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NEET Chemistry MCQ
Quiz 5
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Q.1
In an atom, which has 2K, 8L, 18M and 2N electrons in the ground state. The total number of electrons having magnetic quantum number, m = 0 is
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a) 6
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b) 10
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c) 7
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d) 14
Explanation
2K electron i.e. 2 electron in n =1 ∴ values of l = 0 and m = 0 for m= 0, 2 electrons 8K electron .i.e n = 2 fully filled ∴ values of l = 0,1 for l =0 , m = 0 and l= 1 , m = +1 ,0 , -1 18M electron .i.e n = 3 fully filled ∴ values of l = 0,1, 2 for l =0 , m = 0 and l= 1 , m = +1 ,0 , -1 and l =2 m =+2, +1, 0, -1, -2 2N electron i.e n =4 only two electrons ∴ values of l = 0,1,2,3 have subhsell 4s, thus m =0 Now total m= 0 appear above =7 each m=0 have two electrons = 7×2 = 14 electrons Answer:(d)
Q.2
Assuming the velocity to be same, the wavelength of the waves associated with which of the following particles would be maximum?
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a) An electron
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b) A proton
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c) An α-particle
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d) A deutro
Explanation
λ= h/p, and p = mv As mass of evelctron is smallest wavelength is maximum Answer:(a)
Q.3
Assertion : Energy of electron is taken negative. Reason : Energy of electron at infinity is zero.
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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.4
For the transition from n = 2 → n = 1, which of the following will produce shortest wavelength?
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a) H atom
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b) D atom
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c) He+ ion
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d) Li2+ ion
Explanation
λ ∝ 1/Z H = z = 1, D = z = 1 He+ = z = 2 Li2+ = z = 3 ∴ Li2+ have shorter wavelength Answer:(d)
Q.5
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?
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a) 2.389 × 10⁻¹² ergs
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b) 0.239 × 10⁻¹⁰ ergs
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c) 2.15 × 10⁻¹¹ ergs
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d) 0.1936 × 10⁻¹⁰ ergs
Explanation
Energy of electron = 1312/n2 ( for Hydrogen atom) Answer:(d)
Q.6
If threshold wavelength (λo) for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is
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a) 6 × 10⁻¹⁰ J
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b) 1.2 × 10⁻¹⁸ J
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c) 3 × 10⁻¹⁹ J
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d) 6 × 10⁻¹⁹ J
Explanation
Work functionφ = hc/λo Answer:(d)
Q.7
Which of the following orbital is with the four lobes present on the axis?
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a) dz2
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b) dxy
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c) dyz
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d) dx2-y2
Explanation
Answer:(d)
Q.8
Assertion : Isotopes of an element have almost similar chemical properties. Reason : Isotopes have same electronic configuration.
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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.9
En = –313.6/n2 kcal/mole. If the value of E = –34.84 kcal/mole, to which value does ‘n’ correspond?
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a) 4
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b) 3
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c) 2
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d) 1
Explanation
Substitute E = 34.84 kcal/mole in formula given to find value of n Answer:(b)
Q.10
on Q214) Consider the following sets of quantum number Which of the following sets of quantum number is not possible?
Sr
n
l
m
s
(i)
3
0
0
+1/2
(ii)
2
2
1
+1/2
(iii)
4
3
-2
–1/2
(iv)
1
0
-1
-1/2
(v)
3
2
3
+1/2
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a) (i), (ii), (iii) and (iv)
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b) (ii), (iv) and(v)
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c) (i) and (iii)
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d) (ii), (iii) and (iv)
Explanation
In (ii) n = l not possible In (iv) , n = 1, l = 0 but m =-1 not possible In (v), n =3,l = 2 but m= 3 not possible Answer:(b)
Q.11
How many 3d electrons can have spin quantum number (-1/2)
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a) 5
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b) 7
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c) 8
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d) 10
Explanation
for d, l = 2 and m = +l to -l or m = +2, +1, 0, -1, -2 total 5 suborbvitals each sub-orbital can have two electrons ∴ number of electrons = 10 number half with+1/2 spin = 5 number and half with -1/2 spin = 5 number Answer:(a)
Q.12
If each orbital can hold a maximum of 3 electrons. The number of elements in 2nd period of periodic table
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a) 27
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b) 9
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c) 18
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d) 12
Explanation
For 2nd period electronic configuration = 2s2, 2p6 If each orbital have 3e– then electronic configuration = 2s3, 2px3, 2py3, 2pz3 Total 12 eelectrons will present, and one electron increases when we go from left to right across periodic table Answer:(d)
Q.13
The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)
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a) 4.77
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b) 1.06
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c) 0.13
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d) 2.12
Explanation
Ground state n =1 ,First excited state n = 2 ro = 0.530 Å , n =2 and Z= 1 Answer:(d)
Q.14
Who modified Bohr’s theory by introducing elliptical orbits for electron path?
