### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2008

The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
A
8.51 × 105 J mol−1
B
6.56 × 105 J mol−1
C
7.56 × 105 J mol−1
D
9.84 × 105 J mol−1

## Explanation

Note :

1 eV/atom = 96.485 $\times$ 103 J/mol

$\therefore$ 13.6 eV/atom = 13.6$\times$ 96.485 $\times$ 103 J/mol = 1.312 × 106 J mol−1

Energy required to excite the electron from n1 to n2 is

$\Delta E = 13.6 \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

= 1.312 × 106 × 1$\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

= 1.312 × ${3 \over 4}$ × 106

= 9.84 × 105 J mol−1
2

### 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.
3

### 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.
4

### 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

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