Structure of Atom — Page 7 | JEE Chemistry

Q61

For hydrogen atom, the orbital/s with lowest energy is/are : (A)

\mathrm{4 s}

(B)

3 \mathrm{p}_x

(C)

3 \mathrm{~d}_{x^2-y^2}

(D)

3 \mathrm{~d}_{z^2}

(E)

4 \mathrm{p}_z

Choose the correct answer from the options given below :

A (A) only

B (B) only

C (B), (C) and (D) only

D (A) and (E) only

Correct Answer

Option C

Solution

In hydrogen atom the orbitals in a shell are degenerate means energy depends only on ' $n$ '

\therefore \mathrm{E}_{3 \mathrm{p}_{\mathrm{x}}}=\mathrm{E}_{3 \mathrm{~d}_{x^2-y^2}}=\mathrm{E}_{3 \mathrm{~d}_{z^2}}

Q62

Consider the ground state of chromium atom

(Z=24)

. How many electrons are with Azimuthal quantum number

l=1

and

l=2

respectively ?

A 16 and 5

B 12 and 5

C 12 and 4

D 16 and 4

Correct Answer

Option B

Solution

The ground state electron configuration of chromium (Cr, $Z = 24$ ) is:

\text{Cr: } [\text{Ar}]\,3d^5\,4s^1

Let's break down the configuration: The noble gas core $[\text{Ar}]$ represents: $1s^2$ (with $l=0$ ) $2s^2$ (with $l=0$ ) $2p^6$ (with $l=1$ ) $3s^2$ (with $l=0$ ) $3p^6$ (with $l=1$ ) Electrons with azimuthal quantum number $l = 1$ are found in $p$ orbitals: $2p^6$ : 6 electrons $3p^6$ : 6 electrons Total $p$ -electrons: $6 + 6 = 12$ .

Electrons with azimuthal quantum number $l = 2$ are found in $d$ orbitals: $3d^5$ : 5 electrons Thus, there are 12 electrons with $l = 1$ and 5 electrons with $l = 2$ .

The correct option is Option B.

Q63

The extra stability of half-filled subshell is due to : (A) Symmetrical distribution of electrons (B) Smaller coulombic repulsion energy (C) The presence of electrons with the same spin in non-degenerate orbitals (D) Larger exchange energy (E) Relatively smaller shielding of electrons by one another Indentify the correct statements :

A (B), (D) and (E) only

B

(A),(B),(D)

and (E) only

C (B), (C) and (D) only

D

(A),(B)

and (D) only

Correct Answer

Option B

Solution

The enhanced stability of a half-filled subshell is attributed to the following factors: Symmetrical Distribution of Electrons: Electrons are evenly distributed across the available orbitals, leading to a balanced and stable arrangement.

Large Exchange Energy: The energy due to the exchange of electrons with parallel spins among degenerate orbitals is maximized, contributing to increased stability.

Smaller Coulombic Repulsion: The repulsion between electrons is minimized in a half-filled configuration, reducing the overall energy of the subshell.

Smaller Shielding of Electrons: In a half-filled subshell, electrons are more exposed to the nuclear charge, as there is less shielding effect from other electrons, enhancing stability.

Q64

Which one of the following sets of ions represents the collection of isoelectronic species? (Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)

A K+, Cl-, Mg2+, Sc3+

B Na+, Ca2+, Sc3+, F-

C K+, Ca2+, Sc3+, Cl-

D Na+, Mg2+, Al3+, Cl-

Correct Answer

Option C

Solution

Isoelectronic means those species whose no of electrons are same. .tg .tg Ions Atomic No No of Electron K+ 19 19-1=18 Ca+2 20 20-2=18 Sc+3 21 21-3=18 Cl- 17 17+1=18 So, option

(C)

is correct.

Q65

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen

A 5

\to

2

B 4

\to

1

C 2

\to

5

D 3

\to

2

Correct Answer

Option A

Solution

In Bohr series of lines of hydrogen spectrum :- Red end means end of visible region which is balmer series of lines so n will be 2 Now for third lines with respect to Balmer series means n = (2 + 3) = 5th line $\therefore$ n1 = 2 and n2 = 5 For 2 $\to$ 5 electron will need energy to jump as 2 > 5 so no spectrum will release instead energy will be absorbed.

$\therefore$ 5 $\to$ 2

Q66

The orbital angular momentum for an electron revolving in an orbit is given by

\sqrt {l(l + 1)} {h \over {2\pi }}

. This momentum for an s-electron will be given by

A zero

B

{h \over {2\pi }}

C

\sqrt 2 {h \over {2\pi }}

D

+ {1 \over 2}{h \over {2\pi }}

Correct Answer

Option A

Solution

For s-electron l = 0 $\therefore$

\sqrt {l(l + 1)} {h \over {2\pi }}

=

\sqrt {0(0 + 1)} {h \over {2\pi }}

=

\sqrt 0 {h \over {2\pi }}

= 0 (zero)

Q67

The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are:

A 16, 6 and 3

B 15, 6 and 12

C 16, 5 and 2

D 15, 5 and 3

Correct Answer

Option D

Solution

P[15] = [Ne] 3s23p3 Group number = 10 + valance shell electron = 10 + 5 = 15 number of valence electrons = 5 Valency = 3

Q68

In a hydrogen atom, if energy of an electron in ground state is -13.6 eV, then that in the 2nd excited state is

A -1.51 eV

B -3.4 eV

C -6.04 eV

D -13.6 eV

Correct Answer

Option A

Solution

n = 1 means ground state n = 2 means 1st excited state n = 3 means 2nd excited state Remember: nth excited state means n = n + 1 We know energy of an atom is given by Energy (E) = -13.6 $\times$

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

= -13.6 $\times$

1 \over 9

[ $\because$ Z = atomic number = 1 for hydrogen atom] = -1.51 ev

Q69

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

A 325 nm

B 743 nm

C 518 nm

D 1035 nm

Correct Answer

Option B

Solution

Here a photon is divided into two photons of different wavelength.

According to the law of conservation of energy, Energy of photon before dividing into two parts = Energy of first photon + Energy of second photon

{{hc} \over \lambda } = {{hc} \over {{\lambda _1}}} + {{hc} \over {{\lambda _2}}}

\Rightarrow {1 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}

\Rightarrow {1 \over {355}} = {1 \over {680}} + {1 \over {{\lambda _2}}}

{\lambda _2} = 742.77

nm = 743 nm

Q70

The quantum number of four electrons are given below : n = 4, l = 2, ml =–2, ms = –1/2 n = 3, l = 2, ml = 1, ms = +1/2 n = 4, l = 1, ml = 0, ms = +1/2 n = 3, l = 1, ml = 1, ms = –1/2 The correct order of their increasing enegies will be :

A IV < II < III < I

B I < III < II < IV

C IV < III < II < I

D I < II < III < IV

Correct Answer

Option A

Solution

n = 4, l = 2, 4d orbital, n + l = 6 n = 3, l = 2, 3d orbital, n + l = 5 n = 4, l = 1, 4p orbital, n + l = 5 n = 3, l = 1, 3p orbital, n + l = 4 more is n + l value, more is energy $\therefore$ 3p < 3d < 4p < 4d