MCQ Questions for Class 11 Medical Chemistry Hydrogen Quiz 8

On moving from left to right in a period what is the order of acidic character of hydrides?

0%
$NH_{3} < H_{2}O < HF$
0%
$HF < H_{2}O < NH_{3}$
0%
$H_{2}O < HF < NH_{3}$
0%
$H_{2}O < NH_{3} < HF$

Explanation

On moving from left to right in a period, the acidic character of hydrides increases.

$NH_{3}$ is basic,

$H_{2}O$ is neutral and

$HF$ is acidic.

Electronegativity increases from left to right in a period. The ease of losing a proton increases and acidity increases. Acidity is the ability to donate a proton.

Carbon hydrides of the type,

$C_{n}H_{2n+2}$ do not act as Lewis acid or Lewis base.They behave as normal carbon hydrides because:

0%
carbon hydrides are electron-rich hydrides
0%
carbon hydrides are electron-deficient hydrides
0%
carbon hydrides are electron-precise hydrides
0%
carbon hydrides are non-stoichiometric hydrides.

Explanation

Carbon hydrides with general formula

$C_{n}H_{2n+2}$ are electron-precise hydrides like

$CH_{4}$ . They have a complete octet and they do not have lone pair of electrons. Hence they neither donate nor accept electron pairs. They do not behave as Lewis acid or Lewis base.

Which of the following hydrides is electron deficient?

0%
$NaH$
0%
$CaH_{2}$
0%
$CH_{4}$
0%
$B_{2}H_{6}$

Explanation

$\left (BH_{3} \right )$ is the chemical formula for the Boron atom. It is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet. B-atom in diborane is $sp^{3}$ hybridized. $B_{2}H_{6}$ molecule has a total of 8 covalent bonds but only 12 electrons i.e. 6 pairs. So, it has two electron pairs less than the maximum required number for bonding, which makes it electron deficient.

The answer is option D.

What is the trend of boiling points of hydrides of N, O, and F?

0%
Due to lower molecular masses $NH_{3}, H_{2}O$ and HF have lower boiling points than those of the subsequent group member hydrides
0%
Due to higher electronegativity of N, O and F; $NH_{3}, H_{2}O$ and HF show hydrogen bonding and hence higher boiling points than the hydrides of their subsequent group members.
0%
There is no regular trend in the boiling points of hydrides
0%
Due to higher oxidation states of N, O and F, the boiling points of $NH_{3}, H_{2}O$ and HF are higher than the hydrides of their subsequent group members.

Explanation

The boiling point increases with increase in molecular masses. But in the case of

$N, O$ and

$F$ they show the higher boiling point in molecules.

This is due to the higher electronegativity of

$N, O$ and

$F$ which leads to hydrogen bonding in

$NH_3, H_2O$ and

$HF$ due to which the boiling point of these increases & is higher than the hydrides of their subsequent group members.

In complex hydrides, hydride ions act as ligand and are coordinated to metal ions. These hydrides are good reducing agents. Which of the following hydrides is not a complex hydride?

0%
$LiAIH_{4}$
0%
$NaBH_{4}$
0%
$(AlH_{3})_{n}$
0%
$LiBH_{4}$

Explanation

$(AlH_{3})_{n}$ is a polymeric hydride like

$(BeH_{2})_{n}, (MgH_{2})_{n}$ etc.

Whereas,

$LiAlH_4$ ,

$NaBH_4$ and

$LiBH_4$ are complex co-ordinate hydrides.
Hence option C is correct.

In what respect is the electronic configuration of hydrogen and halogens similar?

0%
Hydrogen and halogens have one electron in their outermost shell
0%
Hydrogen and halogens have one electron less than noble gas configuration
0%
Hydrogen and halogens can lose one electron
0%
Hydrogen and halogens show noble gas configuration

Explanation

Hydrogen has one electron in its only shell. Hence, it needs one more electron to attain a stable duplet electronic configuration. On the other hand, halogens have seven electrons in their valence shell. They also need one more electron to attain a stable octet configuration. Therefore, they are similar in the way that they have one electron less than noble gas configuration.

$H-1s^1; He -1s^2$

$X-ns^2 np^5 : Ne-1s^22s^2 2p^6$

Both have an electron less than the nearest noble gas configuration.

From group 6 only one metal forms hydride. This metal is:

0%
Mo
0%
W
0%
Cr
0%
Co

Explanation

The low affinity of the elements of the group

$6,7,8$ and

$9$ towards hydrogen in their normal oxidation states prevent the formation of hydrides. This is called as hydride gap.

