What is the maximum number of electrons that can be associated with the following set of quantum number?
n=3, l =1 and m=-1.
10
6
4
4
D.
4
The orbital of the electron having =3, l =1 and m= -1 is 3p_{z} (as nl_{m}) and an orbital can have a maximum of two electrons with opposite spins.
therefore, 3p_{z} orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.
Based on equation
E=-2.178 x 10^{-18} J certain conclusions are written. Which of them is not correct?
The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than if would be if the electrons were at the infinite distance from the nucleus.
Larger the value of n, the larger is the orbit radius
Equation can be used to calculate the change in energy when the electron changes orbit
Equation can be used to calculate the change in energy when the electron changes orbit
D.
Equation can be used to calculate the change in energy when the electron changes orbit
If n=1,
E_{1} = - 2.178 x 10^{-18} Z^{2} J
If n=6
From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. it means that electron is more strongly bound in the smallest allowed orbit.
The value of Planck's constant is 6.63 x 10^{-34} Js. The speed of light is 3 x10^{17} nms^{-1} . Which value is closet to the wavelength in nanometer of a quantum of light with frequency of 6 x 10^{15} s^{-1} ?^{}
10
25
50
50
C.
50
Given, Planck's constant,
h= 6.63 x10^{-34}^{}
speed of light, c= 3 x10^{17} nms^{-1}
Frequency of quanta
v=6 x10^{15} s^{-1}
Wavelength, λ =?
We know that,
Which of these is least likely to act as a lewis base?
CO
F^{-}
BF_{3}
BF_{3}
C.
BF_{3}
Electron rich species are called lewis base. Among the given, BF_{3} is an electron deficient species, so have a capacity of electrons accepting instead of donating that's why it is least likely to act as a lewis base. It is a lewis acid.
How many grams of the concentrated nitric solution should be used to prepare 250 mL o 2.0M HNO_{3} ? The concentrated acid is 70% HNO_{3}.
45.0 g conc. HNO_{3}
90.0 g conc. HNO_{3}
70.0 g conc. HNO_{3}
70.0 g conc. HNO_{3}
A.
45.0 g conc. HNO_{3}
Given, molarity of solution = 2
Volume of solution = 250 mL = 250/1000 = 1/4 L
Molar mass of
HNO_{3} = 1+14+3 x 16 = 63 g mol^{-1}
therefore, Molarity
=
Dipole-induced dipole interactions are present in which of the following pair?
H_{2}O and alcohol
Cl_{2} and CCl_{4}
HCl and He atoms
HCl and He atoms
C.
HCl and He atoms
Dipole-induced dipole interaction is present in the pair in which the first species is polar and the other is non-polar.
H_{2}O and alcohol both are non-polar so there exist dipole-dipole interactions in between them.
Cl_{2} and CCl_{4} both are non-polar so there exists induced dipole -induced dipole interactions in between them. Similarly is true for SiCl_{4} and He atoms pair.
Which of the following structure is similar to graphite?
BN
B
B_{4}_{}
B_{4}_{}
A.
BN
Boron nitride, (BN)_{x} rersembles with graphite in structure as,
6.02 x 10^{20} molecules of urea are present in 100 mL of its solution. The concentration of solution is
0.02 M
0.01 M
0.001 M
0.001 M
B.
0.01 M
Given, number of molecules of urea =6.02 x 10^{20}
therefore ,Number of moles
=
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH =10 and by passing hydrogen gas through the platinum wire at 1 atm pressure. The oxidation potential of electrode would be
0.0591 V
0.59 V
0.118 V
0.118 V
B.
0.59 V
For Hydrogen electrode, oxidation half reaction is
A reaction having equal energies of activation for forward and reverse reactions has
C.
ΔH = 0Energy profile diagram for are reaction is as from the figure it is clear that
(E_{a})_{b} = (E_{a})_{f} +ΔH
[Here (E_{a})_{b} = activation energy of backward reaction and (E_{a})_{f} = activation energy of forward reaction].
If (E_{a})_{b} = (E_{a})_{b} = (E_{a})_{f}
then ΔH = 0