\(\underset { 2\ volumes }{ { 2H }_{ { 2 }_{ (g) } } } +\underset { 1\ volumes }{ { O }_{ { 2 }_{ (g) } } } \rightarrow \underset { 2\ volumes }{ { 2H }_{ 2 }{ O }_{ (g) } } \)
Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O (g)
\(\therefore\) 10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O(g).
Thus 10 volumes of H₂O will be produced.

3H₂ + N₂ \(\rightarrow\) 2NH₃
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia = 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = \(\frac { 3 }{ 2 } \times 20=30\) moles.

No.of moles = \(\frac { mass\ of\ the\ substance }{ Molar\ mass\ of\ the\ substanc } =\frac { W }{ M } \)
Molar mass of ethane (C₂H₆) = 24 + 6 = 30
\(\therefore\) Number of moles in 60 g of ethane = \(\frac { 60 }{ 30 } \) = 2 moles.

	Molecule	Compound
i)	A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.	A molecule which contains two or more atoms of different elements are called a compound molecule
ii)	e.g. Na - Monoatomic molecule  O2 - Diatomic molecule P4 - Polyatomic molecule	e.g. CO2 - Carbon dioxid CH4 - Methane H2O- Water

Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.

A matter is anything which has mass and occupies space. Matters exist in all three states such as solid, liquid and gas.

Acidity : The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.
Basicity : The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.

As we know there are three different orientations in space that are possible for a p orbital. All the three p orbitals, namely, px, py and pz have same energies and are called degenerate orbitals. However, in the presence of magnetic or electric field the degeneracy is lost.

For 4p orbital: Number of angular nodes = l
For 4p orbital l = 1
\(\therefore\) Number of angular nodes = 1
Number of radial nodes = n - l - 1
= 4-1-1
= 2
\(\therefore\)Total number of nodes = n -1 = 4 - 1 = 3
1 angular node and 2 radial nodes.
For 4d orbital: Number of angular nodes = 1
For 4d orbital 1= 2
\(\therefore\) Number of angular nodes = 2,
Number of radial nodes = n - l- 1
=4-2-1
= 1
\(\therefore\) Total number of nodes = n - 1 = 4 - 1 = 3
1 radial nodes and 2 angular node.

n = 3, main shell is m.
Total number of orbitals in 3rd energy level =?

When n = 3	l = 0	1	2
Subshell	s	p	d
	3s	3px, 3py, 3pz	3dxz, 3dxy, 3dyz, 3dx2-y2,dz2
	1	3	5

Total number of orbitals = 9.

(i) The theory of electromagnetic radiation states that a moving charged particle should continuously loose its energy in the form of radiation.
(ii) Therefore, the moving electron in an atom should continuously loose its energy and finally collide with nucleus resulting in the collapse of the atom.

(i) C₂H₅OH : (2 x 12) + (5 x 1) + (1 x 16) + (1 x 1) = 46 g
(ii) KMnO₄ : (1 x 39) + (1 x 55) + (4 x 16) = 158 g
(iii) K₂Cr₂O₇ : (2 x 39) + (2 x 52) + (7 x 16) = 294 g
(iv) C₁₂H₂₂O₁₁ : (12 x 12) + (22 x 1) + (11 x 16) = 342 g

Molecular mass = Density x Molar volume
Molar volume of CO₂ = 2.24 x 10^-2m³
Density of CO₂ 1.977 kg m^-3
\(\therefore \text { Molecular mass of } \mathrm{CO}_{2}=1.977 \times 10^{3} \mathrm{~g} \mathrm{~m}^{\not -3} \times 2.24 \times 10^{-2} \mathrm{~m}^{\not -3}\)
= 1.977 X 10¹ x 2.24
= 44g

Molar mass of K₂Cr₂O₇ = 2 x 39 + 2 x 52 + 7 x 16
= 78 + 104 + 112 = 294
In acid medium Cr₂O₇^-2 is reduced to Cr⁺³
Cr₂O₇ ^-2 + 6e + 14H + \(\rightarrow\) 2cr⁺³ + 7H₂O
equivalent mass of K₂Cr₂O₇ = \({Molar \ mass \over No, \ of \ electrons \ taken \ Up }\)
\(={294\over6}=49\)
\(\therefore\) 'n' factor \(={294\over49}=6 eq.mol^{-1}\)

CIO₃^- (CI in CIO₃^-)
Oxidation number of chlorine = x
Oxidation number of oxygen = - 2
x + 3 (-2) = -1
x - 6 = -1 or
x = 5
The oxidation number of chlorine in CIO₃^- is + 5.

Angular momentum of electron in any orbital = \(\sqrt{l(l+1)}\times\frac{h}{2\pi}\)
(i) for 2s orbital, l = 0
Therefore angular momentum = \(\sqrt{0(0+1)}\times\frac{h}{2\pi}=0\)
(ii) for 4f orbital, I = 3
Therefore angular momentum = \(\sqrt{3(3+1)}\times\frac{h}{2\pi}\)
= \(2\sqrt{3}\frac{h}{2\pi}\)
\(=\sqrt{3}\frac{h}{\pi}\)

Similarities:
(i) Both have similar shape.
(ii) Both have same angular momentum = \(\sqrt{l(l+1)} \frac{h}{2\pi}\)
Dissimilarities:
(i) Is orbital has no node while 2s orbital has one node.
(ii) Energy of 2s orbital is greater than Is orbital.
(iii) The size of the 2s orbital is larger than Is orbital.

