The two important points of Bohr's model can be used to derive the given formula are as follows.
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state respectively.

0.2 L(or 200 mL) of 5M solution will contain 
\(\frac { 340\times0.2 }{ 2 } =34g{ H }_{ 2 }{ O }_{ 2 }\)
\(\underset { 2\times 34=68g }{ { 2H }_{ 2 }O_{ 2 } } \rightarrow { 2H }_{ 2 }O+\underset { 2\times 16=32g }{ { O }_{ 2 } } \)
\(\because \ 68g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give 32g\(O_{ 2 }\)
\(\therefore \ 34g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give
\(\frac { 32\times 34 }{ 68 } =16g{ O }_{ 2 }\)

It is LiH because it has significant covalent character due to the smallest alkali metal Li. LiH is very stable. It is almost unreactive towards oxygen and chlorine.
It reacts with \({ Al }_{ 2 }{ Cl }_{ 6 }\) to form lithium aluminium hydride.
\(8LiH+{ Al }_{ 2 }{ Cl }_{ 6 }\longrightarrow 2LiAl{ H }_{ 4 }+6LiCl\)

For the reaction, \({ PCl }_{ 5 }(g)\rightleftharpoons { PCl }_{ 3 }(g)+{ Cl }_{ 2 }(g)\)
At 500 K in a closed container,
\(\left[ { PCl }_{ 5 } \right] =0.8\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\( \left[ { PCl }_{ 5 } \right] =1.2\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\(\left[ { Cl }_{ 2 } \right] \ =1.2\times { 10 }^{ -3 } \ mol \ { L }^{ -1 }\)
\( { K }_{ c } \ =\frac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { { PCl }_{ 5 } } \right] }\)
\( =\frac { (1.2\times { 10 }^{ -3 })\times (1.2\times { 10 }^{ -3 }) }{ (0.8\times { 10 }^{ -3 }) } =1.8\times { 10 }^{ -3 }\)

\(N_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons 2NO_{ 2 }(g)\)
If N₂O₄ is 50% dissociated,d, the mole fraction of both the substances is given by
\(x_{ N_{ 2 }O_{ 4 } }=\frac { 1-0.5 }{ 1+0.5 } \Rightarrow x_{ NO_{ 2 } }=\frac { 2\times 0.5 }{ 1+0.5 }\)
\( p_{ N_{ 2 }O_{ 4 } }=\frac { 0.5 }{ 1.5 } \times 1atm,\ p_{ NO_{ 2 } }=\frac { 1 }{ 1.5 } \times 1atm\)
The equilibrium constant K_p is given by
\(K_{ p }=\frac { (p_{ NO_{ 2 } })^{ 2 } }{ p_{ { N }_{ 2 }{ O }_{ 4 } } } =\frac { 1.5 }{ (1.5)^{ 2 }(0.5) } =1.33atm\)
Since,
\(\Delta _{ r }G^{ \circ }=-RT \ ln \ K_{ p }\)
\( \Delta _{ r }G^{ \circ }=(-8.314 \ JK^{ - } \ mol^{ - })\times (333K)\times (2.303)\times (0.1239)\)
\( =-763.8 \ kJmol^{ -1 }\)

Ionic bond The bond formed, as a result of the electrostatic between the positive and negative ions was termed as the electrovalent bond or ionic bond. e.g. the formation of NaCL from sodium and chlorine can be explained as
\(Na\longrightarrow { Na }^{ + }+{ e }^{ - }\)   
\(\left[ Ne \right] { 3s }^{ 1 } \ \left[ Ne \right] \)
\(Cl \ + { e }^{ - }\longrightarrow { Cl }^{ - }\)
\(\\ \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 5 } \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }or\left[ Ar \right] \)
\({ Na }^{ + }+{ Cl }^{ - }\longrightarrow NaCl \ or \ { Na }^{ + }{ Cl }^{ - }\) 
Similarly, the formation of CaF₂ may be shown as
\( Ca \longrightarrow { Ca }^{ 2+ } +{ 2e }^{ - }\)    
\(\left[ Ar \right] { 4s }^{ 2 } \left[ Ar \right] \)
\(F \ +\ { e }^{ - }\longrightarrow \ { F }^{ - }\) 
\(\\ \left[ He \right] { 2s }^{ 2 }2p^{ 5 } \left[ He \right] { 2s }^{ 2 }2p^{ 6 }or \ \left[ Ne \right] \)
\({ Ca }^{ 2+ }+{ 2F }^{ - }\longrightarrow Ca{ F }_{ 2 } \ or \ { Ca }^{ 2+ }{ 2F }^{ - } \)  
Covalent bond The bond formed between the two atoms by mutual sharing of electrons between them is called covalent bond. e.g. the formation of chlorine molecules can be explained

Similarly, in the formation of HCl
    

The applications of dipole moment are
(a) The dipole moment helps to predict whether a molecule is polar or non-polar. As μ = q × d greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.
(b) The percentage of ionic character can be calculated as Percentage of ionic character \(=\frac { { \mu }_{ observed } }{ { \mu }_{ ionic } } \times 100\)
(c) Symmetrical molecules have zero dipole moment although they have two or more polar bonds.
(d) It helps to distinguish between cis and trans-isomers. Usually cis-isomer has higher dipole moment than trans-isomer.
(e) It helps to distinguish between ortho, meta and para-isomers. Dipole moment of para-isomer is zero. Dipole moment of ortho-isomer is greater than that of.

There are numerous physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, energy of atomisation, etc., which show periodic variations. The cause of periodicity in properties is the repetition of similar outer electronic configuration after certain regular intervals. e.g. all the elements of 1s group (alkali metals) have similar outer electronic configuration, i.e. ns¹.
₃Li = 1s², 2s¹
₁₁Na = 1s², 2s², 2p⁶ , 3s¹
₁₉K = 1s², 2s², 2p⁶ , 3s², 3p⁶, 4s¹
Therefore, due to similar outermost shell electronic configuration all alkali metals have similar properties. e.g., sodium and potassium both are soft and reactive metals. They all form basic oxides and their basic character increases down the group. They all form unipositive ion by the loss of one electron. Similarly, all the elements of 17th group (halogens) have similar outermost shell electronic configuration, i.e. ns²  np⁵ and thus possess similar properties.
₉F = 1s², 2s¹ , 2p⁵
₁₇Cl = 1s², 2s², 2p⁶ , 3s² , 3p⁵
₃₅Br = 1s², 2s², 2p⁶ , 3s², 3p⁶, 3d¹⁰, 4s² , 4p⁵

Given that, mass of Zn = 32.65 g
1 mole of gas occupies =22.7 L volume at STP
Atomic mass of Zn = 65.3\(\mu\) 
The given equation is 
\(\underset { 65.3 \ g }{ Zn } +2HCL\rightarrow { ZnCL }_{ 2 }+\underset { 1mol=22.7 \ L \ at \ STP }{ H_{ 2 } } \)
From the above equation, it is clear that 65.3 gZn, when reacts with HCL, produces = 22.7 of \(H_{ 2 }\) at STP
\(\therefore\) 32.65 g Zn, when reacts with HCL, will produce
\(=\frac { 22.7\times 32.65 }{ 65.3 } =11.35 \ L \ of \ { H }_{ 2 } \ at \ STP\)