The various types of hydrides and examples of each type are given below. Hydride type Compound (A) Electron deficient (i) $$LiH$$ (B) Saline (ii) $$CH_{4}$$ (C) Electrons-precise (iii) $$NH_{3}$$ (D) Interstitial (iv) $$B_{2}H_{6}$$ (E) Electron rich (v) $$CrH$$ Choose the correct matching from the codes given below.

$$(A) - (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)$$

$$(A) - (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)$$

$$(A) - (v), (B) - (iii), (C) - (iv), (D) - (ii), (E) - (i)$$

$$(A) - (iv), (B) - (v), (C) - (i), (D) - (ii), (E) - (iii)$$

Match the reactions of column I with their types given in column II and mark the appropriate choice. Column I Column II A $$H_{2}O + NH_{3} \rightleftharpoons NH^{+}_{4} + OH$$ i Self ionisation of $$H_{2}O$$ B $$FeCl_{3} + 3H_{2}O \rightarrow Fe (OH)_{3} + 3HCl $$ ii Decomposition C $$H_{2}O + H_{2}O \rightleftharpoons H_{3}O^{+} + OH$$ iii Acidic nature of $$H_{2}O$$ D $$2H_{2}O \rightarrow 2H_{2} + O_{2}$$ iv Hydrolysis

(A)$$\rightarrow$$ (iii), (B)$$ \rightarrow $$(iv), (C)$$\rightarrow$$ (i), (D)$$\rightarrow $$(ii)

(A)$$\rightarrow $$(ii), (B) $$\rightarrow $$(i), (C)$$\rightarrow $$(iii), (D)$$\rightarrow$$ (iv)

(A)$$\rightarrow $$(iii), (B)$$ \rightarrow $$(ii), (C)$$\rightarrow$$ (iv), (D)$$\rightarrow $$(i)

(A)$$\rightarrow$$ (i), (B) $$\rightarrow$$ (ii), (C)$$\rightarrow $$(iv), (D)$$\rightarrow$$ (iii)

Match the column I with column II and mark the appropriate choice. column I column II A NaH i Interstitial hydride B $$CH_{4}$$ ii Molecular hydride C $$VH_{0.56}$$ iii Ionic hydride D $$B_{2}H_{6}$$ iv Electron-deficient hydride

$$(A)\rightarrow (iii), (B)\rightarrow (ii), (C)\rightarrow (i) (D)\rightarrow (iv)$$

$$(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii) (D)\rightarrow (i)$$

$$(A)\rightarrow (ii), (B)\rightarrow (iv), (C)\rightarrow (iii) (D)\rightarrow (i)$$

$$(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iv) (D)\rightarrow (iii)$$

MCQ Questions for Class 11 Medical Chemistry Hydrogen Quiz 11

In transforming

$0.01\ mole$ of

$PbS$ to

$PbSO_{4}$ , the volume of

$10$ volume

$H_{2}O_{2}$ required will be:

0%
$11.2\ mL$
0%
$22.4\ mL$
0%
$33.6\ mL$
0%
$44.8\ mL$

Explanation

$PbS + 4H_{2}O_{2}\rightarrow PbSO_{4} + 4H_{2}O$

$0.04\ mol$ of

$H_{2}O_{2}$ is required to react with

$0.01\ mol$

$PbS$ .

Molarity of

$H_{2}O_{2}$ may be calculated as,

Volume strength

$= Molarity \times 11.2$

$10 = M\times 11.2$

$M = 0.892$

Number of moles

$= \dfrac {MV}{1000}$

$0.04 = \dfrac {0.892\times V}{1000}$

$V = 44.8\ mL$ .

Why is water a bent molecule and not linear?

0%
Water has four electron domains repelling each other, causing the molecule to be bent.
0%
There are not enough electrons involved for the molecule to be linear.
0%
Because the oxygen in water can only make bent molecules.
0%
The molecule alternates in between a bent geometry and a linear geometry.

Hydorgen forms binary compounds called hydrides with several elements in the periodic table.Regarding these hydrides the wrong statement:

0%
Among lanthanoids, europioum and ytterbium form ionic hydrides.
0%
The density of ionic hydrides is less, while the density of interstitial hydrides is more than the metals from which they are formed.
0%
The rate of reaction of alkali metals towards hydrogen decreases from lithium to caesium
0%
Hydrogen gas can be purified by the phenomenon of occlusion of hydrogen by palladium metal.

