These are the forms of the same element having the different atomic number but different atomic masses due to different nuclear structure.

These are the forms of the same element having the same atomic number but same atomic masses due to different nuclear structure.

MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 7

An isotope of helium is represented by the symbol

$_{ 2 }^{ 3 }{ He }$ . How many protons, neutrons and electrons are there in an atom of the isotope?

0%
$1$ proton, $2$ neutrons, $1$ electron
0%
$1$ proton, $3$ neutrons, $2$ electrons
0%
$2$ protons, $1$ neutron, $2$ electrons
0%
$2$ protons, $3$ neutrons, $2$ electrons

Explanation

$_{ 2 }^{ 3 }He$ :

Ideal representation of an element :

$_{ Z }^{ A }{ X }$

Here,

$Z=$ atomic number

$=$ number of protons

$A=$ mass number

$=$ number of protons

$+$ number of neutrons

Also, in an atom number of protons

$=$ number of electrons

So, for

$_{2}^{3}{He}$ :

Number of protons

$=$

$Z=$

$2$

Number of electrons

$=$ number of protons

$=$

$2$

number of neutrons

$=$

$A$

$-$

$Z$

$=$

$3$

$-$

$2$

$=$

$1$

Which of the following statements is not true?

0%
All forms of matter have volume
0%
All forms of matter have colour
0%
All forms of matter have mass
0%
Matter can be living or non-living

How many moles of electron weighs one kilogram?

0%
$6.023 \times 10^{23}$
0%
$\dfrac{1}{9.108} \times 10^{32}$
0%
$\dfrac{6.023}{9.108} \times 10^{54}$
0%
$\dfrac{1}{9.108 \times 6.023} \times 10^{8}$

Explanation

Mass of one electron

$\displaystyle =9.108 \times 10^{-31}$ kg.

Mass of one mole of electrons

$\displaystyle =9.108 \times 10^{-31} \times 6.023 \times 10^{23}=9.108 \times 6.023 \times 10^{-8}$ kg.

The number of moles of electrons that weigh one kilogram

$\displaystyle = \dfrac {1}{9.108 \times 6.023 \times 10^{-8}}= \dfrac {1}{9.108 \times 6.023} \times 10^{8}$

Hence, the correct option is

$D$

__________is the nuclear process that transmutes uranium-238 into thorium-234.

0%
Gamma decay
0%
Nuclear fusion
0%
Alpha decay
0%
Positron emission
0%
Nuclear fission

Explanation

$_{92}^{238}U$

$\rightarrow$

$_{90}^{234}Th$

$+$

$_2^4He$

Thus Uranium-238 is converted to Thorium-234 by an Alpha decay. An alpha decay is a radioactive process in which a material with two neutrons and two protons (basically a Helium atom) is obtained from the nucleus of a radioactive atom.

We can also see that when Uranium gets transformed to Thorium, its mass number is decreased by

$4$ units and also the atomic number is decreased by

$2$ units. Which means a material has been produced whose mass number is

$4$ and atomic number is

$2$ . Thus, we can say that Helium has been produced, because Helium has an atomic number of

$2$ and mass number

$4$ .

Hence it is an alpha decay.

Option C is the correct answer.

A sample of water is vaporized and tested in a mass spectrometer. There are

$3$ isotopes of hydrogen,

$^{1}H, ^{2}H$ and

$^{3}H$ . There are

$2$ isotopes of oxygen,

$^{15}O$ and

$^{16}O$ .
How many peaks would be expected to be found on the mass spectrum?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Various isotopes of water with different mass possible are:

$^1H^1H^{ 15 }O(17)$

$^1H^1H^{ 16 }O(18),\ ^1H^2H^{ 15 }O(18)$

$^1H^2H^{ 16 }O(19),\ ^1H^3H^{ 15 }O(19),\ ^2H^2H^{ 15 }O(19)$

$^1H^3H^{ 16 }O(20),\ ^2H^3H^{ 15 }O(20),\ ^2H^2H^{ 16 }O(20)$

$^3H^3H^{ 15 }O(21),\ ^2H^3H^{ 16 }O(21)$

$^3H^3H^{ 16 }O(22)$

Therefore total of 5 peaks will be present in the mass spectrum.

