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

Neutrons are obtained by:

0%
Bombardment of Ra with $\beta$ - particles
0%
Bombardment of Be with $\alpha$ -particles
0%
Radioactive disintegration of uranium
0%
None of these

Explanation

Explanation:

In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron. He discovered it by bombardment of alpha particle on beryllium atom

$\{{}_4^9Be + {}_2^4He \to {}_6^{12}C + {}_0^1n\}$

Hence option B is correct

When a mixture of aluminium powder and iron

$(III)$ oxide is ignited, it produces molten iron and aluminium oxide. In an experiment,

$5.4\ g$ of aluminium was mixed with

$18.5\ g$ of iron

$(III)$ oxide. At the end of the reaction, the mixture contained

$11.2\ g$ of iron,

$10.2\ g$ of aluminium of unreacted iron

$(III)$ oxide. No aluminium was left. What is the mass of the iron

$(III)$ oxide left?

0%
$2.5\ g$
0%
$7.3\ g$
0%
$8.3\ g$
0%
$2.9\ g$

Explanation

$Al+Fe_{2}O_{3}\rightarrow Fe+Al_{2}O_{3}$
Mass of

$Al=5.4\ gm$
Mass of

$Fe_{2}O_{3}=18.5\ gm$
Mass of

$Fe=11.2\ gm$
Mass of

$Al_{2}O_{3}=10.2\ gm$

$\therefore$ Amount of unreacted

$Fe_{2}O_{3}$

$=(5.4+18.5)-(11.2+10.2)\ gm$

$=(23.9-21.4)\ gm$

$=2.5\ gm$
Option

$a$

The commonly used pain reliever, aspirin, has the molecular formula

$C_9H_8O_4.$ If a sample of aspirin contains

$0.968\ g$ of carbon, what is the mass of hydrogen in the sample?

0%
$0.717\ g$
0%
$0.0717\ g$
0%
$8.000\ g$
0%
$0.645\ g$

Explanation

Molar mass of $C_9H_8O_4=9 \times 12+8\times 1+4\times 16$

$=180$ gram/mole

$180$ gram $C_9H_8O_4 \equiv 108$ gram carbon

$\therefore x$ gram $C_9H_8O_4 \equiv 0.968$ gram carbon

$\therefore x=\dfrac{0.968 \times 180}{108}$

$= 1.613$ gram $C_9H_8O_4$

Now

$180$ gram $C_9H_8O_4\equiv 8$ gram Hydrogen

$\therefore 1.613$ gram $C_9H_8O_4 \equiv \dfrac{8\times 1.613}{180}$

$=0.0717$ gram ' $H$ '

Option $(b)$ correct.

Which statement about a

$3$ p orbital is correct?

0%
It can hold a maximum of $6$ electrons
0%
It has the highest energy of the orbitals with principal quantum number $3$
0%
It is at a higher energy level than a $3$ s orbital but has the same shape
0%
It is occupied by one electron in an isolated phosphorous atom

Explanation

Solution:- (A) It can hold a maximum of

$6$ electrons.

Any

$p$ orbital can hold a maximum of

$6$ electrons.

Hence statement

$\left( A \right)$ is correct.

Which of the following is not a fundamental particle?

0%
Electron
0%
Proton
0%
Neutron
0%
$X-$ rays

In the sixth period of the extended from of periodic table, the orbitals are filled as:

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

Which of the following quantity for an electron revolving round the

$H-nucleus$ is independent to the mass of electron ?

0%
distance from nucleus
0%
$K.E.$
0%
$P.E.$
0%
speed

The penultimate and outermost orbit of an element contains

$10$ and

$2$ electrons respectively. If the outermost orbit is fourth orbit, the atomic number of the element should be

0%
$12$
0%
$22$
0%
$32$
0%
$40$

The effective atomic number of cobalt in the complex

$[Co(NH_3)_6]^{3+}$ is:

0%
$36$
0%
$33$
0%
$24$
0%
$30$

$K$ and

$L$ shell of an element are completely filled and there are

$16$ electron in

$M-$ shell and

$2-$ electrons in

$N-$ shell. The atomic number of the element is

0%
$18$
0%
$28$
0%
$22$
0%
$26$

Wireless mouse works on which technology?

0%
IR
0%
Radio frequency
0%
X Rays
0%
Gamma Rays

How many unpaired electrons are present in ground state of chromium

$(Z=24)$ ?

0%
$1$
0%
$5$
0%
$6$
0%
$0$

Explanation

Six.

The electronic configuration for Cr (24) is : 1s

$^2$ 2s

$^2$ 2p

$^6$ 3s

$^2$ 3p

$^6$ 4s

$^1$ 3d

$^5$ .

The no. of unpaired electrons is 6 due to half-filled subshells (4s and 3d) which are more stable than incompletely filled subshells.

In the sixth period, the orbitals are filled as :

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

The ratio of the radii of the first three Bohr orbits is:

0%
$1:5:33$
0%
$1:2:3$
0%
$1:4:9$
0%
$1:8:27$

Explanation

$r_n \propto n^2$

Option C is correct.

