Structure of Atom — Page 6 | NEET Chemistry

Q51

Isoelectronic species are

A CO, CN

-

, NO + , C

{_2^{2 - }}

B CO

-

, CN, NO,

C

C_2^ -

D CO, CN, NO, C 2

Correct Answer

Option A

Solution

No. of electrons CO = CN - = NO + =

C_2^{2 - }

= 14, So these are isoelectronics.

Q52

For given energy, E = 3.03

\times

10

-

19 Joules corresponding wavelength is (h = 6.626

\times

10

-

34 J sec, c = 3

\times

10 8 m/sec)

A 65.6 nm

B 6.56 nm

C 3.4 nm

D 656 nm

Correct Answer

Option D

Solution

E =

{{hc} \over \lambda }

\Rightarrow \lambda

=

{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {3.03 \times {{10}^{ - 19}}}}

= 656 nm

Q53

According to Bohr's theory, the angular momentum of an electron in 5th orbit is

A 10

h/\pi

B 2.5

h/\pi

C 25

h/\pi

D 1.0

h/\pi

Correct Answer

Option B

Solution

Formula of angular momentum of an electron, mvr =

{{nh} \over {2\pi }}

here n = 5 $\therefore$ mvr =

{{5h} \over {2\pi }}

= 2.5

{{h} \over {\pi }}

Q54

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 x 103 ms−1 (Mass of proton = 1.67

\times

10-27 kg and h = 6.63

\times

10-34 Js) :

A 0.40 nm

B 2.5 nm

C 14.0 nm

D 0.32 nm

Correct Answer

Option A

Solution

Wavelength

\left( \lambda \right)

=

{h \over {mv}}

=

{{6.63 \times {{10}^{ - 34}}} \over {1.67 \times {{10}^{ - 27}} \times {{10}^3}}}

= 0.4 $\times$ 10-9 = 0.4 nm

Q55

Which of the following forms of hydrogen emits low energy

\beta

- particles?

A Tritium

_1^3

H

B Proton H+

C Protium

_1^1

H

D Deuterium

_1^2

H

Correct Answer

Option A

Solution

Tritium isotope of hydrogen is radioactive and emits low energy

\beta ^-

particles. It is because of high n/p ratio of tritium which makes nucleus unstable.

Q56

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are respectively

A 16 and 4

B 12 and 5

C 12 and 4

D 16 and 5

Correct Answer

Option B

Solution

Electronic configuration of Cr (Z = 24) = 1s2 2s2 2p6 3s2 3p6 4s1 3d5 For 2p6 and 3p6 , l = 1.

Here in 2p and 3p orbital total 6 + 6 = 12 electrons present.

For 3d5 , l = 2.

Here in 3d orbital total 5 electrons present.

Q57

The radius of the

\mathrm{2^{nd}}

orbit of

\mathrm{Li^{2+}}

is

x

. The expected radius of the

\mathrm{3^{rd}}

orbit of

\mathrm{Be^{3+}}

is

A

\dfrac{16}{27}x

B

\dfrac{4}{9}x

C

\dfrac{9}{4}x

D

\dfrac{27}{16}x

Correct Answer

Option D

Solution

$r_{\mathrm{Li}^{+}}=r_{0} \times \dfrac{2^{2}}{3}=x \Rightarrow r_{0}=\dfrac{3 x}{4}$ $\mathrm{r}_{\mathrm{Be}^{3+}}=\mathrm{r}_{0} \times \dfrac{3^{2}}{4}$ $r_{\mathrm{Be}^{3+}}=\dfrac{3 \mathrm{x}}{4} \times \dfrac{3^{2}}{4}=\dfrac{27 \mathrm{x}}{16}$

Q58

Hydrogen has three isotopes (A), (B) and (C). If the number of neutron(s) in (A), (B) and (C) respectively, are (x), (y) and (z), the sum of (x), (y) an (z) is :

A 3

B 1

C 4

D 2

Correct Answer

Option A

Solution

Hydrogen has three isotopes (A) Protium (

{}_1^1H

) has 0 neutron. (B) Deutrium (

{}_1^2H

) has 1 neutrons. (C) Tritium (

{}_1^3H

) has 2 neutrons. Total number of neutrons in three isotopes of hydrogen = 0 + 1 + 2 = 3

Q59

The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :

A Lyman series

B Balmer series

C Paschen series

D Brackett series

Correct Answer

Option B

Solution

Here electron is coming to the orbital of radius 211.6 pm.

Now we have to find which series have radius of orbital 211.6 pm.

We know, Radius, r =

0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ

Given, r = 211.6 pm = 211.6 $\times$ 10-12 m and Z = 1 for hydrogen atom $\therefore$ 211.6 $\times$ 10-12 =

0.529 \times {{{n^2}} \over 1}

$\times$ 10-10 $\Rightarrow$ n = 2 As n = 2 so the series is Balmer Series.

Q60

Given below are two statements : Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom. Statement II : Bohr's model of hydrogen atom contradicts Heisenberg' uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is false but statement II is true.

B Statement I is true but statement II is false.

C Both statement I and statement II are false.

D Both statement I and statement II are true.

Correct Answer

Option D

Solution

Rutherford's gold foil experiment only proved that electrons are held towards nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds.

Bohr's model gave exact formula for simultaneous calculation of speed & distance of electron from the nucleus, something which was deemed impossible according to Heisenberg.