Practice Start Test

Structure of Atom

NEET Chemistry · 92 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
<p> Match List I with List II :  \text{ Match List I with List II : } </p> <table class=tg><thead> <tr> <th class=tg-c3ow></th> <th class=tg-c3ow colspan=2>List I<br>(Quantum Numbers)</th> <th class=tg-c3ow></th> <th class=tg-c3ow>List II (Orbital)</th> </tr></thead> <tbody> <tr> <td class=tg-baqh></td> <td class=tg-baqh>'n'</td> <td class=tg-baqh>'I'</td> <td class=tg-baqh></td> <td class=tg-baqh></td> </tr> <tr> <td class=tg-c3ow>A.</td> <td class=tg-c3ow>2</td> <td class=tg-baqh>1</td> <td class=tg-c3ow>I.</td> <td class=tg-c3ow>3d</td> </tr> <tr> <td class=tg-c3ow>B.</td> <td class=tg-c3ow>4</td> <td class=tg-baqh>0</td> <td class=tg-c3ow>II.</td> <td class=tg-c3ow>2p</td> </tr> <tr> <td class=tg-c3ow>C.</td> <td class=tg-c3ow>5</td> <td class=tg-baqh>3</td> <td class=tg-c3ow>III.</td> <td class=tg-c3ow>4s</td> </tr> <tr> <td class=tg-c3ow>D.</td> <td class=tg-c3ow>3</td> <td class=tg-baqh>2</td> <td class=tg-c3ow>IV.</td> <td class=tg-c3ow>5f</td> </tr> </tbody></table>Choose the correct answer from the options given below.
A A-IV, B-II, C-III, D-I
B A-II, B-III, C-I, D-IV
C A-II, B-III, C-IV, D-I
D A-I, B-II, C-III, D-IV
Correct Answer
Option C
Solution

Principal Quantum Number (nn): This number tells us the principal energy shell.

Q2
A bulb is rated at 150 watt, converting 8%8 \% energy into light. If energy of one photon is 4.42×10−19 J4.42 \times 10^{-19} \mathrm{~J}, how many photons are emitted by the bulb per second?
A 2.71×10192.71 \times 10^{19}
B 4.06×10194.06 \times 10^{19}
C 27.2×101927.2 \times 10^{19}
D 1.35×10191.35 \times 10^{19}
Correct Answer
Option A
Solution

Energy = Power × Time

=150 watt×1 s= 150~\mathrm{watt} \times 1~\mathrm{s}
=150 J= 150~\mathrm{J}

Only 8% 8\% of this energy is converted into light energy.

Energy as light=150×8100=12 J\text{Energy as light} = \frac{150 \times 8}{100} = 12~\mathrm{J}

We know that total light energy is equal to the number of photons multiplied by the energy of one photon.

E=n×Energy of one photonE = n \times \text{Energy of one photon}

Energy of one photon = 4.42×10−19 J 4.42 \times 10^{-19}~\mathrm{J}

n=EEnergy of one photon=124.42×10−19n = \frac{E}{\text{Energy of one photon}} = \frac{12}{4.42 \times 10^{-19}}
n=2.715×1019n = 2.715 \times 10^{19}

Hence, the bulb emits approximately 2.7×1019 2.7 \times 10^{19} photons per second.

Q3
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n=2→n=3n=2 \rightarrow n=3 and n=4→n=4 \rightarrow n=6\mathrm{n}=6 transitions, respectively, is
A 19\dfrac{1}{9}
B 14\dfrac{1}{4}
C 136\dfrac{1}{36}
D 116\dfrac{1}{16}
Correct Answer
Option B
Solution

To find the ratio of the wavelengths of the light absorbed by a hydrogen atom during the transitions n=2→n=3 n=2 \rightarrow n=3 and n=4→n=6 n=4 \rightarrow n=6 , we start by calculating the change in energy (ΔE\Delta E) for each transition.

The energy of an electron in a hydrogen atom at a given level nn is given by: En=−RHn2 E_n = \dfrac{-R_H}{n^2} where RH R_H is the Rydberg constant.

