### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

Which one of the following groupings represents a collection of isoelectronic species? (At. nos. : Cs : 55, Br : 35)
A
N3-, F-, Na+
B
Be, Al3+, Cl-
C
Ca2+, Cs+, Br
D
Na+, Ca2+, Mg2+

## Explanation

Isoelectronic means those species whose electron number are same.

Species Atomic Number Electron Number
N3- 7 7 + 3 = 10
F- 9 9 + 1 = 10
Na+ 11 11 - 1 = 10
Be 4 4
Al3+ 13 13 - 3 = 10
Cl- 17 17 + 1 = 8
Ca2+ 20 20 - 2 = 18
Cs+ 55 55 - 1 = 54
Br 35 35
Mg2+ 12 12 -2 = 10

$\therefore$ N3-, F-, Na+
2

### AIEEE 2003

The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt {l(l + 1)} {h \over {2\pi }}$. This momentum for an s-electron will be given by
A
zero
B
${h \over {2\pi }}$
C
$\sqrt 2 {h \over {2\pi }}$
D
$+ {1 \over 2}{h \over {2\pi }}$

## Explanation

For s-electron l = 0

$\therefore$ $\sqrt {l(l + 1)} {h \over {2\pi }}$

= $\sqrt {0(0 + 1)} {h \over {2\pi }}$

= $\sqrt 0 {h \over {2\pi }}$ = 0 (zero)
3

### AIEEE 2003

The number of d-electrons retained in Fe2+ (At no of Fe = 26) ion is
A
4
B
5
C
6
D
3

## Explanation

Fe (Z = 26) $\to$ e- = 26

Fe2+ (Z = 26) $\to$ e- = 24

Electronic configuration:

Fe (Z = 26) $\to$ 1s2 2s2 2p6 3s2 3p6 3d6 4s2

Fe2+ (Z = 26) $\to$ 1s2 2s2 2p6 3s2 3p6 3d6 4s0

as 4s is the outermost shell of Fe that's why e- will always removed first from 4s, so retained e- in d-shell is 6
4

### AIEEE 2002

In a hydrogen atom, if energy of an electron in ground state is -13.6 eV, then that in the 2nd excited state is
A
-1.51 eV
B
-3.4 eV
C
-6.04 eV
D
-13.6 eV

## Explanation

n = 1 means ground state

n = 2 means 1st excited state

n = 3 means 2nd excited state

Remember: nth excited state means n = n + 1

We know energy of an atom is given by

Energy (E) = -13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$

= -13.6 $\times$ $1 \over 9$ [ $\because$ Z = atomic number = 1 for hydrogen atom]

= -1.51 ev