Correct statements for an element with atomic number 9 are:A. There can be 5 electrons for which $m_s = +\frac{1}{2}$ and 4 electrons for which $m_s = -\frac{1}{2}$.B. There is only one electron in $p

$$\begin{aligned} &\text { Element with atomic number } 9 \text { is Fluorine }\\ &F(9)=1 s^2 2 s^2 2 p^5 \end{aligned}$$ (A) 5 electrons can be up-spin $\left[\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}\right]$ and 4 electrons can be down spin $\left[\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\right]$ (B) Unpaired electron can be in anyone of $p_x, p_y$ or $p_z$ orbital (C) Last electron is in 2 p subshell with $\mathrm{n}=2, \ell=1$ (D) Angular node for s-orbital $=0$ while of each p -orbital $=1$ S

Structure of Atom — Page 13 | JEE Chemistry

Q121

Which of the following statements are correct? (A) The electronic configuration of Cr is [Ar] 3d5 4s1. (B) The magnetic quantum number may have a negative value. (C) In the ground state of an atom, the orbitals are filled in order of their increasing energies. (D) The total number of nodes are given by n

-

2. Choose the most appropriate answer from the options given below :

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

B (A) and (B) only

C (A) and (C) only

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

Correct Answer

Option D

Solution

(A) Cr (24) = 1s2 2s2 2p6 3s2 3p6 3d5 4s1 = [Ar] 3d5 4s1 (B) Magnetic quantum number (m) values ranging from $-$ l to + l including zero.

$\therefore$ It can have negative value.

(C) According to Aufbau rule, electrons are filled first in these orbitals which have low energy.

$\therefore$ Statement C is correct.

(D) We know, Number of Radial nodes = n $-$ l $-$ 1 and number of Angular nodes = l $\therefore$ Total nodes = n $-$ l $-$ 1 + 1 = n $-$ 1

Q122

Which of the following statements is false ?

A Photon has momentum as well as wavelength.

B Splitting of spectral lines in electrical field is called Stark effect.

C Rydberg constant has unit of energy.

D Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases.

Correct Answer

Option D

Solution

When a black body is heated, black body emit high energy radiation, from higher wavelength to lower wavelength.

Q123

Given below are the quantum numbers for 4 electrons. A.

\mathrm{n}=3,l=2, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2

B.

\mathrm{n}=4,l=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2

C.

\mathrm{n}=4,l=2, \mathrm{~m}_{1}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2

D.

\mathrm{n}=3,l=1, \mathrm{~m}_{1}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2

The correct order of increasing energy is :

A D < B < A < C

B D < A < B < C

C B < D < A < C

D B < D < C < A

Correct Answer

Option B

Solution

The order of increasing energy of electrons in an atom can generally be determined using the principal quantum number ( $n$ ) and the azimuthal quantum number ( $l$ ).

The energy can be compared using the rule: the higher the $(n + l)$ value, the higher the energy.

If two electrons have the same $(n + l)$ value, the electron with the higher $n$ value has the higher energy.

Analyze each electron's quantum numbers: A.

n=3, \ l=2 \quad \Rightarrow \quad n+l = 3+2 = 5

B.

n=4, \ l=1 \quad \Rightarrow \quad n+l = 4+1 = 5

C.

n=4, \ l=2 \quad \Rightarrow \quad n+l = 4+2 = 6

D.

n=3, \ l=1 \quad \Rightarrow \quad n+l = 3+1 = 4

Comparing the values of $(n + l)$ and the principal quantum number $n$ : D: $n+l = 4$ , and $n=3$ .

A and B: $n+l = 5$ , but A has $n=3$ while B has $n=4$ .

Therefore, A has a slightly lower energy than B.

C: $n+l = 6$ .

Thus, the correct order of increasing energy is: $$ D Therefore, Option B is correct.

Q124

Correct statements for an element with atomic number 9 are:A. There can be 5 electrons for which

m_s = +\dfrac{1}{2}

and 4 electrons for which

m_s = -\dfrac{1}{2}

.B. There is only one electron in

p_z

orbital.C. The last electron goes to orbital with

n = 2

and

l = 1

.D. The sum of angular nodes of all the atomic orbitals is 1.Choose the correct answer from the options given below:

A A and B Only

B A, C and D Only

C A and C Only

D C and D Only

Correct Answer

Option C

Solution

\begin{aligned} &\text{ Element with atomic number } 9 \text{ is Fluorine }\\ &F(9)=1 s^2 2 s^2 2 p^5 \end{aligned}

(A) 5 electrons can be up-spin $\left[\mathrm{m}_{\mathrm{s}}=+\dfrac{1}{2}\right]$ and 4 electrons can be down spin $\left[\mathrm{m}_{\mathrm{s}}=-\dfrac{1}{2}\right]$ (B) Unpaired electron can be in anyone of $p_x, p_y$ or $p_z$ orbital (C) Last electron is in 2 p subshell with $\mathrm{n}=2, \ell=1$ (D) Angular node for s-orbital $=0$ while of each p -orbital $=1$ Sum of all angular node $=3$

Q125

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is :

A

{{5A} \over 9}

B

{{9A} \over 5}

C

{{36A} \over 5}

D

{{36A} \over 7}

Correct Answer

Option D

Solution

Note : (1) In Lyman Series, transition happens in n = 1 state from n = 2, 3, . . . . .

