JEE ASPIRANTS

 MCQs on Structure of Atom 

1. How many orbitals can have the following set of quantum numbers, n = 3, l = 1, m1 = 0 ?

(a) 3

(b) 1

(c) 4

(d) 2

Answer: (b)

2. Electronic configuration of the outer shell of the element Gd with atomic number 64 is

(a) 4f4 5d5 6s1

(b) 4f3 5d5 6s2

(c) 4f5 5d4 6s1

(d) 4f7 5d1 6s2

Answer: (d)

3. Maximum number of electrons in a subshell can be

(a) 4l + 2

(b) 4l – 2

(c) 2n2

(d) 2l + 1

Answer: (a)

4. The orientation of atomic orbitals depends on their

(a) spin quantum number

(b) magnetic quantum number

(c) azimuthal quantum number

(d) principal quantum number

Answer: (b)

5. A gas X has Cp and Cv ratio as 1.4, at NTP 11.2 L of gas X will contain_______ number of atoms

(a) 1.2 × 1023

(b) 3.01 × 1023

(c) 2.01 × 1023

(d) 6.02 × 1023

Answer: (d)

6. Number of unpaired electrons in N2+

(a) 3

(b) 1

(c) 2

(d) 0

Answer: (b)

7. The excitation energy of a hydrogen atom from its ground state to its third excited state is

(a) 12.75 eV

(b) 0.85 eV

(c) 10.2 eV

(d) 12.1 eV

Answer: (a)

8. 3p orbital has _ radial nodes

(a) three

(b) two

(c) one

(d) none

Answer: (c)

9. A 0.66 kg ball is moving with a speed of 100 m/s. Find its wavelength

(a) 6.6 × 10-34 m

(b) 6.6 × 10-32 m

(c) 1.0 × 10-32 m

(d) 1.0 × 10-35 m

Answer: (d)

 IIT JEE Physics Math Class 12:

Solved Examples on Some Basic Concepts of Chemistry

Question:1 A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is:(IIT JEE -2013)

1) C3H4

2) C6H5

3) C7H8

4) C2H4

Answer: C

Solution:

General equation for combustion of hydrocarbon:

CxHy + (x+ y/4)O2 → xCO2 + (y/2)H2O

Number of moles of CO2 produced = 3.08/44 = 0.07

Number of moles of H2O produced = 0.72/18 = 0.04

SO, x / (y/2) = 0.07/0.04 =  7/4

The formula of hydrocarbon is C7H8

Hence, the correct option is C.

Question:2)  29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is (IIT JEE 2012)

Solution:

Let mass of the stock solution = 100g

Mass of HCl in 100 g of 29.2 % (w/w) HCl stock solution = 29.2 g

Volume of the stock solution = 100g/1.25 g mL1 = 80 g

Number of moles of HCl in stock solution = 29.2/36.5 = 0.8

Molarity of the stock solution = (0.82/80)×1000 = 10 M

Using,

M1V1 = M2V2

10× V1 =0.4×200

or

V1 = 8mL

Hence, the volume required is 8 mL.

Question:3 Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is  (IIT JEE-2011)

1) 1.78 M

2) 2.00 M

3) 2.05 M

4) 2.22 M

Answer: C

Solution:

For calculating molarity of the solution we require to know two things, (i)number of moles of

urea & (ii) total volume of the solution.

Number of moles of urea dissolved in the solution = 120 g/60g = 2.

Total mass of the solution = 1000g + 120 g = 1120 g

Volume of the total solution = 1120g/1.15 gmL-1 = 974 mL = 0.974 L

Molarity = 2/0.974 = 2.05 M

Hence, the correct option is C.

Question: 4) Energy of an electron is given by E = -2.178×10-18 J Z2/n2. Wavelength of light required to excite an electron in an hydrogen atom from level n =1 to n = 2 will be (h = 6.62 ×10−34Js and c = 3.0 ×108ms−1)   (IIT JEE-2013)

1) 2.816 ×10−7 m

2) 6.500 ×10−7m

3) 8.500 ×10−7m

4) 1.214 ×10−7m

Answer: d

Solution:

We know that energy of an electron can be calculated using Planck relation

E = hc/λ     …………… (i)

But in the question it is given that

E = -2.178×10-18 J Z2 (1/n12 -1/n22) …………(ii)

On comparing both the equations we get,

hc /λ =-2.178×10-18 J Z2(1/n12 -1/n22) ………….(iii)

On substituting the values

h =6.62 ×10−34Js

c = 3.0 ×108ms−1

n1= 1 & n2= 2

In equation (iii) and calculating for the value of λ we get,

λ =1.214 ×10−7m

Hence, the correct option is d.

Question: 5) The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] (IIT JEE 2012)

1) h2/4π2m a02

2) h2/16π2m a02

3) h2/32π2m a02

4) h2/64π2m a02

Answer: c

Solution:

According to Bohr’s postulate,

Electrons can exist only in those orbital around the nucleus, for which the angular momentum is integral multiple of  h/2π i.e.

mvr  = n h/2π

or,

 v = n h/2πmr

So, Kinetic Energy of electrons = ½ m (n h/2πmr)2

Radius of the orbital i.e. rn = (a0×n2)/Z

For, n=2 & z =1

r = 4 a0

So,

Kinetic Energy of electrons = ½ m [4 h2/(4π2m2(4 a0))]2 = h2/32π2m a02

Hence, the correct option is C.