MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 8

The maximum number of electrons in any shell is given by____ rule.

0%
2n
0%
$n^2$
0%
$2n^2$
0%
$4n^2$

Explanation

Maximum no of electrons in a shell

$={2n}^{2}$

this is called

${2n}^{2}$ rule

for example,second shell

$\Rightarrow n=2$

${2n}^{2}$ rule,no of electrons in

${2}^{nd}$ shell=

$2 \left( 2 \right)^2$ =8

The Ionisation potential of Hydrogen is

$2.17\times 10^{-11}erg/ atom$ . The energy of the electron in the second orbit of the hydrogen atom is?

0%
$-\dfrac {2.17\times 10^{-11}}{2}$
0%
$-\dfrac {2.17\times 10^{-11}}{2^{2}}$
0%
$-\dfrac {2.17\times 10^{-7}}{2^{2}}$
0%
$-\dfrac {2.17\times 10^{11}}{2^{2}}$

Explanation

Energy of electron hydrogen atom is given by,

$E_n=-\dfrac{2.178\times 10^{-11}}{n^2}ergatom^{-1}$

Therefore, for second orbit i.e

$n=2$

$E_2=-\dfrac{2.178\times 10^{-11}}{2^2}ergatom^{-1}$

According to the Bohr's atomic model, the relation between principal quantum number (n) and radius of orbit(r) is?

0%
$r\propto n^2$
0%
$r\propto \displaystyle\frac{1}{n^2}$
0%
$r\propto \displaystyle \frac{1}{n}$
0%
$r\propto n$

Explanation

Electron angular momentum about the nucleus is an integer multiple of

$\displaystyle\frac{h}{2\pi}$ , where h is Planck's constant.
Thus,

$I\omega =mvr=\displaystyle \frac{nh}{2\pi}$
Hence

$r\propto n$ .

What does the electronic configuration

$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }$ indicate?

0%
Ground state of fluorine
0%
Excited state of fluorine
0%
Excited state of neon
0%
Excited state of the ${ O }_{ 2 }^{ - }$ ion

Explanation

The atomic number of Neon is

$10$
G.S

$Ne\left[ 10 \right] :1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 }\quad$
E.S

$Ne\left[ 10 \right] :1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }\quad$
Hence

$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }$ electronic configuration indicates the excited state of neon

In Bohr series of lines of hydrogen spectrum, the third line from the red corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

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

Explanation

$\bf{Explanation-}$

Line Spectrum of Hydrogen like atoms are given by:

$\dfrac{1}{\lambda} = RZ^2(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2})$

Where R is called Rhydberg constant,

$R = 1.097 \times 10^7$ ,

Z is atomic number.

$n_1= 1,2 ,3….$

$n_2= n_1+1, n_1+2 ,……$

The electron has minimum energy in the first orbit and its energy increases as

$n$ increases. Here

$n$ represents the number of orbits.

Red line means it is the visible spectra and visible spectra means it is Balmer series that is

$n=2$

So, the third line after

$n=2$ will be

$n=5$ that is Pfund series.

Therefore, the electrons jump from

$n=5$ to

$n=2$ .

$\bf{Correct \ Option}-$ Option

$C$

Which of the following electronic configuration is not possible?

0%
$1{ s }^{ 2 }2{ s }^{ 2 }$
0%
$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 }$
0%
$\left[ Ar \right] 3{ d }^{ 10 },4{ s }^{ 2 },4{ p }^{ 2 }$
0%
$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 2 },3{ s }^{ 1 }$

The electrons, identified by quantum numbers

$n$ and

$l$ (i)

$n=4,l=1$ , (ii)

$n=4,l=0$ (iii)

$n=3,l=2$ (iv)

$n=3,l=1$ can be place in order of increasing energy, from the lowest to highest, as:

0%
(iv)< (ii)< (iii)< (i)
0%
(ii)< (iv)< (ii)< (iii)
0%
(i)< (iii)< (ii)< (ii)
0%
(iii)< (i)< (iv)< (ii)

Explanation

(i)

$4p$ (ii)

$4s$ (iii)

$3d$ (iv)

$3p$

According to Aufbau rule, order of increasing energy is

$3p< 4s< 3d< 4p$

When a hydrogen atom is raised from the ground state to an excited state.