Transition n=2→n=3 n=2 \rightarrow n=3 : ΔE2→3=E3−E2=−RH32−(−RH22) \Delta E_{2 \rightarrow 3} = E_3 - E_2 = \dfrac{-R_H}{3^2} - \left(\dfrac{-R_H}{2^2}\right) =RH(14−19) = R_H\left(\dfrac{1}{4} - \dfrac{1}{9}\right) =RH×536 = R_H \times \dfrac{5}{36} The wavelength λ2→3\lambda_{2 \rightarrow 3} is then: λ2→3=hcΔE2→3=hc⋅36RH⋅5 \lambda_{2 \rightarrow 3} = \dfrac{h c}{\Delta E_{2 \rightarrow 3}} = \dfrac{h c \cdot 36}{R_H \cdot 5} Transition n=4→n=6 n=4 \rightarrow n=6 : ΔE4→6=E6−E4=−RH36+RH16 \Delta E_{4 \rightarrow 6} = E_6 - E_4 = \dfrac{-R_H}{36} + \dfrac{R_H}{16} =RH×2036×16 = \dfrac{R_H \times 20}{36 \times 16} The wavelength λ4→6\lambda_{4 \rightarrow 6} is then: λ4→6=hcΔE4→6=hc×36×16RH⋅20 \lambda_{4 \rightarrow 6} = \dfrac{h c}{\Delta E_{4 \rightarrow 6}} = \dfrac{h c \times 36 \times 16}{R_H \cdot 20} Calculating the ratio: The ratio of the wavelengths λ2→3λ4→6\dfrac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} is: λ2→3λ4→6=hc⋅36RH⋅5hc×36×16RH⋅20 \dfrac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} = \dfrac{\dfrac{h c \cdot 36}{R_H \cdot 5}}{\dfrac{h c \times 36 \times 16}{R_H \cdot 20}} =14 = \dfrac{1}{4} Therefore, the ratio of the wavelengths for the given transitions is 14\dfrac{1}{4}.

Q4
Energy and radius of first Bohr orbit of He+\mathrm{He}^{+}and Li2+\mathrm{Li}^{2+} are [Given RH=2.18×10−18 J,a0=52.9pm\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}, \mathrm{a}_0=52.9 \mathrm{pm} ]
A En(Li2+)=−19.62×10−16 J;rn(Li2+)=17.6pmEn(He+)=−8.72×10−16 J;rn(He+)=26.4pm\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}
B En(Li2+)=−8.72×10−16 J;rn(Li2+)=17.6pmEn(He+)=−19.62×10−16 J;rn(He+)=17.6pm\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}
C En(Li2+)=−19.62×10−18 J;rn(Li2+)=17.6pmEn(He+)=−8.72×10−18 J;rn(He+)=26.4pm\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}
D En(Li2+)=−8.72×10−18 J;rn(Li2+)=26.4pmEn(He+)=−19.62×10−18 J;rn(He+)=17.6pm\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=26.4 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}
Correct Answer
Option C
Solution

The energy and radius of the first Bohr orbit for a hydrogen-like atom is given by: Energy: En=−2.18×10−18×Z2n2 J E_n = \dfrac{-2.18 \times 10^{-18} \times Z^2}{n^2} \, \text{J} Radius: rn=52.9×n2Z pm r_n = \dfrac{52.9 \times n^2}{Z} \, \text{pm} Where: Z Z is the atomic number. n n is the orbit number (or principal quantum number).

Let's calculate these values for the ions He+\text{He}^+ and Li2+\text{Li}^{2+}: For He+\text{He}^+: The atomic number, Z=2 Z = 2 Principal quantum number, n=1 n = 1 Energy, EHe+ E_{\text{He}^+} : EHe+=−2.18×10−18×22=−8.72×10−18 J E_{\text{He}^+} = -2.18 \times 10^{-18} \times 2^2 = -8.72 \times 10^{-18} \, \text{J} Radius, rHe+ r_{\text{He}^+} : rHe+=52.9×122=26.45 pm r_{\text{He}^+} = \dfrac{52.9 \times 1^2}{2} = 26.45 \, \text{pm} For Li2+\text{Li}^{2+}: The atomic number, Z=3 Z = 3 Principal quantum number, n=1 n = 1 Energy, ELi2+ E_{\text{Li}^{2+}} : ELi2+=−2.18×10−18×32=−19.62×10−18 J E_{\text{Li}^{2+}} = -2.18 \times 10^{-18} \times 3^2 = -19.62 \times 10^{-18} \, \text{J} Radius, rLi2+ r_{\text{Li}^{2+}} : rLi2+=52.9×123=17.63 pm r_{\text{Li}^{2+}} = \dfrac{52.9 \times 1^2}{3} = 17.63 \, \text{pm} These calculations show the energy and radius for the first Bohr orbit of He+\text{He}^+ and Li2+\text{Li}^{2+}.