Which of the following hydrolysis reaction is incorrect at moderate temperature?

0%
$4BF_{3} + 3H_{2}O \rightarrow H_{3}BO_{3} + 3HBF_{4}$
0%
$2NF_{3} + 3H_{2}O\rightarrow N_{2}O_{3} + 6HF$
0%
$NCl_{3} + 3H_{2}O\rightarrow NH_{3} + HOCl$
0%
$BiCl_{3} + H_{2}O\rightarrow BiOCl\downarrow + 2HCl$

What is the reaction given below, called?

$H_{2}O_{(l)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + OH^{-}_{(aq)}$

0%
Hydrolysis of water
0%
Hydration of water
0%
Disproportionate of water
0%
Auto-protolysis of water

Explanation

The reaction in which two molecules of the same species react to give ions by proton transfer is known as auto-protolysis or self ionisation. In this reaction, proton from one

$H_{2}O$ molecule is transferred to another

$H_{2}O$ molecule.

$H_{2}O({l}) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}({aq}) + OH^{-}({aq})$

What willl be the mass of oxygen liberated by decomposition of 200 mL hydrogen peroxide solution with a strength of 34 g per litre?

0%
25.5g
0%
3.0g
0%
3.2g
0%
4.2g

Explanation

Strength of the solution

$=34$ g/L.
1 L (or 1000 mL) of the solution contains 34 g of

$H_2O_2$
200 mL of the solution contains

$\dfrac{34}{1000} \times 200= 6.8$ g of

$H_2O_2$

$2H_2O_2\rightarrow2H_2O+O_2$

2 moles of

$H_2O_2$ gives 1 mole of

$O_2$

68g of

$H_2O_2$ gives

$32$ g of

$O_2$

6.8g of

$H_2O_2$ gives

$\dfrac{32}{68} \times 6.8= 3.2$ g of

$O_2$

The various types of hydrides and examples of each type are given below.

	Hydride type		Compound
(A)	Electron deficient	(i)	$LiH$
(B)	Saline	(ii)	$CH_{4}$
(C)	Electrons-precise	(iii)	$NH_{3}$
(D)	Interstitial	(iv)	$B_{2}H_{6}$
(E)	Electron rich	(v)	$CrH$

Choose the correct matching from the codes given below.

0%
$(A) - (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)$
0%
$(A) - (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)$
0%
$(A) - (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)$
0%
$(A) - (v), (B) - (iii), (C) - (iv), (D) - (ii), (E) - (i)$
0%
$(A) - (iv), (B) - (v), (C) - (i), (D) - (ii), (E) - (iii)$

Explanation

(A) Electron deficient

$\rightarrow { B }_{ 2 }{ H }_{ 6 }$ (

$\because$ It has

$3c-2e$ structure, which shows that

$B$ -atom has electron deficient centre)
(B) Saline

$\rightarrow LiH$ is called saline
(C) Electron precise

$\rightarrow C{ H }_{ 4 }$ (

$\because$ It has perfect complete octet

$C$ -atom).
(D) Interstitial

$\rightarrow CrH$ Here, hydrogen atoms are trapped in the interstitial sites of crystal of

$Cr$ .
(E) Electron rich

$\rightarrow \ddot { N } { H }_{ 3 }$ , because nitrogen has lone pair of electrons in its outermost shell.
Hence, correct order is

$(A) - (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)$

The standard potential of the reaction:

$H_2O+e^-\rightarrow\dfrac{1}{2}H_2+OH^-$ at 298 K by using

$K_w(H_2 O)=10^{-14}$ .

0%
-0.828 V
0%
0.828 V
0%
2.0 V
0%
-0.5 V

Explanation

We know

$\Delta G=-RT\ln K$

$\because \Delta G=-nFE$

$\therefore -nFE=-RT\ln K$

$\therefore -1\times 96500\times E=-8.314\times 298\ln (10^{-14})$

$\therefore E=-0.828$ $V$

Hydrogen gas is obtained from:

0%
petroleum
0%
coal
0%
coke
0%
natural gas

Ionic hydride reacts with water to give:

0%
hydride ions
0%
acidic solution
0%
protons
0%
basic solution

Explanation

Ionic hydrides give basic solution when reacts with water.

e.g.

$LiH+{ H }_{ 2 }O\longrightarrow \underset{base}{LiOH}+{ H }_{ 2 }\uparrow$

Which of the following is laboratory preparation of dihydrogen?