0%
Both K.E. and P.E. increase
0%
Both K.E. and P.E. decrease
0%
The P.E. decreases and K.E. increases
0%
The P.E. increases and K.E. decreases

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic.

0%
for all values of E.
0%
for $E < 10.2$ eV.
0%
for $10.2$ eV $<$ E $< 13.6$ eV only.
0%
for $0 <$ E $< 3.4$ eV only.

Explanation

$\begin{array}{l}\text { Energy }=E \\\text { Energy at ground} \text { State for hydrogen like atom}=-13.6\frac{z^{2}}{n^{2}}=V \\\text { }\text {For hydrogen atom, } z=1 \\n=1\end{array}$

$E_{1}=-13.6 \times \frac{1^2}{1^{2}}=-13.6 \mathrm{eV}$

$\text{for 2nd shell}$

$\begin{aligned}E &=-13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \\&=-10.2 \mathrm{eV}\end{aligned}$

$10.2<E<13.6$

$\therefore \text{Opion C is correct}$

2005943_784769_ans_baf5f8317baa48d2b05f5b48130ecaa9.PNG

The Vividh Bharati station of all india Radio, Delhi,broadcasts on a frequency of 1,368kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to ?

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Visible region
0%
Radio wave region
0%
Infrared region
0%
Microwave region

Explanation

$F = \displaystyle \frac{c}{\lambda } \Rightarrow \lambda = \frac{c}{f} = \frac{3\times 10^{8}}{1.368\times 10^{3}} = 2.19 \times 10^{5} m$

Hence, is lies in Radio-Wave Region.

The ratio of the wavelength of photon corresponding to the

$\alpha-$ line of Lyman series in

$H-$ atom and

$\beta-$ line of Balmer series in

$He^{+}$ is:

0%
$1 : 1$
0%
$1 : 2$
0%
$1 : 4$
0%
$3 : 16$

Explanation

$\begin{array}{l} \cfrac { { { \lambda _{ H } } } }{ { { \lambda _{ He } } } } =\cfrac { { { R_{ H } }{ z^{ 2 } }\left( { \cfrac { 1 }{ { { 2^{ 2 } } } } -\cfrac { 1 }{ { { 4^{ 2 } } } } } \right) } }{ { { R_{ H } }\left( { \cfrac { 1 }{ { { 1^{ 2 } } } } -\cfrac { 1 }{ { { 2^{ 2 } } } } } \right) } } \Rightarrow \cfrac { { 4\left( { \cfrac { { 4-1 } }{ { 16 } } } \right) } }{ { \left( { \cfrac { { 4-1 } }{ 4 } } \right) } } =\cfrac { 1 }{ 1 } \\ { \lambda _{ H } }:{ \lambda _{ He } } \\ 1:1 \end{array}$

Hence, the correct option is

$\text{A}$

The energy levels for

$_{Z}A^{(+Z - 1)}$ can be given by:

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$E_{n}$ for $A^{(+Z - 1)} = Z^{2} \times E_{n}$ for $H$
0%
$E_{n}$ for $A^{(+Z - 1)} = Z \times E_{n}$ for $H$
0%
$E_{n}$ for $A^{(+Z - 1)} = \dfrac {1}{Z^{2}} \times E_{n}$ for $H$
0%
$E_{n}$ for $A^{(+Z - 1)} = \dfrac {1}{Z} \times E_{n}$ for $H$

Explanation

According to Bohr's model,

$E = 13.6 \space kcal$ for Hydrogen.

$E = 13.6\dfrac{Z^2}{n}$

Hence

$E_n$ for

$A^{(+Z - 1)} =Z^2 \times E_n$ for

$H.$

What is the atomic number of the element with symbol Uus?

0%
$117$
0%
$116$
0%
$115$
0%
$114$

Which of the following orbitals are nearest to nucleus?