Q5
The energy of an electron in the ground state (n=1)\mathrm{(n=1)} for He+\mathrm{He}^{+} ion is −x J\mathrm{-x} \mathrm{~J}, then that for an electron in (n=2)\mathrm{(n=2)} state for Be3+\mathrm{Be}^{3+} ion in J\mathrm{J} is
A −x-x
B −x9-\dfrac{x}{9}
C −4x-4 x
D −49x-\dfrac{4}{9} x
Correct Answer
Option A
Solution

The energy levels of an electron in a hydrogen-like ion (an atom or ion with only one electron) can be quantified using the formula:

En=−Z2⋅13.6 eVn2E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}

where:

EnE_n

is the energy of the electron in the nth energy level,

ZZ

is the atomic number (number of protons) of the ion,

13.6 eV13.6 \text{ eV}

is the ionization energy of hydrogen,

nn

is the principal quantum number (the energy level).

Since we need to compare this across different ions in different energy states, let's plug in some numbers: For

He+\mathrm{He}^{+}

(Helium ion):

Z=2Z = 2

(as helium has 2 protons)

n=1n = 1

for ground state

E1=−(2)2⋅13.6 eV12=−4⋅13.6 eV=−54.4 eVE_1 = -\frac{(2)^2 \cdot 13.6 \text{ eV}}{1^2} = -4 \cdot 13.6 \text{ eV} = -54.4 \text{ eV}

However, the problem gives the energy in joules, and it's given a constant

xx

. So,

x=54.4 eVx = 54.4 \text{ eV}

(converted to joules as needed). Next, for the

Be3+\mathrm{Be}^{3+}

ion:

Z=4Z = 4

(as beryllium has 4 protons)

n=2n = 2
E2=−(4)2⋅13.6 eV22=−16⋅13.6 eV4=−54.4 eVE_2 = -\frac{(4)^2 \cdot 13.6 \text{ eV}}{2^2} = -\frac{16 \cdot 13.6 \text{ eV}}{4} = -54.4 \text{ eV}

Since initially

x=54.4 eVx = 54.4 \text{ eV}

(or its equivalent in joules) for

He+ at n=1\mathrm{He}^{+} \text{ at } n=1

, and now we have the same energy for

Be3+ at n=2\mathrm{Be}^{3+} \text{ at } n=2

, the energy level in joules would also equate to

xx

, just at a different energy state and ion. So, the answer is: Option A:

−x-x

.

Q6
Incorrect set of quantum numbers from the following is :
A n=4,1=3, m1=−3,−2,−1,0,+1,+2,+3, ms=−1/2\mathrm{n}=4,1=3, \mathrm{~m}_1=-3,-2,-1,0,+1,+2,+3, \mathrm{~m}_{\mathrm{s}}=-1 / 2
B n=5,1=2, m1=−2,−1,+1,+2, ms=+1/2\mathrm{n}=5,1=2, \mathrm{~m}_1=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+1 / 2
C n=4,1=2, m1=−2,−1,0,+1,+2,ms=−1/2\mathrm{n}=4,1=2, \mathrm{~m}_1=-2,-1,0,+1,+2, \mathrm{m}_{\mathrm{s}}=-1 / 2
D n=5,1=3, m1=−3,−2,−1,0,+1,+2,+3,ms=+1/2\mathrm{n}=5,1=3, \mathrm{~m}_1=-3,-2,-1,0,+1,+2,+3, \mathrm{m_s}=+1 / 2
Correct Answer
Option B
Solution
n=5,ℓ=2, m=−2,−1,+1,+2, ms=+12\mathrm{n}=5, \ell=2, \mathrm{~m}=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}
Q7
The relation between nm,(nm=\mathrm{n}_{\mathrm{m}},\left(\mathrm{n}_{\mathrm{m}}=\right. the number of permissible values of magnetic quantum number (m)(\mathrm{m})) for a given value of azimuthal quantum number (l)(l), is
A l=2nm+1l=2 \mathrm{n}_{\mathrm{m}}+1
B nm=2l2+1\mathrm{n}_{\mathrm{m}}=2 l^{2}+1
C nm=l+2\mathrm{n}_{\mathrm{m}}=l+2
D l=nm−12l=\dfrac{\mathrm{n}_{\mathrm{m}}-1}{2}
Correct Answer
Option D
Solution