0%
$3Fe + 4H_{2} O (steam) \rightarrow Fe_{3} O_{4} + 4H_{2}$
0%
$2Na + 2H_{2}O \rightarrow 2NaOH + H_{2}$
0%
$CaH_{2} + 2H_{2}O\rightarrow Ca(OH)_{2} + 2H_{2}$
0%
$Zn + H_{2}SO_{4} (dil.) \rightarrow ZnSO_{4} + H_{2}$

Explanation

In the laboratory, hydrogen gas is prepared by the action of dilute

$H_{2}SO_{4}$ on granulated zinc.

$Zn + H_{2}SO_{4} (dil.) \rightarrow ZnSO_{4} + H_{2}$

$H_{2}$ gas is liberated at cathode and anode both by electrolysis of the given aq. solution except:

0%
$NaH$
0%
$HCOONa$
0%
$CaH_{2}$
0%
$LiH$

Which of the following is most soluble in water?

0%
${ CsClO }_{ 4 }$
0%
${ NaClO }_{ 4 }$
0%
${ KClO }_{ 4 }$
0%
${ LiClO }_{ 4 }$

Explanation

$NaClO_4$ is the inorganic compound. It is a white crystalline, hygroscopic solid that is highly soluble in water and in alcohol. It usually encountered as the monohydrate. The compound is noteworthy as the most soluble of the common perchlorate salts. It decreases from top to bottom

The order of reactivity of halogens towards hydrogen is:

0%
$F_{2} > Cl_{2} > Br_{2} > I_{2}$
0%
$I_{2} > Br_{2} > Cl_{2} > F_{2}$
0%
$Cl_{2} > Br_{2} > I_{2} > F_{2}$
0%
$Br_{2} > Cl_{2} > F_{2} > I_{2}$

Which of the following statements regarding hydrides iis not correct?

0%
Ionic hydrides are crystalline, non-volatile and non-conducting in solid state
0%
Electron-deficient hydrides act as Lewis acids or electron acceptors
0%
Elements of group $-13$ form electron-deficient hydride
0%
Elements of group $15-17$ form electron-precise hydrides

Explanation

Elements of group 15-17 form electron-rich hydrides. Group 14 elements form electron-precise hydrides. Elements of group-13 form electron-deficient hydrides. For example, diborane

$B_2H_6$ is electron deficient, methane

$CH_4$ is electron precise whereas

$NH_3$ ,

$H_2O$ and

$HF$ are electron rich.

Which of the following ions forms a hydroxide that is highly soluble in water?

0%
$K^{+}$
0%
$Zn^{2+}$
0%
$Ni^{2+}$
0%
$Al^{3+}$

Explanation

1. $K^+$ ion will form most water soluble hydroxide because it is the most reactive of these metals and thus forms strong alkali $KOH$ which will undergo complete dissociation in water, thus making it highly soluble.

The production of dihydrogen obtained from coal gasification can be increased by reacting carbon monoxide of syngas mixture with steam in presence of a catalyst iron-chromium. What is this process called?

0%
Hydrogen reaction
0%
Water-gas shift reaction
0%
Coal-gas shift reaction
0%
Syn gasification

Explanation

The production of dihydrogen can be increased by reacting carbon monoxide of syngas with steam. This is called water-gas shift reaction.

$CO({g}) + H_{2}O({g}) \xrightarrow[Catalyst]{673 K} CO_2({g}) + H_2({g})$

Hydrogen peroxide oxidises

$[Fe(CN)_6]^{4-}$ to

$[Fe(CN)_6]^{3-}$ in acidic medium but reduce

$[Fe(CN)_6]^{3-}$ to

$[Fe(CN)_6]^{4-}$ in alkaline medium. The other products formed are, respectively:

0%
$H_2O$ and $(H_2O + O_2)$
0%
$H_2O$ and $(H_2O + OH^-)$
0%
$(H_2O + O_2)$ and $H_2O$
0%
$(H_2O + O_2)$ and $(H_2O + OH^-)$

Explanation

(i)

$[Fe^{+2}(CN)_6]^{4-} +H_2O_2 + 2H^+ \rightarrow [Fe^{+3}(CN)_6]^{3-} +2H_2O$

(ii)

$[Fe^{+3}(CN)_6]^{3-} +H_2O_2 +2OH^- \rightarrow [Fe^{+2}(CN)_6]^{4-} +O_2 + 2H_2O$

From the following reaction it is clear that option A is correct.

