ for hydrogen atom Z = 1 ] =

1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)

$\therefore$ $\lambda$ =

{1 \over {1.097 \times {{10}^7}}}

= 9.11 $\times$ 10-8 m = 91.1 $\times$ 10-9 m = 91.1 nm [ as 1 nm = 10-9 m ]

Q10

The number of subshells associated with n = 4 and m = –2 quantum numbers is

A 8

B 2

C 16

D 4

Correct Answer

Option B

Solution

For n = 4 possible value of l = 0, 1, 2, 3.

Only l = 2 and l = 3 can have m = -2 $\therefore$ 4d & 4f subshell associated with n = 4, m = –2.