MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 10

The Bohr radius of the fifth electron of phosphorus (atomic number = 15) acting as dopant in silicon (relative dielectric constant = 12) is

0%
10.6 $\ { A^0 }$
0%
0.53 $\ { { A ^0} }$
0%
21.2 ${ A^0 }$
0%
None of these

Explanation

${r_n} = { \in _r}\left( {\frac{{{n^2}}}{Z}} \right)$

${a_0} = 12 \times \frac{{{{\left( 5 \right)}^2}}}{{15}} \times 0.53 = 10.6\mathop A\limits^0$

Hence,

option

$(A)$ is correct answer.

Ratio of series limit of Brackett and Pfund series of hydrogen ato spectrum is

0%
$\dfrac{1}{4}$
0%
$\dfrac{4}{9}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{16}{25}$

Explanation

$\begin{array}{c}{\text { Bracket series }} \text { n }_{1}=3, n_{2}=4,5,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{3^{2}}-\dfrac{1}{n_2^2}\right] \\\propto \dfrac{1}{9}\end{array}$

$\begin{array}{l}\text { Pfund series } \quad n_{1}=4, n_{2}=5,6,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{4^{2}}-\dfrac{1}{(n_2)^{2}}\right] \propto \dfrac{1}{16} \\\text { Ratio of series limit }=9/16\end{array}$

The radius of

$He^+$ ion is

$x\ \mathring{A}$ in its ground state. The radius of

$Li^{2+}$ ion in the ground state (

$\mathring{A}$ ) is:

0%
$\dfrac{2x}{3}$
0%
$\dfrac{3x}{2}$
0%
$\dfrac{2}{3x}$
0%
$\dfrac{3}{2x}$

Explanation

Radius of an orbit

$= 0.529\times \dfrac {n^2}{Z}$

$R \propto \dfrac 1Z$

$\dfrac{R_1}{R_2} = \dfrac{Z_2}{Z_1}$

$\dfrac {x}{R_2} = \dfrac 32$

$R_2 = \dfrac{2x}{3}$

Hence, option A is correct.

The heaviest subatomic particle of an atom is___

0%
neutron
0%
positron
0%
proton
0%
electron

Explanation

$\text{Neutron}$ is the heaviest subatomic particle with a mass of

$1.008 \ \text{amu}$ amongst neutron, positron, protons, and electrons.

Hence, option

$A$ is correct.

$M^{3+}$ (mass number

$56$ ) contains

$30.4\%$ more neutrons than electrons. Thus, atomic number of the ion is.

0%
$23$
0%
$24$
0%
$25$
0%
$26$

Which one of the following is considered as the main postulate of Bohr's model of atom

0%
Protons are present in the nucleus
0%
Electrons are revolving around the nucleus
0%
Centrifugal force produced due to the revolving electrons balances the force of attraction between electron and protons
0%
Angular momentum of electron is an integral multiple of $\dfrac{h}{2\pi}$

Which one of the following species will give a series of spectral lines similar to that of

$Mg^{2+}$ ?

0%
$Al^{3+}$
0%
$Na$
0%
$Mg^+$
0%
$F$

Identify the incorrect statement among the following:

0%
$4f$ and $5f$ orbitals are equally shielded
0%
d-Block elements show irregular and erractic chemical properties among themselves.
0%
$La$ and $Lu$ have partially filled d-orbitals and no other partially filled orbitals.
0%
The chemistry of various lanthanoids is very similar.

Explanation

$4f$ orbital is nearer to the nucleus as compared to

$5f$ orbital. Therefore, shielding of

$4f$ is more than

$5f$ orbital. Hence, option (A) is correct.

Equivalent electric current created by electron of a H-atom in its ground using Bohr's model is nearly

0%
$2.14\times {10}^{-35}A$
0%
$4.48\times {10}^{-30}A$
0%
$24\times {10}^{-30}A$
0%
$70\times {10}^{-35}A$

Explanation

$\begin{array}{l}\text { Let, radius of } 1^{\text {St }} \text { orbit Atom be "a_o" } \\\text { velocity of } e^{-} \text { in } 1^{\text {st }} \text { orbit is } v \\\text { Time period, }\left[T=\dfrac{2 \pi \cdot a_{0}}{v}]--(1)\right] \end{array}$

$\begin{array}{l}\text { Electric current, } i=\dfrac{q}{T} \\\text { From } \\\qquad i=\dfrac{q v}{2 \pi a_{0}}\end{array}$

