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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
Which one of the following sets of ions represents the collection of isoelectronic species? (Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
A
K+, Cl-, Mg2+, Sc3+
B
Na+, Ca2+, Sc3+, F-
C
K+, Ca2+, Sc3+, Cl-
D
Na+, Mg2+, Al3+, Cl-

Explanation

Isoelectronic means those species whose no of electrons are same.

Ions Atomic No No of Electron
K+ 19 19-1=18
Ca+2 20 20-2=18
Sc+3 21 21-3=18
Cl- 17 17+1=18


So, option $$(C)$$ is correct.
2

AIEEE 2004

MCQ (Single Correct Answer)
The wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 $$\times$$ 107 m-1)
A
406 nm
B
192 nm
C
91 nm
D
9.1 $$\times$$ 10-8 nm

Explanation

We know Rydberg formula,

$${1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$

[ for hydrogen atom Z = 1 ]

     = $$1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$$

$$ \therefore $$ $$\lambda $$ = $${1 \over {1.097 \times {{10}^7}}}$$

= 9.11 $$ \times $$10-8 m

= 91.1 $$ \times $$10-9 m

= 91.1 nm

[ as 1 nm = 10-9 m ]
3

AIEEE 2004

MCQ (Single Correct Answer)
Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are respectively
A
16 and 4
B
12 and 5
C
12 and 4
D
16 and 5

Explanation

Electronic configuration of Cr (Z = 24) =

1s2 2s2 2p6 3s2 3p6 4s1 3d5

For 2p6 and 3p6 , l = 1. Here in 2p and 3p orbital total 6 + 6 = 12 electrons present.

For 3d5 , l = 2. Here in 3d orbital total 5 electrons present.
4

AIEEE 2004

MCQ (Single Correct Answer)
Which of the following sets of quantum numbers is correct for an electron in 4f orbital?
A
n = 4, l = 3, m = +1, s = + 1/2
B
n = 4, l = 4, m = -4, s = - 1/2
C
n = 4, l = 3, m = +4, s = + 1/2
D
n = 3, l = 2, m = -2, s = + 1/2

Explanation

For 4f orbital,

n = 4 and l = 3

Note : ( for s orbital l = 0, for p orbital l = 1, for d orbital l = 2, for f orbital l = 3)

Value of m = + l to - l

Here m = +3 to -3

Value of s = $$ + {1 \over 2}$$ or $$ - {1 \over 2}$$

Questions Asked from Structure of Atom

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