MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Structure Of Atom
Quiz 14
In an atom $$ 2 \mathrm{K}, 8 \mathrm{L}, 8 \mathrm{M}, 2 \mathrm{N} $$ electrons, the number of electrons present.
If $$ \mathrm{m}=0, \mathrm{s}=1 / 2, $$ are:
Report Question
0%
$$6$$
0%
$$2$$
0%
$$8$$
0%
$$16$$
Which of the following statements in relation to the hydrogen atom is correct?
Report Question
0%
3s orbital is lower in energy than 3p orbital.
0%
3p orbital is lower in energy than 3d orbital.
0%
3s and 3p orbital are of lower energy than 3d orbital.
0%
3s, 3p and 3d orbital all have the same energy.
The energy of an electron in the first Bohr orbit for $$He^+$$ ion is $$-54.4eV.$$ Which one of the following is a possible excited state for electron in Bohr orbit of $$He^+$$ ion ?
Report Question
0%
$$-6.04eV$$
0%
$$-6.08eV$$
0%
$$-1.7eV$$
0%
$$+1.36eV$$
The electrons identified by quantum number $$n $$ and $$l$$:
$$(1)n=4,l=1\quad \quad \quad (2)n=4,l=0\\ (3)n=3,l=2\quad \quad \quad (4)n=3,l=1$$
The correct order of increasing energy is:
Report Question
0%
(4) < (2) < (3) < (1)
0%
(2) < (4) < (1) < (3)
0%
(1) < (3) < (2) < (4)
0%
(3) < (4) < (2) < (1)
In the Bohr's model of the hydrogen atom, let r, v and E represent the orbit radius, speed of the electron, respectively. Which of the following relation is proportional to the orbit number $$n$$?
Report Question
0%
$$v\cdot r$$
0%
$$r/E$$
0%
$$r/V$$
0%
$$r\cdot E$$
The energy of an electron in first Bohr orbit of $$H$$ atom is $$-13.6 eV$$. The possible energy value (s) of excited state(s) for the electron in the Bohr orbits of hydrogen is/are:
Report Question
0%
$$- 3.4\ eV $$
0%
$$- 4.2 \ eV $$
0%
$$6.8 \ eV $$
0%
$$+6.8 \ eV$$
Which one is the ground state?
Report Question
0%
0%
0%
0%
The d orbital(s) involved in the formation of the complex(C) will be?
Report Question
0%
$$d_{z^2}$$
0%
$$d_{x^2-y^2}$$ and $$d_{z^2}$$
0%
$$dxy$$ and $$d_{x^2-y^2}$$
0%
$$d_{x^2-y^2}$$
In what region(s) of the spectrum does this series occur? Visible region is from $$360$$ to $$780\ nm$$.
Report Question
0%
Visible only
0%
Infrarted only
0%
Visible and infrared
0%
Visible and ultraviolet
All of the following possess complete d-shells except:
Report Question
0%
$$Ag^+$$
0%
$$Cu^{2+}$$
0%
$$Ga^{3+}$$
0%
$$Zn^{2+}$$
The time period of revolving in the third orbit of $$Li^{2+}$$ ion is $$x$$ second. The time period of revolution in the second orbit of $$He^+$$ ion should be
Report Question
0%
$$x\ s$$
0%
$$\dfrac {3}{2}x\ s$$
0%
$$\dfrac {2}{3}x\ s$$
0%
$$\dfrac {8}{27}x\ s$$
The atom/ion is
Report Question
0%
$$H$$
0%
$$D$$
0%
$$He^{+}$$
0%
$$Li^{+2}$$
Aluminium chloride exists as a dimer because aluminium has______
Report Question
0%
greater $$I.P.$$
0%
incomplete $$p-$$orbital
0%
larger radius
0%
high nuclear charge
Which element is the end product of each natural radioactive series?
Report Question
0%
Sn
0%
Bi
0%
Pb
0%
C
Explanation
Three of the sets, the thorium series, uranium-series, and actinium series, called natural or classical series.
The thorium series begins with thorium-232 and ends with the stable nuclide lead-208.
The uranium-series begins with uranium-238 and ends with lead-206.
The actinium series, named for its first-discovered member, actinium-227, begins with uranium-235 and ends with lead-207.
So, the end product of each natural radioactive series is lead.
The answer is option $$C$$.
$$\alpha$$ -particles can be detected using:
Report Question
0%
Thin aluminum sheet
0%
Barium sulphate
0%
Zinc sulphide screen
0%
Gold foil
Explanation
Rutherford first of all used zinc sulphide (ZnS) as phosphor in the detection of $$\alpha$$ -particles.
Hence, option C is correct.
Tritium undergoes radioactive decay giving:
Report Question
0%
$$\alpha$$ - particles
0%
$$\beta$$ - particles
0%
Neutrons
0%
None of these
Explanation
Tritium $$(_1H^3 \rightarrow _2He^3 + _{-1}e^0)$$ is a $$\beta$$ - emitter.
Bohr model of an atom could not account for :
Report Question
0%
Emission spectrum
0%
Absorption spectrum
0%
Line spectrum of hydrogen
0%
Fine spectrum
Explanation
Bohr's model breaks down when applied to multi-electron atoms. It does not account for sublevel $$(s\ ,\ p\ ,\ d\ ,\ f)$$ orbitals or electron spin. Bohr's model allows only the classical behavior of an electron (orbiting the nucleus at discrete distances from the nucleus). Thus, the concept of orbitals is thrown out.