$\begin{array}{l}\text { By keeping approximate experimental } \\\text { values in above expression, we get } \\\qquad i=4.489 \times 10^{-30} A\end{array}$

In hydrogen atom, the electron is making

$6.6\times 10^{15}rev/s$ around the nucleus in an orbit of radius

$0.528A^0$ . The equivalent magnetic dipole moment is :

0%
$1\times 10^{-15}Am^2$
0%
$1\times 10^{-10}Am^2$
0%
$1\times 10^{-23}Am^2$
0%
$1\times 10^{-27}Am^2$

According to Bohr's theory radius of

$n^{th}$ shell is given as

$\dfrac {1}{r_{n}}\propto Z^{y}$ (

$Z$ is atomic number). Here

$'y'$ is:

0%
$-1$
0%
$1$
0%
$2$
0%
$-2$

Explanation

According to Bohr's theory radius of any shell

$r_{n}$ is proportional to

$\dfrac{n^{2}}{Z}$ where n= principle quantum no and Z = atomic number.

So,

$1/r_{n}$ is proportional to

$\dfrac{Z}{n^{2}}$ .

So the correct option is B.

The number of electrons having

$n+s=2.5$ in sulphide ion is :-

0%
$3$
0%
$4$
0%
$6$
0%
$8$

Explanation

Solution:

We know that the electronic configuration of sulphide ion i.e.

$S^{2-}=1s^2,2s^2,2p^6,3s^2,3p^6$

Given that

$n+s=2.5$

Where

$n=2,s=0.5$

Means in second shell we have to calculate number of electrons which has

$n=2$ and

$s=+1/2$ , so as per electronic configuration of sulphide ion

second shell has=

$2s^2,2p^6$

$1e^-$ in

$2s^2$ has

$+1/2$ and other has

$-1/2$ spin configuration similarly in

$2p$ subshell also,

Hence on counting there are

$4$ electrons

$1-2s^2$ and

$3-2p^6$

so total

$4$ electrons are there which has

$n=2,s=+1/2$

Option B is correct

$p -$ orbital can accommodate upto:

0%
Four electrons
0%
Six electrons
0%
Two electrons
0%
Eight electons

The radius of the orbit of an electron in a Hydrogen-like atom is

$3{ a }_{ 0 }$ , where

${ a }_{ 0 }$ is the Bohr radius. Its orbital angular momentum is

$\frac { 3h }{ 2\pi }$ . It is given that h is Planck's constant and R is Rydberg constant. The possible wavelength, when the atom de-excites is :

0%
$\dfrac { 4 }{ 5R }$
0%
$\dfrac { 4 }{ 9R }$
0%
$\dfrac { 1 }{ 2R }$
0%
$\dfrac { 9 }{ 32R }$

Explanation

solution:

given ;

$4.5a_0 = a_0 \dfrac{n^2}{Z}----(i)$

$\dfrac{nh}{2\pi} = \dfrac{3h}{2\pi} ----(ii)$

so , n= 3 and z=2

so , possible wavelength are:

$\dfrac{1}{\lambda _1}= RZ^2[\dfrac{1}{1^2} - \dfrac{1}{3^2}] \Rightarrow \lambda _1= \dfrac{9}{32R}$

hence the correct option:D

Any

$nf-$ orbital can accommodate upto:-

0%
$14$ electron
0%
Six electrons
0%
Two electrons with parallel spin
0%
Two electrons with opposite spin

The velocity of the electron in 3rd orbit of hydrogen atom in

$cms^{-1}$

0%
${ 4.88\times 10 }^{ 4 }$
0%
${ 2.44\times 10 }^{ 8 }$
0%
${ 7.29\times 10 }^{ 7 }$
0%
${ 2.44\times 10 }^{ 12 }$

Explanation

For an electron,

$e = 4.803 \times 10^{-10}esu$

$m = 9.1\times 10^{-28}g$

$z = 1$

Radius of

$3rd$ orbit =

$r_1\times n^2$ =

$0.529 \times 10^{-8} \times 9$

$V_n$ =

$[ \dfrac{ze^2}{mr_n}]^{1/2}$

$V_n$ =

$7.29\times10^7cm sec^{-1}$

Hence,option C is correct.

In a hydrogen like sample electron is in 2nd excited state,the energy of the 4th state of this sample is -13.6 eV,then the incorrect statement is:

0%
Atomic number of element is 4.
0%
3 different types of spectral lines will be observed if electrons make transition upto ground state from the 2nd excited state.
0%
A 25 eV photon can set free the electron from the 2nd excited state of this sample
0%
2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atom

Which one of the following complex ions is expected to absorb visible light?