Given below are two reaction of water with sodium and carbon dioxide, what is the nature of water in these reaction?
(i)

$2Na+2H_2O\rightarrow 2NaOH +H_2$
(ii)

$6CO_2+12H_2O\rightarrow C_6H_12_6+6H_2O+6O_2$

0%
In (ii) water acts as an oxidising agent and in (i) it acts as a reducing agent
0%
In (i) water acts as an oxidising agent and in (ii) it acts as a reducing agent
0%
In both (i) and (ii) hydrogen acts as a reducing agent
0%
In both (i) and (ii) hydrogen acts as an oxidizing agent

Explanation

With sodium metal water acts as an oxidizing agent gets reduced to

$H_2$ and with

$CO_2$ water acts as a reducing agent and is oxidized to oxygen

Which of the following is not a process of preparation of hydrogen peroxide?

0%
Auto-oxidation of 2-ethylanthraquinol.
0%
By passing oxygen through boiling water
0%
By oxidation of isopropyl alcohol
0%
By reaction of barium peroxide with dil. $H_2SO_4$

Explanation

Preparation of 2-ethylanthraquinol

preparation of

$H_{2}O_{2}$

(i) 2- ethylanthrquinol

$\xrightarrow[H_{2}/pd]{O_{2}}H_{2}O_{2}+$ oxidized product

(ii) By reaction of barium oxide (per)with

dif.

$H_{2}SO_{4}$

$BaO_{2}.8H_{2}O(s)+H_{2}SO_{4}\rightarrow BaSO_{4}+H_{2}O_{2}+8H_{2}O$

(iii) By oxidation of is propyl (alcohol)

Option B will not give

$H_{2}O_{2}$

1091514_927273_ans_a37e0c4531ed4605b8e1efa1233b343e.png

Study the following reactions and mark the correct properties shown by water :
(i)

$SO_{3} + H_{2}O \rightarrow H_{2}SO_{4}$
(ii)

$Cl_{2}O_{7} + H_{2}O \rightarrow 2HClO_{4}$
(iii)

$Na_{2}O + H_{2}O \rightarrow 2NaOH$
(iv)

$CaO + H_{2}O \rightarrow Ca (OH)_{2}$

0%
All oxides react with water to give hydroxides
0%
Acidic oxides are formed by metals and basic oxides by non-metals
0%
Non-metal oxides combine with water to form acids while metallic oxides combine with water to form alkalies
0%
Acidic oxides are stronger than basic oxides since they form strong acids

Explanation

(i)

$SO_{3}+H_{2}O\rightarrow H_{2}SO_{4}$

(ii)

$Cl_{2}O_{7}+H_{2}O\rightarrow 2HClO_{4}$

$SO_{3}$ and

$Cl_{2}O_{7}$ are non metallic oxides form acid on reacting with water

(iii)

$Na_{2}O+H_{2}O \rightarrow 2NaOH$

(iv)

$CaO+H_{2}O \rightarrow Ca(OH)_{2}$

$Na_{2}O$ and

$CaO$ are metallic oxide form base on reacting with water

Ans (C) Non metal oxides combines with water to form acids while metallic oxides combine with water to form alkalies.

Match the reactions of column I with their types given in column II and mark the appropriate choice.

	Column I		Column II
A	$H_{2}O + NH_{3} \rightleftharpoons NH^{+}_{4} + OH$	i	Self ionisation of $H_{2}O$
B	$FeCl_{3} + 3H_{2}O \rightarrow Fe (OH)_{3} + 3HCl$	ii	Decomposition
C	$H_{2}O + H_{2}O \rightleftharpoons H_{3}O^{+} + OH$	iii	Acidic nature of $H_{2}O$
D	$2H_{2}O \rightarrow 2H_{2} + O_{2}$	iv	Hydrolysis

0%
(A) $\rightarrow$ (ii), (B) $\rightarrow$ (i), (C) $\rightarrow$ (iii), (D) $\rightarrow$ (iv)
0%
(A) $\rightarrow$ (iii), (B) $\rightarrow$ (ii), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (i)
0%
(A) $\rightarrow$ (i), (B) $\rightarrow$ (ii), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (iii)
0%
(A) $\rightarrow$ (iii), (B) $\rightarrow$ (iv), (C) $\rightarrow$ (i), (D) $\rightarrow$ (ii)

Explanation

(A)