Bohr's model failed in explaining the line spectrum of multi-electron species. Hence, Bohr's model could not explain the emission spectrum.
Hence, option (A) is correct.
Choose the element which is not radioactive.
Report Question
0%
Cm
0%
No
0%
Mo
0%
Md
Explanation
Radioactive elements are unstable isotopes that release subatomic particles or energy as they decay.
Curium(Cm), Nobelium (No), Mendelevium(Md) are transuranic radioactive chemical elements. But molybdenum (Mo) is not a radioactive element.
Hence, option $$C$$ is correct.
The French physicist Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength $$\lambda$$ of a material particle, its linear momentum p and Planck constant h. $$\lambda=\dfrac {h}{p}=\dfrac{h}{mv}$$. The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have a shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves. These waves differ from the electromagnetic waves as they (i) have lower velocities (ii) have no electrical and magnetic fields and (iii) are not emitted by the particle under consideration. The experimental confirmation of the de Broglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave as proposed by de Broglie.
Using Bohr's theory, the transition, so that the electrons de-Broglie wavelength becomes 3 times of its original value in $$He^{+}$$ ion will be
Report Question
0%
$$2 \rightarrow 6$$
0%
$$2 \rightarrow 4$$
0%
$$1 \rightarrow 4$$
0%
$$1 \rightarrow 6$$
Radioactivity was discovered by
Report Question
0%
Henry Becquerel
0%
Rutherford
0%
J. J. Thomson
0%
Madam Curie
Explanation
It was Henri Becquerel that discovered radioactivity, but it was Marie Curie who coined the term.
Hence, option $$A$$ is correct.
If radium and chlorine combine to from radium chloride the compound is:
Report Question
0%
No longer radioactive
0%
Twice as radioactive as radium
0%
Half as radioactive as radium
0%
As radioactive as radium
Explanation
Radioactivity is a nuclear property.
The formation of a compound from a radioactive element does not involve any change in the structure of nucleus . Thus, radium chloride will be as radioactive as radium itself.
Hence, option $$D$$ is correct.
The spectrum produced from an element is:
Report Question
0%
atomic spectrum
0%
line spectrum
0%
absorption spectrum
0%
any one of the above
Explanation
Hence option D is correct
Two elements X and Y have
(i) X has 17 protons and 18 neutrons
(ii) Y has 17 protons and 20 neutrons Both X and Y are
Report Question
0%
Isobars
0%
Isotopes
0%
Isotones
0%
None of the above
Which of the following is discrete in Bohr's theory ?
Report Question
0%
Potential energy
0%
Kinetic energy
0%
Velocity
0%
Angular momentum
Explanation
Bohr's hypothesis:
1) Electrons revolves round the nucleus in water orbits.
2) Orbit of the electron around the nucleus can take only some special values of radius.
3) The energy of the atom as a definite value in these orbits.
4) In this Orbits, Angular momentum (e) of the electron is integral multiple of the plank's constant h divided by 2n i.e. $$l=n \dfrac{h}{2\pi}$$.
Hence option D is correct
The probability of finding electrons in $$d_{xy}$$ orbital is:
Report Question
0%
along X- and Y-axis
0%
along X- and Z-axis
0%
along Y- and Z-axis
0%
at an angle of $$45^\circ$$ with X-axis
Explanation
The maximum probability of finding an electron in the $$d_{xy}$$ orbital is at an angle of $$45^o$$ from the $$x$$ and $$y$$ axis.
Hence Option D is correct
The figure shows a graph between $$\ln \left|\dfrac {A_n}{A_1}\right|$$ and $$1n |n|$$ where $$A_n$$ is the area enclosed by the $$n^{th}$$ orbit in a hydrogen like atom. The correct curve is
Report Question
0%
$$4$$
0%
$$3$$
0%
$$2$$
0%
$$1$$
Explanation
$$A_n=\pi r_n^2 \Rightarrow \dfrac {A_n}{A_1} =\left(\dfrac {r_n}{r_1}\right)^2 =\left (\dfrac {n}{1}\right)^4 \quad (\because r_n \propto n^2)$$
Taking $$\log_e $$ both side $$\log_e \dfrac {A_n}{A_1}=4\log_e(n)$$
Comparing it with $$y=mx+c$$, graph $$(4)$$ is correct.
Radioactivity is affected by:
Report Question
0%
temperature
0%
pressure
0%
electric and magnetic field
0%
none of these
Explanation
$$ \begin{array}{l} \text { Temperature, pressure, electric and magnetic field } \\ \text { does not affect Radioactivity. } \\ \text { The correct is option D. } \end{array} $$
One or more answers is/are correct.
In electron capture (radioactive process)
Report Question
0%
a neutron is formed
0%
a proton is consumed
0%
$$\gamma$$-ray emission takes place
0%
X-ray emission takes place
Explanation
$$_{1}^{1}P + _{-1}e^{0} \rightarrow _{0}^{1}n + X-rays$$
Hence option (A) and (B) is correct.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Medical Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page