0%
$[Cu(NH_3)_4]^+$
0%
$[Cu(NH_3)_4]^{+2}$
0%
$[Zn(H_2O)_6]^{+2}$
0%
$[Zn(H_2O)_6]^{+3}$

A stationary

$He^+$ ion emitted a photon corresponding to the first line of the Lyman series. The photon liberates a photo electron from a stationary hydrogen atom in ground state. The kinetic energy?

0%
$10.2 eV$
0%
$13.6 eV$
0%
$20.4 eV$
0%
$27.2 eV$

The maximum number of electrons that can be contained in

0%
$\text{4f subshell}$
0%
$3^{rd}\ \text{shell}$
0%
$\text{Both (A) and (B)}$
0%
$\text{Neither (A) nor (B)}$

The electron present in penultimate orbit of coinage metal atom are:-

0%
8
0%
18
0%
2
0%
32

Consider the electronic configuration

$1s^{2}\ 2s^{2}\ 2p_{x}^{2}$

0%
It violates Aufbau principle.
0%
It violates Hund's rule
0%
It violates Pauli's exclusion principle.
0%
It is correct.

Explanation

The given electronic configuration violates the $Hunds \ Rule$ according to which every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.
So Here $2p_x^2$ has $2$ electrons it must have $1$ electron and another electron should be in $2p_y$ orbital.

Option B is correct

If the velocity of an electron moving in first Bohr's orbit of hydrogen has to be reduced to

$7.27\times 10^{5}ms^{-1}$ it should be shifted from its existing position by

0%
$4.23\mathring{A}$ away from nucleus
0%
$4.23\mathring{A}$ towards nucleus
0%
$4.76\mathring{A}$ away from nucleus
0%
$4.76\mathring{A}$ towards nucleus

In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr's orbit in atom of hydrogen

0%
$4\rightarrow 1$
0%
$2\rightarrow 5$
0%
$3\rightarrow 2$
0%
$5\rightarrow 2$

Explanation

We see red end means it is a part of the visible region and obviously only Balmer series corresponds to the visible region for the Balmer series

$n_1=2$ and red end means low energy or third end from this means

$n_2=-5$

$2(r)5. 3$

$2(r)4. 2$

$2(r)3. 1$

The correct answer is (D)

$5 \to 2$

The

$71^{st}$ electron of an element

$X$ with an atomic number of

$71$ enters into the orbital:

0%
$4f$
0%
$6p$
0%
$6s$
0%
$5d$

Explanation

Electronic Configuration of Element X with atomic number 71 is

$[Xe] 4{f}^{14} 6{s}^{2} 5{d}^{1}$ .

The Last electron will enter in 5d orbital.

Hence, the correct option is D.

The electron in hydrogen atom is in the first Bohr's orbit (n=1). The ratio of electronic transition energies,

$\Delta E(n=1\rightarrow n=3)\ to \ \Delta E(n=1\rightarrow n=2)\ is$ :

0%
32/27
0%
16/27
0%
32/9
0%
8/9

Electronic transition in

${\text{H}}{{\text{e}}^{\text{ + }}}$ would emit light in which region when

${{\text{e}}^{\text{ - }}}$ jumps from fourth orbit to second orbit :

0%
UV
0%
Visible
0%
IR
0%
No spectral region cab be defined

The first three radius ratio of Bohr orbits

0%
$1:0.5:0.5$
0%
$1:2:3$
0%
$1:4:9$
0%
$1:8:27$

The ratio of the radii of the first three Bohr orbit in H atom is

0%
$1:\dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 }$
0%
$1:2:3$
0%
$1:4:9$
0%
$1:8:27$

Explanation

Atomic number

$Z$ is equal to

$1$ .

Hence the radius of

$n th$ orbit,

$r_n=0.529 n^2 A$

For first three orbits,

$n$ values are

$1,2$ and

$3$

ratio of radii of first three orbit

${r_1}:{r_2}:{r_3} = n_1^2:n_2^2:n_3^2$

$\begin{array}{l} \Rightarrow { 1^{ 2 } }:{ 2^{ 2 } }:{ 3^{ 2 } } \\ \Rightarrow 1:4:9 \end{array}$

How many carbon atoms are present in palmitic acid?