$H_{2}O+NH_{3}\rightleftharpoons NH_{4}+OH^{-}$

Here

$NH_{3}$ is Lewis base and

$H_{2}O$

acid as acid to it represents

acidic nature of

$H_{2}O$

(b)

$F_{e}Cl_{3}+ H_{2}O \rightarrow F_{e}(OH)_{3}+3HCl$

it is are example of hydrolysis

reaction which produces acid and base

(C)

$H_2O+H_2o \rightarrow H_{3}O+OH^{-}$

self ionisation of water

(D)

$2H_{2}O\rightarrow 2H_{2}+O_{2}$

Decomposition of water in their

corresponding molecule

(A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)

1090723_927152_ans_4615fc09db7e44a28dbc3bd8b1805c0a.png

Which of the following reagents cannot be used for the preparation of hydrogen peroxide?

0%
Sodium peroxide
0%
2- Ethylanthraquinol
0%
Sodium thiosulphate
0%
Barium Peroxide

Explanation

Sodium thiosulphate cannot be used for the preparation of hydrogen peroxide. Instead, electrolysis of ammonium hydrogen sulphate (ammonium sulphate +

$H_2SO_4$ ) can be carried out.

A commercial sample of hydrogen peroxide is labelled as

$10$ volume. Its percentage strength is nearly ?

0%
$3\%$
0%
$1\%$
0%
$90\%$
0%
$10\%$

Explanation

Percent concentration

$= \dfrac {68}{224} \times$ volume strength

Percent concentration

$= \dfrac {68}{224} \times 10=3$ %.

Match the column I with column II and mark the appropriate choice.

	column I		column II
A	NaH	i	Interstitial hydride
B	$CH_{4}$	ii	Molecular hydride
C	$VH_{0.56}$	iii	Ionic hydride
D	$B_{2}H_{6}$	iv	Electron-deficient hydride

0%
$(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii) (D)\rightarrow (i)$
0%
$(A)\rightarrow (ii), (B)\rightarrow (iv), (C)\rightarrow (iii) (D)\rightarrow (i)$
0%
$(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iv) (D)\rightarrow (iii)$
0%
$(A)\rightarrow (iii), (B)\rightarrow (ii), (C)\rightarrow (i) (D)\rightarrow (iv)$

Explanation

NaH is an ionic compound and is made of sodium cations

$\left ( Na^{+} \right )$ and hydride anions

$\left ( H^{-} \right )$ . It has the octahedral crystal structure with each sodium ion surrounded by six hydride ions.

$CH_{4}$ forms molecular or covalent hydrides. These are mainly formed by p - block elements and some s - block elements.

The third compound has an oxidation number of hydrogen which is zero. So it belongs to Interstitial hydride.

$B_{2}H_{6}$ the Boron atom is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet.

The correct option is D.

Water plays a key role in the biosphere, It is due to certain properties of

$H_2O$ as compared to other liquids. These are except:

0%
higer specific heat
0%
lesser thermal conductivity
0%
high dielectric constant
0%
high surface tension

Which of the following represents the chemical equation involved in the preparation of

$H_2O_2$ from barium peroxide?

0%
$BaO_2.8H_2O+H_2SO_4\rightarrow BaSO_4+H_2O_2+8H_2O$
0%
$CH_3CHOHCH_3\rightarrow CH_3COOCH_3+H_2O_2$
0%
$BaO_2+CO_2+H_2O\rightarrow BaCO_3+H_2O_2$
0%
$Ba_3(PO_4)_2+3H_2SO_4\rightarrow 3BaSO_4+2H_3PO_4$

Explanation

Barium peroxide react with sulfuric acid to produce Hydrogen peroxide and Barium sulfate.

Isopropyl Alcohol oxidises to Acetone and water, and not hydrogen peroxide.

$CO_{2}$ reacts with water to form carbonic acid. It then reacts with barium peroxide to form barium carbonate and hydrogen peroxide. Barium carbonate is insoluble and is easy to remove by filtration. This process is widely used to manufacture $H_{2}O_{2}$ .

Barium phosphate reacts with Sulphuric acid to form Barium sulfate and Phosphoric acid.

The correct option is C.

What will be the strength of 20 vol of

$H_2O_2$ in terms of gram per litre?