0%
4f
0%
5d
0%
6s
0%
7p

An element with atomic number 117 is known as:

0%
nihonium
0%
flerovium
0%
tennessine
0%
roentgenium

Meitnerium is IUPAC official name of an element with atomic number?

0%
113
0%
118
0%
104
0%
109

The matter can not be perceived through our senses.

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True
0%
False

Which of the following statements is correct for an electron having azimuthal quantum number

$l = 2$ ?

0%
The electron may be in the lowest energy shell
0%
The electron is in a spherical orbital
0%
The electron must have spin $m_{s} = + \dfrac {1}{2}$
0%
None

Explanation

Azimuthal quantum number is also called Angular momentum quantum number. It is represented by letter

$l$ . This represents the sub-shell to which the electron belongs. It defines the shape of the orbital occupied by the electron and the angular momentum of the electron. For

$l$ =2 corresponding magnetic quantum numbers(

$m_s$ )=

$-2,-1,0,1,2$ and shape of the orbital is dumbbell shape that corresponds to the the

$d$

$orbital$ as

$E_N$

$\alpha$

$\dfrac {1}{n^2}$ . Value of n for

$l=2$ ,

$n=3$ . Therefore the energy of the subshell is lowest for

$l=2$ .

If EAN of a central metal ion

${ X }^{ +2 }$ in a complex isand atomic number of

$X$ isThe number of monodentate ligands present in complex are:

0%
3
0%
4
0%
6
0%
2

Explanation

EAN=(Atomic Number)-(Charge)+2(No. of monodentate ligands)

$34=28-(2)+2 \times$ (No. of monodentate ligands)

$8=2 \times$ (No. of monodentate ligands)

$\therefore$ No. of monodentate ligands

$=4$

What is the name and symbol of the element with atomic number 112?

0%
Ununbium, Uub
0%
Unnilbium, Unb
0%
Ununnillum, Uun
0%
Ununtrium, Uut

The electron in Bohr's model of hydrogen atom is pictured as revolving around the nucleus in order for it to :

0%
emit photons
0%
keep from being pulled into the nucleus
0%
keep from being repelled by the nucleus
0%
possess energy

If the radius of first Bohr orbit is x pm, then the radius of the third orbit would be :

0%
$(3\times x )pm$
0%
$(6\times x )pm$
0%
$(\frac{1}{2}\times x )pm$
0%
$(9\times x )pm$

Explanation

The radii of the

$n^{th}$ stationary states of Hydrogen and Hydrogen like atom is given by:

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

where

$a_0 = 52.9\ pm, n=n^{th}$ Bohr orbit, Z=atomic number .

Given: The radius of the first stationary state i.e Bohr orbit=x pm

$r_1=a_0 \frac{1^2}{Z}=x$ pm

$x=a_0 \frac{1}{Z}$

$Z= \frac{a_0}{x}$

Radius of third orbit:

$r_3=a_0 \frac{3^2}{Z}$

substitute the above relation:

Radius of third orbit:

$r_3=a_0 \frac{9x}{a_0}=9x$ pm

The atoms in a molecule do not remain stationary. They vibrate around their mean position, by stretching or bending out of plane. We gain useful knowledge about these vibrations and the energy they carry by studying:

0%
$X-$ ray spectra
0%
ultraviolet spectra
0%
infrared spectra
0%
visible spectra

The radius of a hydrogen atom in the ground state is

$0.53 \mathring {A}$ . What will be the radius of

$Li^{2+}$ in the ground state?

0%
$1.06\mathring{A}$
0%
$0.265\mathring{A}$
0%
$0.17\mathring{A}$
0%
$0.53\mathring{A}$

Explanation

The radii of the

$n^{th}$ stationary state for a hydrogen-like specie is expressed as :

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

where

$Z=$ atomic number and

$a_0=$ radius of Bohr orbit.