The radius of the first Bohr orbit of hydrogen atom is

$r$ . The radius of the

$3^{rd}$ orbit would be:

0%
$3r$
0%
$9r$
0%
$27r$
0%
$none\ of\ these$

Explanation

$r_n =r_0 \dfrac {n^2}{Z}\Rightarrow r_3 =r(3)^2 =9r \left(A / q, \dfrac {r_0}{Z}=r \right)$

Option B is correct.

An element X belongs to fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of X? It has:

0%
partially filled d-orbitals and completely filled s-orbital
0%
completely filled s-orbital and completely filled p-orbitals
0%
completely filled s- orbital and half filled p-orbitals
0%
half filled d-orbitals and completely filled s-orbital
0%
completely filled s-, p- and d-orbitals

Explanation

Answer C.

Since, element 'X' belongs to the fourth period, n=4. For the fifteenth group, the general outer electronic configuration is

$ns^2np^3$ .

The electronic configuration of X can be written as

$X=1s^22s^22p^63s^23p^64s^23d^{10}4p^3$ . So, element 'X' has completely filled s- and d-orbitals and half-filled p-orbitals.

The

$_{88}Ra^{226}$ is:

0%
n-mesons
0%
u-mesons
0%
Radioactive
0%
Non-radioactive

Explanation

$_{88}Ra^{226}$ is radioactive because

$\dfrac{n}{p}$ ratio for it is 1.56 which is greater than 1.5.

The expression for Bohr's radius of an atom is

0%
$r = \dfrac{n^2h^2}{4\pi^2me^4z^2}$
0%
$r = \dfrac{n^2h^2}{4\pi^2me^2z}$
0%
$r = \dfrac{n^2h^2}{4\pi^2me^2z^2}$
0%
$r = \dfrac{n^2h^2}{4\pi^2m^2e^2z^2}$

Explanation

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

Hence, option B is correct.

Nuclear reactivity of

$Na$ and

$Na^+$ is same because both have:

0%
Same electron and proton
0%
Same proton and same neutron
0%
Different electron and proton
0%
Different proton and neutron

Explanation

Nuclear reactivity depends upon

$\frac{n}{p}$ ratio and

$\frac{n}{p}$ ratio is same for both

$Na$ and $$Na^+$$, so they both have same nuclear reactivity.

Hence, option

$B$ is correct.

The

$\alpha$ -particle is identical with:

0%
Helium nucleus
0%
Hydrogen nucleus
0%
Electron
0%
Proton

Explanation

$\alpha$ - particle is identical with

$_2He^4$ helium nucleus. Alpha particles are subatomic fragments consisting of two neutrons and two protons.

$_6C^{14}$ is formed from

$_7N^{14}$ in the upper atmosphere by the action of the fundamental particle:

0%
Positron
0%
Neutron
0%
Electron
0%
Proton

Explanation

The radioactive isotope

$_6C^{14}$ is produced in the atmosphere by the action of cosmic ray neutrons on

$_7N^{14}$ .

$_7N^{14} + _0n^1 \rightarrow _6C^{14} + _1H^1$

An element

$_{96}X^{227}$ emits

$4\alpha$ and

$5\beta$ particles to form new element Y. Then atomic number and mass number of Y are:

0%
93; 211
0%
211; 93
0%
212; 88
0%
88; 211

Explanation

$_{96}X^{227} \rightarrow _aY^b + 4\alpha + 5 \beta$

On equating mass number:

$227 = a + 4 \times 4 + 0, \ a = 211$

On equating atomic number:

$96 = b + 2 \times 4 - 5, \ b = 93$

Hence, option A is correct.

Which of the following is radioactive element?

0%
Sulphur
0%
Polonium
0%
Tellurium
0%
Selenium

Alpha particles are ...... times heavier (approximately) than neutrons.

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

Explanation

Since an alpha particle has 2 protons and 2 neutrons which are literally the nucleus of a helium atom, the atomic weight of helium which is about 4.031882 amu and a neutron is about 1.008665 amu.

Alpha particle is about 3.99 (=4) times heavier than a neutron.

Hence, option

$B$ is correct.

Radioactivity is due to:

0%
Stable electronic configuration
0%
Unstable electronic configuration
0%
Stable nucleus
0%
Unstable nucleus

Explanation

Radioactive decay (radioactivity) occurs in unstable atomic nuclei that don't have enough binding energy to hold the nucleus together due to an excess of either protons or neutrons.

Hence, option

$D$ is correct.

The appreciable radioactivity of uranium minerals is mainly due to:

0%
An uranium isotope of mass number 235
0%
A thorium isotope of mass number 232
0%
Actinium
0%
Radium

A radium

$_{88}Ra^{224}$ isotope, on emission of an

$\alpha$ -particle gives rise to a new element whose mass number and atomic number will be:

0%
220 and 86
0%
225 and 87
0%
228 and 88
0%
224 and 86

Explanation

When an

$\alpha$ -particle is emitted by any nucleus than atomic weight decreases by four units and atomic number decreases by two units,

$_{88}Ra^{224} \xrightarrow []{-\alpha} \ _{86}X^{220}$ .