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a) Rutherford
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b) Thomson
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c) Hund
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d) Sommerfield
Explanation
Answer:(d)
Q.15
The ion that is isoelectronic with CO is
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a) CN–
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b) N2+
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c) O2–
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d) N2–
Explanation
Isoelectronic means same number of electrons CO = Number of electrons = 14 CN– = 6 + 7 + 1 = 14 Answer:(a)
Q.16
Which of the following has maximum number of unpaired d-electrons?
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a) N3+
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b) Fe2+
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c) Zn+
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d) Cu+
Explanation
N3+ = 4 = 1s2, 2s2 Zero unpaired Fe2+ = 24 = 1s2, 2s2, 2p6, 3s2, 3p6, 3d6, Four unpaired Zn+ = 29 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10 1 unpaired Cu+ = 28 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 Zero unpaired Answer:(b)
Q.17
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at
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a) (5R/36) cm-1
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b) (3R/4) cm-1
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c) (7R/144) cm-1
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d) (9R/400) cm-1
Explanation
1st line in the Balmer series means n = 2, n = 3 Answer:(a)
Q.18
Which of the following is not among shortcomings of Bohr’s model?
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a) Bohr theory could not account for the fine lines in the atomic spectrum
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b) Bohr theory was unable to account for the splitting of the spectral lines in the presence of magnetic field
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c) Bohr theory failed for He atom
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d) It did not give information about energy level
Explanation
Bohr's model explain the energy level i.e. Energy of electron in each orbital is quantized. Answer:(d)
Q.19
According to Bohr’s theory angular momentum of an electron in 6th orbit is
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a) 2.5h/π
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b) 6h/π
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c) 3h/π
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d) 2.5h/2π
Explanation
Angular momentum= nh/2π n = 6 Answer:(c)
Q.20
Assertion : In third energy level there is no f-subshell. Reason : For n = 3, the possible values of l are 0, 1, 2 and for f-subshell l = 3.
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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.21
For which one of the following set of quantum numbers an electron will have the highest energy?
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a) 3, 2, 1, 1/2
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b) 4, 2, 1, 1/2
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c) 4, 1, 0, -1/2
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d) 5, 0, 0, 1/2
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.22
In the ground state, an element has 13 electrons in its M-shell. The element is
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a) Manganese
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b) Cobalt
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c) Nickel
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d) Iron
Explanation
M shell means 3rd orbit Mn = 25 = 1s2 , 2s2 , 2p6 , 3s2 , 3p6 , 4s2 , 3d5 3s, 3p and 3d have total 18 electrons Answer:(a)
Q.23
Assertion : Orbital angular momentum of (1s, 2s, 3s etc.) all s electrons is same. Reason : Orbital angular momentum depends on orientation of orbitals.
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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
Angular momentum, depends on the angular (azimuthal) quantum number (L) not the magnetic quantum number(m) therefore statement of Reason is incorrect Answer:(c)
Q.24
For principal quantum number n = 5, the total number of orbitals having l = 3 is
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a) 7
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b) 14
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c) 9
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d) 18
Explanation
For l = 3 m = –3, –2, –1, 0, +1, +2, +3 i.e., 7 orbitals are present Answer:(a)
Q.25
The following quantum numbers are possible for how many orbitals : n = 3, l = 2, m = +2?
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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
As the value of m = + 2 i.e. one value Therefore one orbital is represented Answer:(a)
Q.26
A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly [AIPMT (Prelims)-2008]
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a) 20 × 1014 Hz
0%
b) 10 × 1014 Hz
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c) 5 × 1014 Hz
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d) 1 × 1014 Hz
Explanation
Answer:(c)
Q.27
Which transition of Li2+ is associated with same energy change as n = 6 to n = 4 transition in He+?