For ground state of hydrogen atom,

$n=1$ and

$Z=1$

thus for Hydrogen

$r_1(H)=a_0=0.53 \mathring A$ (given)

Thus the radius of the first stationary state of hydrogen atom i.e. in ground state, called the Bohr orbit, is

$0.53 \mathring A$ .

for

$_3Li^{2+}$ , Z=3 and ground state means

$n=1$

thus

$r_1(Li^{2+})=\frac{1^2 \times 0.53}{3}$

$r_1(Li^{2+})=0.176 \mathring A$

The spectrum of a molecule is so detailed that a molecule can be identified by its spectrum.

0%
Only if it is small
0%
Only by test-tube analysis
0%
With great difficulty
0%
Easily, like a fingerprint

If an electron jumps from orbit

$A$ to orbit

$B$ it loses energy while when it jumps from

$C$ to

$B$ it gains energy. Arrange the orbits in the increasing order of distance from the nucleus.

0%
$A < B < C$
0%
$C < A < B$
0%
$C < B < A$
0%
$A < C < B$

Explanation

When an electrons goes from lower debit to higher orbit it gains energy while when it comes to lower orbit from higher orbit it releases energy.So,the correct order will be.

$C<B<A$

Hence, option

$C$ is correct answer.

Which of the sequences given below shows the correct increasing order of energy?

0%
$3s, 3p, 4s, 3d, 5s, 5p, 4d$
0%
$3s, 3p, 4s, 4p, 4d, 5s, 5p$
0%
$3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p$
0%
$3s, 3p, 4s, 5s, 3d, 4d, 5p$

Explanation

The energy of an orbital depends on the (n+l) value. Higher is the (n+l) higher is the energy of the orbital.

Value of n majorly decides the energy of an orbital. For two orbitals with the same value for (n+l), the orbital with a higher value of n will have higher energy.

Thus increasing order of energy for orbitals is:

$1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p$

For the given set of orbitals the order is:

$3s<3p<4s<3d<4p<5s<4d<5p$

The radius of the stationary state which is also called Bohr radius is given by the expression

$r_n= n^2a_0$ where the value of

$a_0$ is:

0%
52.9 pm
0%
5.29 pm
0%
529 pm
0%
0.529 pm

Explanation

The radii of the

$n^{th}$ stationary states of Hydrogen and Hydrogen like atom is given by:

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

where

$a_0 = 52.9\ pm, n=n^{th}$ Bohr orbit,

$Z$ =atomic number .

The radius of the first stationary state of Hydrogen atom

$(n=1)$ , is called as Bohr radius that has a value of

$52.9\ pm$ . As n increases the value of r will increase. In other words, the electron will be present away from the nucleus.

Consider a

$p_{y}$ -orbital of an atom and identify correct statement.

0%
$s-$ orbital of another atom produces $\pi$ -bond when y is the bond formation axis.
0%
$p_{y}$ -orbital of another atom produces $\Sigma$ -bond when x is the bond formation axis.
0%
$p_{z}$ -orbital of another atom produces $\Sigma$ -bond when x is the bond formation axis.
0%
$d_{xy}$ -orbital of another atom produces $\Sigma$ -bond when x is the bond formation axis.

According to Bohr's theory, the angular momentum for an electron of

${5}^{th}$ orbit is:

0%
$2.5{h}/{\pi}$
0%
$5{h}/{\pi}$
0%
$25{h}/{\pi}$
0%
$6{h}/{2\pi}$

Explanation

$\bf{Explanation-}$

According to Bohr's theory the angular momentum of an electron is given as;

$mvr = \dfrac{nh}{2\pi}$

For

$5th$ orbit,

$n = 5$

Putting the value of

$n$ in the above equation we get;

Angular momentum

$= \dfrac{5h}{2\pi} = \dfrac{2.5h}{\pi}$

$\bf{Conclusion-}$ Hence, option A is correct.

An electron with energy

$12.09$ eV strikes hydrogen atom in ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atom is excited to __________ state.

0%
Fourth
0%
Third
0%
Second
0%
First

Explanation

We know that energy of Hydrogen atom in it's orbit is given by formula

$\dfrac{-13.6}{n^2} eV$ .

