Structure of Atom — Page 10 | JEE Chemistry

Q91

Given below are two statements : Statement I : Bohr's theory accounts for the stability and line spectrum of Li+ ion. Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field. 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 Both statement I and statement II are true.

C Statement I is true but statement II is false.

D Both statement I and statement II are false.

Correct Answer

Option A

Solution

Bohr’s theory is applicable for unielectronic species only Li+ has two electrons.

Bohr’s theory could not explain the splitting of spectral lines in the presence of external magnetic field (Zeeman effect) Statement I – false Statement II – true

Q92

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are :

A Paschen and Pfund

B Balmer and Brackett

C Lyman and Paschen

D Brackett and Pfund

Correct Answer

Option C

Solution

Shortest wavelength can found when n2 = $\infty$

{\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}

Here n1 = series number. $\Rightarrow$

{\lambda _{shortest}} =

{{{R{Z^2}} \over {n_1^2}}}

=

{{{R{{\left( 1 \right)}^2}} \over {n_1^2}}}

$\therefore$

{{{\lambda _A}} \over {{\lambda _B}}} = {\left( {{{{n_B}} \over {{n_A}}}} \right)^2}

Given,

{\left( {{{{n_B}} \over {{n_A}}}} \right)^2} = 9

$\Rightarrow$

{{{n_B}} \over {{n_A}}} = 3

$\therefore$ Series number of B = 3 $\times$ Series number of A So, If series number of A is Lyman( n = 1 ) then series number of B is Paschen( n = 3 ).

Q93

In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in absence of electric and magnetic fields? A.

\mathrm{n}=1, \mathrm{l}=0, \mathrm{~m}_1=0

B.

\mathrm{n}=2, \mathrm{l}=0, \mathrm{~m}_1=0

C.

\mathrm{n}=2, \mathrm{I}=1, \mathrm{~m}_1=1

D.

\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}_1=1

E.

\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}_1=0

Choose the correct answer from the options given below:

A B and C Only

B C and D Only

C A and B Only

D D and E Only

Correct Answer

Option D

Solution

.tg .tg orbital $\mathrm{A}: \mathrm{n}=1, \ell=0, \mathrm{~m}_{\ell}=0$ 1s $\mathrm{B}: \mathrm{n}=2, \ell=0, \mathrm{~m}_{\ell}=0$ 2s $\mathrm{C}: \mathrm{n}=3, \ell=1, \mathrm{~m}_{\ell}=1$ 3p $\mathrm{D}: \mathrm{n}=3, \ell=2, \mathrm{~m}_{\ell}=1$ 3d $\mathrm{E}: \mathrm{n}=3, \ell=2, \mathrm{~m}_{\ell}=0$ 3d In absence of electric and magnetic fields, all orbitals of 3d are degenerate

Q94

Ejection of the photoelectron from metal in the photoelectric experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is :

A 4 eV

B 4.5 eV

C 5 eV

D 5.5 eV

Correct Answer

Option B

Solution

$\lambda$ = 250 nm = 2500

\mathop A\limits^ \circ

E =

{{hc} \over \lambda }

=

{{12400} \over {2500}}

= 4.96 eV K. E = 0.5 eV As,

\,\,\,\,\,\,

K. E = E $-$ $\omega$ 0 $\Rightarrow$

\,\,\,

0.5 = 4.96 $-$

{\omega _0}

$\Rightarrow$

\,\,\,

{\omega _0}

= 4.46 eV

\simeq

4.5 eV

Q95

The correct set of four quantum numbers for the valence elections of rubidium atom (Z= 37) is:

A 5, 1, 1, + 1/2

B 5, 1, 0, + 1/2

C 5, 0, 0, + 1/2

D 5, 0, 1, + 1/2

Correct Answer

Option C

Solution

The electronic configuration of Rubidium (Rb = 37): 1s22s22p63s23p63d104s24p65s1 As you can see last electron or valence electron enterin 5s subshell So, the quantum numbers are n = 5,

l

= 0, m = 0, s =

\pm {1 \over 2}

Q96

Which one of the following sets of ions represents a collection of isoelectronic species? (Given : Atomic Number :

\mathrm{F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21}

)

A

\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{S}^{2-}

B

\mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}

C

\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}

D

\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}

Correct Answer

Option D

Solution

Isoelectronic species have same number of electrons.

\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}

all have 18 electrons, hence these are isoelectronic.

Q97

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.

Q98

The group having isoelectronic species is:

A O2- , F- , Na+, Mg2+

B O2- , F- , Na, Mg2+

C O- , F- , Na+, Mg2+

D O- , F- , Na, Mg+

Correct Answer

Option A

Solution

O2- , F- , Na+ and Mg2+, all have 10 electrons each.

Q99

Outermost electronic configurations of four elements A, B, C, D are given below : (A)

3 s^{2}

(B)

3 s^{2} 3 p^{1}

(C)

3 s^{2} 3 p^{3}

(D)

3 s^{2} 3 p^{4}

The correct order of first ionization enthalpy for them is :

A (A) < (B) < (C) < (D)

B (B) < (A) < (D) < (C)

C (B) < (D) < (A) < (C)

D (B) < (A) < (C) < (D)

Correct Answer

Option B

Solution

Orbitals with fully filled and half-filled electronic configuration are stable, and require more energy for ionization Elements with greater electronegativity require more energy for ionisation Hence the correct order is $C>D>A>B$

Q100

Outermost electronic configuration of a group 13 element, E, is 4s2, 4p1. The electronic configuration of an element of p-block period-five placed diagonally to element, E is :

A [Kr] 3d10 4s2 4p2

B [Ar] 3d10 4s2 4p2

C [Xe] 5d10 6s2 6p2

D [Kr] 4d10 5s2 5p2

Correct Answer

Option D

Solution

The element E is Ga and the diagonal element of 5th period is

{}_{50}Sn

having outer electronic configuration will be [Kr] 4d10 5s2 5p2.