Structure of Atom — Page 5 | JEE Chemistry

Q41

The number of radial nodes and total number of nodes in 4p orbital respectively are :

A 2 and 3

B 2 and 2

C 3 and 4

D 4 and 4

Correct Answer

Option A

Solution

For $4 p$

n=4, l=1

Total number of nodes $=n-1=4-1=3$ Total radial nodes $=n-l-1=4-1-1=2$

Q42

Consider the following statements : (A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3, ... (B) The azimuthal quantum number 'l' for a given 'n' (principal quantum number) can have values as 'l' = 0, 1, 2, ...... n (C) Magnetic orbital quantum number 'ml' for a particular 'l' (azimuthal quantum number) has (2l + 1) values. (D)

\pm

1/2 are the two possible orientations of electron spin. (E) For l = 5, there will be a total of 9 orbital Which of the above statements are correct?

A (A), (B) and (C)

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

C (A), (C) and (D)

D (A), (B), (C) and (D)

Correct Answer

Option C

Solution

(A) Principle quantum number n is a positive integer and it's possible values are n = 1, 2, 3 ........

$\therefore$ A is correct.

(B) Azimuthal quantum number 'l' for a given 'n' can have values as l = 0, 1, 2 ....... (n $-$ 1) $\therefore$ Statement B is wrong.

(C) Magnetic orbital quantum number ml for particular l has values from $-$ l to + 1 including zero means 2l + 1 values.

$\therefore$ Statement C is correct.

(D)

\pm \,\,{1 \over 2}

are the two possible orientation of electron spin.

$\therefore$ Statement D is correct.

(E) For l = 5, there will be a total of 11 orbitals.

l

= 0 $\Rightarrow$ 5 subshell

l

= 1 $\Rightarrow$ p subshell

l

= 2 $\Rightarrow$ d subshell

l

= 3 $\Rightarrow$ f subshell

l

= 4 $\Rightarrow$ g subshell

l

= 5 $\Rightarrow$ h subshell We know, Number of orbital in any subshell = 2

l

+ 1. $\therefore$ For h subshell, number of orbitals = 2 $\times$ 5 + 1 = 11

Q43

Given below are two statements. Statement I : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus. Statement II : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option C

Solution

Velocity of electron in Bohr's atom is given by

V \propto {Z \over n}

Z = atomic number of atom, corresponds to +ve charge so as Z increase velocity increases so statement I is wrong and as 'n' decreases velocity increases so statement II is correct.

Q44

The pair, in which ions are isoelectronic with AI3+ is :

A Br

-

and Be2+

B Cl

-

and Li+

C S2

-

and K+

D O2

-

and Mg2+

Correct Answer

Option D

Solution

O2–, Mg2+ and Al3+ are isoelectronic. All have 10 electrons.

Q45

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.

Q46

Given below are two statements: One is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion

\mathbf{A}

: Zero orbital overlap is an out of phase overlap. Reason

\mathbf{R}

: It results due to different orientation / direction of approach of orbitals. In the light of the above statements, choose the correct answer from the options given below

A Both A and R are true and R is the correct explanation of A

B Both A and R are true but R is NOT the correct explanation of A

C A is true but R is false

D A is false but R is true

Correct Answer

Option A

Solution

Zero overlapping is something in which there is no overlapping between two orbitals.

The first condition is that the two orbitals should not be symmetrical and the second condition is that both orbitals should be in different planes.

Q47

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

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

Q48

The number of s-electrons present in an ion with 55 protons in its unipositive state is

A 10

B 9

C 12

D 8

Correct Answer

Option A

Solution

The number of protons in an element determines its atomic number, which is the number of electrons in a neutral atom of the element.

The unipositive state of an ion indicates that it has lost one electron.

Therefore, the number of electrons in a unipositive ion is one less than the atomic number.

The element with 55 protons is cesium (Cs), which is in group 1 of the periodic table.

Group 1 elements have one valence electron in the s sublevel of their outermost energy level.

In a neutral atom of cesium, there are 55 electrons, with the electron configuration [Xe] 6s1.

When cesium loses one electron to form a unipositive ion (Cs+), the resulting ion has 54 electrons.

Out of this 54 electrons s subshell has 10 electrons in 1s2, 2s2, 3s2, 4s2, 5s2.

Q49

The wave function

(\Psi)

of

2 \mathrm{~s}

is given by

\Psi_{2 \mathrm{~s}}=\dfrac{1}{2 \sqrt{2 \pi}}\left(\dfrac{1}{a_0}\right)^{1 / 2}\left(2-\dfrac{r}{a_0}\right) e^{-r / 2 a_0}

At

r=r_0

, radial node is formed. Thus,

r_0

in terms of

a_0

A

r_0=a_0

B

r_0=4 a_0

C

r_0=2 a_0

D

r_0=\dfrac{a_0}{2}

Correct Answer

Option C

Solution

For radial node

{\psi _{2s}} = 0

$\therefore$

r = 2{a_0}

Q50

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

A Ti

^{2+}

B V

^{2+}

C Cr

^{2+}

D Mn

^{2+}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{Cr}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^4, 4 \mathrm{~s}^0 \mathrm{n}=4, \mu=\sqrt{4(4+2)}=\sqrt{24} =4.89 \mathrm{BM} \\\\ & \mathrm{Mn}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^5, 4 \mathrm{~s}^0 \mathrm{n}=5, \mu=\sqrt{5(5+2)}=\sqrt{35} =5.91 \mathrm{BM} \\\\ & \mathrm{V}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^3, 4 \mathrm{~s}^0 \mathrm{n}=3, \mu=\sqrt{3(3+2)}=\sqrt{15} =3.87 \mathrm{BM} \\\\ & \mathrm{Ti}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^2, 4 \mathrm{~s}^0 \mathrm{n}=2, \mu=\sqrt{2(2+2)}=\sqrt{8} =2.82 \mathrm{BM} \end{aligned}