Hydrogen and It's Compounds

The problem asks for the approximate difference between the temperature of maximum density of heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$ and ordinary water $\left(\mathrm{H}_2 \mathrm{O}\right)$.

The temperature of maximum density of $\mathrm{H}_2 \mathrm{O}$ is approximately $3.98^{\circ} \mathrm{C}$.

The temperature of maximum density of $\mathrm{D}_2 \mathrm{O}$ is approximately $11.2^{\circ} \mathrm{C}$.

The difference in temperature in Celsius is

$ \Delta T_C=11.2^{\circ} \mathrm{C}-3.98^{\circ} \mathrm{C}=7.22^{\circ} \mathrm{C} $

The problem specifies the temperatures in Kelvin, with $y$ being the temperature for $\mathrm{H}_2 \mathrm{O}$ and $x$ for $\mathrm{D}_2 \mathrm{O}$. The difference is $(x-y)$ in Kelvin.

The magnitude of a temperature difference is the same whether measured in Celsius or Kelvin.

$ \Delta T_K=\Delta T_C $

So, the difference in Kelvin is 7.22 K . The value 7.22 K is closest to the option 7.0 K .

$\mathrm{LiAlH}_4=$ Total charge $=0$

Li oxidation number $=+1$

Al oxidation number $=+3$

$ \begin{aligned} 1+3+x \times 4 & =0 \\ x & =-1 \\ \mathrm{H}_2 \mathrm{O}=2 \times x-2=0, x & =+1 \\ \text { Hence, } X=\mathrm{LiAlH}_4, Y & =\mathrm{H}_2 \mathrm{O} \end{aligned} $

$ \mathrm{BeH}_{2} $ and $ \mathrm{MgH}_{2} $ have polymeric structure. Polymeric structure are those materials/compounds that are made up of long chain repeating molecules.

Same structure is for $ \mathrm{MgH}_{2} $.

Hydrogen peroxide $ \left(\mathrm{H}_{2} \mathrm{O}_{2}\right) $ is decomposed by the rough surface of glass, alkali oxide present in it and light. Therefore, to prevent its decomposition $ \mathrm{H}_{2} \mathrm{O}_{2} $ is usually stored in waxed lined plastic bottle and kept in dark.

The complete reaction is as follows.

$ 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \longrightarrow \underset{\substack{\text { (Product } X \text { ) } \\\\ \text { Diborane }}}{2 \mathrm{~B}_2 \mathrm{H}_6 }+3 \mathrm{AlF}_3+3 \mathrm{LiF} $

$\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaH} \longrightarrow \underset{\substack{\text { (Product } Y \text { ) } \\\\ \text { Sodium tetrahydroborate }}}{2 \mathrm{NaBH}_4}$

It is a good reducing agent.

Oxidation state of H in $ \mathrm{NaBH}_{4} $ is -1 .

[ $ \mathrm{Na}=+1, \mathrm{~B}=+3 $ oxidation state ]

$ [1+3+4 x=0 $ then $ x=-1] $

Hence, option (a) is incorrect about compound $ Y $,

$\mathrm{H}_2 \mathrm{O}_2$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

$\mathrm{H}_2 \mathrm{O}_2$ structure in solid phase, dihedral angle is $90.2^{\circ}$.

The type of hydrides that have the exact number of electrons to form a covalent bond is called electron precise. These types of compounds are usually formed by group-14 elements.

Group 13 elements forms electron deficient hydrides.

Group 15 elements formed electron excess.

• In saline hydrides oxidation state of hydrogen is -1 .

So, statement I is incorrect.

• LaH is an interstitial hydride. So, statement II is correct.

• $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ have the tendency to form hydrogen bonds. So, statement III is also correct.

  Electrolysis of molten sodium hydride librates hydrogen gas at anode and sodium is formed at cathode. So, statement IV is incorrect.

  Thus, statements II and III are correct about the hydrides.

Statements II and IV are correct, while statements I and III are incorrect. Here are the corrected versions of statements I and III:

I. Sodium hydride reacts with water to release hydrogen gas:

$ \mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(a q) \longrightarrow \mathrm{NaOH}(a q) + \mathrm{H}_2(g) $

III. Ammonia and water molecules are examples of electron-rich compounds.

The correct match is $\mathrm{A}-\mathrm{III}, \mathrm{B} \cdot \mathrm{IV}$, C-I, D-II..

$ \begin{aligned} & \text { } \mathrm{CaC}_2+2 \mathrm{D}_2 \mathrm{O} \longrightarrow \mathrm{C}_2 \mathrm{D}_2+\mathrm{Ca}(\mathrm{OD})_2 \\ & \therefore P \text { is } \mathrm{C}_2 \mathrm{D}_2 \end{aligned} $

Deionised or demineralised water is obtained by passing hard water through both cation and anion exchanger one after other.

