Structure of Atom Practice Questions - DPP for JEE

 Page 1


1. (c) The species CO, NO
+
, CN
–
 and C
2
? 2–
 contain 14 electrons each.
2. (d) NaCl : No. of e
– 
in Na
+
= At. No. of Na–1
= 11 – 1 = 10
No. e
–
 in Cl
–
= At. No. of Cl + 1
= 17 + 1 = 18
CsF  : No. of e
–
 in Cs
+
= 55 – 1 = 54
No. of e
–
 in F
–
= 9 + 1 = 10
NaI : No. of e
–
 in Na
+
= 11 – 1 = 10
No. of e
–
 in I
–
= 53 + 1 = 54
K
2
S : No. of e
–
 in K
+
= 19 – 1 = 18
No. of e
–
 in S
2–
= 16 + 2 = 18
3. (a) For He
+
,
For H,
For same frequency,
= 
?
? n
1
 = 1 & n
2
 = 2
4. (c) Series limit is the last line of the series, i.e. n
2
 = 8.
Page 2


1. (c) The species CO, NO
+
, CN
–
 and C
2
? 2–
 contain 14 electrons each.
2. (d) NaCl : No. of e
– 
in Na
+
= At. No. of Na–1
= 11 – 1 = 10
No. e
–
 in Cl
–
= At. No. of Cl + 1
= 17 + 1 = 18
CsF  : No. of e
–
 in Cs
+
= 55 – 1 = 54
No. of e
–
 in F
–
= 9 + 1 = 10
NaI : No. of e
–
 in Na
+
= 11 – 1 = 10
No. of e
–
 in I
–
= 53 + 1 = 54
K
2
S : No. of e
–
 in K
+
= 19 – 1 = 18
No. of e
–
 in S
2–
= 16 + 2 = 18
3. (a) For He
+
,
For H,
For same frequency,
= 
?
? n
1
 = 1 & n
2
 = 2
4. (c) Series limit is the last line of the series, i.e. n
2
 = 8.
  The line belongs to Paschen series.
5. (d) de Broglie wavelength, 
KE 
6. (c) Fe (III) = [Ar] unpaired electrons = 5;
Magnetic moment = ;
Ratio = 
Co(II) = [Ar] unpaired electrons = 3;
Magnetic moment =
Ratio = 
7. (b)
Page 3


1. (c) The species CO, NO
+
, CN
–
 and C
2
? 2–
 contain 14 electrons each.
2. (d) NaCl : No. of e
– 
in Na
+
= At. No. of Na–1
= 11 – 1 = 10
No. e
–
 in Cl
–
= At. No. of Cl + 1
= 17 + 1 = 18
CsF  : No. of e
–
 in Cs
+
= 55 – 1 = 54
No. of e
–
 in F
–
= 9 + 1 = 10
NaI : No. of e
–
 in Na
+
= 11 – 1 = 10
No. of e
–
 in I
–
= 53 + 1 = 54
K
2
S : No. of e
–
 in K
+
= 19 – 1 = 18
No. of e
–
 in S
2–
= 16 + 2 = 18
3. (a) For He
+
,
For H,
For same frequency,
= 
?
? n
1
 = 1 & n
2
 = 2
4. (c) Series limit is the last line of the series, i.e. n
2
 = 8.
  The line belongs to Paschen series.
5. (d) de Broglie wavelength, 
KE 
6. (c) Fe (III) = [Ar] unpaired electrons = 5;
Magnetic moment = ;
Ratio = 
Co(II) = [Ar] unpaired electrons = 3;
Magnetic moment =
Ratio = 
7. (b)
8. (d) 
Hence, 
9. (a) We know 
since ?p = ?x (given)
? 
or m?v    [ ? ?p= m?v]
or 
or 
10. (b) I. E = ...(i)
or ...(ii)
Given I
1
 = – 19.6 × 10
–18
 , Z
1
 = 2,
n
1
 = 1 , Z
2
 = 3 and n
2
 = 1
Substituting these values in equation (ii).
– 
or 
= – 4.41 × 10
–17
 J/atom
11. (c) As per Bohr’s postulate,
mvr = 
Page 4


