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NEET Chemistry MCQ
Quiz 10
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Q.1
The number of photons of light having wave number x in 1J of energy source is ...
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a) hcx
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b) hc/x
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c)x/hc
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d)1/hcx
Explanation
Energy=nhc/λ here n is number of photonsbut wave number x=1/λE=nhcxE=1 given ∴ n=1/hcxAnswer: (d)
Q.2
Number of nodel planes in px orbital is [ IIT 2000]
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a) one
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b) two
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c)three
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d)zero
Explanation
Number of planar nodes or nodel planes in an orbital=l ∴ number of nodal plane in p x=1 Answer:(a)
Q.3
the polarizing ability of which one of the following is highest? [ DEC 2003]
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a) large -ve ion
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b) small, highly -ve ion
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c) large +ve ion
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d) small, highly +ve ion
Explanation
According bto fajan's rule, small cations with high charge have highest polarising ability Answer: (d)
Q.4
Which of the following molecules has pyramidal shape ? [ DEC 2004]
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a)CO2-
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b) SPO3
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c)PCl3
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d)NO3-
Explanation
PCl3 hybridization is sp3 , pyramidal structure,loan pair occupy one vertices.Answer: (c)
Q.5
Assertion : The radius of second orbit of He+ is equal to that of first orbit of hydrogen. Reason : The radius of an orbit in hydrogen like species is directly proportional to n and inversely proportional to Z.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion
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c) Assertion is correct but Reason is wrong
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d) Both Assertion and Reason are incorrect.
Explanation
Answer:(d)
Q.6
The orbital angular momentum for an electron revolving in an 's' orbit is given by .. [ AIEEE 2003]
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a)h/4π
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b) zero
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c)h/2π
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d)h/(√2π)
Explanation
Since value of 'l'' for 's' orbit is zero Answer:(b)
Q.7
uggest two transitions in the atomic spectrum hydrogen for which the wave number ratio is 108 :7
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a) First Lyman and first Bracket transition
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b) First and second Lyman series transition
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c) First Lyman and first Balmer transition
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d) First Lyman and first Paschen transition
Explanation
Use formula and try and error for for all the options Answer: (d)
Q.8
The threshold energy is given as E0 and radiation of energy E falls on metal, then K.E. is given as
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a) (E- E0)/2
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b) E- E0
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c) E0 - E
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d) E/E0
Explanation
E = Incident energy E0= Threshold energy E = E0 + K.E. K.E. = E - E0 Answer:(b)
Q.9
Which of the following is not linear molecule? [DEC 2003]
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a) BeCl2
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b) CS2
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c)CO2
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d)SO3
Explanation
SO3 is sp2 have planar triangular structureAnswer: (d)
Q.10
Correct energy value order is ... [ Orisa JEE 2004]
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a) ns, np, nd, (n-1)f
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b) ns, np, (n-1)d, (n-2)f
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c)ns, np, (n-1)d, (n-1)f
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d)ns, (n-1)d, np, (n-2)f
Explanation
Apply (n+l) rule Answer:(d)
Q.11
The number of waves in the third orbit of H atom is
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a) 1
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b) 2
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c) 3
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d) 4
Explanation
Number of waves = Number of shell Can be drived as follows Answer:(c)
Q.12
The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove electron from three lowest orbits of hydrogen atom are
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a) 13.6, 6.8, 8.4
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b) 13.6, 10.2, 3.4
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c) 13.6, 27.2, 40.8
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d) 13.6, 3.4, 1.51
Explanation
Energy of orbitalsis gien byformula E1 = -13.6 eV Z= 1 and n = 1, 2 3 Energy ofInonization=E∞ - En Answer:(d)
Q.13
Any f-orbital can accommodate upto
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a) 2 electrons with parallel spin
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b) 6 electrons
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c) 2 electrons with opposite spin
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d) 14 electrons
Explanation
Any orbital have maximum of two electrons with opposite spin. Answer:(c)
Q.14
Probability can be assessed by
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a) ψ2
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b) ψψ*
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c) both above
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d)ψ*
Explanation
Answer: (c)
Q.15
The correct order of increasing energy of atomic orbital is
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a) 5p < 4f < 6s < 5d
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b) 5p < 6s < 4f < 5d
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c) 4f < 5p < 5d < 6s
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d) 5p < 5d < 4f < 6s
Explanation
More the (n + l) value ; more will be the energy Two (n+l) have same values then higher then higher the value Answer:(b)
Q.16
Assertion : Wavelength for first line of any series in hydrogen spectrum is biggest among all other lines of the same series. Reason : Wavelength of spectral line for an electronic transition is inversely related to difference in the energy levels involved in the transition.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion
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c) Assertion is correct but Reason is wrong
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d) Both Assertion and Reason are incorrect.