Initial Energy of Electron =

$-13.6 eV$ (Given that atom in i's ground state i.e n=1)

Energy after striking =

$(-13.6 + 12.09) eV = -1.51 eV$

$\Rightarrow$

$\dfrac{-13.6}{n^2} eV = 1.51eV$

$\Rightarrow$

$n^2= 9$

$\Rightarrow$

$n=3.$

So, hydrogen atom is excited to third state.

Therefore, B is correct option.

The number of orbital having values of n = 4 and l = 3 :

0%
3
0%
7
0%
5
0%
9

Explanation

The number of orbital having values of

$n=4$ and

$l=3$

$\Rightarrow 4f$

f subshell

7 orbitals

$\begin{bmatrix} f-value & 0 & 1 & 2 & 3 \\ orbital & s & p & d & f \end{bmatrix}$

Which of the following contains maximum number of atoms?

0%
$0.16\ g\ CH_3$
0%
$0.28\ g \ N_2$
0%
$0.36\ g\ C_6H_{12}O_6$
0%
$1.1\ g \ CO_2$

Explanation

No. of atoms in

$CH_3$ =

$N_a \times mole$

$N_a \times \dfrac{0.16}{15}$

No. of atoms in

$N_2$ =

$N_a \times mole$

$N_a \times \dfrac{0.28}{28}$

No. of atoms in

$C_6H_{12}O_6 = N_a \times \dfrac{0.36}{180}$

No. of atoms in

$CO_2 = N_a\times \dfrac{1.1}{44}$

Hence

$1.1$ g of

$CO_2$ has the maximum no. of atoms.

Thus D is the correct option.

The correct order of increasing energy of atomic orbital is:

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

One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in:

0%
visible region
0%
infrared region
0%
ultraviolet region
0%
microwave region

Explanation

$\begin{array}{l}h\nu = 1.6 \times {10^{ - 19}}\\\nu = \dfrac{{1.6 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}} = 2.4 \times {10^{14}}Hz\end{array}$

In the Bohr model of the hydrogen atom.

0%
The radius of the $n^{th}$ orbit is proportional to $n^{2}$
0%
The total energy of the electron in the $n^{th}$ orbit is inversely proportional to $'n'$
0%
The angular momentum of the electron in an orbit is an integral multiple of $h/2\pi$
0%
The magnitude of the potential energy of the electron in any orbit is greater than its kinetic energy

Explanation

Radius of Bohr's orbit of hydrogen atom is given by

$r=\cfrac{n^2h^2}{4\pi^2 mKze^2}$ or

$r=(0.59 \mathring { A } )\cfrac { n^{ 2 } }{ z }$

So from expression we found

$r \propto n^2$

And according to Bohr's theory

Angular momentum of elctron in an orbit will be Integral multiple of

$(h/2\pi)$

Magnitude of potential energy is twice of kinetic energy of electron in an orbit

$\left| P.E \right| =2\left| K.E \right|$

$K.E=(13.6ev)\cfrac{(z^2)}{(n^2)}$

If the diameter of two different nuclei are in the ratio 1 : 2, then their mass number are in the ratio:

0%
1 : 2
0%
8 : 1
0%
1 : 8
0%
1 : 4

Explanation

Given:

Ratio of diameter of $2$ nuclei = $1:2$

To find the Ratio of their mass number.

As we know the - Radius of the nucleus is given as:

$R=r_0A^{1/3}$

$A= Mass\ number$ , $R=Radius\ of\ the\ nucleus$ , $r_0=Nucleon\ number=1.25\ fm$

As, $\frac{D_1}{D_2}=\frac{1}{2}$

Radius is just half of diameter as $Radius\ \times \ 2=Diameter$

So ratio of Radius $=\frac{1}{2}$

$\frac{R_1}{R_2}=\frac{r_0\ A_1^{1/3}}{r_0\ A_2^{1/3}}$

So, $R^3\propto \ A$

So Ratio of mass number i.e. $\frac{A_1}{A_2}=\frac{1}{8}$

optiin C 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 or absorbed in erg?