1. (c) The species CO, NO
+
, CN
–
 and C
2
? 2–
 contain 14 electrons each.
2. (d) NaCl : No. of e
– 
in Na
+
= At. No. of Na–1
= 11 – 1 = 10
No. e
–
 in Cl
–
= At. No. of Cl + 1
= 17 + 1 = 18
CsF  : No. of e
–
 in Cs
+
= 55 – 1 = 54
No. of e
–
 in F
–
= 9 + 1 = 10
NaI : No. of e
–
 in Na
+
= 11 – 1 = 10
No. of e
–
 in I
–
= 53 + 1 = 54
K
2
S : No. of e
–
 in K
+
= 19 – 1 = 18
No. of e
–
 in S
2–
= 16 + 2 = 18
3. (a) For He
+
,
For H,
For same frequency,
= 
?
? n
1
 = 1 & n
2
 = 2
4. (c) Series limit is the last line of the series, i.e. n
2
 = 8.
  The line belongs to Paschen series.
5. (d) de Broglie wavelength, 
KE 
6. (c) Fe (III) = [Ar] unpaired electrons = 5;
Magnetic moment = ;
Ratio = 
Co(II) = [Ar] unpaired electrons = 3;
Magnetic moment =
Ratio = 
7. (b)
8. (d) 
Hence, 
9. (a) We know 
since ?p = ?x (given)
? 
or m?v    [ ? ?p= m?v]
or 
or 
10. (b) I. E = ...(i)
or ...(ii)
Given I
1
 = – 19.6 × 10
–18
 , Z
1
 = 2,
n
1
 = 1 , Z
2
 = 3 and n
2
 = 1
Substituting these values in equation (ii).
– 
or 
= – 4.41 × 10
–17
 J/atom
11. (c) As per Bohr’s postulate,
mvr = 
So, 
KE = 
So,  KE = 
Since, 
So, for 2
nd
 Bohr orbit
KE = 
KE = 
12. (c) Not more than two electrons can be present in same atomic orbital.
This is Pauli's exclusion principle.
13. (a) 2
nd
 excited state will be the 3
rd
 energy level.
 or  
14. (a)
In Balmer series n
1
 = 2 & n
2
 = 3, 4, 5.... Last line of the spectrum is
called series limit.
Limiting line is the line of shortest wavelength and high energy when n
2
= 8
 
15. (c) r
n
 = a
0
n
2
Page 5


1. (c) The species CO, NO
+
, CN
–
 and C
2
? 2–
 contain 14 electrons each.
2. (d) NaCl : No. of e
– 
in Na
+
= At. No. of Na–1
= 11 – 1 = 10
No. e
–
 in Cl
–
= At. No. of Cl + 1
= 17 + 1 = 18
CsF  : No. of e
–
 in Cs
+
= 55 – 1 = 54
No. of e
–
 in F
–
= 9 + 1 = 10
NaI : No. of e
–
 in Na
+
= 11 – 1 = 10
No. of e
–
 in I
–
= 53 + 1 = 54
K
2
S : No. of e
–
 in K
+
= 19 – 1 = 18
No. of e
–
 in S
2–
= 16 + 2 = 18
3. (a) For He
+
,
For H,
For same frequency,
= 
?
? n
1
 = 1 & n
2
 = 2
4. (c) Series limit is the last line of the series, i.e. n
2
 = 8.
  The line belongs to Paschen series.
5. (d) de Broglie wavelength, 
KE 
6. (c) Fe (III) = [Ar] unpaired electrons = 5;
Magnetic moment = ;
Ratio = 
Co(II) = [Ar] unpaired electrons = 3;
Magnetic moment =
Ratio = 
7. (b)
8. (d) 
Hence, 
9. (a) We know 
since ?p = ?x (given)
? 
or m?v    [ ? ?p= m?v]
or 
or 
10. (b) I. E = ...(i)
or ...(ii)
Given I
1
 = – 19.6 × 10
–18
 , Z
1
 = 2,
n
1
 = 1 , Z
2
 = 3 and n
2
 = 1
Substituting these values in equation (ii).
– 
or 
= – 4.41 × 10
–17
 J/atom
11. (c) As per Bohr’s postulate,
mvr = 
So, 
KE = 
So,  KE = 
Since, 
So, for 2
nd
 Bohr orbit
KE = 
KE = 
12. (c) Not more than two electrons can be present in same atomic orbital.
This is Pauli's exclusion principle.
13. (a) 2
nd
 excited state will be the 3
rd
 energy level.
 or  
14. (a)
In Balmer series n
1
 = 2 & n
2
 = 3, 4, 5.... Last line of the spectrum is
called series limit.
Limiting line is the line of shortest wavelength and high energy when n
2
= 8
 
15. (c) r
n
 = a
0
n
2
r = a
0
 × (3)
2
 = 9a
0
 
16. (c) l = 2 represent d orbital for which
17. (b) de – Broglie wavelength is given by :
... (i)
K.E. 
Substituting this in equation (i)
...(i)
i.e. 
? when KE become 4 times wavelength become 1/2.
18. (a) The electronic configuration of Rubidium (Rb = 37) is
Since last electron enters in 5s orbital