In the group,

$6$ , Chromium

$(Cr)$ is an exception as it is the only group

$6$ element to form a hydride.

Answer:

$(Cr)$

Peroxodisulphate on hydrolysis yields:

0%
water
0%
dihydrogen
0%
hydrogen peroxide
0%
deuterium

Explanation

Peroxodisulphate on hydrolysis yield hydrogen peroxide.

$HO_3SO-OSO_3H(aq) +2H_2O (l)\xrightarrow{hydrolysis}2HSO^-_4(aq)+2H^+(aq)+H_2O_2(aq)$

Non-stoichiometric hydrides are produced by:

0%
palladium, vanadium
0%
manganese, lithium
0%
nitrogen, fluorine
0%
carbon, nickel

Explanation

The hydrogen deficient compounds formed by the reaction of

$d-block$ and

$f-block$ elements with dihydrogen are called Non-stoichiometric compounds.

The d-block and f-block element form non-stoichiometric hydride because of the vacant d- and f-orbitals along with the small size.

Their elemental composition proportions cannot be represented in integers. They disobey the law of constant composition. Among the elements given, only vanadium and palladium form non-stoichiometric hydrides.

Answer: (A) palladium, vanadium

The maximum number of hydrogen bonds formed by a water molecule in ice is:

0%
$4$
0%
$1$
0%
$2$
0%
$3$

Explanation

Explanation:

Part: 1

In a molecule of $H_2O$ , both the $H$ atoms are involved in the H-bonding with $O$ atom of the other $H_2O$ molecule.
Due to the presence of two lone pairs on the oxygen atom, it is capable of forming two more H-bonds with the hydrogen atoms of the nearby $H_2O$ molecule.

Part: 2 Structure:

Final answer:

Therefore, the correct answer is option $A$ , as there is a total of four H-bonds formed.

Which of the statements given below are true for the structure of water molecule?
(i) Oxygen undergoes

$sp^{3}$ hybridization.
(ii) Due to the presence of two lone pairs of electrons on oxygen, the

$H-O-H$ bond angle is

$118.4^{0}$
(iii) Due to angular geometry the net dipole moment of water is not zero, u = 1.84 D.

0%
(i) and (ii)
0%
(ii) and (iii)
0%
(i) and (iii)
0%
only (ii)

Explanation

$H-O-H$ angle in water is slightly less than the typical tetrahedral angle. It is

$104.5^o$ . If all the substituents were the same (such as in

$CH_4$ ), the bond angle would be

$109.5^o$ . In water,

$H-O-H$ bond angle decreases from

$109.5^o$ to

$104.5^o$ as lone pair-lone pair repulsion is more than bond pair-bond pair repulsion and lone pair-bond pair repulsion.

Liquid water is denser than ice due to:

0%
higher surface tension
0%
hydrogen bonding
0%
van der Waals forces
0%
covalent bonding

Which is not a property of water ?

0%
It is a colorless and tasteless liquid
0%
There is no hydrogen bonding in solid state of water
0%
It is an excellent solvent for transportation of ions in plants and animals
0%
Frozen water is lighter than liquid water.

When

$CO_2$ is bubbled through a solution of barium peroxide in water:

0%
carbonic acid is formed
0%
$H_2O_2$ is formed
0%
$H_2O$ is formed
0%
barium hydroxide is formed

Explanation

When

$CO_2$ is bubbled through a cold solution of barium peroxide in water

$H_2O_2$ is obtained

$BaO_2+CO_2+H_2O \rightarrow BaCO_3+H_2O_2$

During hydrate formation from aqueous solution water can be associated in different forms. Indicate the wrong combination.
Coordinated water -

$[Cr(H_{2}O)_{6}]^{3+}3Cl^{-}$
Interstitial water -

$BaCl_{2}.2H_{2}O$
Hydrogen bonded water --

$[Cu(H_{2}O)_{4}]^{2+}SO_{4}^{2-}.H_{2}O$

0%
(i)
0%
(ii)
0%
(iii)
0%
none of these

Which of the following properties of water is related to the hydrogen bonding?

0%
High boiling point
0%
High heat of vaporizatiom
0%
Low density of ice compared to water
0%
None of these

When a substance reacts with water and breaks into acids and bases the process is called:

0%
dialysis
0%
solvolysis
0%
hydrolysis
0%
dissolution

Elements of which of the following group (s) of periodic table do not form hydrides?

