 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields?
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0
A
(D) and (E)
B
(C) and (D)
C
(B) and (C)
D
(A) and (B)

Explanation

As here is no electric and magnetic field so ignore m to calculate the energy of orbital.

As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.

In (A), n + l = 1 + 0 = 1

In (B), n + l = 2 + 0 = 2

In (C), n + l = 2 + 1 = 3

In (D), n + l = 3 + 2 = 5

In (E), n + l = 3 + 2 = 5

So (D) and (E) will be of equal energy.
2

AIEEE 2005

Of the following sets which one does NOT contain isoelectronic species?
A
$BO_3^{3 - }$, $CO_3^{2 - }$, $NO_3^{- }$
B
$SO_3^{2 - }$, $CO_3^{2 - }$, $NO_3^{-}$
C
$CN^{- }$, $N_2$, $C_2^{2 - }$
D
$PO_4^{3 - }$, $SO_4^{2 - }$, $ClO_4^{ - }$

Explanation

In $SO_3^{2 - }$ no of electrons = 16 + 24 + 2 = 42

In $CO_3^{2 - }$ no of electrons = 6 + 24 + 2 = 32

In $NO_3^{-}$ no of electrons = 7 + 24 + 1 = 32

So they are not isoelectronic.
3

AIEEE 2004

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.
4

AIEEE 2004

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 ]