0%
$16$
0%
$20$
0%
$18$
0%
$12$

Imagine an atom made up of proton and a a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength

$\lambda$ (give in term of the Rydberg constant R for the hydrogen atom) equal to

0%
9/5 R
0%
36/5 R
0%
18/5 R
0%
4/ R

Explanation

Energy is related to mass

${R_n} \propto m$

The longest wavelength λmaxλmax photon will correspond to the transition of particle from

$n=3$ to

$n=2$

$\begin{array}{l} \dfrac { 1 }{ { { \lambda _{ \max } } } } =2R\left( { \dfrac { 1 }{ { { 2^{ 2 } } } } -\dfrac { 1 }{ { { 3^{ 2 } } } } } \right) \\ { \lambda _{ \max } }=\dfrac { { 18 } }{ 5 } R \end{array}$

Hence, Option

$C$ is correct.

The effective atomic number of Mn,in

$Mn_{2}(CO)_{10}$ is _____.

0%
36
0%
38
0%
32
0%
34

If the Thomson model is correct what should be the observation in the

$\alpha - ray$ scattering experiment

0%
All the $\alpha - ray$ should pass through the gold foil
0%
Only few $\alpha - ray$ should pass through the gold foil
0%
98% of $\alpha - ray$ should get reflected back
0%
Both B & C

Explanation

According to Thomson model, positive charge is uniformly distributed with electrons embedded in it like seeds of watermelon.

$\alpha$ ray is much heavier than proton as

$\alpha=^{4}_{2}He^{2+}$ . Alpha particle has mass four times that of a proton. So since the positive charge is not concentrated in one place. So alpha ray will pass through the gold foil as there is no concentrated positive charge to stop it.

Hence, option

$A$ would be the answer, if the Thomson model is correct.

In a Bohr's model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted?

0%
$2.15\times10^{-11}$ ergs
0%
$2.389\times10^{-12}$ ergs
0%
$0.239\times10^{-10}$ ergs
0%
$0.1936\times10^{-10}$ ergs

The distance between

$3^{rd}$ &

$2^{nd}$ Bohr orbit

$He^{+}$ is

0%
$2.645\ A^o$
0%
$1.322\ A^o$
0%
$0.2645\ A^{o}$
0%
$None\ of\ these$

Explanation

Dis\tance between the

${ 2^{ nd } }$ and the

${ 3^{ rd } }$ orbits=distance of

${ 3^{ rd } }$ orbit-distance of second orbit from the nucleus

${ d_{ 3 } } = 0.529 \times \dfrac { n^{ 2 }}{ z } \times { 10^{ -10 } } m=0.529 \times 9 \times { 10^{ -10 } } m$

${ d_{ 2 } }=0.529\times 4\times { 10^{ -10 } }\, m \\ { d_{ 3 } }-{ d_{ 2 } }=0.529\times 5\times { 10^{ -10 } }\, m =2.645\times { 10^{ -10 } }\, m=2.645A^{ 0 }$

Hence,

option

$(A)$ is correct answer.

According to Bohr's model, if the kinetic energy of an electron in

$2^{nd}$ orbit of

$He^{+}$ is

$x$ , then what should be the ionisaion energy of the electron revolving in

$3^{rd}$ orbit of

$m^{5+} ions>$

0%
$X$
0%
$4X$
0%
$x/4$
0%
$2X$

Explanation

$\begin{array}{l} X=13\cdot 6\frac { { { { \left( 2 \right) }^{ 2 } } } }{ { { { \left( 2 \right) }^{ 2 } } } } \\ =13\cdot 6 \\ X'=13\cdot \frac { { { { \left( 6 \right) }^{ 2 } } } }{ { { { \left( 3 \right) }^{ 2 } } } } \\ =13\cdot 6\times \frac { { 36 } }{ 9 } \\ =4X \\ Hence,\, option\, B\, is\, correct\, answer. \end{array}$

The atomic number of an element represents :

0%
number of neutrons in the nucleus
0%
number of protons in the nucleus
0%
atomic mass of element
0%
valency of element

Neutrons is a fundamental particle carrying

0%
A charge of + 1 unit and a mass of 1 unit
0%
no charge and a mass of 1 unit
0%
no charge and no mass
0%
a charge of - 1 and a mass of 1 unit

If an electron moves away from nucleus, its potential energy:

0%
decreases
0%
not changes
0%
increases
0%
none of these

$E_n=-\dfrac{313.6}{n^2}$ , if the value of

$E_i=-34.84$ to which value '

$n$ ' corresponds:

0%
$2$
0%
$4$
0%
$1$
0%
$3$

Explanation

Solution:- (D)

$3$

${E}_{n} = - \cfrac{313.6}{{n}^{2}}$

For

${E}_{i} = -34.84; \quad n = ?$

$-34.84 = - \cfrac{313.6}{{n}^{2}}$

$\Rightarrow n = \sqrt{\cfrac{313.6}{34.84}} = 3$

Hence the value of

$n$ is

$3$ .

