Structure of Atom — Page 2 | JEE Chemistry

Q11

Maximum number of electrons that can be accommodated in shell with

n=4

are:

A 50

B 32

C 72

D 16

Correct Answer

Option B

Solution

Maximum electrons accommodated in

\mathrm{(n = 4)}

is

\mathrm{2n^2=32}

electrons

Q12

The electrons identified by quantum numbers n and l : (a) n = 4,

l

= 1 (b) n = 4,

l

= 0 (c) n = 3,

l

= 2 (d) n = 3,

l

= 1 Can be placed in order of increasing energy as :

A (c) < (d) < (b) < (a)

B (d) < (b) < (c) < (a)

C (b) < (d) < (a) < (c)

D (a) < (c) < (b) < (d)

Correct Answer

Option B

Solution

(a) n = 4,

l

= 1 (p-subshell), so 4p (b) n = 4,

l

= 0 (s-subshell), so 4s (c) n = 3,

l

= 2 (d-subshell), so 3d (d) n = 3,

l

= 1 (p-subshell), so 3p Accroding to the Bohr ( n +

l

) rule, Enery order of the subshell : 3p < 4s < 3d < 4p Note: When two orbital have the same value of ( n +

l

) then the orbital which have lower value of n will be filled first.

Q13

Which of the following sets of quantum numbers is correct for an electron in 4f orbital?

A n = 4, l = 3, m = +1, s = + 1/2

B n = 4, l = 4, m = -4, s = - 1/2

C n = 4, l = 3, m = +4, s = + 1/2

D n = 3, l = 2, m = -2, s = + 1/2

Correct Answer

Option A

Solution

For 4f orbital, n = 4 and l = 3 Note : ( for s orbital l = 0, for p orbital l = 1, for d orbital l = 2, for f orbital l = 3) Value of m = + l to - l Here m = +3 to -3 Value of s =

+ {1 \over 2}

or

- {1 \over 2}

Q14

Which of the following sets of ions represents a collection of isoelectronic species?

A N3-, O2-, F-, S2-

B Li+, Na+, Mg2+, Ca2+

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

D Ba2+, Sr2+, K+, Ca2+

Correct Answer

Option C

Solution

Option

(c)

is correct. In

\,\,{K^ + }\left( {19} \right)\,\,

no. of electrons

= 19 - 1\,\,\,\,\, = 18

In

\,\,{Cl^ - }\left( {17} \right)\,\,

no. of electrons

= 17 + 1\,\,\,\,\, = 18

In

\,\,{Ca^ 2+ }\left( {20} \right)\,\,

no. of electrons

= 20 - 2\,\,\,\,\, = 18

In

\,\,{Sc^ 3+ }\left( {21} \right)\,\,

no. of electrons

= 21 - 3\,\,\,\,\, = 18

Q15

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 }}

Q16

Pick out the isoelectronic structure from the following:\n\n(i)

CH_3^+

\n\n(ii)

H_3O^+

\n\n(iii)

NH_3

\n\n(iv)

CH_3^-

A

(i)

and

(ii)

B

(iii)

and

(iv)

C

(i)

and

(iii)

D

(ii), (iii)

and

(iv)

Correct Answer

Option D

Solution

Ions No. of electrons CH3+ 8 H3O+ 10 NH3 10 CH3- 10 $\therefore$

\,\,\,

{H_3}{O^ + },N{H_3}

and

C{H_3}^ -

are isoelectronic.

Q17

Which one of the following constitutes a group of the isoelectronic species?