0%
Groups 7, 8, 9
0%
Group 13
0%
Groups 15, 16, 17
0%
Group 14

Explanation

$\bf{Hint-}$ Hydrogen having negative charge is called as hydride

$H^-$ .

$\bf{Explanation-}$

The first elements of group

$7, 8, 9, 13, 14, 15, 16,$ and

$17$ are

$Mn, Fe, Co, B, C, N, O,$ and

$F$ from the periodic table.

$\bullet$ Consider group-13, Boran forms more hydrides with the general formula

$B_nH_{n+4}$ .

For example: diborane

$(B_2H_6 )$ , pentaborane

$(B_5H_9 )$ , etc.

$\bullet$ In group-14, carbon forms hydrides with hydrogen.

For example: methane

$(CH_4)$ , etc.

$\bullet$ In group-15, ammonia is an important nitrogen hydride.

$\bullet$ In group-16, oxygen forms hydrides are water

$(H_2O)$ , hydrogen peroxide

$(H_2O_2)$ , etc.

$\bullet$ In group-17, fluorine forms hydrides as

$HF$ .

$\bullet$ Group

$7, 8, 9$ does not form hydrides out of the periodic table. The property does not form hydrides of these

$7, 8, 9$ groups are known as the hydride gap.

$\bf{Conclusion-}$ Hence, option A is correct.

Which of the following hydrides is electron-precise hydride?

0%
$B_2H_6$
0%
$NH_3$
0%
$H_2O$
0%
$CH_4$

Explanation

Methane

$CH_4$ is electron-precise hydride.

Diborane

$B_2H_6$ is electron-deficient hydride

Ammonia

$NH_3$ and water

$H_2O$ are electron-rich hydrides.

Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is the most important in this respect?

0%
Its tendency to lose an electron to form a cation
0%
It tendency to gain a single electron in its valence shell to attain stable electronic configuration
0%
Its low negative electron gain enthalpy value
0%
Its small size

Explanation

Hydrogen resembles halogens as it has similar tendency to gain a single electron to attain stable noble gas configuration. The electronic configuration of hydrogen is

$1s^1$ . It needs one more electron to attain stable electronic configuration of hellium, a noble gas. The electronic configuration of halogen is

$ns^2np^5$ . It needs one more electron to attain stable electronic configuration of a noble gas.

The oxidising property of

$H_2O_2$ is best explained by assuming that two oxygen atoms in its molecule are_______

0%
bonded differently.
0%
bonded similarly
0%
bonded covalently
0%
bonded by hydrogen bonds

Name the phenomenon used in dissolution of water.

0%
solubility
0%
crystallization
0%
polarization
0%
surface tension

Calculate the ebullioscopic for water. The heat of vapourisation is

$40.685 \ kJ\ mol^{-1}$ .

0%
$0.512\ K\ kg\ mol^{-1}$
0%
$1.86\ K\ kg\ mol^{-1}$
0%
$5.12\ K\ kg\ mol^{-1}$
0%
$3.56\ K\ kg\ mol^{-1}$

Explanation

$K_b = \dfrac{MRT_b\times T_b}{1000\times DH_v}$

$K_b =\dfrac{18\times 8.314\times 273\times 273}{1000\times 40.685\times 1000}$

$K_b= \dfrac{20820933.108}{40.685\times 1000000}$

$K_b=0.512$ M is molecular mass of water

$T_b$ is boiling point of water

Out of the following metals which will give

$H_2$ reaction with

$NaOH$ ?

$I: Zn$

$II: Mg$

$III: Al$

$IV: Be$

0%
$I, II, III, IV$
0%
$I, III, IV$
0%
$II, IV$
0%
$I, III$

Explanation

The reactions of following metals with

$NaOH$ can be given as:-

$Zn+2NaOH\longrightarrow Na_2ZnO_2+H_2$

$Mg+2NaOH\longrightarrow 2Na+Mg(OH)_2$

$2Al+2NaOH+2H_2O\longrightarrow 2NaAlO_2+3H_2$

$Be+NaOH+H_2O\longrightarrow Na_2[Be(OH)_4]+H_2$

From the above reactions, it is clear that

$Zn,Al$ and

$Be$ react with

$NaOH$ to give

$H_2$ while

$Mg$ does not.

Which of the hydrated species can exist?
I.