Hence, option

$C$ is correct.

Beer's Law allows the measurement of the progress of a reaction by measuring the change of which of the following?

0%
Temperature
0%
Surface Area
0%
Concentration
0%
All of the above

Explanation

Beer's law stated that absorbance is proportional to the concentrations of the attenuating species in the material sample.

$\therefore$ Beer's law allows the measurement of the progress of a reaction by measuring the change of concentration.

Identify the error(s) in the orbital notation diagram for oxygen in the in the image

$a$ .

0%
The electrons are not drawn with the appropriate spins
0%
This is not a picture of oxygen. It is nitrogen
0%
The electrons are not filled out according to Aufbau's Principle
0%
The electrons are not filled out according to Hund's Rule

Explanation

The correct orbital rotation for oxygen is given by oxygen.

Thus given represent is wrong.

$\therefore$ Option B

Statement I: The

$5s$ orbital is filled after the

$4d$ orbital
BECAUSE
Statement II: The

$5s$ orbital has a lower energy than the

$4d$ orbital

0%
Statement $1$ and Statement $2$ are correct and Statement 2 is the correct explanation of Statement $1$
0%
Both the Statement $1$ and Statement $2$ are correct and Statement $2$ is NOT the correct explanation of Statement $1$
0%
Statement $1$ is correct but Statement $2$ is not correct
0%
Statement $1$ is not correct but Statement $2$ is correct
0%
Both the Statement $1$ and Statement $2$ are not correct

Explanation

The

$5s$ orbital has a lower energy than the

$4d$ orbital. Lower energy orbitals are filled first. Hence

$5s$ will be filled before

$4d$ .

Option D is the correct answer.

Given there are

$3$ isotopes of hydrogen

$(^{1}H, ^{2}H$ and

$^{3}H$ ) and

$2$ isotopes of oxygen

$(^{15}O$ and

$^{16}O)$ , how many peaks would be expected on a mass spectrum for the hydronium ion,

$H_{3}O^{+}$ ?

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Peaks would be expected on a mass spectrum for different masses of compound.

For hydronium ion production by isotope of oxygen

$O^{15},O^{16}$ and isotope of hydrogen

$H^1,H^2,H^3$ there are 8 different masses of hydronium ion produced,so 8 peaks are generated on mass spectrum.

What are isotopes?

0%
These are the forms of the same element having the same atomic number but different atomic masses due to different nuclear structure.
0%
These are the forms of the same element having the different atomic number but different atomic masses due to different nuclear structure.
0%
These are the forms of the same element having the same atomic number but same atomic masses due to different nuclear structure.
0%
none of these

Explanation

Isotopes are the forms of same element having the same atomic number but different atomic masses due to different nuclear structure.

So, the correct option is

$A$

Scientists previously thought an atom looked like plum pudding. In this model, it was assumed that protons and electrons were distributed throughout the atom almost randomly. Which of the following contributions DISPROVED the Plum Pudding Model?

0%
Rutherford's Gold Foil Experiment
0%
J.J. Thomson's Cathode Ray Experiment
0%
Millikan's Oil Drop Experiment
0%
Bohr's Model of the Atom

Explanation

In Rutherford's gold foil experiment, few positively charged

$\alpha$ - particles were deflected. The deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed in his plum-pudding model. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged

$\alpha$ -particles.

Thomson's experiment gave the existence of electron and Millikan's oil drop experiment tells about charge of electron.

Bohr's model explained the stability of atom and distribution of electron in orbit.

Thus, option A is correct.

Which experiment was developed on the plum pudding model of the atom?

0%
Gold foil
0%
Cathod ray tube
0%
Emission spectra of hydrogen
0%
Oil drop

Maximum number of electrons that an M-shell contain is/are _____________.