0%
$60.71g L^{-1}$
0%
$5.6g L^{-1}$
0%
$30.62g L^{-1}$
0%
$17g L^{-1}$

Explanation

Concentration (in g/L)

$= \dfrac {680}{224} \times$ volume strength

Concentration (in g/L)

$= \dfrac {680}{224} \times 20= 60.71 \ g/L$

$5.0 cm^3 of H_2O_2$ liberates 0.508 g of iodine from an acidified KI solution. The strength of

$H_2O_2$ solution in terms of volume strength at STP is:

0%
6.48 volumes
0%
4.48 volumes
0%
7.68 volumes
0%
none of these

Explanation

Moles of iodine=

$\cfrac {0.508}{254}$ moles

Now,

$M_1V_1=M_2V_2$

$M_1\longrightarrow$ Molarity of

$H_2O_2$ solution

$\therefore$

$V_1\longrightarrow$ Volume of

$H_2O_2$

$M_2\longrightarrow$ Molarity of iodine

$V_2\longrightarrow$

$1000$

$ml$

$\therefore$

$M_1\times 5=\cfrac {0.508}{254}\times 1000$

$\therefore$

$M_1=0.4$

$M$

The reaction can be given as,

$H_2O_2+2I^{-}+2H^{+}\longrightarrow2H_2O+I_2$

$\therefore$ n-factor for

$H_2O_2=2$

$\therefore$ Normality of

$H_2O_2=2\times 0.4=0.8N$

$\therefore$ Volume strength of

$H_2O_2=0.8\times 5.6$

$4.48$ volumes

Metal hydrides are ionic, covalent or molecular in nature. Among

$LiH,\ NaH,\ KH,\ RbH$ and

$CsH$ , the correct order of increasing ionic character is:

0%
$LiH > NaH > CsH > KH > RbH$
0%
$LiH < NaH < KH < RbH < CsH$
0%
$RbH > CsH > NaH > KH > LiH$
0%
$NaH > CsH > RbH > LiH > KH$

Explanation

According to Fajan’s rule, as the size of cation increases, the polarisation of anion decreases; the percentage covalent character decreases and the percentage ionic character increases.
And as we move down the group, the size of elements increases .

Thus, the order of atomic radii of the alkali metals is

$Li < Na< K< Rb< Cs$ .
Thus, order of increasing percentage ionic character in alkali metal hydrides will be:

$LiH< NaH<KH< RbH<CsH$

Two structure of

$H_2O_2$ are drawn below. Identify the phase

$(X)$ and

$(Y)$ of

$H_2O_2$ .

0%
$(X)$ is the structure of $H_2O_2$ in gas phase and $(Y)$ in solid phase
0%
$(X)$ is structure of $H_2O_2$ in solid phase and $(Y)$ in gas phase
0%
$(X)$ and $(Y)$ are structures of $H_2O_2$ in gas phase
0%
$(X)$ and $(Y)$ are structures of $H_2O_2$ in solid phase

Explanation

$(X)$ is the structure of

$H_2O_2$ in the gas phase and

$(Y)$ in the solid phase.

$O-O$ single bond length is higher in the gas phase.

$H-O-O$ bond angle is higher in the solid phase.

Hence, the correct answer is option

$\text{A}$ .

When

$NaBH_{4}$ is dissolves in water

0%
It decomposes with the evolution of $H_{2}$
0%
$Na^{+}$ and $BH^{-}_{4}$ are formed which are stable
0%
$BH^{-}_{4}$ ions formed initially decompose to produce $OH^{-}$ ions, which prevent future decomposition
0%
$NaH$ and $B_{2}H_{6}$ are produced

Explanation

When

$NaBH_{4}$ is dissolves in water it decomposes with the evolution of

$H_{2}$ and the reaction is as follows:

$NaBH_4+2H_2O\to NaBO_2+4H_2$

Which of the following reactions increases production of dihydrogen from synthesis gas?

0%
$CH_4(g) + H_2O(g) \xrightarrow[Ni]{1270K} CO(g) + 3H_2(g$ )
0%
$C(s) + H_2O(g) \overset{1270K}{\rightarrow} CO(g) + H_2(g)$
0%
$CO(g) + H_2O(g) \xrightarrow[Catalyst]{673 K} CO_2(g) + H_2(g)$
0%
None of the above

Explanation

Correct Answer: Option C

Explanation:

By reacting carbon monoxide from syngas mixture with steam in the presence of iron chromate as a catalyst, the production of dihydrogen gas can be boosted.