$H_5O^+_2$
II.

$H_3O^+$
III.

$H_3O^-_2$
IV.

$H_7O^+_3$
Select alternate from.

0%
II only
0%
I and II
0%
I, II and IV
0%
I, II, III and IV

Explanation

Water on self ionization form

$H_3O^+$ . Water in strong acid also lead to the formation of

$H_5O_2^+$ ,

$H_7O_3^+$ and water in strong alkali also lead to the formation of

${H_3O_2}^-$ . So all the hydrated species can exist.

Industrial hydrogen is manufactured by passing an electric current through a solution of?

0%
Alumina
0%
Sodium hydroxide
0%
Carbon dioxide
0%
Magnesium hydroxide

Explanation

Electrolysis of

$NaOH$ gives hydrogen

$A^{+}$ cathode

$Na^{+}+e^{-}\rightarrow Na$

At anode

$20H^{-}\rightarrow H_{2}+O_{2}+2e^{-}$

Which of the following is not true about

${ H }_{ 2 }O$ molecule?

0%
The molecule has $\mu =0$
0%
The molecule can act as a base
0%
Shows abnormally high boiling point in comparison to hydrogen oxygen group
0%
The molecule has a bent shape

Explanation

Solution:- (A) The molecule has

$\mu = 0$ .

The dipole moment of water is

$1.85 D$ .

The dipole moment arises because oxygen is more electronegative than hydrogen; the oxygen pulls in the shared electrons and increases the electron density around itself.

This creates an electric dipole moment vector, with the partial negative charge on the oxygen atom.

There are two

$O-H$ bonds, so the dipole moment of the molecule is the resultant vertical vector projection of the individual diagonal dipole vectors.

Hence, (A) is not true.

The sources of dissolved oxygen in water is ________.

0%
photosynthesis
0%
natural aeration
0%
mechanical aeration
0%
all of the above

Hydrogen exists in atomic state in which of the following compounds?

0%
Metallic hydrides
0%
Ionic Hydride
0%
Molecular hydrides
0%
${ H }_{ 2 }O$

Which of the following cannot be oxidized by

$H_2O_2$ ?

0%
$KI + HCl$
0%
$O_3$
0%
$PbS$
0%
$Na_2SO_3$

$H_2$ gas can not be prepared by :

0%
$Be$ $+$ $NaOH$
0%
$Na$ $+$ $NaOH$
0%
$Mg$ $+$ $NaOH$
0%
both A and B

Explanation

For (A):-

$Be+NaOH\longrightarrow Na_2BeO_2+H_2$

For (B):-

$Na+NaOH\longrightarrow Na_2O+H_2$

For (C):-

$Mg+2NaOH\longrightarrow Mg(OH)_2+2Na$

From the above three reactions, it can be concluded that (A) and (B) react with

$NaOH$ to give

$H_2$ gas but (C) does not give

$H_2$ gas.

Therefore, option D is correct.

Which combination do not produce

${ H }_{ 2 }$ except:-

0%
Au + cold water
0%
Cu + aq. ${ H }_{ 2 }{ SO }_{ 4 }$
0%
Pt + $dil.{ HNO }_{ 3 }$
0%
All of these

Explanation

Because gold (Au) is the least reactive metal, even less than hydrogen, now as Hydrogen is more reactive, gold cannot displace it, hence water and gold will not react.

$Cu$ can't liberate

$H_2$ from

$H_2SO_4$ and same case with

$Pt$ and

$HNO_3$ .

Hence option D is correct.

Which of the following has maximum purity

0%
River water
0%
Distilled water
0%
Rain water
0%
none

Explanation

$\begin{array}{l} \text { Distilled water is water that has been boiled} \\ \text { into vapour and condensed back into liquid in a } \\ \text { seperate container. Impurities in the original } \\ \text { water that do not boil below or near the} \\ \text { boiling point of water remain in the original } \\ \text { container. } \\ \text { So, Distilled water has maximum purity. } \\ \text { (B) Distilled water. } \end{array}$

Very pure hydrogen

$(99.9\%)$ can be made by which of the following processes?

0%
Mixing natural hydrocarbons of high molecular weight
0%
Electrolysis of water
0%
Reaction of salt like hydrides with water
0%
Reaction of methane with steam.

Explanation

Electrolysis of water gives

$99.9$ % pure hydrogen.

It is the decomposition of water into Oxygen

$(O_2)$ and Hydrogen gas

$(H_2)$ , due to an electric current being passed through the water.