This is because, according to this principle, if one electron of an atom has same particular values, for the four quantum numbers, then all the other electrons in that atom are excluded from having the same set of values.

(iii) \(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \)  Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O

 (ii) \({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)   MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)  2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ +2KOH

(i) Atoms consist of huge positively charged centers called nuclei.
(ii) Most of the space inside the atom is empty.

According to this theory, atom was assumed to consist of a sphere of uniform distribution of about 10^-10 m positive charge with electrons embedded in it such that the number of electrons equal to the number of positive charges and the atom as a whole is electrically neutral.

(i) ground state :

\(​​\upharpoonleft \)

\(​​\upharpoonleft \)

\(​​\upharpoonleft \)

\(​​\upharpoonleft \)

\(​​\upharpoonleft \)

This type of electronic configuration have ten possible arrangement (i.e. Half filled configuration is More).

a) Principal quantum number defines energy and size of an orbital.
b) Azimuthal quantum number defines shape of an orbital
c) Magnetic quantum number defines spatial orientation (direction) of an orbital.

Orbital	n	1	Radial node n-1-1	Angular node 1
2s	2	0	1	0
4p	4	1	2	1
5d	5	2	2	2
4f	4	3	0	3

Energy electrons symmetry
This is due to the symmetrical distribution and exchange energy of given d- electrons. Symmetry leads to stability.
Exchange energy:
If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy. If more number of exchanges are possible, more exchange energy in released. More number of exchanges are possible only in case of half filled and fully filled configurations.
For example, in chromium the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges as shown in figure. On the other hand only six exchanges are possible for [Ar]3d⁴ 4s² configuration. Hence, exchange energy for the half filled configuration is more. This increases the stability of half filled 3d orbitals.

The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.

Average atomic mass
= \(\frac { (78.99\times 23.99)+(10\times 24.99)+(11.01\times 25.98) }{ 100 } \)
= \(\frac { 2430.9 }{ 100 } \)
= 24.31 u

Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62

i) urea [CO(NH₂)₂]
Mol.mass = 1 (C) + 2(N) + 4(H) + 1(0)
= 1(12) + 2(14) + 4(1) + 1(16)
 = 12 + 28 + 4 + 16 = 60  
ii) Acetone [CH₃ COCH₃]
Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58
iii) Boric Acid [H₃ BO₃]
Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62
iv) Sulphuric  Acid [H₂ SO₄]
Mol.mass = 2(H) + 1(S) + 4(0)
= 2(1) + 1(32) + 4(16)
= 2 + 32 + 64 = 98

Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.

CH_4(g) + 2O₂ \(\rightarrow\) CO₂ + 2H₂O
16g                          (2x18)g
As per stoichiometric equation,
16 g of methane produces 36 g of H₂O
\(\therefore\) 32 g of methane will produce = \(\frac { 36 }{ 16 } \times 32=72\) g of water.

\(\mathrm{E}_{n}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}\)
Second excited state
\(\therefore E_{3}=\frac{-13.6}{9} \mathrm{eV}\)
n = 3
E₃ = -1.51 eV

i) ₂₅Mn - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 4s², 3d⁵
₂₃Mn²⁺ - 1s², 2S^2, 2p⁶, 3s², 3p⁶, 3d⁵
ii) ₂₄Cr - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 4s¹, 3d³
₂₁Cr³⁺ - 1s^2, 2S²,2p⁶, 3s², 3p⁶, 3d³

n	1	Sub Energy	m1values	Number of orbitals
4	2	4d	-2,-1,0+1,+2	Five 4d orbitals
5	3	5f	-3,-2,-1,0,+1,+2,+3,	seven 5f orbitals
7	0	7s	0	one 7s orbitals

De Broglie wavelength of the tennis ball equal to 5400 \(\overset { 0 }{ A } \).
m = 54 g 
V = ?
\(\lambda=\frac{h}{mV}\)
\(V=\frac{h}{m\lambda}\)
\(\mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{JS}}{54 \times 10^{-3} \mathrm{~kg} \times 5400 \times 10^{-10} \mathrm{~m}}=2.27 \times 10^{-26} \mathrm{~ms}^{-1}\)

(i) no. of electrons: 35 (given)
no. of protons : 35
(ii) Electronic configuration
1s² 2S² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
(iii) Last electron:

\(\downharpoonleft\upharpoonright\)

\(\upharpoonleft\downharpoonright\)

\(\upharpoonleft\)

4P_x        4P_y     4p_z
last electron present in 4P_y orbital y
n = 4, l = 1 m₁ = either + 1 or -1 and s = -1/2

half reaction \(\Rightarrow \) \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { S }_{ 4 }{ O }_{ 6 }^{ 2- }\)
\({ I }_{ 2 }\longrightarrow { I }^{ - }\)
 

The solution to Schrodinger equation gives the permitted total energy values called eigen values and the corresponding wave function represent atomic orbitals.

Orbital	n	l
3px	3	1
4dx2-y2	4	2