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a) n = 3 to n = 1
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b) n = 8 to n = 6
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c) n = 9 to n = 6
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d) n = 2 to n = 1
Explanation
Energy of nth orbitalis given by For He+ , Z = 2 and n=6 ,Have same energy of Li2+ let n = n1 For He+ , Z = 2 and n=4 ,Have same energy of Li2+ let n = n2 Answer:(c)
Q.28
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon? [AIPMT (Mains)-2011]
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a) n = 6 to n = 5
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b) n = 5 to n = 3
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c) n = 6 to n = 1
0%
d) n = 5 to n = 4
Explanation
Because (E2– E1) > (E3 – E2) > (E4 – E3) > (E5– E4) > (E6– E5) As the difference is of one energy levels ∴ (E6– E5) have less energy {Alternatively value of Δ E [difference between two successive energy level decreases] as the distance from the nucleus increases.} Answer:(a)
Q.29
The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is [AIPMT (Prelims)-2012]
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a) 5, 0, 0, 1/2
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b) 5, 1, 0, 1/2
0%
c) 5, 1, 1, 1/2
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d) 6, 0, 0, 1/2
Explanation
Rb = 37 = [Ar] 4s2, 3d10, 4p6 5s1 5s1 is last electron For s orbitals , l=0, m=0, Answer:(a)
Q.30
In a hydrogen atom, if the energy of electron in the ground state is –x eV., then that in the 2nd excited state of He+ is
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a) –x eV
0%
b) -4x/9 eV
0%
c) +2x eV
0%
d) -9x/4 eV
Explanation
Second excited state n=3 and Z = 2 Now E1 = -x eV Answer:(b)
Q.31
The orientation of an atomic orbital is governed by [AIPMT (Prelims)-2006]
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a) Azimuthal quantum number
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b) Spin quantum number
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c) Magnetic quantum number
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d) Principal quantum number
Explanation
Answer:(c)
Q.32
The orbital angular momentum of a p-electron is given as [AIPMT (Mains)-2012]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Angular momentum = l (l+1) ħ For p orbit l= 1 Answer:(a)
Q.33
In hydrogen atom, energy of first excited state is –3.4 eV. Then find out KE of same orbit of hydrogen atom
0%
a) +3.4 eV
0%
b) +6.8 eV
0%
c) –13.6 eV
0%
d) +13.6 eV
Explanation
K= −TotalEnergy(E)=3.4eV Negative kinetic energy equals half the potential energy (−K = ½U). Potential energy equals twice the total energy (U = 2E). Total energy equals negative kinetic energy (E = −K). Twice the kinetic energy plus the potential energy equals zero (2K + U = 0). This is a key relationship for a larger problem in orbital mechanics known as the virial theorem. Answer:(a)
Q.34
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = –1 [NEET-2013]
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a) 6
0%
b) 4
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c) 2
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d) 10
Explanation
Value of m = –1 represents one orbital. Therefore maximum number of electrons will be two Answer:(c)
Q.35
With which of the following configuration an atom has the lowest ionization enthalpy ? [AIPMT (Prelims)-2007]
0%
a) 1s2 2s2 2p6
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b) 1s2 2s2 2p5
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c) 1s2 2s2 2p3
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d) 1s2 2s2 2p5 3s1
Explanation
Answer:(d)
Q.36
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?
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a) 3 → 1
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b) 5 → 2
0%
c) 2 → 5
0%
d) 3 → 2
Explanation
Third line means third excited state i.e. n1= 2 Balmer series (visible region) n2 = 5 Third line ∴ Third line will appear when electron comes from 5th energy level to 2nd level. Answer:(b)
Q.37
The angular momentum of electron in 'd' orbital is equal to ... [ AIMPMT 2015]
0%
a) 0ħ
0%
b) √6 ħ
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c) √2 ħ
0%
d) 2√3 ħ
Explanation
Formula for angular momnetum = √l(l+1)× ħ For p orbital l = 2 Answer:(b)
Q.38
The number of photons of light of wavelength 7000Å equivalent to 1 J are
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b) 3.52 × 1018
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c) 50,000
0%
d) 10,0000
0%
a) 3.52 × 10⁻¹⁸
Explanation
According to Plank's quantum theory energy od single photon Answer:(b)
Q.39
Which of the following contains maximum number of electrons in the antibonding molecular orbitals
0%
a) O2
0%
b) O22-
0%
c)O2-
0%
d)O2+
Explanation
Number of electrons in the antibonding MO=4. We will have 6 in O22- 5 in O2- and 3 in O2+Answer: (b)
Q.40
Number of spectral lines in Balmer series when an electron return from 7th orbit to 1st orbit of hydrogen atom
0%
a) 5
0%
b) 6
0%
c) 21
0%
d) 15
Explanation
As only visible lines have to be calculated i.e. Balmer lines n = 7 ∴ Visible lines when ground state = 2 Possible arrangements 7 → 2, 6 → 2, 5 → 2, 4 → 2, 3 → 2 Answer:(a)
Q.41
Which of the following transition will emit maximum energy in hydrogen atom?
