JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2008

Which one of the following constitutes a group of the isoelectronic species?
A
$C_2^{2 - }$, $O_2^{-}$, CO, NO
B
$NO^{+}$, $C_2^{2 - }$, CN-, $N_2$
C
CN-, $N_2$, $O_2^{2-}$, $C_2^{2 - }$
D
$N_2$, $O_2^{-}$, $NO^{+}$, CO

Explanation

Species No. of electrons
C22- 12+2 = 14
O2- 16+1 = 17
CO 6+8 = 14
NO 7+8 = 15
NO+ 7+8-1 = 14
C22- 12+2 = 14
CN- 6+7+1 = 14
N2 14
CN- 6+7+1 = 14
N2 14
O22- 16+2 = 18
C2 12
N2 14
O2- 16+1 = 17
NO+ 7+8-1 = 14
CO 6+8 = 14

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

AIEEE 2007

Which of the following sets of quantum numbers represents the highest energy of an atom?
A
n = 3, l = 0, m = 0, s = +1/ 2
B
n = 3, l = 1, m = 1, s = +1/ 2
C
n = 3, l = 2, m = 1, s = +1/ 2
D
n = 4, l = 0, m = 0, s = +1/ 2

Explanation

In which quantum number have highest ( n + l ) value, will represent highest energy of an atom.

In option (C), n + l = 3 + 2 = 5 = maximum.
3

AIEEE 2006

Which of the following sets of ions represents a collection of isoelectronic species?
A
N3-, O2-, F-, S2-
B
Li+, Na+, Mg2+, Ca2+
C
K+, Cl-, Ca2+, Sc3+
D
Ba2+, Sr2+, K+, Ca2+

Explanation

Option $(c)$ is correct.

In $\,\,{K^ + }\left( {19} \right)\,\,$ no. of electrons

$= 19 - 1\,\,\,\,\, = 18$

In $\,\,{Cl^ - }\left( {17} \right)\,\,$ no. of electrons

$= 17 + 1\,\,\,\,\, = 18$

In $\,\,{Ca^ 2+ }\left( {20} \right)\,\,$ no. of electrons

$= 20 - 2\,\,\,\,\, = 18$

In $\,\,{Sc^ 3+ }\left( {21} \right)\,\,$ no. of electrons

$= 21 - 3\,\,\,\,\, = 18$
4

AIEEE 2006

Uncertainty in the position of an electron (mass = 9.1 $\times$ 10-31 kg) moving with a velocity 300 ms-1, accurate upto 0.001% will be (h = 6.63 $\times$ 10-34 Js)
A
1.92 $\times$ 10-2 m
B
3.84 $\times$ 10-2 m
C
19.2 $\times$ 10-2 m
D
5.76 $\times$ 10-2 m

Explanation

% error in velocity = ${{\Delta V} \over V} \times 100$

$\therefore$ 0.001 = ${{\Delta V} \over {300}} \times 100$

$\Rightarrow$ $\Delta$V = 3 $\times$ 10-3

According to Heisenberg uncertainty principle,

$\Delta x.m\Delta V \ge {h \over {4\pi }}$

$\Rightarrow$ $\Delta x = {h \over {4\pi m\Delta V}}$

$\Rightarrow$ $\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 3}}}}$

= 1.92 $\times$ 10-2 m