If uranium (mass no. 238 and atomic no. 92) emits

$\alpha$ -particle, the product has mass number and atomic number:

0%
234, 90
0%
236, 92
0%
238, 90
0%
236, 90

Explanation

The required reaction is:

$_{92}U^{238} \rightarrow \ _{90}Th^{234} + \ _2He^4$

Atomic number: 90

Mass number: 234

The maximum sum of the number of neutrons and protons in an isotope of hydrogen is:

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

Explanation

$_1^3H$ , there are 1 proton and 2 neutrons.

The amount of energy which is released due to addition of extra electron to the outermost orbit of gaseous atom is called

0%
Electron capacity
0%
Electron affinity
0%
ionisation potential
0%
Electronegativity

Which of the following property displays progressive increase down a group in the Bohr's periodic table

0%
Electronegativity
0%
Electron affinity
0%
Ionization potential
0%
Size of the atom

The number of elements in the

$5^{th}$ period of the periodic table are

0%
$8$
0%
$10$
0%
$18$
0%
$32$

Explanation

$5^{th}$ period has 18 elements.

Which property is different for neutral atoms of the two isotopes of the same element?

0%
Number of protons
0%
Atomic number
0%
Number of neutrons
0%
None of these

The symbol of an isotope is

$_{32} X^{65}$ , this reveals that:

0%
Its atomic number is 32 and atomic weight is 65
0%
Its atomic number is 65
0%
It has 65 electrons
0%
It has 32 neutrons

Explanation

In isotope

$_{32}X^{65}$ , 32 is atomic number and 65 is atomic weight.

Which character is different of the two isotopes of an element?

0%
Atomic mass
0%
Atomic number
0%
Number of electrons
0%
Number of protons

The atomic number of bromine is 35 and its atomic weight is 79. Two isotopes of bromine are present in equal amounts. Which of the following statements represents the correct number of neutrons?

0%
First isotope - 34, Second isotope - 36
0%
First isotope - 44, Second isotope - 46
0%
First isotope - 45, Second isotope - 47
0%
First isotope - 79, Second isotope - 81

Explanation

Two isotopes of bromine are

$_{35}Br^{79}, \ _{35}Br^{81}$ .

No. of neutrons in

$_{35}Br^{79} = 79 - 35 = 44$

No. of neutrons in

$_{35}Br^{81} = 81 - 35 = 46$

Atomic weight of the isotope of hydrogen which contains 2 neutrons is the nucleus would be:

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

Explanation

Atoms of different elements having different atomic no. but same mass no. are called isobars.

Let isotope of hydrogen

$= \ _1H^y$

No of neutron

$= 2$

Atomic mass

$=1+2 =3$

An element in +3 oxidation state has the electronic configuration

$(Ar)3d^3$ , Its atomic number is

0%
24
0%
23
0%
22
0%
21

Explanation

If element in +3 oxidation state has its configuration as

$(Ar)3d^3$ . Then in ground state, its electronic configuration will be

$(Ar)3d^44s^2$ .

Hence, the atomic no. is 24.

Hence, Option "B" is the correct answer.

Difference in

$_{17}Cl^{35}$ and

$_{17}Cl^{37}$ is of:

0%
Atomic number
0%
Number of protons
0%
Number of neutrons
0%
Number of electrons

Which of the following properties are different for neutral atoms of isotopes of the same element?

0%
Mass
0%
Atomic number
0%
General chemical reactions
0%
Number of electrons

Isotopes are those which contain:

0%
Same number of neutrons
0%
Same physical properties
0%
Same chemical properties
0%
Different atomic mass

The elements with a atomic number 26 is

0%
A non-metal
0%
Krypton
0%
Iron
0%
Manganese

Out of all the known elements the number of transitional elements is

0%
80
0%
61
0%
43
0%
38

In transition elements, the orbitals partially filled by electrons are

0%
s- orbitals
0%
p- orbitals
0%
d- orbitals
0%
f- orbitals

All those elements belong to f- block whose atomic number are

0%
58 to 71
0%
90 to 103
0%
Both (A) and (B)
0%
None

Those elements whose two outermost orbitals are incompletely filled with electrons are

0%
p - block elements
0%
s - block elements
0%
transitional elements
0%
Both s and p - block elements

How are isotopes of the same element similar?

0%
They have the same particles in the nucleus
0%
They have the same number of neutrons
0%
They are equally abundant in nature
0%
They have the same number of protons
0%
They have the same mass

Which of the following in figure do not represent Bohr’s model of an atom correctly?

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

Tick the appropriate answer.
Which one of the following scientists put forward the theory regarding the extra-nuclear structure of the atom ?

0%
James Chadwick
0%
John Dalton
0%
Rutherford
0%
Niels Bohr

Identify the pairs which are not isotopes?

0%
$^{12}_{6}X\ ^{13}_{6}Y$
0%
$^{35}_{17}X\ ^{37}_{17}Y$
0%
$^{34}_{6}X\ ^{14}_{7}Y$
0%
$^{8}_{4}X\ ^{8}_{2}Y$

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

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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