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

Explanation

The general formula for number of electrons in nth shell is given by

$2(n^2)$ .

Now M shell is 3rd shell i.e, n = 3.

Therefore maximum number of electron

$=2(n^2)$

$=2\times 3^2$

$=2\times 9=18$ electrons.

Rutherford model could not explain:

0%
maxwell's theory of elctrodynamics
0%
line spectra of atom
0%
both A and B
0%
None of these

Watermelon model of atom was proposed by:

0%
Rutherford
0%
Thomson
0%
Bohr
0%
Sommerfeld

Assertion (A) : Atoms with completely filled and half-filled subshells are stable.

Reason (R) : Completely filled and half-filled subshells have the symmetrical distribution of electrons and have maximum exchange energy.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Assertion (A) : Atoms with completely filled and half filled sub-shells are stable.
Reason (R) : Completely filled and half filled sub-shells have symmetrical distribution of electrons and have maximum exchange energy.
For example, He with completely filled

$1s$ sub-shell and

$N$ with half filled

$2p$ sub-shell are stable.
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Hence, the option (A) is the correct answer.

$n=6$ , the correct sequence for filling of electrons will be:

0%
$ns\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f\longrightarrow np$
0%
$ns\longrightarrow \left( n-2 \right) f\longrightarrow np\longrightarrow \left( n-1 \right) d$
0%
$ns\longrightarrow np\longrightarrow \left( n-1 \right) d\longrightarrow \left( n-2 \right) f$
0%
$ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$

Explanation

$n=6$ , the correct sequence for filling of electrons will be

$\displaystyle ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$ .

It will be

$\displaystyle 6s\longrightarrow 4 f \longrightarrow5 d\longrightarrow 6 p$

The energy of

$4 f$ and

$5 d$ orbitals are in between the energies of

$4s$ and

$6 p$ orbitals.

Note: Higher is the

$(n+l)$ , higher will be the energy of the orbital.

For n = 6

6s = 6+0 = 6

6p = 6+1 = 7

5d = 5+2 = 7

4f = 4+3 = 7

Hence, correct order is:

$ns\longrightarrow \left( n-2 \right) f\longrightarrow \left( n-1 \right) d\longrightarrow np$

Hence, the correct option is

$D$

According to the drawback of Rutherford's structure of atom electron should fall into the :

0%
space
0%
outer enviroment
0%
shell
0%
nucleus

$E_P$ and

$E_k$ represent potential energy and kinetic energy respectively, of an orbital electron, then according to Bohr's theory:

0%
$E_k= -E_p/2$
0%
$E_k= -E_p$
0%
$E_k= -2E_p$
0%
$E_k= 2E_p$

Explanation

Potential energy of an orbital electron

$E_p=-\dfrac{Ze^2}{r}$

Kinetic energy of an electron

$E_k=\dfrac{Ze^2}{2r}$ , where Z=atomic number

So,

$E_k=-\dfrac{E_p}{2}$

Which of the following statement in relation to the hydrogen atom is correct?

0%
$3s,3p$ and $3d$ orbitals all have the same energy
0%
$3s$ and $3p$ -orbitals are of lower energy than $3d$ -orbitals
0%
$3p$ -orbital is lower in energy than $3d$ - orbital
0%
$3s$ -orbital is lower in energy than $3p$ -orbital

Explanation

Hydrogen atom is in

$1{ s }^{ 1 }$ and the

$3s,3p$ and

$3d-$ orbitals will have the same energy w.r.t

$1s$ orbital

The ionization energy of

${ Li }^{ 2+ }$ is equal to

0%
$9 hcR$
0%
$6 hcR$
0%
$2 hcR$
0%
$hcR$

Explanation

Ionization energy of an atom

$=hc R{ Z }^{ 2 }$
where

$R$ is the Rydberg constant and

$Z$ is the atomic number.
We know,

$Z=3$ for

${ Li }^{ 2+ }$
Ionization energy of

$Li^{2+}$

$={ \left( 3 \right) }^{ 2 }hcR$

$=9 hcR$

The density of water and ethanol at room temperature is

$1.0$ g

$/$ mL and

$0.789$ g

$/$ mL respectively. What volume of ethanol contains the same number of molecules as are present in

$100$ mL of water?