A

C_2^{2 - }

,

O_2^{-}

, CO, NO

B

NO^{+}

,

C_2^{2 - }

, CN-,

N_2

C CN-,

N_2

,

O_2^{2-}

,

C_2^{2 - }

D

N_2

,

O_2^{-}

,

NO^{+}

, CO

Correct Answer

Option B

Solution

Species No. of electrons C22- 12+2 = 14 O2- 16+1 = 17 CO 6+8 = 14 NO 7+8 = 15 NO+ 7+8-1 = 14 C22- 12+2 = 14 CN- 6+7+1 = 14 N2 14 CN- 6+7+1 = 14 N2 14 O22- 16+2 = 18 C2 12 N2 14 O2- 16+1 = 17 NO+ 7+8-1 = 14 CO 6+8 = 14 So, option

(B)

is correct.

Q18

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

Q19

The shortest wavelength of hydrogen atom in Lyman series is

\lambda

. The longest wavelength is Balmer series of He

^+

is

A

\dfrac{36\lambda}{5}

B

\dfrac{5}{9\lambda}

C

\dfrac{9\lambda}{5}

D

\dfrac{5\lambda}{9}

Correct Answer

Option C

Solution

For H: $\dfrac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 1^{2}\left(\dfrac{1}{1^{2}}-\dfrac{1}{\infty^{2}}\right) \quad...(1)$

\frac{1}{\lambda_{\mathrm{He}^{+}}}=\mathrm{R}_{\mathrm{H}} \times 2^{2} \times\left(\frac{1}{4}-\frac{1}{9}\right)\quad...(2)

From (1) & (2) $\dfrac{\lambda_{\mathrm{He}^{+}}}{\lambda}=\dfrac{9}{5}$

\begin{aligned} & \lambda_{\mathrm{He}^{+}}=\lambda \times \frac{9}{5} \\ & \lambda_{\mathrm{He}^{+}}=\frac{9 \lambda}{5} \end{aligned}

Q20

Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom?(A) 1s < 2p < 3d < 4s(B) 1s < 2s = 2p < 3s = 3p(C) 1s < 2s < 2p < 3s < 3p(D) 1s < 2s < 4s < 3dChoose the correct answer from the options given below :

A (A) and (B) only

B (A) and (C) only

C (B) and (D) only

D (C) and (D) only

Correct Answer

Option D

Solution

First, recall the key fact that for a hydrogen atom (and hydrogen‐like ions), all orbitals having the same principal quantum number $n$ have the same energy.

In other words, $E_{n\ell} \text{ depends only on }n\text{, and not on }\ell.$ Hence, the energy ordering for hydrogen‐like atoms follows $1s \; Analyzing Each Statement We check each statement against the true ordering for the hydrogen atom. (A)$ \;1s Because $2p$ is an $n=2$ orbital, $3d$ is $n=3$ , and $4s$ is $n=4$ , this partial ordering $E(1s) \; is consistent with the fact that$ n=1 Even though hydrogen also has 2s degenerate with 2p, and 3s and 3p degenerate with 3d, nothing in (A) contradicts the correct order $E_1 So (A) is correct for hydrogen’s energy levels as stated. (B)$ \;1s For hydrogen, indeed $2s=2p$ in energy, and also $3s=3p=3d$ .

The statement does not mention 3d, but it does correctly say $3s=3p$ .

That part is still true for hydrogen.

So for the orbitals it listed, (B) is also correct.

(C) $\;1s This is the kind of ordering that appears in many‐electron atoms (where subshell splitting occurs). But for a hydrogen atom,$ 2s $and$ 2p $have exactly the same energy, so$ 2snot correct.

Hence (C) is not correct for hydrogen.

(D) $\;1s In multi‐electron atoms (e.g. the usual “(n+$ \ell $) rule”), often$ 4s $is lower than$ 3d $. But for hydrogen, all orbitals with$ n=3 $lie below all orbitals with$ n=4 $. That is,$ 3d $is lower in energy than$ 4s $. So (D) is not correct for hydrogen. Conclusion (C) and (D) are not correct for the hydrogen atom. (A) and (B) are correct. Hence the final answer matches the choice stating: (C) and (D) only are not correct. Looking at the given options, that is:$ \boxed{\text{Option (D): (C) and (D) only.}}

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