n = 4; f orbital l = 3 \(\Rightarrow\) m₁= - 3, -2,-1, 0, +1, +2, +3
out of two electrons, one electron occupies 4f orbital with m₁ = -3 and another electron occupies 4f orbital with m₁ = -2.
All the four quantum numbers for the two electrons are

Electronic configuration of Fe³⁺ 1s²2s²2p⁶3s²3p⁶3d⁶4s²

Electronic configuration of mn²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
Five unpaired electrons
Electronic configuration of Ar is 1s² 2s² 2p⁶ 3s² 3p⁶ 
no unpaired electrons.

\(\lambda =\frac{h}{\sqrt{2KEm}}\ \lambda=\frac{h}{mv}or \frac{h}{p}\)
E = Kinetic energy
m = mass of particle               
\(\lambda\) = 4.8 pm = 4.8 x 10^-12 m     
h = 6.63 x 10^-34 J.S.               
m = 9.11 x 10^-31 kg             
 4.8 x 10^-12 = \(\frac{6.63\times 10^{-34}}{\sqrt{12\times E\times 9.11\times 10^{-31}}}\)
E = 1.047 x 10^-14 J.

Uncertainty in speed = 2% of 40 ms^-1
i.e., \(\Delta v=\frac{2}{100}\times 45\) = 0.9 ms^-1
Applying uncertainty principle,
\(\Delta x(m\times \Delta v)=\frac{h}{4\pi}\)
\(\Delta x=\frac{h}{4\pi.m.\Delta v}\)
= \(\frac{6.626\times 10^{-34}kgm^2 s^{-1}}{4\times 3.14\times (40\times 10^{-3}kg)(0.9ms^{-1})}\)
= 1.46 x 10^-33 m.

The shape of an atomic orbital is found by finding the probability (\({ \Psi }^{ 2 }\) ) of the electron at different points around the nucleus and representing the density of the nucleus. Same orbitals are found to have a region of space where the probability of finding electron is zero. This is called a node.
(i) All 's' orbitals are spherical in shape.
(ii) All 'p' orbitals are dumb-bell in shape.
The three 'p' orbitals differ in orientation. They lie along the axes and called P_x, P_y and P_z orbitals.
(iii) All 'd' orbitals have clover- leaf shaped with four lobes. Three have lobes between axes and are called d_xy, d_yz, d_xz orbitals.
The fourth lobe is along the axes and is called \({ d }_{ { x }^{ 2 }-{ y }^{ 2 } }\) the fifth has two lobes along 'z' axis and dough-nut shape in the centre and is called \({ d }_{ { z }^{ 2 } }\)

If uncertainty principle is applied to an object of mass say about a milligram (10^-6 kg), then
Δv.Δx = \(\frac { h }{ 4\pi m } \)
Δv.Δx = \(\frac { 6.626\times { 10 }^{ -34 }kg\quad { m }^{ 2 }{ s }^{ -1 } }{ 4\times 3.14\times 10^{ -6 }kg } \)
= 0.52 \(\times\) 10^-28 m²s^-1
The value of Δv.Δx obtained is extremely small and is insignificant. Therefore, for milligram sized or heavier objects, the associated uncertainties are hardly of any real consequence.

No, for a stationary electron, velocity = 0 and thus, position can be measured accurately.

Bohr's theory does not consider the de-Broglie concept of dual nature of electron and also contradicts with the Heisenberg's uncertainty principle, while the Schrodinger equation is based on quantum mechanics which deals with the microscopic objects having both the particle as well as wave like character.