So, option (B) is correct.

The physical state of water is liquid at which of the following temperatures?

0%
$25^o C$
0%
$0^oC$
0%
$100^oc$
0%
none of these

Explanation

At 0

$^{\circ}C$ , water exists as ice.

At 100

$^{\circ}C$ , it exists as steam.

At 25

$^{\circ}C$ , water exists in liquid form.

Very pure hydrogen (99.9%) can be made by 'which of the following processes?

0%
Reaction of methane with steam
0%
Mixing natural hydrocarbons of high molecular weight
0%
Elecrolysis of water
0%
Reaction of salt like hydrides with water

Explanation

Highly pure hydrogen can be obtained by electrolysis of water.

It is the decomposition of water into

$O_2$ &

$H_2$ gas by passing electric current.

At anode,

$6H_2O(l) \longrightarrow O_2(g)+4H_3O^++4e^-$

cathode,

$4e^-+4H_2O(l) \longrightarrow 2H_2(g)+4OH^-(aq)$

Overall

$\longrightarrow 2H_2O(l) \longrightarrow 2H_2(g)+O_2(g)%$

Which of the following compound will evolve

$H_{2}$ gas when reacts with alkali metals?

0%
$H_{2}C =CH_{2}$
0%
$R-O-R$
0%
$R-CH_{3}$
0%
$R-OH$

Explanation

Alcohols when reacting with sodium metal evolution of hydrogen gas takes place. For example ethyl alcohol (ethanol) when reacting with sodium metal, the formation of sodium ethoxide and hydrogen gas takes place. It can be shown by the following reaction.

$\underset{alcohol}{2R-OH} + 2Na \longrightarrow \underset{sodium \ alkoxide}{2R-O^{-}Na^{+}} + H_{{2}{(g)}}\uparrow$

$\underset{Ethanol}{2R-OH} + 2Na \longrightarrow \underset{sodium \ ethoxide}{2R-O^{-}Na^{+}} + H_{{2}{(g)}}\uparrow$

Hence, option

$D$ is correct.

At what temperature, water have maximum density?

0%
$0^oC$
0%
$4^oC$
0%
$30^oC$
0%
$25^oC$

Explanation

$\bf{Hint:}$ On freezing the volume of ice increases.

$\bf{Correct \ answer:}$ Option

$B$

$\bf{Explanation:}$

Water has a freezing point

$0^{o}C$ , at this temperature its molecules get arranged in a crystalline structure, thereby attaining a defined shape and the overall volume increases, and water 'expands'.

$0^{o}C$ temperature the hydrogen bonds became in crystalline form which keeps the molecules far enough apart to make ice less dense but at

$4^{o}C$ these crystalline hydrogen bonds breaks, therefore water molecules comes closer and its density increases.

Why is the maximum density of water at 4°C? | Socratic

Why is the maximum density of water at 4°C? | Socratic

Which of the following reaction will not produce hydrogen gas?

0%
Reaction between $Fe$ and dil. $HCl$
0%
Reaction between $Zn$ and conc. $H_{2}SO_{4}$
0%
Reaction between $Zn$ and $NaOH$
0%
Electrolysis of $NaCl$ in Nelson's cell

Explanation

$a)$

$Fe_{(s)}+2HCl_{(aq)}\longrightarrow FeCl_{2(aq)}+H_{2(g)}$

$b)$

$Zn+conc.\text{ }2H_2SO_4\longrightarrow ZnSO_4+SO_2+2H_2O$

$c)$

$Zn+2NaOH\longrightarrow Na_2ZnO_2+H_2$

$d)$ In Nelson's cell, at cathode:

$2H_2O+2e^-\rightleftharpoons H_2+2OH^-$

Hence,

$B$ is the correct answer.

${ H_{ 2 }O }_{ 2 }$ is:

0%
oxidising agent
0%
both oxidising and reducing agent
0%
reducing agent
0%
none of the above

Explanation

$H_2O_2$ is both oxidizing as well as reducing agent.

These two chemical equation shows different behaviour of

$H_2O_2$ :-

$\longrightarrow H_2O_2+2I^-+2H^+ \longrightarrow I_2+2H_2O$

$\longrightarrow 5H_2O_2+2MnO^-_4+6H^+ \longrightarrow 2Mn^{2+}$

In first reaction

$H_2O_2$ act as an oxidizing agent as oxidation states change from

$-1$ to

$-2$ .