$CO\left( g \right)+{{H}_{2}}O\left( g \right)\xrightarrow[catalyst]{673K}C{{O}_{2}}\left( g \right)+{{H}_{2}}\left( g \right)$

This is called the water-gas shift reaction.

Final Answer:

The correct answer is option $(C).$

Which of the following information is true for structure of

$H_2O_2$ in solid phase?

Explanation

$H_2O_2$ has an open book structure with

$H-O-O$ bond angle equal to 101.9^0 in solid phase. The

$O-O$ single bond distance is 145.8 pm. The

$O-H$ single bond distance is 98.8 pm.

Strength of 10 volume hydrogen peroxide solution is:

0%
$30.35\ g\ L^{-1}$
0%
$17\ g\ L^{-1}$
0%
$34\ g\ L^{-1}$
0%
$68\ g\ L^{-1}$

Explanation

Concentration (in g/L)

$= \dfrac {680}{224} \times$ volume strength

Concentration (in g/L)

$= \dfrac {680}{224} \times 10= 30.35 \ g/L$

HOTS
The various types of hydrides and examples of each type are given below :
Hydride type Compound
A Electron deficient i LiH
B Saline ii

$CH_{4}$
C Electron-precise iii

$NH_{3}$
D Interstitial iv

$B_{2}H_{6}$
E Electron rich v CrH
Choose the correct matching from the code given below :

0%
(A)- (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)
0%
(A)- (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)
0%
(A)- (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)
0%
(A)- (v), (B) - (iii), (C) - i(v), (D) - (ii), (E) - (i)

Explanation

(A) electron deficient

$\rightarrow B_{2}H_{6}$

(which can accept a love pair of electron)

(B) saline (Ionic hydrides)

$\rightarrow LiH$

$\rightarrow CH_{4}$

(which can neither accept nor give pair of

$e^{-} 14^{th}$ group element makes this hydride)

(D) Interstitial

$\rightarrow CrH$

(d block element from intersfical hydride)

(E) Electron rich

$\rightarrow NH_{3}$

molecule which can donate pair of electron

(A)-iv, (B)-(i), (C)-(ii), (D)-(v), (E)-(iii)

1091267_927638_ans_be49b27966ab4eec99ffb7dbab17b851.jpg

Match list I with list II. Choose the correct matching codes from the choices given.
List I List II
(Hydride) (Type of hydride)
A.

$BeH_{2}$ 1. Complex
B..

$AsH_{3}$ Lewis acid
C.

$B_{2}H_{6}$ Interstitial
D.

$LaH_{3}$ Covalent
E.

$LiAlH_{4}$ Intermediate

0%
A-6, B-2, C-4, D-5, E-1
0%
A-6, B-2, C-4, D-3, E-1
0%
A-6, B-4, C-2, D-3, E-15
0%
A-5, B-4, C-2, D-3, E-1

Explanation

$A.$

$BeH_2\rightarrow$ Intermediate

$B.$

$AsH_3\rightarrow$ Covalent

$C.$

$B_2H_6\rightarrow$ Lewis acid

$D.$

$LaH_3\rightarrow$ Interstitial

$E.$

$LiAlH_4\rightarrow$ Complex

The enthalpy change on freezing of 1 mol of water at

$5^oC$ to ice at

$-5^oC$ is:

$[{ Given:\quad \triangle }_{ fus }H=6kJ{ mol }^{ -1 }at\ 0^oC,\ { C }_{ p }\left( { H }_{ 2 }O,l \right) =75.3J{ mol }^{ -1 }{ K }^{ -1 },{ C }_{ p }\left( { H }_{ 2 }O,s \right)$

$=36.8J{ mol }^{ -1 }{ K }^{ -1 }]$

0%
$-6.56kJ{ mol }^{ -1 }$
0%
$-5.81kJ{ mol }^{ -1 }$
0%
$-6.00kJ{ mol }^{ -1 }$
0%
$-5.44kJ{ mol }^{ -1 }$

Which of the following will give

${H}_{2}$ gas with dilute

${HNO}_{3}$ ?

0%
$Mg$
0%
$Zn$
0%
$Cu$
0%
$Hg$

Explanation

$Mg+2HN{ O }_{ 3 }\rightarrow Mg{ \left( N{ O }_{ 3 } \right) }_{ 2 }+{ H }_{ 2 }\left( g \right)$

$'Mg'$ being a strong reducing agent compared to zinc, copper and mercury produce hydrogen gas by reducing nitric acid.