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a) 4f → 2s
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b) 4d → 2p
0%
c) 4p → 2s
0%
d) All have same energy
Explanation
Transition energy depends upon the shell number i.e. value of principle quantum number 'n' in all the case transition is between 4th energy level to 2nd level ∴ All have same energy Answer:(d)
Q.42
The set of quantum numbers not applicable to an electron
0%
a) 1, 1, 1, 1/2
0%
b) 1, 0, 0, 1/2
0%
c) 1, 0, 0 , -1/2
0%
d) 2, 0, 0, 1/2
Explanation
The value of l can never be equal to n If n = 1, l = 1 which is not possible Answer:(a)
Q.43
The frequency of a wave is 6 × 1015 s–Its wave number would be
0%
a) 105 cm–1
0%
b) 2 × 107 m–1
0%
c) 2 × 107 cm–1
0%
d) 2 × 105 cm–1
Explanation
Wave number = 1/wave length λ = c/ν (ν is frequency) Wave number = ν /c Wave number = 6 × 1015 / 3×108 Wave number =2 × 107 m–1 Answer:(b)
Q.44
Atomic number and mass number of an element M are 25 and 52 respectively. The number of electrons, protons and neutrons in M2+ ion are respectively
0%
a) 25, 25 and 27
0%
b) 25, 27 and 25
0%
c) 27, 25 and 27
0%
d) 23, 25 and 27
Explanation
Atomic number = 25 Mass number = 52 Number of proton = Atomic number = 25 Number of neutron = (Mass number – Atomic number) 52 – 25 = 27 Number of e– = Number of protons, but M2+ ion means 2e– are removed Number of e– = Number of protons – 2 = 25 – 2 = 23 Answer:(d)
Q.45
The maximum number of electrons in an atom which can have n = 4 is
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a) 4
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b) 8
0%
c) 16
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d) 32
Explanation
Number of electrons = 2n2 (n = shell number) For 4th shell = 2 × (4)2 = 32 electrons Answer:(d)
Q.46
Maximum number of electrons in a subshell with l = 3 and n = 4 is
0%
a) 10
0%
b) 12
0%
c) 14
0%
d) 16
Explanation
n =4, l = 3 means 4f for l = 3, m = –3, –2, –1, 0, 1, 2, 3 = 7 orbital Therefore, maximum 14 electrons are present. Answer:(c)
Q.47
Assertion : Zn(II) salts are diamagnetic. Reason : Zn2+ ion has one unpaired electron.
0%
a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
0%
b) Both are correct but Reason is not explanation of Assertion
0%
c) Assertion is correct but Reason is wrong
0%
d) Both Assertion and Reason are incorrect.
Explanation
Answer:(c)
Q.48
Which of the following has maximum number of unpaired electrons?
0%
a) Mg2+
0%
b) Ti3+
0%
c) Fe2+
0%
d) Mn2+
Explanation
Mg2+ = 10 = 1s2, 2s2, 2p6 number of unpaired electrons = 0 Ti3+ = 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d1 number of unpaired electrons =1 Fe2+ = 24 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d6 number of unpaired electrons =4 Mn2+ = 23 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d5 number of unpaired electrons =5 maximum number Answer:(d)
Q.49
A certain metal when irradiated with light (ν = 3.2 × 1016 Hz) emits photo electrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light (ν = 2.0 × 1016 Hz). Calculate ν0 ofelectron?
0%
a) 1.2 × 1014 Hz
0%
b) 8 × 1015 Hz
0%
c) 1.2 × 1016 Hz
0%
d) 4 × 1012 Hz
Explanation
K. E. = h(ν – ν0 ) K.E. of photoelectrons when ν = 3.2 × 1016 Hz K.E1= h (3.2 × 1016 – ν0) K E. of photoelectron when ν = 2.0 × 1016 Hz KE2 = h(2.0 × 1016 – ν0) According to question KE1 = 2KE2 ∴ h(3.2 × 1016 – ν0) = 2h(2.0 × 1016 – ν0) 3.2 × 1016 – ν0 = 4.0 × 1016 – 2ν0 ν0= 4.0 ×1016 – 3.2 × 1016 = 0.8 × 1016 Hz = 8 × 1015 Hz Answer:(b)
Q.50
The principal and azimuthal quantum number of electrons in 4f orbitals are
0%
a) 4, 2
0%
b) 4, 4
0%
c) 4, 3
0%
d) 3, 4
Explanation
for s l= 0 for p, l = 1 for d. l = 2 for f, l = 3 Thus for 4f electron n =4 and l=3 Answer:(c)
0 h : 0 m : 1 s
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