0%
$307.8$ mL
0%
$168.7$ mL
0%
$323.9$ mL
0%
$253.3$ mL

Explanation

An equal number of moles contain equal number of molecules.

$\displaystyle\dfrac{wt. \ of \ C_2H_5OH}{mol. \ wt. \ of \ C_2H_5OH}=\frac{wt. of H_2O}{mol. \ wt. \ of \ H_2O}$

$\displaystyle\frac{0.789\times \ volume \ of \ C_2H_5OH}{46}=\dfrac{1.0\times 100}{18}$

Volume of

$C_2H_5OH=\displaystyle\frac{100}{18}\times \dfrac{46}{0.789}$

$=323.9$ mL.

Hence, option

$C$ is correct.

Choose one of the following definitions that give the correct meaning of matter :

0%
anything that occupies space and has mass is known as matter.
0%
anything that has definite shape and volume is known as matter.
0%
anything that has mass but not definite shape is known as matter.
0%
anything that is intangible is known as matter.

In Bohr's model, if the atomic radius of the first orbit

${r}_{1}$ , then radius of fourth orbit will be:

0%
$4{r}_{1}$
0%
$6{r}_{1}$
0%
$16{r}_{1}$
0%
$\dfrac{{r}_{1}}{16}$

Explanation

In Bohr's model, if the atomic radius of the first orbit

$\displaystyle r_1$ , then the radius of the fourth orbit will be

$\displaystyle 16 \times r_1$ .

In general,

$\displaystyle r_n = n^2 \times r_1$

$\therefore \displaystyle r_4 = 4^2 \times r_1$

$\therefore \displaystyle r_4 = 16 \times r_1$

Hence, option

$C$ is correct.

The energy liberated when an excited electron returns to its ground state can have:

0%
any value from zero to infinity
0%
only negative values
0%
only specified positive values
0%
none of the above

Positronium is the name given to an atom-like combination formed between_____________.

0%
a positron and a proton
0%
a positron and a neutron
0%
a positron and an $\alpha$ -particle
0%
a positron and an electron

According to classical theory, if an electron is moving in a circular orbit around the nucleus:

0%
it will continue to do so for sometime
0%
its orbit will continuously shrink
0%
its orbit will continuously enlarge
0%
it will continue to do so for all the time

On Bohr stationary orbits:

0%
electrons do not move
0%
electrons move emitting radiations
0%
energy of the electron remains constant
0%
angular momentum of the electron is $\dfrac{h}{2 \pi}$

Explanation

Hint: Bohr orbit is the hypothetical path of an electron around the nucleus of the Bohr atom.

Explanation:

Electrons revolve in fixed stationary orbits where they do not emit radiation.
The angular momentum of electrons is $nh/2\pi$ .
Each stationary state is associated with a definite amount of energy.
The energy of an electron remains constant.

Hence, the correct answer is option $C.$

The fundamental particles present in equal numbers in neutral atoms (atomic number

$71$ ) are____________.

0%
protons and electrons
0%
neutrons and electrons
0%
protons and neutrons
0%
protons and positrons

Any

$p$ -orbital can accommodate up to:

0%
$4$ electrons
0%
$2$ electrons with parallel spins
0%
$6$ electrons
0%
$2$ electrons with opposite spins

Explanation

Any p-orbital can accommodate up to 6 electrons.
p-orbital has three sub-shells

$\displaystyle p_x$ ,

$\displaystyle p_y$ and

$\displaystyle p_z$ each of which can accommodate 2 electrons.

Which of the properties of the elements is a whole number?

0%
Atomic mass
0%
Atomic number
0%
Atomic radius
0%
Atomic volume

Atomic number of an element represents__________.