In an atom the wave function \(\Psi \) for an electron has no physical significance as such. However, its square i.e.\(\Psi ^{ 2 }\) at any point gives the intensity of the electron wave at that point.
In view of Heisenberg's uncertainty principle, it shows the probability of finding the electron at that point and therefore, termed as probability density

It is impossible to accurately determine both the position as well as the momentum of a microscopic particle simultaneously.
\(\Delta x.\Delta p\ge h/4\pi \)
where, Δx and Δp are uncertainties in determining the position and momentum, respectively

Schrodinger Wave Equation:
\(\frac { \partial ^{ 2 }\Psi }{ \partial x^{ 2 } } +\frac { \partial ^{ 2 }\Psi }{ \partial y^{ 2 } } +\frac { \partial ^{ 2 }\Psi }{ \partial z^{ 2 } } +\frac { 8\pi ^{ 2 }m }{ { h }^{ 2 } } (E-V)\Psi =0\)
\(\Psi \) = amplitude of wave; E = total energy of electron
V = potential energy; m = mass of electron

Isoelectronic - Species having same number of electrons
₁₁Na⁺ = 11 - 1 = 10e^-; 19K⁺ = 19 - 1 = 18e^-
₁₂Mg²⁺= 12 - 2 = 10e^-; .₂₀Ca²⁺ = 20 - 2 = 18e^-
₁₆S^2- 16 + 2 = 18e^-; _I8Ar = 18e^-
Hence the isoelectronic species are
(i) Na⁺ and Mg2+
(ii) K⁺, Ca²⁺, S^2- and Ar

Mass of the ball = 150 g = 0.150 kg
Uncertainty in velocity (Δv) = 3.52 10^-24 ms^-1
Δx.mΔv ≥ \(\frac { h }{ 4\pi } \)
Δx  = \(\frac { 6.626\times { 10 }^{ -34 }{ kg }\quad m^{ 2 }{ s }^{ -1 } }{ 4\times 3.142\times (0.150\quad kg)\times (3.52\times { 10 }^{ -24 }{ ms }^{ -1 }) } \)
Δx = 10^-10 m
Δx = 1Å (1Å = 10^-10 m) 

We know, mass of electron  = 9.1\(\times\) 10^-31 kg;
  h  = 6.626 \(\times\) 10^-34 Js
∴ λ = \(\frac { 6.626\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 2.19\times { 10 }^{ 6 } } \)
= 3.32\(\times\)10^-10 m

According to de-Broglie relation
λ_A = \(\frac { h }{ { P }_{ A } } \) ;λ_A = \(\frac { h }{ { P }_{ B } } \)
λ_A = 5 10-8 m ; λ_B =  ?
P_B = 2\(\times\) P_A
λ_A P_A = λ_B P_B
λ_A P_A =  λ_B (2\(\times\) P_A)
λ_B = \(\frac { { \lambda }_{ A }{ P }_{ A } }{ 2{ P }_{ A } } =5\times { 10 }^{ -8 }\times \frac { { P }_{ A } }{ 2{ P }_{ A } } \)
Wavelength of B = 2.\(\times\)5 10^-8 m 

Any form of matter that exhibits both particle as well as wave characteristics is said to show dual character.

The statement is true. According to Bohr, as long as an electron remains in a particular orbit, it does not lose or gain energy. This means that energy of an electron in a particular path remains constant. Therefore, these orbits are also called stationary states.

According to Bohr, electrons revolve round the nucleus only in certain selected circular paths called orbits.. These orbits are associated with definite energies and so are called energy shells or energy levels or quantum levels.

If a substance which gives a line emission spectrum is placed in an external electric field, its lines get split into a number of closely spaced lines. This phenomenon is known as Stark effect.

If a substance which gives a line emission spectrum, is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines. This phenomenon is known as Zeeman effect.

The composition of any atom can be represented by using the normal element symbol X with superscript on the left side as the mass number A and subscript Z on the left side as atomic number (ie) \(_{ Z}^{ A }{ X }\)
Hence, the symbols \(_{ 79 }^{ 35 }{ Br }\) and ³⁵Br are not acceptable.

Chlorine atom on accepting an electron becomes CI- and aluminium atom after donating an electron becomes AI3^+.
These changes will affect their number of protons, neutrons and electrons.
CI^- :
No. of electrons 17 + 1 = 18
No. of protons = 17
No. of neutrons = 35 - 17 = 18
AI3^+.
No. of electrons 13 - 3 = 10
No. of protons 13
No. of neutrons 27 -13 = 14.