0%
$2.15\times { 10 }^{ -11 }$
0%
$0.191\times { 10 }^{ -10 }$
0%
$2.389\times { 10 }^{ -12 }$
0%
$0.239\times { 10 }^{ -10 }$

Consider

$\alpha-$ particles,

$\beta-$ particles &

$\gamma-$ rays, each having an energy of

$0.5\ MeV$ . Increasing order of penetrating of the radiations are:

0%
$\alpha < \beta< \gamma$
0%
$\alpha< \gamma< \beta$
0%
$\beta< \gamma< \alpha$
0%
$\gamma< \beta< \alpha$

Explanation

$\alpha$ -particles are essentially helium ions, and these are Massive and are Poorly penetrating.

$\beta$ -radiation are fast-moving electrons, and these are slightly more penetrating with respect to $α$ -particles

$\gamma$ -radiation are faster moving particles, on the photonic scale. And these have the most penetrating power. They require lead shielding of several mm for safe containment.

Hence answer is Option A

Atomic Radius is of the order

$10^{-8}$ cm & nucleus order of

$10^{-13}$ cm. Calculate what fraction of atom is occupied by nucleus?

0%
$10^{-14} \times V_{atom}$
0%
$10^{-13} \times V_{atom}$
0%
$10^{-15} \times V_{atom}$
0%
$10^{-11} \times V_{atom}$

Explanation

We know the volume of a sphere varies to 3rd power of the radius and also nucleus and atoms are spherical in shape

$\dfrac{V_{atom}}{V_{nucleus}}=\dfrac{(10^{-8})^3}{(10^{-13})^3}$

$\dfrac{V_{atom}}{V_{nucleus}}=\dfrac{1}{(10^{-5})^3}$

$V_{nucleus} = 10^{-15}\times V_{atom}$

Hence, the correct option is

$C$

Atomic radius is of the order of

$10^{-8}\ cm$ and nuclear radius is of the order of

$10^{-13}\ cm$ . The fraction of atom occupied by nucleus is:

0%
$10^{-5}$
0%
$10^{5}$
0%
$10^{-15}$
0%
none of these

Explanation

A nucleus is the positively charged center of the atom consisting of protons and neutrons. It's also known as the "atomic nucleus". Nearly all the mass of an atom is contained within the nucleus, since protons and neutrons have much more mass than electrons

according to qes.

Volume of nucleus =

$\dfrac{4}{3}{\pi}r^3$ =

$\dfrac{4}{3}\pi({10^{-3}})^3 cm^3$

Volume of atom =

$\dfrac{4}{3}\pi({10^{-8}})^3cm^3$

$\dfrac{V_N}{V_{atom}}$

$=$

$\dfrac{10^{-39}}{10^{-24}}$

$\dfrac{V_N}{V_{atom}}$

$=$

$10^{-15}$

A certain electronic transition from an excited state to the ground state of the

$H$ atom in one or more steps gives rise to four lines in the ultraviolet region of the spectrum, how many lines does this transition produce in the infrared region of the spectrum.

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

Explanation

Let the four lines are observed in the ultraviolet region of the spectrum.

These correspond to the transitions

$5 \to 1,4 \to 1,3 \to 1$ and

$2 \to 1$

The possible transitions in the IR region are

$5 \to 4,5 \to 3$ and

$4 \to 3$

Therefore the infrared region of the spectrum,

$3$ lines are produced.

Hence, te correct option is

$\text{C}$

The orbital diagram in which the Aufbau principle is violated is:

Explanation

$\bf{Hint-}$ The Aufbau principle states that electrons fill lower-energy atomic orbitals before filling higher-energy atomic orbitals.

$\bf{Explanation-}$

Aufbau's principle states that electrons are filled into atomic orbitals in the increasing order of orbital energy level.
That means an electron will occupy an orbital of lower energy level followed by higher energy levels.

Aufbau principle is violated in option B as $2s$ is not completely filled, yet electrons are filled in $2p$ orbital which has higher energy than $2s$ .
In the case of options $A, C$ and $D$ the electrons are filled according to their energy level. So, the Aufbau principle is not violated here.

$\bf{Conclusion-}$ Hence, option B is correct.

Which of the following orbitals are in the increasing order of energy.