Explanation
Answer:(a)
Q.17
Consider the following statements,. the d-orbitals have(i) four lobes and two nodes(ii) four lobes and one node(iii)same sign in opposite lobs(iv)opposite sign in the opposite lobs
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a)1 alone
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b) 1 and 3
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c)2 and 3
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d)1 and 4
Explanation
d-orbital have four lobes and two nodes and opposite lobes have the same sign Answer:(b)
Q.18
If r1 is the radius of the first orbit of hydrogen atom, then the radii of second, third and fourth orbitals in terms of r1 are
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a) r12, r13 , r14
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b) 8r1, 27r1, 64r1
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c) 4r1, 9r1, 16r1
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d) 2r1, 6r1, 8r1
Explanation
rn=r1 n2 Answer: (c)
Q.19
For a particle it is found that uncertainty in velocity (in ms-1) is reciprocal of the uncertainty in position (in m). the mass of the particle will have
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a)a fixed value
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b) a variable value
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c)a value not grater than a particular value
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d)a value not less than a particular value
Explanation
Mass is fixedAnswer: (a)
Q.20
The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar state is
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a) 0.53 Å
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b) 1.06 Å
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c) 0.16 Å
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d) 0.265 Å
Explanation
Answer:(c)
Q.21
the maximum number of electrons in a subshell having the same value of spin quantum number is given by
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a) l+2
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b) 2l+1
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c)l(l+1)
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d)√[l(l+1)]
Explanation
Number of orbitla in subshell=2l+1. each of these can cave one electron having same value of spin Answer:(b)
Q.22
Which of the following statement is false?
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a) The optical microscope makes use of the wave property of light, while the electron microscope uses the particle property of electrons
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b) Raman spectrometer uses the wave property of light, while the electron microscope uses the wave property of electrons
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c) Optical microscope uses the wave property of light, electron microscope uses the wave property of electrons
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d) Optical microscope uses the wave property of light, while Raman spectrometer uses the particle property of light
Explanation
Answer: (a)
Q.23
The energies of orbitals of H-atom are in the order
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a) 3s < 3p < 4s < 3d < 4p
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b) 3s < 3p < 3d < 4s < 4p
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c) 3s = 3p = 3d < 4s = 4p
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d) 3s = 3p = 3d < 4s < 4p
Explanation
In the hydrogen atom the energy of the electron depends only on the value of the principle quantum number n as shown by the solution to the Schrodinger Wave Equation and experimentally by the emission spectrum. ∴ 3s = 3p = 3d < 4s = 4p Answer:(c)
Q.24
Assertion : Bohr’s orbits are also called stationary states. Reason : Electrons are stationary in an orbit.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion
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c) Assertion is correct but Reason is wrong
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d) Both Assertion and Reason are incorrect.
Explanation
Answer:(c)
Q.25
For a particles having same kinetic energy, the deBroglie wavelength is
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a)directly proportional to its velocity
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b) inversely proportional to its velocity
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c)independent of velocity and mass
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d)meaningless
Explanation
thus option "a" is correct Answer:(a)
Q.26
The number of d-electrons in Fe2+ ( Z=26) is not equal to that of the
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a) p-electrons in Ne ( at. no.=10)
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b) s-electrons in Mg 9 at no=12)
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c) d-electrons in Fe
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d) p-electrons in Cl- ( at no of Cl=17)
Explanation
26Fe=[Ar] 3d6 4s2 Fe2+=[Ar] 3d6d electron=6, p-electrons in Cl-=12 Answer: (d)
Q.27
The electron configuration of P in H3PO4 is
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a) 1s2 2s2 2p6 3s2 3p3
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b) 1s2 2s2 2p6 3s2
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c) 1s2 2s2 2p6
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d) 1s2 2s2 2p6 3s2 3p6 4s1
Explanation
Oxidation state of P in H3PO415 P=1s2 2s2 2p63s2 3p3P+5=1s2 2s2 2p6 Answer: (c)
Q.28
Which of the following has sp3 hybridisation? [ DEC 2001]
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a) BBr3
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b) XeF4
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c)BCl3
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d)XeO3
Explanation
Answer: (d)
Q.29
In a ground state of Cu+, the number of shells occupied, subshells occupied, filled orbitals and unpaired electrons respectively are
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a) 4, 8, 15, 0
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b) 3, 6, 15, 1
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c)3, 6, 14, 0
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d)4, 7, 14, 2
Explanation
Cu+=1s2 , 2s2 , 2p6, 3s2, 3p63d10 Answer:(c)
Q.30
Which of the following has non-spherical shell of electron?