$\dfrac { weight\quad of\quad one\quad atom\quad of\quad an\quad element }{ Its\quad atomic\quad weight } =x.$

$\dfrac { weight\quad of\quad one\quad molecule\quad of\quad a\quad compound }{ Its\quad molecular\quad weight } =y.$
Then, x : y is

0%
$1:\frac { 1 }{ 12 }$
0%
$2:1$
0%
$1:2$
0%
$1:1$

Which of the following pairs is not correctly matched

0%
Effective ato,ic number of Pt in ${ \left[ { PtCl }_{ 6 } \right] }^{ 2- }$
0%
Absorption peak for ${ \left[ { Cr }^{ m }{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] }^{ +3 }=21680{ cm }^{ -1 }$
0%
Crystal field stabilization energy of ${ d }^{ 2 }$ in weak ligand field $=\left( - \right) 0.8{ \Delta }_{ 0 }$
0%
Example of weak ligand field for ${ d }^{ 5 }$ configuration ${ d }^{ 5 }$

The energy of electron in the excited state of hydrogen atom is

$-0.85 eV$ . If 'h' is Plank's constant, the angular momentum of electron in the excited state is :

0%
$\dfrac{4h}{2 \pi}$
0%
$h$
0%
$\dfrac{3 h}{2 \pi}$
0%
$\dfrac{h}{\pi}$

Explanation

Energy of the

$nth$ state of H-atom

$E_n = \dfrac{-13.6}{n^2} \ eV$

So,

$-0.85 = -\dfrac{13.6}{n^2}$

$n^2 = \dfrac{13.6}{0.85} = 16$

$\implies \ n = 4$

Angular momentum

$L = \dfrac{nh}{2\pi} = \dfrac{4h}{2\pi}$

How many spectral line of balmer series present in visible region?

0%
$5$
0%
$4$
0%
$2$
0%
$3$

Explanation

Solution:- (B)

$4$

Four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths between

$400 \; nm - 700 \; nm$ .

The average atomic mass of a sample of an element X is 16.2u. What are the percentage of isotope

$^{16}_8X$ and

$^{18}_8X$ in the sample respectively?

0%
$90$ %, $10$ %
0%
$80$ %, $20$ %
0%
$10$ %, $90$ %
0%
$20$ %, $80$ %

General electronic configuration of outermost and penultimate shell is

$(n-1)s^{2}(n-1)p^{6}(n-1)d^{x}ns^{2}$ If n = 4 and x = 5, then number of protons in the nucleus will be :-

0%
> 25
0%
< 24
0%
25
0%
30

The process of successive addition of protons to the nucleus followed by an addition of the same number of electrons to the available orbitals in the sequence of increasing energy to obtain the electronic configuration of many electron atom is known

0%
Pauli's exclusion principle
0%
Hund's rule
0%
Heisenberg's uncertainty principle
0%
Aufbau principle

Which of the following has the correct combination considering List-I and List-II

0%
$(III),(S)$
0%
$(IV),(Q)$
0%
$(III),(P)$
0%
$(IV),(U)$

Explanation

The kinetic energy of the electron in the

$n^{th}$ orbit is given as:

$K.E.=\dfrac{me^4}{8{{\epsilon}_0}^2n^2h^2}$

$\therefore \ K.E.\propto n^{-2}$

Similarly,

radius of

$n^{th}$ orbit

$\quad \propto \dfrac{n^2}{Z}$

Angular momentum

$|L|\propto n$

and Potential energy

$P.E. \propto \dfrac {-Z^2}{n^2}$

Hence, the correct option is

$A$

What are isotopes of an element?

0%
Atoms of a different element with a different number of neutrons
0%
Atoms of a different element with a different number of protons
0%
Atoms of the same element with a different number of neutrons
0%
Atoms of the same element with a different number of protons

Twenty molecular of

$SO_3$ will weight as much as _____ molecular of oxygen.

0%
$100$
0%
$50$
0%
$15$
0%
$8$

Explanation

The weight of one

$SO_3$ molecule

$= 32+( 3\times16 )$

$= 32+48=80u$

Weight of twenty

$SO_3$ molecule

$= 20\times80=1600u$

Weight of each molecule of oxygen

$=2\times16=32u$

Therefore, the no. of oxygen molecules

$= \dfrac{1600}{32}$

$=50$

So, the correct option is

$B$

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.