0%
$1s, 2s, 2p,3s$
0%
$4s, 3s, 3p, 4d$
0%
$5p, 4d, 5d, 4f, 6s$
0%
$4p, 2s, 3s, 2p$

Explanation

Energy of the orbital

$\propto$ Principal Quantum number

Energy of the orbital

$\propto$

$(n+l)$ rule

$(n+l)$ for two orbitals is same then the energy is judged by

$'n'$ . So order using this rule is and according to Aufbau's principle as shown in picture is:

$1s<2s<2p<3s$

Hence, option

$A$ is correct.

1054813_1066388_ans_4f08d1b21a1d43f089cd0969ad7879e0.PNG

Which series have highest energy in hydrogen spectrum?

0%
Balmer
0%
Bracket
0%
Pfund
0%
Lyman

Explanation

Energy of electron varies with

$\dfrac{1}{n^2}$ i.e principal quantum number. For lyman series principal quantum number

$(n)$ is 1. Hence, Lyman series have highest energy in hydrogen spectrum.

An element has three electrons in the 4th shell, the atomic number of the element is:

0%
13
0%
21
0%
27
0%
31

Explanation

The configuration is

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^1$ .

Three electrons in the 4th shell is the element Gallium (Ga), atomic number 31.

Hence, option D is correct.

The electronic configuration,

$1s^22s^22p^63s^23p^63d^9$ , represents a:

0%
metal atom
0%
non-metal atom
0%
non-metallic anion
0%
metallic cation

Explanation

The electronic configuration,

$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}$ , represents a metallic cation because the

$e$ ' of

$4s$ Sulosshell are filled before

$3d$ subshell.

In terms of Bohr radius

${a}_{0}$ , the radius of the second Bohr orbit of a hydrogen atom is given by:

0%
$8{a}_{0}$
0%
$4{a}_{0}$
0%
$2{a}_{0}$
0%
$\sqrt {2}{a}_{0}$

Explanation

Since,

$r \propto {n^2}$

Radius of

$2^{nd}$ Bohr's orbit

$= 4a_0$ .

$1.00$ g of

$BaCl_2$ is treated with excess of aqueous

$AgNO_3$ and all chlorine is recovered as

$1.38$ g of AgCl. What is the atomic weight of

$Ba$ ? (

$Cl$

$=35.5$ ,

$Ag$

$=108$ )

0%
$137$
0%
$172.5$
0%
$33$
0%
$68.5$

Explanation

$BaCl_2 + 2AgNO_3 ==> 2AgCl + Ba(NO_3)_2$

$1.38\, g\, AgCl \times \frac{1 \,mole\, AgCl}{143.3 \,g\, AgCl} \times\frac{1\, mole\,BaCl_2}{2\, moles\, AgCl} = 0.00482\, moles\,BaCl2$

$moles \,BaCl_2 = \frac{grams\,BaCl_2 }{formula\, wt.\,BaCl_2 }$

$formula \,wt.\, BaCl_2 = \frac{grams\, XCl_2}{moles\, BaCl_2} = \frac{1.00}{0.00482} = 208$

$BaCl_2 = 208\, amu$

$Ba = 208 - 2Cl = 208 - 2(35.5) = 137 \,amu$

So atomic weight of Ba is 137

Radius of

$2nd$ shell of

$He^+$ (where

$a_0$ - Bohr radius)

0%
$3a_0$
0%
$a_0$
0%
$\dfrac{3}{2}a_0$
0%
$2a_0$

Explanation

Radius of the Bohr's orbit is given by,

$r=\dfrac{h^2n^2}{4\pi^2 m Z e^2}$

Substituting values of h, m for helium ion and the electronic charge r may be calculated as

$r=\dfrac{0.529n^2}{Z}A^0$

Where

$0.529A^0$ is is the Bohr's radius

$a_0=0.529$

Substituting

$Z=2$ ,

$n=2$

$r=2a_0A^0$

Atomic number	Name	Symbol
104	Rutherfordium	Rf
109	Meitnerium	Mt
113	Ununtrium	Uut
118	Ununoctium	Uuo

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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