(a) $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ are polymeric hydrides, a large number of $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ units join together by hydrogen bonding to produce a long chain.

(b) LiH is by no means unreactive towards air at moderate temperature because it is a powerful reducing agent.

(c) $\mathrm{BeH}_2$ and $\mathrm{MgH}_2$ are covalent in nature due to small size of $\mathrm{Mg}^{2+}$ and $\mathrm{Be}^{2+}$ ions and $\mathrm{H}^{-}$ions is large anion.

(d) The statement (d )is incorrect. The correct stability order of alkali hydrides is

$ \mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH} $

Hence, (a), (b), (c) are correct statements.

The extra neutron in the deuterium isotope makes heavy water denser. Also, the freezing point of heavy water is higher than the ordinary water. The freezing point of heavy water at 1 atm pressure is $3.8^{\circ} \mathrm{C}$.

The most effective water softening method is ion exchange process because in this method the dissolved calcium and magnesium ions that cause hardness of water are removed easily with the help of an ion exchange process. Also, in this method relatively large number of calcium and magnesium ions displace smaller sodium ions.

This method is cheaper method and the chemicals used in it are safe to work with and easy to handle.

Consider LiH as the molten ionic hydrides of s-block,

$ \mathrm{LiH} \longrightarrow \mathrm{Li}^{+}+\mathrm{H}^{-} $

Upon passing electric current, the oxidation reaction that take place at anode.

At anode, $2 \mathrm{H}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g})+2 e^{-}$

So, $\mathrm{H}_2$ gas is liberated at anode.

∵ (I) $\mathrm{Zn}+2 \mathrm{HCl}$ (dil.) $\longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2$ and ∵ (II)

$ \mathrm{Zn}+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 $

while calcium hydrogen carbonate, i.e. calcium bicarbonate $\left[\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2\right]$. On heating give

$ \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \xrightarrow{\Delta} \mathrm{CaCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} $

As it only exsist in solution and will not give $\mathrm{H}_2$.

Hence, (I) and (II) only give $\mathrm{H}_2$, i.e. dihydrogen and option (c) is the correct answer.

$\because \mathrm{H}_2 \mathrm{O}_2$ show oxidising as well as reducing action.

Oxidising nature (i.e. act as oxidising agent)

(i) In acidic solution,

$ \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} $

(ii) In alkaline solution,

$ \mathrm{H}_2 \mathrm{O}_2+2 e^{-} \longrightarrow 2 \mathrm{OH}^{-} $

Reducing nature (i.e. act as reducing agent)

$ \begin{aligned} &\mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}^{+}+\mathrm{O}_2+2 e^{-}\\ &\text { (in acidic medium) } \end{aligned} $

$ \begin{aligned} &\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+2 e^{-}\\ &\text { (in alkaline medium) } \end{aligned} $

Hence, (i), (ii), (iii) and (iv) all are possible and (c) is the correct answer.

Primary components by synthesis gas are hydrogen and carbon monoxide.

It is also known as syn gas. It is used as a fuel gas. Hence, option (a) is the correct option.

$22.4 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2$ means I mL of $\mathrm{H}_2 \mathrm{O}_2$

$ \begin{array}{ll} 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & +\mathrm{O}_2 \\ 2 \mathrm{~mol} & 1 \mathrm{~mol} \\ 68 \mathrm{~g} & 22400 \mathrm{~mL} \text { at NTP } \end{array} $

$\therefore 22400 \mathrm{~mL} \mathrm{O}_2$ is obtained by 68 g $\mathrm{H}_2 \mathrm{O}_2$

$\therefore 22.4 \mathrm{~mL} \mathrm{O}_2$ is obtained by

$ \frac{68 \times 22.4}{22400}=0.068 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2 $

or $\mathrm{l} \mathrm{mL} \mathrm{H}_2 \mathrm{O}_2$ solution $=0.068 \mathrm{~g}$

$\therefore 100 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2$ solution

$ =0.0680 \times 100=6.8 \% $

Hard water does not produce lather with soap is called hard water. Hardness is due to the presence of bicarbonate, chlorides and sulphates of Ca and Mg .

$ \begin{aligned} & M^{2+}+2 \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa} \longrightarrow \\ & \left(M=\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\right) \\ & \left(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}\right)_2 M+2 \mathrm{Na}^{+} \\ & \text {Sodium stearate (soap) } \end{aligned} $

$\mathrm{Ca}^{2+}$ or $\mathrm{Mg}^{2+}$ ion present in hard water react with soap to form precipitate of Ca and Mg salt of fatty acids and hence, no lather is produced.