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

Explanation

Atomic number of an element represents the number of protons in the nucleus. It is also equal to the number of electrons in neutral atom. The number of neutrons is equal to the difference between mass number and atomic number.
For example, for

$\displaystyle _{17}^{37}Cl$ , the atomic number is 17, the mass number is 37, there are 17, electrons, 17 protons and

$\displaystyle 37-17=20$ neutrons.

Bohr advanced the ideal of

0%
stationary electrons
0%
stationary nucleus
0%
stationary orbits
0%
elliptical orbits

Krypton (Atomic number

$36$ ) has the electronic configuration

$\left[Ar\right] 4{s}^{2} 3{d}^{10} 4{p}^{6}$ . The

$37$ th electron will go into which one of the following sub-levels?

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

Explanation

The 37th electron will go into 5s sub-level.

According to aufbau principle, the correct order of energy of 4p and 5s-orbitals is

$\displaystyle 4p<5s$

Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.

The 4p sub-energy level is at a lower energy than the 5s sub-energy level
For 4p,

$\displaystyle n=4$ and

$\displaystyle l=1$ . Hence,

$\displaystyle (n+l) = 4+1=5$
For 5s,

$\displaystyle n=5$ and

$\displaystyle l=0$ . Hence,

$\displaystyle (n+l) = 5+0=5$
As the value of

$\displaystyle (n+l)$ for 4p orbital is same as that of 5s orbital, 4p orbital is filled before 5s orbital as 4p orbital has lower value of n than 5s orbital.

Note: If two orbitals have same value of

$\displaystyle (n+l)$ , then the orbital with lower value of n will be filled first.

According to Aufbau principle, the correct order of energy of

$3d, 4s$ and

$4p$ -orbitals is:

0%
$4p < 3d < 4s$
0%
$4s < 4p < 3d$
0%
$4s < 3d < 4p$
0%
$3d < 4s < 4p$

Explanation

According to aufbau principle, the correct order of energy of 3d,4s and 4p-orbitals is

$\displaystyle 4s<3d<4p$

Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.

The 4s sub-energy level is at a lower energy than the 3d sub-energy level
For 3d,

$\displaystyle n=3$ and

$\displaystyle l=2$ . Hence,

$\displaystyle (n+l) = 3+2=5$
For 4s,

$\displaystyle n=4$ and

$\displaystyle l=0$ . Hence,

$\displaystyle (n+l) = 4+0=4$
As the value of

$\displaystyle (n+l)$ for 4s orbital is less than that of 3d orbital, 4s orbital is filled before 3d orbital.

For 4p,

$\displaystyle n=4$ and

$\displaystyle l=1$ . Hence,

$\displaystyle (n+l) = 4+1=5$
As the value of

$\displaystyle (n+l)$ for 4p orbital is same as that of 3d orbital, 3d orbital is filled before 4p orbital as 3d orbital has lower value of n than 4p orbital.

Note: If two orbitals have the same value of

$\displaystyle (n+l)$ , then the orbital with the lower value of n will be filled first.

Hence, the correct option is

$\text{C}$

After

$np$ -orbitals are filled, the next orbital filled will be:

0%
$\left(n+1\right)s$
0%
$\left(n+2\right)p$
0%
$\left(n+1\right)d$
0%
$\left(n+2\right)s$

Explanation

After np-orbitals are filled, the next orbital filled will be

$\displaystyle (n+1)s$ .
According to Aufbau principle, orbitals are filled in order of increasing value of

$\displaystyle (n+l)$ .
If two orbitals have same value of

$\displaystyle (n+l)$ , then the orbital with lower value of n will be filled first.
For np orbital,

$\displaystyle (n+l) = (n+1)$
For

$\displaystyle (n+1)s$ orbital,

$\displaystyle (n+l) = (n+1+0)=(n+1)$
For

$\displaystyle (n+2)p$ orbital,

$\displaystyle (n+l) = (n+2+1)=(n+3)$
For

$\displaystyle (n+1)d$ orbital,

$\displaystyle (n+l) = (n+1+2)=(n+3)$
For

$\displaystyle (n+2)s$ orbital,

$\displaystyle (n+l) = (n+2+0)=(n+2)$

In which of the following orbital diagrams are both Pauli exclusion principle and Hund's rule violated?