Hydrogen peroxide is obtained by the electrolysis of:

0%
water
0%
sulphuric acid
0%
hydrochloric acid
0%
fused sodium peroxide

Explanation

$H_2O_2$ is obtained by the electrolysis of a cold 50% solution of

$H_2O_2$

$2H_2SO_4\rightarrow 2H^+ + 2HSO^-_4$
The electrolysis of

$HSO^-_4$ gives

$H_2O_2$ .

Industrially,

${ H }_{ 2 }{ O }_{ 2 }$ is obtained by the following cyclic process :

$\begin{matrix} 2-Ethylanthraquinol \\ (P) \end{matrix}\overset { { O }_{ 2 } }{ \underset { { H }_{ 2 }/Pd }{ \rightleftharpoons } } \begin{matrix} 2-Ethylanthraquinone \\ (Q) \end{matrix}+{ H }_{ 2 }{ O }_{ 2 }$
Which one of the following properties is wrong about the mixture of organic solvents in which the reaction is carried out?

0%
It must resist oxidation.
0%
It must be miscible with water.
0%
It must dissolve both (P) and (Q).
0%
It is generally a mixture of ester and hydrocarbon.

Explanation

Industrially

$H-20_2$ is obtained by this reaction

$2$ ethylanthraquinol

$\frac {o_2} {H_2/pd} 2-$ ethylanthraquinone

It is auto-oxidation of

$2$ - alkylanthraquints. The wrong statement is that, organic solvents should be miscible with water because it dissolves just like that, means the organic solvent wouldn't be miscible in water.

Volume strength of

$500ml$ solution containing

$3.4gm$ of

${H_2}{O_2}$ is

0%
$11.2$
0%
$6.8$
0%
$1.12$
0%
$2.24$

A commercial sample of

${H_2}{O_2}$ marked as 100 volume hydrogen peroxide, it means that

0%
${\rm{1}}\,\,{\rm{mL}}\,{\rm{of}}\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\,{\rm{give}}\,{\rm{100}}\,{\rm{mL}}\ {{\rm{O}}_{\rm{2}}}\,{\rm{at}}\,{\rm{STP}}$
0%
${\rm{1}}\,\,{\rm{L}}\,{\rm{of}}\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\,{\rm{will}}\,{\rm{give}}\,{\rm{100}}\,{\rm{mL}}\ {{\rm{O}}_{\rm{2}}}\,{\rm{at}}\,{\rm{STP}}$
0%
${\rm{1}}\,\,{\rm{L}}\,{\rm{of}}\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\,{\rm{will}}\,{\rm{give}}\,\ {\rm{22}}{\rm{.4L}}\,{\rm{}}\ {{\rm{O}}_{\rm{2}}}\,{\rm{at}}\,{\rm{STP}}$
0%
None of these

$H_2O_2$ can be estimated by

0%
Titration with acidified $K_2Cr_2O_7$
0%
Titration with $NaOH$ solution
0%
Titration with acidified $KMnO_4$ solution
0%
Titration with $Na_2CO_3$ solution

Which one of the following metals cannot evolve

$H_{2}$ from acids or

$H_{2}O$ or from its compounds?

0%
Hg
0%
Al
0%
Pb
0%
Fe

Explanation

$Hg$ has greater reduction potential than that of

$H^+$ and hence cannot displace hydrogen from acid.

When barium chloride solution is added to sodium sulphate solution, a white precipitate is formed. The white precipitate is:

0%
Barium chloride
0%
Sodium sulphate
0%
Barium sulphate
0%
Sodium chloride

Hydrogen does not combine with ______.

0%
Sb
0%
Na
0%
He
0%
Zn

Enthalpy of vapourisation for water is

$186.5kJ$

${mol}^{-1}$ . The entropy change during vapourisation is _____

$kJ{K}^{-1}{mol}^{-1}$

0%
$0.5$
0%
$1.0$
0%
$1.5$
0%
$2.0$

Which is correct order of BP of VA , VIA and VIIA hydrides:

0%
$SbH_3 > NH_3 > AsH_3 > PH_3$
0%
$H_2 O > H_2 Te > H_2Se > H_2S$
0%
$HF > HI > HBr > HCl$
0%
All correcct

Industrially

$H_2O_2$ is produced

0%
By oxidation of 2 ethyl anthraquinone in and cyclic process
0%
By partial oxidation of propane2 ol
0%
Both
0%
None

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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