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a) He
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b) B
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c) Be
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d) Li
Explanation
B has 2p shell which is not spherical Answer: (b)
Q.31
An isotone of 76 Ge32 is
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a) 77Ge32
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b) 77As33
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c) 77Se34
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d) 78Se74
Explanation
Isotone means same number of neutrons Number of neutrons = Mass number = Atomic number Number of neutron in Ge = 76 – 32 = 44 Number of neutron in As = 77 – 33 = 44 Answer: (b)
Q.32
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × 10⁻¹⁸ g cm s–The uncertainty in electron velocity is, (Mass of an electron is 9 × 10⁻²⁸g) [AIPMT (Prelims)-2008]
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a) 1 × 1011 cm s–1
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b) 1 × 109 cm s–1
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c) 1 × 106 cm s–1
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d) 1 × 105 cm s–1
Explanation
Answer:(b)
Q.33
Uncertainty in position of an electron (mass = 9.1 × 10⁻²⁸g) moving with a velocity of 3 × 104 cm/s accurate upto 0.001% will be (Use h/(4π) in uncertainty expression where h = 6.626 × 10⁻²⁷ erg-s)
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a) 5.76 cm
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b) 7.68 cm
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c) 1.93 cm
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d) 3.84 cm
Explanation
Answer:(c)
Q.34
The frequency in sec⁻¹ of a radiation having a wavelength of 1000Å is ....
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c)1 ×105
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d)3.0 ×1015
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a) 3.3 ×10⁻¹⁵
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b) 3.0 ×10⁻¹³
Explanation
Answer:(d)
Q.35
The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within 5.0 × 10⁻²⁶ kg ms–The minimum uncertainty in the measurement of the momentum of the helium atom is
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b) 80 kg ms–1
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c) 50 kg ms–1
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a) 8.0 × 10⁻²⁶ kg ms–1
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d) 5.0 × 10⁻²⁶ kg ms–1
Explanation
Δx × Δp = h/4π ( constant irrespective of any particle) Now it is given "position of both, an electron and a helium atom is known within 1.0 nm" or Δx is same for electron and a helium atom ∴ Δp must be same for electron and helium atom Answer:(d)
Q.36
Given : The mass of electron is 9.11×10⁻³¹ kg. Planck’s constant is 6.626×10⁻³⁴ Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is [AIPMT (Prelims)-2006]
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a) 5.79 × 106 ms–1
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b) 5.79 × 107 ms–1
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c) 5.79 × 108 ms–1
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d) 5.79 × 105 ms–1
Explanation
Δx × Δp = h/4π Answer:(a)
Q.37
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × 10⁻³⁴ Js) [AIPMT (Mains)-2010]
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a) 6.6 × 10⁻³² m
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b) 6.6 × 10⁻³⁴ m
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c) 1.0 × 10⁻³⁵ m
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d) 1.0 × 10⁻³² m
Explanation
Use formula λ = h/mv Answer:(c)
Q.38
The value of Planck's constant is 6.63 × 10⁻³⁴ Js. The speed of light is 3 × 1017 nm s–Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s–1? [NEET-2013]
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a) 25
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b) 50
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c) 75
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d) 10
Explanation
Use formula λ = C/ν here ν if frequency As velocity oflightis given in nm/s and wave length asked is also in nm no need to convert in m Answer:(b)
Q.39
The de-Broglie wavelength of an electron travelling with 10% of velocity of light is equal to (mass of electron =9.1× 10⁻³¹ kg)
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a) 242.4 pm
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b) 24.2 pm
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c) 2.42 pm
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d) 2.424 pm
Explanation
Velocity of electron is 10%of velocity of light = 3×107 Answer:(b)
Q.40