Sol. Number of permissible values of magnetic quantum number for a given value of azimuthal quantum

(l)(l)
⇒nm=2ℓ+1\Rightarrow \mathrm{n}_{\mathrm{m}}=2 \ell+1
⇒ℓ=nm−12\Rightarrow \ell=\frac{\mathrm{n}_{m}-1}{2}
Q8
When electromagnetic radiation of wavelength 300 nm falls on the surface of a metal, electrons are emitted with the kinetic energy of 1.68 ×\times 10 5 J mol −-1 . What is the minimum energy needed to remove an electron from the metal? (h = 6.626 ×\times 10 −-34 Js, c = 3 ×\times 10 8 ms −-1 , N A = 6.022 ×\times 10 23 mol −-1 )
A 2.31 ×\times 10 5 J mol −-1
B 2.31 ×\times 10 6 J mol −-1
C 3.84 ×\times 10 4 J mol −-1
D 3.84 ×\times 10 −-19 J mol −-1
Correct Answer
Option A
Solution

Energy of one photon

=hcλ= {{hc} \over \lambda }

(λ\lambda = 300 nm) For one mole photons,

E=hcλ×NAE = {{hc} \over \lambda } \times {N_A}
E=6.626×10−34×3×108×6.023×1023300×10−9E = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.023 \times {{10}^{23}}} \over {300 \times {{10}^{ - 9}}}}
E=3.99×105E = 3.99 \times {10^5}

J mol −-1 Kinetic energy

=1.68×105= 1.68 \times {10^5}

J mol −-1

W0=E−K.E.{W_0} = E - K.E.
=3.99×105−1.68×105= 3.99 \times {10^5} - 1.68 \times {10^5}
=2.31×105= 2.31 \times {10^5}

J mol −-1

Q9
Identify the incorrect statement from the following.
A All the five 5d orbitals are different in size when compared to the respective 4d orbitals.
B All the five 4d orbitals have shapes similar to the respective 3d orbitals.
C In an atom, all the five 3d orbitals are equal in energy in free state.
D The shapes of d xy , d yz and d zx orbitals are similar to each other; and d x 2 −- y 2 and d z 2 are similar to each other.
Correct Answer
Option D
Solution

∙\bullet In an atom, all the five 3d orbitals are equal in energy in free state i.e., degenerate.

∙\bullet The shape of d x 2 −- y 2 is different then shape of d z 2 ∙\bullet The size of orbital depends on principal quantum number 'n' therefore all the five 3d orbitals are different in size when compared to the respective 4d orbitals.

∙\bullet Shape of orbitals depends on azimuthal quantum number 'I' therefore shapes of 4d orbitals are similar to the respective 3d orbitals.

Q10
If radius of second Bohr orbit of the He + ion is 105.8 pm, what is the radius of third Bohr orbit of Li 2+ ion?
A 158.7 pm
B 15.87 pm
C 1.587 pm
D 158.7 Ao\mathop A\limits^o
Correct Answer
Option A
Solution
rn∝n2Z{r_n} \propto {{{n^2}} \over Z}
r3(Li2+)r2(He+)=(n3)2Z(Li2+)×Z(He+)(n2)2{{{r_3}(L{i^{2 + }})} \over {{r_2}(H{e^ + })}} = {{{{({n_3})}^2}} \over {Z(L{i^{2 + }})}} \times {{Z(H{e^ + })} \over {{{({n_2})}^2}}}
r3(Li2+)105.8=(3)23×2(2)2{{{r_3}(L{i^{2 + }})} \over {105.8}} = {{{{(3)}^2}} \over 3} \times {2 \over {{{(2)}^2}}}
=105.8×32= 105.8 \times {3 \over 2}
r3(Li2+)=158.7{r_3}(L{i^{2 + }}) = 158.7

pm

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