The electronic configuration of an atom/ion can be defined by which of the following?

0%
Aufbau principle
0%
Pauli's exclusion principle
0%
Hund's rule of maximum multiplicity
0%
All of the above

Which among the following is incorrect of

$_{5}{B}$ in normal state?

Explanation

The option (C) is correct of

$_5B$ in normal state. It is against Pauli exclusion principle and not Hund's rule of maximum multiplicity.
Pauli exclusion principle : Two electrons in an atom cannot have the same set of all four quantum numbers.
In the given diagram, two electrons in s subshell have same spin. This violets Pauli exclusion principle.

Hund's rule : When several orbitals of equal energy (degenerate orbitals) are available, the electrons first fill all the orbitals singly before pairing in any of these orbitals.

According to Aufbau principle, orbitals are filled in order of increasing value of

$(n+l)$ .
If two orbitals have same value of

$(n+l)$ , then the orbital with lower value of

$n$ will be filled first.

After filling the

$4d$ -orbitals, an electron will enter in:

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

Explanation

After filling the 4d-orbital, an electron will enter in 5p orbital.

The 4d sub-energy level is at a lower energy than the 5p sub-energy level
For 4d,

$\displaystyle n=4$ and

$\displaystyle l=2$ . Hence,

$\displaystyle (n+l) = 4+2=6$

For 5p,

$\displaystyle n=5$ and

$\displaystyle l=1$ . Hence,

$\displaystyle (n+l) = 5+1=6$

As the value of

$\displaystyle (n+l)$ for 4d orbital is same as that of 5p orbital, 4d orbital is filled before 5p orbital as 4d orbital has lower value of n than 5 p orbital.

Note: If two orbitals have same value of

$\displaystyle (n+l)$ , then the orbital with lower value of n will be filled first.

The orbital diagram which violates the 'aufbau principle' is:

Explanation

The orbital diagram in which 'aufbau principle' is violated, is represented by the option (B).

Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.

The 2s sub-energy level is at a lower energy than the 2p sub-energy level
For 2s,

$\displaystyle n=2$ and

$\displaystyle l=0$ . Hence,

$\displaystyle (n+l) = 2+0=2$
For 2p,

$\displaystyle n=2$ and

$\displaystyle l=1$ . Hence,

$\displaystyle (n+l) = 2+1=3$
As the value of

$\displaystyle (n+l)$ for

$2s$ orbital is less than that of

$2p$ orbital,

$2s$ orbital is filled before

$2p$ orbital.

What is ratio of time periods

$\left( { { T }_{ 1 } }/{ { T }_{ 2 } } \right)$ in second orbit of hydrogen atom to third orbit of

${He}^{+}$ ion?

0%
$\dfrac { 8 }{ 27 }$
0%
$\dfrac { 32 }{ 27 }$
0%
$\dfrac { 27 }{ 32 }$
0%
$\dfrac { 27 }{ 8 }$

Explanation

we know that

time period

$\propto \cfrac { { n }^{ 3 } }{ { t }^{ 2 } }$

n=energy level

t=z=valan enly

$\cfrac { { T }_{ H } }{ { T }_{ He } } =\cfrac { { n }_{ 1 }^{ 3 }{ t }_{ 2 }^{ 2 } }{ { n }_{ 2 }^{ 3 }{ t }_{ 1 }^{ 2 } }$

$\cfrac { \left( 8 \right) \times \left( 4 \right) }{ \left( 27 \right) \times \left( 1 \right) }$

ratio=

$\cfrac { 32 }{ 17 }$

Binding energy per nucleon of three nuclei

$A$ ,

$B$ and

$C$ are

$5.5$ ,

$8.5$ and

$7.5$ respectively. Which one of the following nuclei is most stable?