Uncertainty in the momentum of an electron is 1.0×10⁻⁵ kg m-the uncertainty in its position will be ( h=6.62 ×10⁻³⁴ kg m2 s-1) ... [ AFMC 1998]
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a) 1.05 ×10⁻²⁸ m
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b) 1.05 ×10⁻²⁶ m
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c) 5.27 ×10⁻³⁰ m
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d)5.25 ×10⁻²⁸ m
Explanation
use formula Answer:(c)
Q.41
The energy absorbed by each molecule (A2) of a substance is 4.4 × 10⁻¹⁹J and bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be [AIPMT (Prelims)-2009]
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a) 2.2 × 10⁻¹⁹ J
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b) 2.0 × 10⁻¹⁹ J
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c) 4.0 × 10⁻²⁰J
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d) 2.0 × 10⁻²⁰J
Explanation
KE = Energy observed by molecule – Energy required to break one bond Now A2 have two atoms Thus Energy per atom= KE/2 Answer:(d)
Q.42
The uncertainty in momentum of an electron is 1 × 10⁻⁵ kg-m/s. The uncertainty in its position will be (h = 6.62 × 10⁻³⁴ kg-m2/s)
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a) 5.27 × 10⁻³⁰ m
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b) 1.05 × 10⁻²⁶ m
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c) 1.05 × 10⁻²⁸ m
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d) 5.25 × 10⁻²⁸ m
Explanation
Use equation Δx × Δp = h/4π Answer:(a)
Q.43
Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck's constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 108 ms–1) [AIPMT-2014]
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a) 6.67 × 1015
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b) 6.67 × 1011
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c) 4.42 × 10⁻¹⁵
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d) 4.42 × 10⁻¹⁸
Explanation
Use formula E = hc/λ Answer:(d)
Q.44
if the planck's constant h=6.6×10⁻³⁴ Js, the de Broglie wave length of a particle having momentum of 3.3×10⁻²⁴ kg m s-1 will be ... [ Pb CET 1998]
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a) 0.002 Ao
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b)0.5 Ao
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c)2.0 Ao
0%
d)500 Ao
Explanation
Answer:(c)
Q.45
Two particles A and B are in motion. if the wavelength associated with the particle A is 5.0 ×10⁻⁸ m, the wavelength of particle B having momentum half of A is
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a) 2.5×10⁻⁸ m
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b) 1.25×10⁻⁸ m
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c) 1.0×10⁻⁷ m
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d) 1.0×10⁻⁸ m
Explanation
λ ∝ (1/p) Answer: (c)
Q.46
The energy of a photon is given as ΔE/atom=3.03×10⁻¹⁹ J atom-thus the wavelength (λ) of photon is .. [CBSE PMT 2000]h=6.63 ×10⁻³⁴ Js, c=3.0 ×108 m/s
0%
a) 65.5 nm
0%
b) 656 nm
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c)0.656 nm
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d)6.56 nm
Explanation
E=hc/λAnswer: (b)
Q.47
The uncertainty in the position of an electron ( mass 9.1×10⁻²⁸ g) moving with velocity of 3.0×104 cm s-1 accurate up to 0.11% will be ... [ CBSE PMT 1995]
0%
a)1.92 cm
0%
b) 7.68 cm
0%
c)0.175 cm
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d)3.84 cm
Explanation
Here Δv=0.11% of 3.0×104 cm s-1=3.3 cm s-1Answer: (c)
Q.48
the position of both, electron and helium atom is known with in 0.1nm. Further the momentum of electron is known with in 5.0×10⁻²⁶ kg ms-1The minimum uncertainty in the measurement of momentum of helium atom is
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a) 50 kg m s-1
0%
c) 80 kg m s-1
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b) 5.0 ×10⁻²⁶ kg m s-1
0%
d) 80.0 ×10⁻²⁶ kg m s-1
Explanation
Δxe Δpe=ΔxHe ΔpHe As Δxe=ΔxHeΔpe=ΔpHe Answer: (b)
Q.49
The wavelength associated with a ball of 200 g and moving with a speed of 5 m/hour is of the order of
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a) 10–10 m
0%
b) 10–20 m
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c) 10–30 m
0%
d) 10–40 m
Explanation
v = 5 m/hour =5/3600 m/s m =200 g = 0.2 kg h =6.6× 10⁻³⁴ J-s Use forula λ = h/p, p=mv Answer:(c)
Q.50
If the energy difference between the ground state and excited state of an atom is 4.4 × 10⁻¹⁹ J. The wavelength of photon required to produce this transition is
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a) 4.5 × 10⁻⁷ m
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b) 4.5 × 10⁻⁷ nm
0%
c) 4.5 × 10⁻⁷ Å
0%
d) 4.5 × 10⁻⁷ cm
Explanation
ΔE = [Excited state - ground state] = 4.4 × 10⁻¹⁹ J According to Plank's quantum theory energy od single photon Answer:(a)
0 h : 0 m : 1 s
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