1. Heavy atoms like gold, platinum have heavy nucleus. Heavy nucleus contains large amount of positive charge.
2. When a beam of \(\alpha\)-particles is shot at a thin gold foil, most of them pass through without much effect.
3. Some are deflected back due to enormous repulsive force of heavy nucleus.
4. If thin foil is made of light aluminium, then the number of \(\alpha\) -particles deflected back will be negligible.

(1) Phosphorous atom:
Atomic number (Z) No. of protons = 15
= No. of electrons = 15
Number of neutrons = Mass number - Atomic Number
= 31 - 15
= 16
(ii) Phosphate ion:( \({ PO }_{ 4 }^{ 3- }\))
Number of electron = 15+ 4 x 8 + 3 = 50
Number of protons = 15+ 4 x 8 = 47
Number of neutrons = 10+ 4 x 8 = 48.

K.E = 2.5 \(\times\)10^-24 J
Kinetic energy = \(\frac { 1 }{ 2 } \)mv² = 2.5 \(\times\)10^-24
m = 9.1\(\times\) 10-31 kg
v = \(\left( \frac { 2K.E }{ m } \right) ^{ \frac { 1 }{ 2 } }=\left[ \frac { 2\times 2.5\times { 10 }^{ -24 } }{ 9.1\times { 10 }^{ -31 } } \right] ^{ \frac { 1 }{ 2 } }\)
= 2.34 \(\times\)10³ ms^-1
\(\boxed{v = 2.34 103\ {ms}^{-1}}\)
Wavelength , \(\lambda =\frac { h }{ mv } \)
= \(\frac { 6.626\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 2.34\times { 10 }^{ 3 } } =311.1\times { 10 }^{ -9 }m\)
\(\boxed{\lambda =311.1\times 10^{ -9 }m(or)311.1\ nm}\)

(i) Both have identical shape, consisting of four lobes
(ii) Lobes of d_x²-y² lie along the x and y-axes where those of d_xy lie between x and y-axes.
(iii) 45^o

f-orbitals :
For 'f' orbital, l = 3 and the m values are -3, -2, -1, 0, + 1, +2, +3 corresponding to seven f orbitals  \({ f }_{ { z }^{ 3 }, }{ f }_{ { xz }^{ 2 }, }{ f }_{ { yz }^{ 3 }, }{ f }_{ { xz }y },{ f }_{ { z({ x }^{ 2 }-y^{ 2 }) } },{ f }_{ { x({ x }^{ 2 }-3y^{ 2 }) } },{ f }_{ { y({ 3x }^{ 2 }-y^{ 2 }) } }\)
There are 3 nodal planes in the f-orbitals.

d-orbitals :
(a) For 'd' orbital I= 2 and the corresponding m values are -2, -1,0 +1,+2.
(b) The shape of the d orbital looks like a clover leaf.
(c) The five m values give rise to five d orbitals namely d_xy,d_yz,d_zx,d_x² -y² and d_z2.
(d) The 3d orbitals contain two nodal planes.

p-orbitals :
(a) For p orbitals l = 1 and the corresponding m values are -1,0 and + 1. The angular distribution functions are quite complex
(b) The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x,p_y and p_z and the angular distribution for these orbitals shows that the lobes are along the x, y and z axis respectively.
(c) The 2p orbitals have one nodal plane

n = 4 l = 0,1,2,3
4 sub shells s, p, d & f.
I = 0 m₁ = 0 + one 4s orbital.
I = 1 m₁ = -1, 0, + 1 \(\Rightarrow\) three 4p orbitals.
I = 2 m₁ = -2,:1, 0, +1, +2 \(\Rightarrow\) five 4d orbitals.
I = 3 m₁ = -3, -2, -1,0, +1, +2, +3 \(\Rightarrow\) seven 4f orbitals.
Over all 16 orbitals are possible.