0%
$A$
0%
$C$
0%
$B$
0%
Cannot be predicted

Arrangement of the following group of orbitals in which they fill with electrons :
(5f, 6p, 4p, 6s, 4d, 4f )

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

Explanation

Electrons will first fill orbitals by the sum of the quantum numbers n and l. Orbitals with equal values of (n+l) will fill with the lower n values first.

So energies of

$4p, 4d, 6s, 4f, 6p, 5f$ are 5,6,6,7,7,8 respectively.
Hence option A is correct.

Which of the following sets of quanta numbers could represent the last electron added to complete the electronic configuration for a ground state atom of Br (Z = 35)?

0%
$4, 0, 0, - \cfrac {1}{2}$
0%
$4, 1, 1, - \cfrac {1}{2}$
0%
$3, 1, 1, - \cfrac {1}{2}$
0%
$4, 1, 2, + \cfrac {1}{2}$

Explanation

The valence node shell configuration is

$4s^2\,\, 4p^5$ for the last electron

$n=4, l=1$ .

$m=-1, 0$ or

$+1, s=+\cfrac {1}{2}$ or

$-\cfrac {1}{2}$ .

Hence,

$4, 1, 1, -\cfrac {1}{2}$

Therefore, option B is correct.

Which of the following pairs is correctly matched ?

0%
Experimental observation - (a) X-ray spectra

Phenomenon - Charge on the nucleus
0%
Experimental observation - (b) a- particle scattering

Phenomenon - Quantized electron orbit
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Experimental observation - (c) Emission spectra

Phenomenon - Quantization of energy
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Experimental observation - (d) The photoelectric effect

Phenomenon - The nuclear atom

The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of hydrogen atom is

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

Explanation

The kinetic energy of the electron in the nth state

$K=\dfrac {mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$
The total energy of the electron in the nth state

$T=\dfrac {-mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$

$\therefore \dfrac {K}{T}=-1$

In the Bohr's model of the atom:

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

Explanation

According to the Bohr model of atom

Energy of electron in the

$n^{th}$ orbit of hydrogen like species

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

Radius of

$n^{th}$ orbit of hydrogen like species

$R_n = R_0 \cfrac{n^2}{Z}$

Angular momentum of electron in the

$n^{th}$ orbit

$mvr = \cfrac{nh}{2\pi}$ integral multiple of

$\cfrac{h}{2\pi}$

Relation between energies

$-KE = TE= \cfrac{PE}{2}$

Comparing magnitudes =

$2|KE| = |PE|$

$\implies |PE| >|KE|$

The magnitude of potential energy is greater than that of kinetic energy.

$0.25 g$ of an element

$M$ reacts with excess fluorine to produce

$0.547 g$ of the hexafluoride

$MF_6$ . What is the element?

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$Cr$
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$Mo$
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$S$
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$Te$

Explanation

We have

$M + F_2 \rightarrow MF_6$

moles of M = moles of $MF_6$

let mass of element M = $x g$

therefore we have

$\dfrac{0.25}{x} = \dfrac{0.547}{x+114}$

so, solving we get , $x = 95.9 g$

therefore, the elemnt is $Mo$

Which electron,orbitals are designated by
(i) n= 2, l=1, m= 0;
(ii)n= 3, l= 2, m= 0;
(iii)n = 4, l = 2, m = 1 and
(iv) n = 5, I= 3, m =3 respectively ?

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2p, 3d, 4d and 5f, respectively
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2p, 3d, 4d and 5d, respectively
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2p, 3d, 4f and 5f, respectively
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2p, 3p, 4d and 5f, respectively

Explanation

For

$(i) n = 2, l= 1, m =.0$ ; the orbital occupied by electron is 2p.

For

$(ii) n = 3, l= 2, m = 0$ ;the orbital occupied by electron is 3d.

For

$(iii) n = 4, l= 2, m =1$ ; the orbital occupied by electron is 4d.

For

$(iv) n = 5, l = 3, m = 3$ ; the orbital occupied by electron is 5f.

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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