$\propto$ (2) In Balmer Series, transition happens in n = 2 state from n = 3, 4, . . . . .

$\propto$ (3) In Paschen Series, transition happens in n = 3 state from n = 4, 5, . . . . .

$\propto$ (4) In Bracktt Series, transition happens in n = 4 state from n = 5, 6 . . . . . .

$\propto$ (5) In Pfund Series, transition happens in n = 5 state from n = 6, 7, . . . .

$\propto$ We know,

{1 \over \lambda }

= Rz2 (

{1 \over {n_1^2}}

$-$

{1 \over {n_2^2}}

) The shortest wavelength of hydrogen atom in Lyman series is from n1 = 1 to n2 = $\propto$

\therefore\,\,\,

{1 \over A}

= 12 R (

{1 \over {{1^2}}}

$-$

{1 \over {{ \propto ^2}}}

) [for hydrogen, z = 1] $\Rightarrow$

\,\,\,

{1 \over A}

= R The longest wavelength in pascal series of He+ is from n1 = 3 to n2 = 4 For He+, z = 2.

{1 \over \lambda }

= RZ2 (

{1 \over {n_1^2}}

$-$

{1 \over {n_2^2}}

) $\Rightarrow$

\,\,\,

{1 \over \lambda }

=

{1 \over A}

(2)2 (

{1 \over {{3^2}}}

$-$

{1 \over {{4^2}}}

) $\Rightarrow$

\,\,\,

{1 \over \lambda }

=

{4 \over A}

\left( {{7 \over {16 \times 9}}} \right)

=

{7 \over {36A}}

$\Rightarrow$

\,\,\,

$\lambda$ =

{{36A} \over 7}

Q126

Which one of the following about an electron occupying the 1 s orbital in a hydrogen atom is incorrect? (Bohr's radius is represented by

\mathrm{a}_0

)

A The probability density of finding the electron is maximum at the nucleus

B The total energy of the electron is maximum when it is at a distance

a_0

from the nucleus

C The electron can be found at a distance

2 a_0

from the nucleus

D The 1 s orbital is spherically symmetrical

Correct Answer

Option B

Solution

1.

$\Psi^2=$ Probability density is maximum at nucleus.

2.

Electron can exist upto infinity from nucleus.

3.

True 4.

Energy of electron is maximum at infinite distance from nucleus.

Q127

Choose the Incorrect Statement about Dalton's Atomic Theory

A Compounds are formed when atoms of different elements combine in any ratio.

B Matter consists of indivisible atoms.

C chemical reactions involve reorganization of atoms

D All the atoms of a given element have identical properties including identical mass.

Correct Answer

Option A

Solution

According to Dalton’s theory, compounds are formed when atoms of different elements combine in fixed ratio.

Q128

Radius of the first excited state of Helium ion is given as :

\mathrm{a}_0 \rightarrow

radius of first stationary state of hydrogen atom.

A

\mathrm{r}=\dfrac{\mathrm{a}_0}{4}

B

\mathrm{r}=2 \mathrm{a}_0

C

\mathrm{r}=4 \mathrm{a}_0

D

\mathrm{r=\dfrac{a_0}{2}}

Correct Answer

Option B

Solution

r_n = \frac{n^2\, a_0}{Z}

For the helium ion, which is hydrogen-like with a nuclear charge of

Z = 2

, the first excited state corresponds to

n = 2

. Substituting these values:

r_2 = \frac{(2)^2\, a_0}{2} = \frac{4\, a_0}{2} = 2\, a_0

Thus, the radius of the first excited state of the helium ion is

2\, a_0

.

Q129

Of the following sets which one does NOT contain isoelectronic species?

A

BO_3^{3 - }

,

CO_3^{2 - }

,

NO_3^{- }

B

SO_3^{2 - }

,

CO_3^{2 - }

,

NO_3^{-}

C

CN^{- }

,

N_2

,

C_2^{2 - }

D

PO_4^{3 - }

,

SO_4^{2 - }

,

ClO_4^{ - }

Correct Answer

Option B

Solution

In

SO_3^{2 - }

no of electrons = 16 + 24 + 2 = 42 In

CO_3^{2 - }

no of electrons = 6 + 24 + 2 = 32 In

NO_3^{-}

no of electrons = 7 + 24 + 1 = 32 So they are not isoelectronic.