$2^{nd}$ reaction

$H_2O_2$ act as reducing agent as oxidation state of

$Mn$ changes from

$+7$ to

$+2$ by

$H_2O_2$ (itself gets oxidized).

Hydrogen peroxide is now generally prepared to industrial scale by the:

0%
action of $H_{2}SO_{4}$ on barium peroxide
0%
action of $H_{2}SO_{4}$ on sodium peroxide
0%
electrolysis of $50$ % $H_2SO_4$
0%
burning hydrogen in excess of oxygen

Explanation

Hydrogen peroxide is prepared from barium peroxide by the following methods.

By the action of dilute sulphuric acid. A paste of hydrated barium peroxide ( $BaO_2.8H_2O$ ) is prepared in ice – cold water and then added slowly to an ice-cold solution of 20% $H_2SO_4.$

Water released from dams is measured in?

0%
Feet per second or CFS
0%
Liters
0%
Cusesc
0%
Feet

The following factor are responsible for this property of water:

0%
Water molecule is polar in nature.
0%
Water molecule can form hydrogen bonds with the molecules of other compounds.
0%
water molecules can enter in a chemical reaction with molecules of other compounds.
0%
Water has a high value of dielectric constant.

On an industrial scale

${ H }_{ 2 }{ O }_{ 2 }$ is prepared by auto oxidation of ______.

0%
2- ethylanthraquinol
0%
2-ethylanthraquinone
0%
1- ethylanthraquinol
0%
1- ethylanthraquinone

Explanation

(I) 2-Ethyl anthraquinone is reduced to 2-ethyl anthraquinol by

$H_2$ in the presence of palladium.

(II) This 2-ethyl anthraquinol is oxidised by air to 2-ethyl anthrequinone again. Then

$H_2O_7$ (20% solution) forms.

1056429_1117339_ans_abd5140d463c4c8eac0e2344420238dc.png

Hydrogen oxidised is reduced by

0%
$H_2O$
0%
Acidic solution of $KMnO_{4}$
0%
Berium oxide
0%
None of the above

Hydrogen peroxide is obtained by the electrolysis of _________

0%
water
0%
sulpharic acid
0%
hydrochoric acid
0%
fused sodium

Explanation

$H_2O_2$ is offained by the electrolysis of sulphuric acid.

$H_2SO_4\longrightarrow H^++HSO_4^-$

$2HSO_4^-\longrightarrow HO_3SOOSO_3H+2e^-$

$HO_3SOOSO_3H+2H_2O\longrightarrow 2H_2SO_4+H_2O_2$

Intramolecular H-bonding is present in:

0%
$O-nitrophenol$
0%
$HCl$
0%
$HBr$
0%
$HF$

Explanation

Intramolecular hydrogen bonding takes place in ortho-nitrophenol because of the $NO_2$ and OH molecules close proximity to each other and this disrupts the possibility of molecular association.

Hence, the correct option is A.

1100372_1136214_ans_60db3cbb1ddc4b35906efd75bdbcc32d.gif

The structure of

$H_2O_2$ is :

0%
Half open book like
0%
Linear
0%
Closed book
0%
Pyramidal

The amount of water recommended by united nations for drinking, washing, cooking and maintaning, proper hygiene is a minimum__________ (i) litre per person/day. This amount is about _____(ii) bucket of water Per Person/day. The upper limit of layer, where all the space between particle of soil and gaps between rocks are filled with water is called ______ (iii) and and it can be recharge by _____(v)._________(vi) is a technique of watering plants by making use of narrow deliver water directly A the base of plants.

0%
(i) 60 litre, (iii) water cycle, (v) Bawaris
0%
(i) Two and half (iii) water table (vi) drip irrigation
0%
(i) 50 litre, (ii) two, (v) seeping
0%
(iii) Water cycle (v) infiltration (vi) ground water

Correct order of ionic character is:

0%
$CaH_2 < BaH_2 < CH_4$
0%
$CH_4 < CaH_2 < BaH_2$
0%
$CH_4 < BaH_2 < CaH_2$
0%
$BaH_2< CaH_2 < CH_4$

Explanation

$BeH_2 < CaH_2<BaH_2$

Smaller the size of cation more will be its polarising power. Hence,

$BeH_2$ will be the least iconic.

"or"

On moving down the group metallic character of metals increases so the ionic character of metal hydride increases.

Hence option B is correct.

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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